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Case Study Question for Class 12 Physics Chapter 1 Electric Charges and Fields

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There is Case Study Questions in class 12 Physics in session 2020-21. The first two questions in the board exam question paper will be based on Case Study and Assertion & Reason. Case Study Questions will have 5 MCQs out of which students will have to attempt any 4 questions. Here are the questions based on case study.

Case Study Question 1:

Electric field strength is proportional to the density of lines of force i.e., electric field strength at a point is proportional to the number of lines of force cutting a unit area element placed normal to the field at that point. As illustrated in given figure, the electric field at P is stronger than at Q.

(i) Electric lines of force about a positive point charge are (a) radially outwards (b) circular clockwise (c) radially inwards (d) parallel straight lines

(ii) Which of the following is false for electric lines of force? (a) They always start from positive charge and terminate on negative charges. (b) They are always perpendicular to the surface of a charged conductor. (c) They always form closed loops. (d) They are parallel and equally spaced in a region of uniform electric field.

(iii) Which one of the following patterns of electric line of force is not possible in field due to stationary charges?

(iv) Electric field lines are curved (a) in the field of a single positive or negative charge (b) in the field of two equal and opposite charges. (c) in the field of two like charges. (d) both (b) and (c)

(v) The figure below shows the electric field lines due to two positive charges. The magnitudes E A , E B and E C of the electric fields at point A, B and C respectively are related as

(a) E A >E B >E C (b) E B >E A >E C (c) E A =E B >E C (d) E A >E B =E C

Case Study Question 2:

Smallest charge that can exist in nature is the charge of an electron. During friction it is only the transfer of electron which makes the body charged. Hence net charge on any body is an integral multiple of charge of an electron (1.6 x 10 -19 C) i.e., q=±ne where r= 1, 2, 3, 4 …. Hence no body can have a charge represented as 1.8e, 2.7e, 2e/5, etc. Recently, it has been discovered that elementary particles such as protons or neutrons are elemental units called quarks.

(i) Which of the following properties is not satisfied by an electric charge? (a) Total charge conservation. (b) Quantization of charge. (c) Two types of charge. (d) Circular line of force.

(ii) Which one of the following charges is possible? (a) 5.8 x 10 -18 C (b) 3.2 x 10 -18 C (c) 4.5 x 10 -19 C (d) 8.6 x 10 -19 C

(iii) If a charge on a body is 1 nC, then how many electrons are present on the body? (a) 6.25 x 10 27 (b) 1.6 x 10 19 (c) 6.25 X 10 28 (d) 6.25 X 10 9

(iv) If a body gives out 10 9 electrons every second, how much time is required to get a total charge of 1 from it? (a) 190.19 years (b) 150.12 years (c) 198.19 years (d) 188.21 years

(v) A polythene piece rubbed with wool is found to have a negative charge of 3.2 x 10 -7 C. Calculate the number of electrons transferred. (a) 2 x 10 12 (b) 3 x 10 12 (c) 2 x 10 14 (d) 3 x 10 14

Case Study Question 3:

When electric dipole is placed in uniform electric field, its two charges experience equal and opposite forces, which cancel each other and hence net force on electric dipole in uniform electric field is zero. However these forces are not collinear, so they give rise to some torque on the dipole. Since net force on electric dipole in uniform electric field is zero, so no work is done in moving the electric dipole in uniform electric field. However some work is done in rotating the dipole against the torque acting on it.

(i) The dipole moment of a dipole in a uniform external field Ē is B. Then the torque τ acting on the dipole is (a) τ=p x E (b) τ = P. Ē (c) τ = 2(p + Ē) (d) τ = (P + E)

(ii) An electric dipole consists of two opposite charges, each of magnitude 1.0 μC separated by a distance of 2.0 cm. The dipole is placed in an external field of 10 5 NC -1 . The maximum torque on the dipole is (a) 0.2 x 10 -3 Nm (b) 1x 10 -3 Nm (c) 2 x 10 -3 Nm (d) 4x 10 -3 Nm

(iii) Torque on a dipole in uniform electric field is minimum when θ is equal to (a) 0° (b) 90° (c) 180° (d) Both (a) and (c)

(iv) When an electric dipole is held at an angle in a uniform electric field, the net force F and torque τ on the dipole are (a) F= 0, τ = 0 (b) F≠0, τ≠0 (c) F=0, τ ≠ 0 (d) F≠0, τ=0

(v) An electric dipole of moment p is placed in an electric field of intensity E. The dipole acquires a position such that the axis of the dipole makes an angle with the direction of the field. Assuming that potential energy of the dipole to be zero when 0 = 90°, the torque and the potential energy of the dipole will respectively be (a) pEsinθ, -pEcosθ (b) pEsinθ, -2pEcosθ (c) pEsinθ, 2pEcosθ (d) pEcosθ, – pEsinθ

Case Study Question 4:

A charge is a property associated with the matter due to which it experiences and produces an electric and magnetic field. Charges are scalar in nature and they add up like real number. Also, the total charge of an isolated system is always conserved. When the objects rub against each other charges acquired by them must be equal and opposite.

(i) The cause of a charging is: (a) the actual transfer of protons. (b) the actual transfer of electrons. (c) the actual transfer of neutrons. (d) none the above

(ii) Pick the correct statement. (a) The glass rod gives protons to silk when they are rubbed against each other. (b) The glass rod gives electrons to silk when they are rubbed against each other. (c) The glass rod gains protons from silk when they are rubbed against each other. (d) The glass rod gains electrons when they are rubbed against each other.

(iii) If two electrons are each 1.5 × 10 –10 m from a proton, the magnitude of the net electric force they will exert on the proton is (a) 1.97 × 10 –8 N (b) 2.73 × 10 –8 N (c) 3.83 × 10 –8 N (d) 4.63 × 10 –8 N

(iv) A charge is a property associated with the matter due to which it produces and experiences: (a) electric effects only (b) magnetic effects only (c) both electric and magnetic effects (d) none of these.

(v) The cause of quantization of electric charges is: (a) Transfer of an integral number of neutrons. (b) Transfer of an integral number of protons. (c) Transfer of an integral number of electrons. (d) None of the above.

Case Study Question 5:

Surface Charge Density. Surface charge density is defined as the charge per unit surface area the surface (Arial) charge symmetric distribution and follow Gauss law of electro statics mathematical term of surface charge density σ=ΔQ/ΔS

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite sign (± s). Having magnitude 8.8 × 10 –12 cm –2 as shown here. The intensity of electrified at a point is E =σ/ε 0   and flux is Φ=E.ΔS, where ΔS = 1 m 2 (unit arial plate)

(i) E in the outer region (I) of the first (A) plate is (a) 1.7 × 10 –22 N/C (b) 1.1 × 10 –12 V/m (c) Zero (d) Insufficient data

(ii) E in the outer region (III) of the second plate (B) is (a) 1 N/C (b) 0.1 V/m (c) 0.5 N/C (d) zero

(iii) E between (II) the plate is (a) 1 N/C (b) 0.1 V/m (c) 0.5 N/C (d) None of these

(iv) The ratio of E from left side of plate A at distance 1 cm and 2 m respectively is (a) 1 : 2 (b) 10 : 2 (c) 1 : 1 (d) 20 : 1

(v) In order to estimate the electric field due to a thin finite plane metal plate the Gaussian surface considered is (a) Spherical (b) Linear (c) Cylindrical (d) Cybic

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CBSE Class 12 Physics Case Study Questions PDF Download

Case Study questions for the Class 12 Physics board exams are available here. You can read the Class 12 Physics Case Study Questions broken down by chapter. Subject matter specialists and seasoned teachers created these quizzes. You can verify the right response to each question by referring to the answer key, which is also provided. To achieve success on your final exams, practice the following questions.

CBSE (Central Board of Secondary Education) is a renowned educational board in India that designs the curriculum for Class 12 Physics with the goal of promoting a scientific temperament and nurturing critical thinking among students. As part of their Physics examination, CBSE includes case study questions to assess students’ ability to apply theoretical knowledge to real-world scenarios effectively.

Chapter-wise Solved Case Study Questions for Class 12 Physics

Before the exams, students in class 12 should review crucial Physics Case Study issues. They will gain a better understanding of the kinds of Case Study questions that may be asked in Physics exams for Grade 12. These questions were created by our highly qualified standard 12 Physics staff based on the questions that appeared most frequently in last year’s exams. The solutions have been written in a way that will make them simple to grasp and will aid students in grade 12 in understanding the topics.

Best Books for Class 12 Physics Exams

Strictly in accordance with the new term-by-term curriculum for the class 12 board exams to be held in the academic year 2023–2024, which will include multiple choice questions based on new board typologies including stand-alone MCQs and case-based MCQs with an assertion–reason. Included are inquiries from the official CBSE Question Bank that was released in April 2024. What changes have been made to the book: strictly in accordance with the term-by-term syllabus for the board exams that will be held during the 2024 academic year? Chapter- and topic-based Questions with multiple choices that are based on the unique evaluation method used for the Class 12th Physics Final Board Exams.

Key Benefits of Solving CBSE Class 12 Physics Case Study Questions

• Application of Concepts: Case study questions demand the application of theoretical knowledge in practical scenarios, preparing students for real-world challenges and professional pursuits.
• Critical Thinking: By evaluating and analyzing case studies, students develop critical thinking abilities, enabling them to approach complex problems with a logical mindset.
• In-Depth Understanding: Addressing case study questions necessitates a thorough understanding of physics concepts, leading to a more profound comprehension of the subject matter.
• Holistic Evaluation: CBSE adopts case study questions as they provide a holistic evaluation of a student’s aptitude and proficiency in physics, moving beyond rote memorization.
• Preparation for Competitive Exams: Since competitive exams often include similar application-based questions, practicing case study questions equips students for various entrance tests.

How to Approach CBSE Class 12 Physics Case Study Questions

• Read and Analyze Thoroughly: Carefully read the case study to grasp its context and identify the underlying physics principles involved.
• Identify Relevant Concepts: Highlight the physics theories and concepts applicable to the given scenario.
• Create a Systematic Solution: Formulate a step-by-step solution using the identified concepts, explaining each step with clarity.
• Include Diagrams and Charts: If relevant, incorporate diagrams, charts, or graphs to visually represent the situation, aiding better comprehension.
• Double-Check Answers: Always review your answers for accuracy, ensuring that they align with the principles of physics.

Tips for Excelling in CBSE Class 12 Physics

• Conceptual Clarity: Focus on building a strong foundation of physics concepts, as this will enable you to apply them effectively to case study questions.
• Practice Regularly: Dedicate time to solving case study questions regularly, enhancing your proficiency in handling real-world scenarios.
• Seek Guidance: Don’t hesitate to seek guidance from teachers, peers, or online resources to gain additional insights into challenging concepts.
• Time Management: During exams, practice efficient time management to ensure you allocate enough time to each case study question without rushing.
• Stay Positive: Approach case study questions with a positive mindset, embracing them as opportunities to showcase your skills and knowledge.

CBSE Class 12 Physics case study questions play a pivotal role in promoting practical understanding and critical thinking among students. By embracing these questions as opportunities for growth, students can excel in their physics examinations and beyond.

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• NCERT Solutions for Class 12 Physics Chapter 1 - Electric Charges And Fields
• NCERT Solutions

NCERT Class 12 Physics Chapter 1: Electric Charges And Fields

The very first chapter of Class 12 NCERT Physics introduces students to the concept of electric charges and how it exerts a force on other materials by attracting or repelling them. The NCERT Solutions for Class 12 Chapter 1 are available on Vedantu for students preparing for the CBSE 12th board examinations along with other competitive exams. Skilled teachers with years of experience have prepared these solutions to facilitate a strong understanding of the topics and the best techniques for fetching more marks. Students can download the NCERT Class 12 Physics Chapter 1 PDF for free from Vedantu.

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List of Important Topics Covered under NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges And Fields

Below are The Important Topics Discussed in The Chapter 1 of Class 12 Physics

Deleted syllabus of class 12 electromagnetic waves (2024-25), list of important formulas in class 12 physics chapter 1 electric charges and fields.

While preparing a chapter, it is important for students to memorize the formula of a particular topic. This will help them to boost their score. Some important formulas are listed below.

Coulomb’s Law

The electric force between the two-point charges are directly proportional to the product of the charges and inversely proportional to the square of the distance between them. 

F = K$\frac{q_{1}q_{2}}{r2}$

Here, 

F represents the electric force.

K represents Coulomb’s constant.

$q_{1}q_{2}$ represents the two charges.

$\lambda$ represents the distance between two charges.  

Electric field intensity 

Electric field intensity is the vector quantity.             

E $\frac{F}{q_{1}}$

F is the force experienced by the test charge.

• $q_1$ is the test charge.

Electric Flux 

dФ = E.da cos𝛉 

Go through the below link to recap all the important formulas in the Class 12 Physics Chapter 1 Electric Charges And Fields.

Would you like to view a summarized version of this chapter? Check out the 'Chapter at a glance' section below the PDF of NCERT Solutions.

Related chapters.

Chapter at a Glance for Electric Charges & Fields NCERT Solutions

From simple experiments on frictional electricity, one can infer that there are two types of charges in nature; and that like charges repel and unlike charges attract. By convention, the charge on a glass rod rubbed with silk is positive; that on a plastic rod rubbed with fur is then negative.

Charge is not only a scalar (or invariant) under rotation; it is also invariant for frames of reference in relative motion. This is not always true for every scalar. For Example-, kinetic energy is a scalar under rotation, but is not invariant for frames of reference in relative motion.

Conductors allow movement of electric charge through them, insulators do not. In metals, the mobile charges are electrons; in electrolytes both positive and negative ions are mobile.

Coulomb’s Law: The mutual electrostatic force between two point charges q 1 and q 2 is proportional to the product q 1 q 2 and inversely proportional to the square of the distance r 21 separating them. 

$F_{21}=\text{force}\;q_{2} \;\text{du to}\;q_{1}\;=\frac{k(q_{1}q_{2})}{r^2_{21}}\;r_{21}$

Where $r_{21}$ is a unit vector in the direction from q 1 to q 2 and $k=\frac{1}{4\pi \varepsilon _{0}}$ is the constant of proportionality.

In SI units, the unit of charge is coulomb. The experimental value of the constant $\varepsilon _{0}$ is 

$\varepsilon _{0}=8.854\times 10^{-12}C^{2}N^{-}M^{-2}$

The approximate value of k is- 9\times $10^{9}Nm^2C^{-2}$

Superposition Principle: The principle is based on the property that the forces with which two charges attract or repel each other are not affected by the presence of a third (or more) additional charge(s). For an assembly of charges Q 1 , Q 2 , Q 3 ..., the force on any charge, say Q 1 , is the vector sum of the force on  Q 1 due to Q 2 , the force on Q 1 due to Q 3 , and so on. For each pair, the force is given by Coulomb's law for two charges stated earlier.

An electric field line is a curve drawn in such a way that the tangent at each point on the curve gives the direction of the electric field at that point. The relative closeness of field lines indicates the relative strength of electric field at different points; they crowd near each other in regions of strong electric field and are far apart where the electric field is weak. In regions of constant electric field, the field lines are uniformly spaced parallel straight lines.

Some of the important properties of field lines are: (i) Field lines are continuous curves without any breaks. (ii) Two field lines cannot cross each other. (iii) Electrostatic field lines start at positive charges and end at negative charges (iv) They cannot form closed loops.

Electric flux is the measure of the field lines crossing a surface. It is scalar quantity, with SI unit $\frac{N}{C}$-m 2 or V-m. “The number of field lines passing through perpendicular unit area will be proportional to the magnitude of Electric Field there” 

Gauss’s law: The flux of electric field through any closed surface S is $1/\varepsilon _0$ times the total charge enclosed by S. The law is especially useful in determining electric field E, when the source distribution has simple symmetry.

An electric dipole is a pair of equal and opposite charges q and –q separated by some distance 2a. Its dipole moment vector p has magnitude 2qa and is in the direction of the dipole axis from –q to q.

Field of an electric dipole in its equatorial plane (i.e., the plane perpendicular to its axis and passing through its centre) at a distance r from the centre:

$E=\frac{-P}{4\pi \varepsilon _0}\frac{1}{(\alpha^2+r^2)^{3/2}}\cong \frac{-P}{4\pi \varepsilon _0r^3}.\,\text{for}\,r\gg \alpha $

Dipole electric field on the axis at a distance r from the centre:

$E=\frac{2pr}{4\pi\varepsilon_0(r^2-\alpha^2)^3}\cong \frac{2{p}}{4\pi\varepsilon_0{r}^3}\,\,\text{for}\,\,r\gg\alpha$

The 1/r 3 dependence of dipole electric fields should be noted in contrast to the 1/r 2 dependence of electric field due to a point charge.

In a uniform electric field E, a dipole experiences a torque  τ given by: τ = p × E

but experiences no net force.

Electric Field Due to Various Uniform Charge Distribution

Mastering Class 12 Physics Chapter 8: Electromagnetic Waves - MCQs, Question and Answers, and Tips for Success

1. What is the force between two small charged spheres of charges 

$2\times {{10}^{-7}}C$ and $3\times {{10}^{-7}}C$ placed $30cm$ apart in air?

Ans: We are given the following information:

Repulsive force of magnitude,$F=6\times {{10}^{-3}}N$

Charge on the first sphere, ${{q}_{1}}=2\times {{10}^{-7}}C$

Charge on the second sphere, ${{q}_{2}}=3\times {{10}^{-7}}C$

Distance between the two spheres, $r=30cm=0.3m$

Electrostatic force between the two spheres is given by Coulomb’s law as,

$F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$

Where, ${{\varepsilon }_{0}}$is the permittivity of free space and, 

$\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$

Now on substituting the given values, Coulomb’s law becomes, 

\[F = \frac{9\times 10^{9}\times 2\times 10^{-7}\times 3\times 10^{-7}}{(0.3)^{2}}\]

Therefore, we found the electrostatic force between the given charged spheres to be $F=6\times {{10}^{-3}}N$. Since the charges are of the same nature, we could say that the force is repulsive.

2. The electrostatic force on a small sphere of charge $0.4\mu C$ due to another small sphere of charge $-0.8\mu C$ in air is 0.2N.

a) What is the distance between the two spheres?

Ans: Electrostatic force on the first sphere is given to be, $F=0.2N$

Charge of the first sphere is, ${{q}_{1}}=0.4\mu C=0.4\times {{10}^{-6}}C$

Charge of the second sphere is, ${{q}_{2}}=-0.8\mu C=-0.8\times {{10}^{-6}}C$

We have the electrostatic force given by Coulomb’s law as,

$\Rightarrow r=\sqrt{\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\varepsilon }_{0}}F}}$

Substituting the given values in the above equation, we get, 

$\Rightarrow r=\sqrt{\frac{0.4\times {{10}^{-6}}\times 8\times {{10}^{-6}}\times 9\times {{10}^{9}}}{0.2}}$

$\Rightarrow r=\sqrt{144\times {{10}^{-4}}}$

$\therefore r=0.12m$

Therefore, we found the distance between charged spheres to be $r=0.12m$.

b) What is the force on the second sphere due to the first?

Ans: From Newton’s third law of motion, we know that every action has an equal and opposite reaction. 

Thus, we could say that the given two spheres would attract each other with the same force. 

So, the force on the second sphere due to the first sphere will be $0.2N$. 

3. Check whether the ratio $\frac{k{{e}^{2}}}{G{{m}_{e}}{{m}_{p}}}$is dimensionless. Look up a table of physical constants and hence determine the value of the given ratio. What does the ratio signify?

Ans: We are given the ratio, $\frac{k{{e}^{2}}}{G{{m}_{e}}{{m}_{p}}}$.

Here, G is the gravitational constant which has its unit $N{{m}^{2}}k{{g}^{-2}}$;

${{m}_{e}}$and ${{m}_{p}}$ are the masses of electron and proton in $kg$ respectively;

$e$ is the electric charge in $C$; 

$k$ is a constant given by $k=\frac{1}{4\pi {{\varepsilon }_{0}}}$

In the expression for k, ${{\varepsilon }_{0}}$ is the permittivity of free space which has its unit $N{{m}^{2}}{{C}^{-2}}$. 

Now, we could find the dimension of the given ratio by considering their units as follows: 

$\frac{k{{e}^{2}}}{G{{m}_{e}}{{m}_{p}}}=\frac{\left[ N{{m}^{2}}{{C}^{-2}} \right]{{\left[ C \right]}^{2}}}{\left[ N{{m}^{2}}k{{g}^{-2}} \right]\left[ kg \right]\left[ kg \right]}={{M}^{0}}{{L}^{0}}{{T}^{0}}$

Clearly, it is understood that the given ratio is dimensionless. 

Now, we know the values for the given physical quantities as, 

$e=1.6\times {{10}^{-19}}C$

$G=6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$

${{m}_{e}}=9.1\times {{10}^{-31}}kg$

${{m}_{p}}=1.66\times {{10}^{-27}}kg$

Substituting these values into the required ratio, we get, 

$\frac{k{{e}^{2}}}{G{{m}_{e}}{{m}_{p}}}=\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{6.67\times {{10}^{-11}}\times 9.1\times {{10}^{-3}}\times 1.67\times {{10}^{-22}}}$

$\Rightarrow \frac{k{{e}^{2}}}{G{{m}_{e}}{{m}_{p}}}\approx 2.3\times {{10}^{39}}$

We could infer that the given ratio is the ratio of electrical force to the gravitational force between a proton and an electron when the distance between them is kept constant. 

a) Explain the meaning of the statements ‘electric charge of a body is quantized’. 

Ans: The given statement ‘Electric charge of a body is quantized’ means that only the integral number $(1,2,3,...,n)$ of electrons can be transferred from one body to another. 

That is, charges cannot be transferred from one body to another in fraction.

b) Why can one ignore quantization of electric charge when dealing with macroscopic i.e., large scale charges?

Ans: On a macroscopic scale or large-scale, the number of charges is as large as the magnitude of an electric charge. 

So, quantization is considered insignificant at a macroscopic scale for an electric charge and electric charges are considered continuous.

5. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge. 

Ans: Rubbing two objects would produce charges that are equal in magnitude and opposite in nature on the two bodies. 

This happens due to the reason that charges are created in pairs. This phenomenon is called charging by friction. 

The net charge of the system however remains zero as the opposite charges equal in magnitude annihilate each other. 

So, rubbing a glass rod with a silk cloth creates opposite charges of equal magnitude on both of them and this observation is found to be consistent with the law of conservation of charge. 

6. Four point charges ${{q}_{A}}=2\mu C$, ${{q}_{B}}=-5\mu C$, ${{q}_{C}}=2\mu C$and ${{q}_{D}}=-5\mu C$ are located at the corners of a square ABCD with side 10cm. What is the force on the $1\mu C$ charge placed at the centre of this square?

Ans: Consider the square of side length $10cm$ given below with four charges at its corners and let O be its centre.

From the figure we find the diagonals to be, 

$AC=BD=10\sqrt{2}cm$

$\Rightarrow AO=OC=DO=OB=5\sqrt{2}cm$

Now the repulsive force at O due to charge at A, 

${{F}_{AO}}=k\frac{{{q}_{A}}{{q}_{O}}}{O{{A}^{2}}}=k\frac{\left( +2\mu C \right)\left( 1\mu C \right)}{{{\left( 5\sqrt{2} \right)}^{2}}}$…………………………………………… (1)

And the repulsive force at O due to charge at D, 

${{F}_{DO}}=k\frac{{{q}_{D}}{{q}_{O}}}{O{{D}^{2}}}=k\frac{\left( +2\mu C \right)\left( 1\mu C \right)}{{{\left( 5\sqrt{2} \right)}^{2}}}$………………………………………….. (2) 

And the attractive force at O due to charge at B, 

${{F}_{BO}}=k\frac{{{q}_{B}}{{q}_{O}}}{O{{B}^{2}}}=k\frac{\left( -5\mu C \right)\left( 1\mu C \right)}{{{\left( 5\sqrt{2} \right)}^{2}}}$……………………………………………. (3)

And the attractive force at O due to charge at C, 

${{F}_{CO}}=k\frac{{{q}_{C}}{{q}_{O}}}{O{{C}^{2}}}=k\frac{\left( -5\mu C \right)\left( 1\mu C \right)}{{{\left( 5\sqrt{2} \right)}^{2}}}$……………………………………………… (4)

We find that (1) and (2) are of same magnitude but they act in the opposite direction and hence they cancel out each other. 

Similarly, (3) and (4) are of the same magnitude but in the opposite direction and hence they cancel out each other too. 

Hence, the net force on charge at centre O is found to be zero. 

a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?  

Ans: An electrostatic field line is a continuous curve as the charge experiences a continuous force on being placed in an electric field. 

As the charge doesn’t jump from one point to the other, field lines will not have sudden breaks. 

b) Explain why two field lines never cross each other at any point?

Ans: If two field lines are seen to cross each other at a point, it would imply that the electric field intensity has two different directions at that point, as two different tangents (representing the direction of electric field intensity at that point) can be drawn at the point of intersection. 

This is however impossible and thus, two field lines never cross each other.

8. Two point charges ${{q}_{A}}=3\mu C$and ${{q}_{B}}=-3\mu C$are located 20cm apart in vacuum. 

a) What is the electric field at the midpoint O of the line AB joining the two charges?

Ans: The situation could be represented in the following figure. Let O be the midpoint of line AB.

We are given:

\[AB=20cm\]

\[AO=OB=10cm\]

Take E to be the electric field at point O, then, 

The electric field at point O due to charge $+3\mu C$would be, 

${{E}_{1}}=\frac{3\times {{10}^{-6}}}{4\pi {{\varepsilon }_{0}}{{\left( AO \right)}^{2}}}=\frac{3\times {{10}^{-6}}}{4\pi {{\varepsilon }_{0}}{{\left( 10\times {{10}^{-2}} \right)}^{2}}}N{{C}^{-1}}$along OB

The electric field at point O due to charge $-3\mu C$would be, 

${{E}_{2}}=\left| \frac{3\times {{10}^{-6}}}{4\pi {{\varepsilon }_{0}}{{\left( OB \right)}^{2}}} \right|=\frac{3\times {{10}^{-6}}}{4\pi {{\varepsilon }_{0}}{{\left( 10\times {{10}^{-2}} \right)}^{2}}}N{{C}^{-1}}$along OB 

The net electric field,

$\Rightarrow E={{E}_{1}}+{{E}_{2}}$

$\Rightarrow E=2\times \frac{9\times {{10}^{9}}\times 3\times {{10}^{-6}}}{{{\left( 10\times {{10}^{-2}} \right)}^{2}}}$

$\Rightarrow E=5.4\times {{10}^{6}}N{{C}^{-1}}$

Therefore, the electric field at mid-point O is $E=5.4\times {{10}^{6}}N{{C}^{-1}}$ along OB. 

b) If a negative test charge of magnitude $1.5\times {{10}^{-19}}C$ is placed at this point, what is the force experienced by the test charge?

Ans: We have a test charge of magnitude $1.5\times {{10}^{-9}}C$ placed at mid-point O and we found the electric field at this point to be $E=5.4\times {{10}^{6}}N{{C}^{-1}}$.

So, the force experienced by the test charge would be F, 

$\Rightarrow F=qE$

$\Rightarrow F=1.5\times {{10}^{-9}}\times 5.4\times {{10}^{6}}$

$\Rightarrow F=8.1\times {{10}^{-3}}N$

This force will be directed along OA since like charges repel and unlike charges attract.

9. A system has two charges ${{q}_{A}}=2.5\times {{10}^{-7}}C$and ${{q}_{B}}=-2.5\times {{10}^{-7}}C$located at points $A:\left( 0,0,-15cm \right)$ and $B:\left( 0,0,+15cm \right)$ respectively. What are the total charge and electric dipole moment of the system?

Ans: The figure given below represents the system mentioned in the question:

The charge at point A, ${{q}_{A}}=2.5\times {{10}^{-7}}C$

The charge at point B, ${{q}_{B}}=-2.5\times {{10}^{-7}}C$

Then, the net charge would be, $q={{q}_{A}}+{{q}_{B}}=2.5\times {{10}^{-7}}C-2.5\times {{10}^{-7}}C=0$

The distance between two charges at A and B would be,

\[d=15+15=30cm\]

The electric dipole moment of the system could be given by,

$P=\mathop{q}_{A}\times d=\mathop{q}_{B}\times d$

$\Rightarrow P=2.5\times {{10}^{-7}}\times 0.3$

$\therefore P=7.5\times {{10}^{-8}}Cm$ along the \[+z\]axis.

Therefore, the electric dipole moment of the system is found to be $7.5\times {{10}^{-8}}Cm$ and it is directed along the positive \[z\]-axis.

10. An electric dipole with dipole moment $4\times {{10}^{-9}}Cm$ is aligned at $30{}^\circ $ with direction of a uniform electric field of magnitude $5\times {{10}^{4}}N{{C}^{-1}}$. Calculate the magnitude of the torque acting on the dipole.

Ans: We are given the following:

Electric dipole moment, $\overrightarrow{p}=4\times {{10}^{-9}}Cm$ 

Angle made by $\overrightarrow{p}$ with uniform electric field, $\theta =30{}^\circ $ 

Electric field, $\overrightarrow{E}=5\times {{10}^{4}}N{{C}^{-1}}$ 

Torque acting on the dipole is given by

$\tau =pE\sin \theta $ 

Substituting the given values we get, 

$\Rightarrow \tau =4\times {{10}^{-9}}\times 5\times {{10}^{4}}\times \sin 30{}^\circ $

$\Rightarrow \tau =20\times {{10}^{-5}}\times \frac{1}{2}$

$\therefore \tau ={{10}^{-4}}Nm$

Thus, the magnitude of the torque acting on the dipole is found to be ${{10}^{-4}}Nm$.

11. A polythene piece rubbed with wool is found to have a negative charge of $3\times {{10}^{-7}}C$ 

a) Estimate the number of electrons transferred (from which to which?)

Ans: When polythene is rubbed against wool, a certain number of electrons get transferred from wool to polythene. 

As a result of which wool becomes positively charged on losing electrons and polythene becomes negatively charged on gaining them.

Charge on the polythene piece, $q=-3\times {{10}^{-7}}C$ 

Charge of an electron, $e=-1.6\times {{10}^{-19}}C$ 

Let n be the number of electrons transferred from wool to polythene, then, from the property of quantization we have, 

$q=ne$ 

$\Rightarrow n=\frac{q}{e}$

Now, on substituting the given values, we get, 

$\Rightarrow n=\frac{-3\times {{10}^{-7}}}{-1.6\times {{10}^{-19}}}$

$\therefore n=1.87\times {{10}^{12}}$

Therefore, the number of electrons transferred from wool to polythene would be$1.87\times {{10}^{12}}$.

b) Is there a transfer of mass from wool to polythene?

Ans: Yes, during the transfer of electrons from wool to polythene, along with charge, mass is transferred too. 

Let $m$ be the mass being transferred in the given case and ${{m}_{e}}$ be the mass of the electron, then,

$m={{m}_{e}}\times n$ 

$\Rightarrow m=9.1\times {{10}^{-31}}\times 1.85\times {{10}^{12}}$

$\therefore m=1.706\times {{10}^{-18}}kg$

Thus, we found that a negligible amount of mass does get transferred from wool to polythene.

a) Two insulated charged copper spheres $A$ and $B$ have their centres separated by a distance of $50cm$. What is the mutual force of electrostatic repulsion if the charge on each is $6.5\times {{10}^{-7}}C$? The radii of A and B are negligible compared to the distance of separation. 

Ans: We are given:

Charges on spheres $A$ and $B$ are equal,

${{q}_{A}}={{q}_{B}}=6.5\times {{10}^{-7}}C$

Distance between the centres of the spheres is given as,

$r=50cm=0.5m$

It is known that the force of repulsion between the two spheres would be given by Coulomb’s law as,

$F=\frac{{{q}_{A}}{{q}_{B}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$

Where, ${{\varepsilon }_{o}}$ is the permittivity of the free space

Substituting the known values into the above expression, we get,

$F=\frac{9\times {{10}^{9}}\times {{(6.5\times {{10}^{-7}})}^{2}}}{{{(0.5)}^{2}}}=1.52\times {{10}^{-2}}N$

Thus, the mutual force of electrostatic repulsion between the two spheres is found to be$F=1.52\times {{10}^{-2}}N$.

b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Ans: It is told that the charges on both the spheres are doubled and the distance between the centres of the spheres is halved. That is,

${{q}_{A}}'={{q}_{B}}'=2\times 6.5\times {{10}^{-7}}=13\times {{10}^{-7}}C$

$r'=\frac{1}{2}(0.5)=0.25m$

Now, we could substitute these values in Coulomb’s law to get,

$F'=\frac{{{q}_{A}}'{{q}_{B}}'}{4\pi {{\varepsilon }_{0}}r{{'}^{2}}}$

$\Rightarrow F=\frac{9\times {{10}^{9}}\times {{(13\times {{10}^{-7}})}^{2}}}{{{(0.25)}^{2}}}$

$\Rightarrow F=0.243N$

The new mutual force of electrostatic repulsion between the two spheres is found to be $0.243N$.

13. Suppose the spheres $A$ and $B$ in question $12$ have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between $A$ and $B$?

Distance between the spheres $A$ and $B$ is $r=0.5m$

The charge on each sphere initially is found to be ${{q}_{A}}={{q}_{B}}=6.5\times {{10}^{-7}}C$

Now, when an uncharged sphere $C$ is made to touch the sphere $A$, a certain amount of charge from $A$ will get transferred to the sphere $C$, making both $A$ and $C$ to have equal charges in them. So,

${{q}_{A}}'={{q}_{C}}=\frac{1}{2}(6.5\times {{10}^{-7}})=3.25\times {{10}^{-7}}C$

Now, when the sphere $C$ is made to touch the sphere $B$, there is a similar transfer of charge making both $C$ and $B$ to have equal charges in them. So,

${{q}_{C}}'={{q}_{B}}'=\frac{3.25\times {{10}^{-7}}+6.5\times {{10}^{-7}}}{2}=4.875\times {{10}^{-7}}C$

Thus, the new force of repulsion between the spheres $A$ and $B$ would now become,

$F'=\frac{{{q}_{A}}'{{q}_{B}}'}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$

$\Rightarrow F'=\frac{9\times {{10}^{9}}\times 3.25\times {{10}^{-7}}\times 4.875\times {{10}^{-7}}}{{{(0.5)}^{2}}}$

$\Rightarrow F'=5.703\times {{10}^{-3}}N$

14. Figure below shows tracks taken by three charged particles in a uniform electrostatic field. Give the signs of the three charges and also mention which particle has the highest charge to mass ratio?

Ans: From the known properties of charges, we know that the unlike charges attract and like charges repel each other. 

So, the particles 1 and 2 that move towards the positively charged plate while repelling away from the negatively charged plate would be negatively charged and the particle 3 that moves towards the negatively charged plate while repelling away from the positively charged plate would be positively charged.

Now, we know that the charge to mass ratio (which is generally known as emf) is directly proportional to the displacement or the amount of deflection for a given velocity. 

Since the deflection of particle 3 is found to be maximum among the three, it would have the highest charge to mass ratio.

15. Consider a uniform electric field $E=3\times {{10}^{3}}\hat{i}N/C$. 

a) Find the flux of this field through a square of side $10cm$whose plane is parallel to the y-z plane. 

Electric field intensity, $\overrightarrow{E}=3\times {{10}^{3}}\hat{i}N/C$

Magnitude of electric field intensity, $\left| \overrightarrow{E} \right|=3\times {{10}^{3}}N/C$

Side of the square, $a=10cm=0.1m$ 

Area of the square, $A={{a}^{2}}=0.01{{m}^{2}}$ 

Since the plane of the square is parallel to the y-z plane, the normal to its plane would be directed in the x direction. So, angle between normal to the plane and the electric field would be, $\theta =0{}^\circ $ 

We know that the flux through a surface is given by the relation, 

$\phi =\left| E \right|\left| A \right|\cos \theta $

Substituting the given values, we get, 

$\Rightarrow \phi =3\times {{10}^{3}}\times 0.01\times \cos 0{}^\circ $

$\therefore \phi =30N{{m}^{2}}/C$

Thus, we found the net flux through the given surface to be $\phi =30N{{m}^{2}}/C$.

b) What would be the flux through the same square if the normal to its plane makes $60{}^\circ $ angle with the x-axis? 

Ans: When the plane makes an angle of $60{}^\circ $ with the x-axis, the flux through the given surface would be,

$\Rightarrow \phi =3\times {{10}^{3}}\times 0.01\times \cos 60{}^\circ $

$\Rightarrow \phi =30\times \frac{1}{2}$

$\Rightarrow \phi =15N{{m}^{2}}/C$

So, we found the flux in this case to be, $\phi =15N{{m}^{2}}/C$.

16. What is the net flux of the uniform electric field of exercise $1.15$ through a cube of side $20cm$ oriented so that its faces are parallel to the coordinate planes?

Ans: We are given that all the faces of the cube are parallel to the coordinate planes. 

Clearly, the number of field lines entering the cube is equal to the number of field lines entering out of the cube. As a result, the net flux through the cube would be zero.

17. Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is $8.0\times {{10}^{3}}N{{m}^{2}}/C$ . 

a) What is the net charge inside the box? 

Ans: We are given that:

Net outward flux through surface of the box,

$\phi =8.0\times {{10}^{3}}N{{m}^{2}}/C$ 

For a body containing of net charge $q$, flux could be given by,

$\phi =\frac{q}{{{\varepsilon }_{0}}}$ 

Where, ${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}=$ Permittivity of free space 

Therefore, the charge $q$ is given by

$q=\phi {{\varepsilon }_{0}}$

$\Rightarrow q=8.854\times {{10}^{-12}}\times 8.0\times {{10}^{3}}$

$\Rightarrow q=7.08\times {{10}^{-8}}$

$\Rightarrow q=0.07\mu C$

Therefore, the net charge inside the box is found to be $0.07\mu C$.

b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?

Ans: No, the net flux entering out through a body depends on the net charge contained within the body according to Gauss’s law. 

So, if the net flux is given to be zero, then it can be inferred that the net charge inside the body is zero. 

However, the net charge of the body being zero only implies that the body has equal amount of positive and negative charges and thus, we cannot conclude that there were no charges inside the box.

18. A point charge $+10\mu C$ is a distance $5cm$ directly above the centre of a square of side $10cm$, as shown in Figure below. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge $10cm$) 

Ans: Consider the square as one face of a cube of edge length $10cm$ with a charge $q$ at its centre, according to Gauss's theorem for a cube, total electric flux is through all its six faces.

${{\phi }_{total}}=\frac{q}{{{\varepsilon }_{0}}}$

The electric flux through one face of the cube could be now given by,  \[\phi =\frac{{{\phi }_{total}}}{6}\].

$\phi =\frac{1}{6}\frac{q}{{{\varepsilon }_{0}}}$

${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}=$ Permittivity of free space

The net charge enclosed would be, $q=10\mu C=10\times {{10}^{-6}}C$

Substituting the values given in the question, we get, 

$\phi =\frac{1}{6}\times \frac{10\times {{10}^{-6}}}{8.854\times {{10}^{-12}}}$

$\therefore \phi =1.88\times {{10}^{5}}N{{m}^{2}}{{C}^{-1}}$

Therefore, electric flux through the square is found to be $1.88\times {{10}^{5}}N{{m}^{2}}{{C}^{-1}}$.

19. A point charge of $2.0\mu C$ is kept at the centre of a cubic Gaussian surface of edge length $9cm$. What is the net electric flux through this surface? 

Ans: Let us consider one of the faces of the cubical Gaussian surface considered (square).

Since a cube has six such square faces in total, we could say that the flux through one surface would be one-sixth the total flux through the gaussian surface considered. 

The net flux through the cubical Gaussian surface by Gauss’s law could be given by, 

So, the electric flux through one face of the cube would be, \[\phi =\frac{{{\phi }_{total}}}{6}\]

$\Rightarrow \phi =\frac{1}{6}\frac{q}{{{\varepsilon }_{0}}}$……………………………….. (1)

But we have, 

Charge enclosed, $q=10\mu C=10\times {{10}^{-6}}C$

Substituting the given values in (1) we get, 

$\Rightarrow \phi =1.88\times {{10}^{5}}N{{m}^{2}}{{C}^{-1}}$

Therefore, electric flux through the square surface is $1.88\times {{10}^{5}}N{{m}^{2}}{{C}^{-1}}$.

20. A point charge causes an electric flux of $-1.0\times {{10}^{3}}N{{m}^{2}}/C$ to pass through a spherical Gaussian surface of $10cm$ radius centred on the charge. 

a) If the radius of the Gaussian surface were doubled, how much flux could pass through the surface? 

Electric flux due to the given point charge, $\phi =-1.0\times {{10}^{3}}N{{m}^{2}}/C$ 

Radius of the Gaussian surface enclosing the point charge,$r=10.0cm$ 

Electric flux piercing out through a surface depends on the net charge enclosed by the surface according to Gauss’s law and is independent of the dimensions of the arbitrary surface assumed to enclose this charge. 

Hence, if the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., $-{{10}^{3}}N{{m}^{2}}/C$ .

b) What is the magnitude of the point charge?

Ans: Electric flux could be given by the relation,

Where,$q=$ net charge enclosed by the spherical surface

$\Rightarrow q=\phi {{\varepsilon }_{0}}$

Substituting the given values,

$\Rightarrow q=-1.0\times {{10}^{3}}\times 8.854\times {{10}^{-12}}=-8.854\times {{10}^{-9}}C$

$\Rightarrow q=-8.854nC$

Thus, the value of the point charge is found to be $-8.854nC$.

21. A conducting sphere of radius $10cm$ has an unknown charge. If the electric field at a point $20cm$ from the centre of the sphere of magnitude $1.5\times {{10}^{3}}N/C$ is directed radially inward, what is the net charge on the sphere?

Ans: We have the relation for electric field intensity $E$ at a distance \[\left( d \right)\] from the centre of a sphere containing net charge $q$ is given by,

$E=\frac{q}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}$ ……………………………………………… (1)

Where, 

Net charge, $q=1.5\times {{10}^{3}}N/C$

Distance from the centre, $d=20cm=0.2m$ 

From (1), the unknown charge would be, 

$q=E\left( 4\pi {{\varepsilon }_{0}} \right){{d}^{2}}$

$\Rightarrow q=\frac{1.5\times {{10}^{3}}\times {{\left( 0.2 \right)}^{2}}}{9\times {{10}^{9}}}=6.67\times {{10}^{-9}}C$

$\Rightarrow q=6.67nC$

Therefore, the net charge on the sphere is found to be$6.67nC$.

22. A uniformly charged conducting sphere of $2.4m$ diameter has a surface charge density of $80.0\mu C/{{m}^{2}}$ .

a) Find the charge on the sphere. 

Ans: Given that,

Diameter of the sphere, $d=2.4m$. 

Radius of the sphere, $r=1.2m$. 

Surface charge density, 

\[\sigma =80.0\mu C/{{m}^{2}}=80\times {{10}^{-6}}C/{{m}^{2}}\] 

Total charge on the surface of the sphere,

$Q=\text{Charge density }\times \text{ Surface area}$ 

$\Rightarrow \text{Q}=\sigma \times \text{4}\pi {{\text{r}}^{2}}=80\times {{10}^{-6}}\times 4\times 3.14\times {{\left( 1.2 \right)}^{2}}$

$\Rightarrow Q=1.447\times {{10}^{-3}}C$

Therefore, the charge on the sphere is found to be $1.447\times {{10}^{-3}}C$.

b) What is the total electric flux leaving the surface of the sphere?

Ans: Total electric flux $\left( {{\phi }_{total}} \right)$ leaving out the surface containing net charge $Q$ is given by Gauss’s law as, 

${{\phi }_{total}}=\frac{Q}{{{\varepsilon }_{0}}}$…………………………………………………. (1)

Where, permittivity of free space,

${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$

We found the charge on the sphere to be, 

$Q=1.447\times {{10}^{-3}}C$

Substituting these in (1), we get, 

${{\phi }_{total}}=\frac{1.447\times {{10}^{-3}}}{8.854\times {{10}^{-12}}}$

$\Rightarrow {{\phi }_{total}}=1.63\times {{10}^{-8}}N{{C}^{-1}}{{m}^{2}}$

Therefore, the total electric flux leaving the surface of the sphere is found to be $1.63\times {{10}^{-8}}N{{C}^{-1}}{{m}^{2}}$.

23. An infinite line charge produces a field of magnitude $9\times {{10}^{4}}N/C$ at a distance of $2cm$. Calculate the linear charge density.

Ans: Electric field produced by the given infinite line charge at a distance $d$having linear charge density$\lambda $ could be given by the relation,

$E=\frac{\lambda }{2\pi {{\varepsilon }_{0}}d}$ 

$\Rightarrow \lambda =2\pi {{\varepsilon }_{0}}Ed$…………………………………….. (1)

$d=2cm=0.02m$  

$E=9\times {{10}^{4}}N/C$ 

Permittivity of free space,

${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$ 

Substituting these values in (1) we get, 

$\Rightarrow \lambda =2\pi \left( 8.854\times {{10}^{-12}} \right)\left( 9\times {{10}^{4}} \right)\left( 0.02 \right)$

$\Rightarrow \lambda =10\times {{10}^{-8}}C/m$

Therefore, we found the linear charge density to be $10\times {{10}^{-8}}C/m$.

24. Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $17.0\times {{10}^{-22}}C{{m}^{-2}}$. What is $E$ in the outer region of the first plate? What is $E$ in the outer region of the second plate? What is E between the plates?

Ans: The given nature of metal plates is represented in the figure below: 

Here, A and B are two parallel plates kept close to each other. The outer region of plate A is denoted as $I$, outer region of plate B is denoted as $III$, and the region between the plates, A and B, is denoted as $II$.

It is given that:

Charge density of plate A, $\sigma =17.0\times {{10}^{-22}}C/{{m}^{2}}$

Charge density of plate B, $\sigma =-17.0\times {{10}^{-22}}C/{{m}^{2}}$

In the regions $I$and$III$, electric field E is zero. This is because the charge is not enclosed within the respective plates.

Now, the electric field $E$ in the region $II$ is given by

$E=\frac{|\sigma |}{{{\varepsilon }_{0}}}$ 

Permittivity of free space ${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$ 

$E=\frac{17.0\times {{10}^{-22}}}{8.854\times {{10}^{-12}}}$

$\Rightarrow E=1.92\times {{10}^{-10}}N/C$

Thus, the electric field between the plates is $1.92\times {{10}^{-10}}N/C$.

25. An oil drop of $12$ excess electrons is held stationary under a constant electric field of $2.55\times {{10}^{4}}N{{C}^{-1}}$ in Millikan's oil drop experiment. The density of the oil is $1.26gm/c{{m}^{3}}$ . Estimate the radius of the drop. $\left( g=9.81m{{s}^{-2}},e=1.60\times {{10}^{-19}}C \right)$.

Ans: It is given that:

The number of excess electrons on the oil drop, \[n=12\] 

Electric field intensity, $E=2.55\times {{10}^{4}}N{{C}^{-1}}$ 

The density of oil, $\rho =1.26gm/c{{m}^{3}}=1.26\times {{10}^{3}}kg/{{m}^{3}}$ 

Acceleration due to gravity, $g=9.81m{{s}^{-2}}$ 

Charge on an electron $e=1.60\times {{10}^{-19}}C$

Radius of the oil drop $=r$ 

Here, the force (F) due to electric field E is equal to the weight of the oil drop (W).

$\Rightarrow Eq=mg$

$\Rightarrow Ene=\frac{4}{3}\pi {{r}^{2}}\rho \times g$ 

$q$ is the net charge on the oil drop $=ne$

$m$ is the mass of the oil drop $=\text{Volume of the oil drop}\times \text{Density of oil}$\[=\frac{4}{3}\pi {{r}^{3}}\times p\]

Therefore, radius of the oil drop can be calculated as 

\[r=\sqrt{\frac{3Ene}{4\pi \rho g}}\]

\[\Rightarrow r=\sqrt{\frac{3\times 2.55\times {{10}^{4}}\times 12\times 1.6\times {{10}^{-19}}}{4\times 3.14\times 1.26\times {{10}^{3}}\times 9.81}}\]

\[\Rightarrow r=\sqrt{946.09\times {{10}^{-21}}}\]

\[\Rightarrow r=9.72\times {{10}^{-10}}m\]

Therefore, the radius of the oil drop is $9.72\times {{10}^{-10}}m$.

26. Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?

Ans: The field lines shown in (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor which is a characterizing property of electric field lines.

Ans: The lines shown in (b) do not represent electrostatic field lines because field lines cannot emerge from a negative charge and cannot terminate at a positive charge since the direction of the electric field is from positive to negative charge. 

Ans: The field lines shown in (c) do represent electrostatic field lines as they are directed outwards from positive charge in accordance with the property of electric field.

Ans: The field lines shown in (d) do not represent electrostatic field lines because electric field lines should not intersect each other.

Ans: The field lines shown in (e) do not represent electrostatic field lines because electric field lines do not form closed loops

27. In a certain region of space, the electric field is along the z-direction throughout. The magnitude of the electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of ${{10}^{5}}N{{C}^{-1}}$ per meter. What are the force and torque experienced by a system having a total dipole moment equal to ${{10}^{-7}}Cm$ in the negative z-direction?

Ans: We know that the dipole moment of the system, $P=q\times dl=-{{10}^{-7}}Cm$.

Also, the rate of increase of electric field per unit length is given as

 $\frac{dE}{dl}={{10}^{5}}N{{C}^{-1}}$

Now, the force (F) experienced by the system is given by $F=qE$

$F=q\frac{dE}{dl}\times dl$

$\Rightarrow F=P\frac{dE}{dl}$

$\Rightarrow F=-{{10}^{-7}}\times {{10}^{5}}$

$\Rightarrow F=-{{10}^{-2}}N$ 

Clearly, the force is equal to $-{{10}^{-2}}N$ in the negative z-direction i.e., it is opposite to the direction of the electric field. 

Thus, the angle between the electric field and dipole moment is equal to \[180{}^\circ \].

Now, the torque is given by $\tau =PE\sin \theta $

$\tau =PE\sin 180{}^\circ =0$

Therefore, it can be concluded that the torque experienced by the system is zero.

a) A conductor A with a cavity as shown in the Fig. 1.36(a) is given a charge $Q$. Show that the entire charge must appear on the outer surface of the conductor.

Now, let \[q\] be the charge inside the conductor and ${{\varepsilon }_{0}}$, the permittivity of free space.

According to Gauss's law,

Flux is given by

 $\phi =\overrightarrow{E}.ds=\frac{q}{{{\varepsilon }_{0}}}$ 

Here, $\phi =0$  as $E=0$ inside the conductor

Clearly, 

$0=\frac{q}{8.854\times {{10}^{-12}}}$

$\Rightarrow q=0$ 

Therefore, the charge inside the conductor is zero.

And hence, the entire charge $Q$ appears on the outer surface of the conductor.

b) Another conductor B with charge $q$ is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is $Q+q$ [Fig. 1.36 (b)].

Ans: The outer surface of conductor A has a charge of $Q$.

It is given that another conductor B, having a charge \[+q\] is kept inside conductor A and is insulated from conductor A.

Clearly, a charge of \[-q\] will get induced in the inner surface of conductor A and a charge of \[+q\] will get induced on the outer surface of conductor A.

Therefore, the total charge on the outer surface of conductor A amounts to $Q+q$.

c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

Ans: A sensitive instrument can be shielded from a strong electrostatic field in its environment by enclosing it fully inside a metallic envelope. 

Such a closed metallic body provides hindrance to electrostatic fields and thus can be used as a shield.

29. A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $\left[ \frac{\sigma }{2{{\varepsilon }_{0}}} \right]\overset{\wedge }{\mathop{n}}\,$, where $\overset{\wedge }{\mathop{n}}\,$ is the unit vector in the outward normal direction, and $\sigma $ is the surface charge density near the hole.

Ans: Firstly, let us consider a conductor with a cavity or a hole as shown in the diagram below. It is known that the electric field inside the cavity is zero. 

Let us assume E to be the electric field just outside the conductor, $q$ be the electric charge, $\sigma $ be the charge density, and ${{\varepsilon }_{0}}$, the permittivity of free space.

We know that charge $\left| q \right|=\sigma \times d$ 

Now, according to Gauss's law,

$\phi =E.ds=\frac{\left| q \right|}{{{\varepsilon }_{0}}}$

$\Rightarrow E.ds=\frac{\sigma \times d}{{{\varepsilon }_{0}}}$

$\Rightarrow E=\frac{\sigma }{{{\varepsilon }_{0}}}\overset{\wedge }{\mathop{n}}\,$

where $\overset{\wedge }{\mathop{n}}\,$ is the unit vector in the outward normal direction.  

Thus, the electric field just outside the conductor is $\frac{\sigma }{{{\varepsilon }_{0}}}\overset{\wedge }{\mathop{n}}\,$. 

Now, this field is actually a superposition of the field due to the cavity ${{E}_{1}}$ and the field due to the rest of the charged conductor ${{E}_{2}}$. 

These electric fields are equal and opposite inside the conductor whereas equal in magnitude as well as direction outside the conductor. 

${{E}_{1}}+{{E}_{2}}=E$

$\Rightarrow {{E}_{1}}={{E}_{2}}=\frac{E}{2}=\frac{\sigma }{2{{\varepsilon }_{0}}}\overset{\wedge }{\mathop{n}}\,$

Therefore, the electric field in the hole is  $\frac{\sigma }{2{{\varepsilon }_{0}}}\overset{\wedge }{\mathop{n}}\,$. 

Hence, proved.

30. Obtain the formula for the electric field due to a long thin wire of uniform linear charge density $\lambda $ without using Gauss's law. (Hint: Use Coulomb's law directly and evaluate the necessary integral)

Ans: Firstly, let us take a long thin wire XY as shown in the figure below. This wire is of uniform linear charge density $\lambda $ .

Now, consider a point A at a perpendicular distance l from the mid-point O of the wire as shown in the figure below:

Consider E to be the electric field at point A due to the wire.

Also consider a small length element $dx$ on the wire section with $OZ=x$ as shown.

Let $q$ be the charge on this element.

Clearly, $q=\lambda dx$ 

Now, the electric field due to this small element can be given as

$dE=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\lambda dx}{{{\left( AZ \right)}^{2}}}$ 

However, $AZ=\sqrt{{{1}^{2}}+{{x}^{2}}}$ 

\[\Rightarrow dE=\frac{\lambda dx}{4\pi {{\varepsilon }_{0}}\left( {{1}^{2}}+{{x}^{2}} \right)}\]

Now, let us resolve the electric field into two rectangular components. Doing so, $dE\cos \theta $ is the perpendicular component and $dE\sin \theta $ is the parallel component.

When the whole wire is considered, the component $dE\sin \theta $ gets cancelled and only the perpendicular component $dE\cos \theta $  affects the point A.

Thus, the effective electric field at point A due to the element $dx$ can be written as 

$d{{E}_{1}}=\frac{\lambda dx\cos \theta }{4\pi {{\varepsilon }_{0}}\left( {{l}^{2}}+{{x}^{2}} \right)}$    ....(1) 

Now, in $\Delta AZO$, we have

$\tan \theta =\frac{x}{l}$

$x=l\tan \theta \text{          }......\text{(2)}$ 

On differentiating equation (2), we obtain

$dx=l{{\sec }^{2}}d\theta \text{    }......\text{(3)}$ 

From equation (2)

${{x}^{2}}+{{l}^{2}}={{l}^{2}}+{{l}^{2}}{{\tan }^{2}}\theta $

$\Rightarrow {{l}^{2}}\left( 1+{{\tan }^{2}}\theta  \right)={{l}^{2}}{{\sec }^{2}}\theta $

$\Rightarrow {{x}^{2}}+{{l}^{2}}={{l}^{2}}{{\sec }^{2}}\theta \text{         }.....\text{(4)}$

Putting equations (3) and (4) in equation (1), we obtain

$d{{E}_{1}}=\frac{\lambda l{{\sec }^{2}}d\theta }{4\pi {{\varepsilon }_{0}}\left( {{l}^{2}}{{\sec }^{2}}\theta  \right)}\cos \theta $

\[\Rightarrow d{{E}_{1}}=\frac{\lambda \cos \theta d\theta }{4\pi {{\varepsilon }_{0}}l}\text{       }.....\text{(5)}\]

Now, the wire is taken so long that it ends from $-\frac{\pi }{2}$ to $+\frac{\pi }{2}$.

Therefore, by integrating equation (5), we obtain the value of field ${{E}_{1}}$ as

\[\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{d{{E}_{1}}}=\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\frac{\lambda }{4\pi {{\varepsilon }_{0}}l}\cos \theta d\theta }\]

\[\Rightarrow {{E}_{1}}=\frac{\lambda }{4\pi {{\varepsilon }_{0}}l}\times \text{2}\]

\[\Rightarrow {{E}_{1}}=\frac{\lambda }{2\pi {{\varepsilon }_{0}}l}\]

Thus, the electric field due to the long wire is derived to be equal to \[\frac{\lambda }{2\pi {{\varepsilon }_{0}}l}\].

31. It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called 'up quark (denoted by $u$)  of charge $\left( +\frac{1}{2} \right)e$ and the 'down' quark (denoted by $d$) of charge $-\left( \frac{1}{3} \right)e$ together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.                        

Ans: It is known that a proton has three quarks. Let us consider \[n\] up quarks in a proton, each having a charge of $+\left( \frac{2}{3}e \right)$. 

Now, the charge due to \[n\] up quarks $=\left( \frac{2}{3}e \right)n$ 

The number of down quarks in a proton $=3-n$ 

Also, each down quark has a charge of $-\frac{1}{3}e$ 

Therefore, the charge due to $\left( 3-n \right)$ down quarks $=\left( -\frac{1}{3}e \right)\left( 3-n \right)$

We know that the total charge on a proton $=+e$ 

$e=\left( \frac{2}{3}e \right)n+\left( -\frac{1}{3}e \right)\left( 3-n \right)$

$\Rightarrow e=\left( \frac{2ne}{3} \right)-e+\frac{ne}{3}$

$\Rightarrow 2e=ne$

$\Rightarrow n=2$

Clearly, the number of up quarks in a proton, $n=2$

Thus, the number of down quarks in a proton $=3-n=3-2=1$ 

Therefore, a proton can be represented as $uud$.

A neutron is also said to have three quarks. Let us consider \[n\] up quarks in a neutron, each having a charge of $+\left( \frac{2}{3}e \right)$ .

It is given that the charge on a neutron due to \[n\] up quarks $=\left( +\frac{3}{2}e \right)n$ 

Also, the number of down quarks is $\left( 3-n \right)$, each having a charge of $=\left( -\frac{3}{2} \right)e$

Thus, the charge on a neutron due to $\left( 3-n \right)$ down quarks $=\left( -\frac{1}{3}e \right)\left( 3-n \right)$

Now, we know that the total charge on a neutron $=0$ 

$0=\left( \frac{2}{3}e \right)n+\left( -\frac{1}{3}e \right)\left( 3-n \right)$

$\Rightarrow 0=\left( \frac{2ne}{3} \right)-e+\frac{ne}{3}$

$\Rightarrow e=ne$

$\Rightarrow n=1$

Clearly, the number of up quarks in a neutron, $n=1$

Thus, the number of down quarks in a neutron $=3-n=2$ 

Therefore, a neutron can be represented as $udd$.

a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where \[\mathbf{E}=\mathbf{0}\]) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.

Ans: Firstly, let us assume that the small test charge placed at the null point of the given setup is in stable equilibrium. 

By stable equilibrium, it means that even a slight displacement of the test charge in any direction will cause the charge to return to the null point as there will be strong restoring forces acting around it. 

This further suggests that all the electric lines of force around the null point act inwards and towards the given null point. 

But by Gauss law, we know that the net electric flux through a chargeless enclosing surface is equal to zero. This truth contradicts the assumption which we had started with. Therefore, it can be concluded that the equilibrium of the test charge is necessarily unstable.

b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed at a certain distance apart.

Ans: When we consider this configuration setup with two charges of the same magnitude and sign placed at a certain distance apart, the null point happens to be at the midpoint of the line joining these two charges. 

As per the previous assumption, the test charge, when placed at this mid-point will experience strong restoring forces when it tries to displace itself. 

But when the test charge tries to displace in a direction normal to the line joining the two charges, the test charge gets pulled off as there is no restoring force along the normal to the line considered. 

Since stable equilibrium prioritizes restoring force in all directions, the assumption in this case also gets contradicted. 

33. A particle of mass $m$ and charge $\left( -q \right)$ enters the region between the two charged plates initially moving along x- axis with speed $vx$ (like particle 1 in Fig 1.33). The length of plate is $L$ and a uniform electric field $E$ is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is $\frac{qE{{L}^{2}}}{2m{{v}_{x}}^{2}}$.

Compare this motion with motion of a projectile in the gravitational field discussed in section 4.10 of class XI textbook of Physics.

The charge on a particle of mass $m=-q$

Velocity of the particle $=vx$ 

Length of the plates $=L$ 

Magnitude of the uniform electric field between the plates $=E$ 

Mechanical force, $F=\text{ Mass }(m)\times \text{Acceleration }(a)$ 

Thus, acceleration, $a=\frac{F}{m}$

However, electric force, $F=qE$ 

Therefore, acceleration, $=\frac{qE}{m}$  .........(1)

Here, the time taken by the particle to cross the field of length $L$ is given by,

$t=\frac{\text{Length of the plate}}{\text{Velocity of the plate}}=\frac{L}{{{v}_{x}}}$  ......(2)

In the vertical direction, we know that the initial velocity, $u=0$ 

Now, according to the third equation of motion, vertical deflection $s$ of the particle can be derived as

$s=ut+\frac{1}{2}a{{t}^{2}}$

$\Rightarrow s=0+\frac{1}{2}\left( \frac{qE}{m} \right){{\left( \frac{L}{{{v}_{x}}} \right)}^{2}}$ 

$\Rightarrow s=\frac{qE{{L}^{2}}}{2m{{v}_{x}}^{2}}$   .....(3)

Thus, the vertical deflection of the particle at the far edge of the plate is $\frac{qE{{L}^{2}}}{2m{{v}_{x}}^{2}}$. 

In comparison, we can see that this is similar to the motion of horizontal projectiles under gravity.

34. Suppose that the particle in Exercise in 1.33 is an electron projected with velocity ${{v}_{x}}=2.0\times {{10}^{6}}m{{s}^{-1}}$. If $E$between the plates separated by $0.5cm$ is $9.1\times {{10}^{2}}N/C$, where will the electron strike the upper plate? ($\left| e \right|=1.6\times {{10}^{-19}}C,{{m}_{e}}=9.1\times {{10}^{-31}}kg$ )

Ans: We are given the velocity of the particle, ${{v}_{x}}=2.0\times {{10}^{6}}m{{s}^{-1}}$.

Separation between the two plates, $d=0.5cm=0.005m$ 

Electric field between the two plates, $E=9.1\times {{10}^{2}}N/C$ 

Charge on an electron, $e=1.6\times {{10}^{-19}}C$ 

mass of an electron, ${{m}_{e}}=9.1\times {{10}^{-31}}kg$ 

Let$s$be the deflection when the electron strikes the upper plate at the end of the plate $L$, then, we have the deflection given by, 

$s=\frac{qE{{L}^{2}}}{2m{{v}_{x}}}$ 

$\Rightarrow L=\sqrt{\frac{2dm{{v}_{x}}}{qE}}$

Substituting the given values, 

$\Rightarrow L=\sqrt{\frac{2\times 0.005\times 9.1\times {{10}^{-31}}\times {{\left( 2.0\times {{10}^{6}} \right)}^{2}}}{1.6\times {{10}^{-19}}\times 9.1\times {{10}^{2}}}}=\sqrt{0.025\times {{10}^{-2}}}=\sqrt{2.5\times {{10}^{-4}}}$

$\Rightarrow L=1.6\times {{10}^{-2}}=1.6cm$

Therefore, we found that the electron will strike the upper plate after travelling a distance of $1.6cm$.

NCERT Class 12 Physics Chapter 1 PDF

The NCERT Class 12 Physics Chapter 1 PDF deals with a wide range of concepts. The solutions consist of various objective type questions, including numerical problems, theoretical answers, short answer type questions as well as long-type questions. The thorough discussion of each question guides students to grasp all the concepts well. It helps them to ace in competitive exams like JEE as well as in their 12th board exams. For better guidance and assistance, download and refer to the PDF of NCERT Solution for 12 Physics Chapter 1 from Vedantu.

NCERT Class 12 Physics Solutions

Chapter 1 - Electric Charges and Fields

Chapter 2 - Electrostatic Potential and Capacitance

Chapter 3 - Current Electricity

Chapter 4 - Moving Charges and Magnetism

Chapter 5 - Magnetism And Matter

Chapter 6 - Electromagnetic Induction

Chapter 7 - Alternating Current

Chapter 8 - Electromagnetic Waves

Chapter 9 - Ray Optics and Optical Instruments

Chapter 10 - Wave Optics

Chapter 11 - Dual Nature of Radiation and Matter

Chapter 12 - Atoms

Chapter 13 - Nuclei

Chapter 14 - Semiconductor Electronic: Material, Devices And Simple Circuits

Chapter 15 - Communication Systems

NCERT Solutions for Electric Charges and Fields Class 12: Question-Wise Explanations

NCERT Class 12 Physics Chapter 1 introduces students to one of the most important laws of Physics, Coulomb’s Law. Students will get to learn about insulators, conductors, and other materials in this chapter. Also, NCERT Solutions to chapter 1 broadly discusses the various sub-topics covered in this chapter. The PDF solution comprises a total of 34 questions, and their solutions will definitely guide all students to understand the applications of Coulomb’s law and other concepts.

Question 1: Physics Class 12 Chapter 1

The first question of the exercise helps students to identify the force between two charged spheres suspended at certain distances apart in the air.

Question 2:

The second solution gives students an idea about the electrostatic force on different charged spheres (due to one another) in air. Students will get to learn about the distance between two charged spheres and the force on the second sphere due to first.

Question 3:

The solution to this question provides an explanation of the ratio ke2 / G me mp is dimensionless and its significance.

Question 4:

Here, students get an elaborate explanation of the statement “electric charge of a body is quantized”. Also, students learn that the quantization of electric charges is of no use on a macroscopic scale.

Question 5:

This solution enlightens students about the phenomena of charging by friction.

Question 6:

Question 6 provides students with the answer of charge at the center of a 10 cm square with four charged points.

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Question 7:

The solution to this question broadly discusses why electrostatic field line cannot have sudden breaks. Also, students get to know why two electric field lines never intersect at any point.

Question 8:

In this solution, students are provided with an answer to the electric field at the midpoint of two charged points suspended in the vacuum. Also, students learn about the force experienced by test charge at the midpoint.

Question 9:

Question 9 offers students with an explanation of the total charge and electric dipole moment of a system that has two charges.

Question 10:

The solution to this question elaborately discusses the magnitude of torque acting on the dipole.

Apart from these answers, class 12 Physics Chapter 1 NCERT solutions come with an extensive explanation of 24 other solutions. Some of the topics covered in the solutions are as follows.

Coulomb’s law: applications.

Calculation of linear charge density.

Gaussian Surface: applications.

Important Questions on 12th Physics Chapter 1

‘Electric charge of a body is quantised’ - Explain the meaning of this statement. 

Give the reason behind one being able to ignore the quantisation of an electric charge while dealing with macroscopic or large scale charges?

When you rub a glass rod with a silk cloth, charges appear on both these materials. A similar phenomenon is also seen in the case of many other pairs of bodies. Explain the consistency of this observation with the law of conservation of charge.

Why is an electrostatic field line always a continuous curve, that is, a field line does not have any sudden breaks? 

Provide the reason behind two field lines never crossing each other at any point.

Determine the net flux of the uniform electric field of Exercise 1.15 through a cube having a side equal to 20 cm, that is oriented in a manner such that its faces are parallel to the coordinate planes.

Consider the configuration of an arbitrary electrostatic field. Let a small test charge be placed at the null point, that is, E = 0, in this configuration. Prove that the equilibrium of this test charge is unstable.

If two charges of the same magnitude and sign are placed at a certain distance apart, then verify the result of this basic configuration.

Physics Class 12 Chapter 1: Marks Weightage

The chapter carries a 15-point weightage. Students who are having difficulty grasping the issues discussed in this chapter can go to Vedantu's answers for assistance.

Advantages of NCERT Solutions for Class 12 Physics Chapter 1

Students who seek the NCERT solutions for Class 12 Physics Chapter 1 find it helpful for various reasons.

The solutions are updated according to the latest NCERT guidelines.

All the solutions to the theoretical and numerical questions are provided in a simple and logical manner.

Each and every question is explained by highly experienced teachers.

All the solutions are designed to help students to develop a better understanding of the concepts of Electric Charges and Fields. Students who want to excel in this field of Physics can download and find assistance in the NCERT Solutions for Class 12 Physics Chapter 1 for free.

Why Students of Class 12 Opt for Vedantu?

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Quality online education, expert guidance, and personalized attention for 12th class students

Vedantu's NCERT Solutions for Class 12 Physics Chapter 1 provide a comprehensive and reliable resource for students studying this subject. The solutions are well-structured, covering all the key topics and exercises from the chapter. With Vedantu's solutions, students can enhance their understanding of physics concepts and improve their problem-solving skills. The solutions are designed to be easily accessible and user-friendly, enabling students to grasp complex concepts with ease. Whether it's for exam preparation or concept clarification, Vedantu's NCERT Solutions for Class 12 Physics Chapter 1 serve as a valuable tool for students seeking academic success in physics.

FAQs on NCERT Solutions for Class 12 Physics Chapter 1 - Electric Charges And Fields

1. What are the Benefits of Class 12 Physics Chapter 1 NCERT Solutions?

NCERT Solutions of the chapter of Electric Charges and Fields class 12 come with an elaborate discussion of the topic and subtopics. Moreover, the solutions are explained in a simple manner for a better understanding of the concepts.

Students who want to ace in the CBSE 12th board examination as well as in competitive exams like JEE can seek the guidance of NCERT solutions. The solutions are created in light of the latest CBSE guidelines that help students to secure better marks. Students can completely rely on the solutions provided by Vedantu, as these are all solved by the subject matter expert of Physics.

2. How can I get the NCERT Class 12 Physics Chapter 1 PDF?

The NCERT solutions of Physics class 12 chapter 1 PDF can be downloaded from the Vedantu website and mobile application. Therefore, students who are facing problems in solving the very first chapter of class 12 Physics can seek guidance in the NCERT solutions. The way every solution is described has been considered helpful. Interestingly, students can download this PDF for absolutely free.

3. What are the Subtopics of the Chapter Electric Charges and Fields?

The first chapter of Physics Class 12 Electric Charges and Fields consists of several sub-topics under it, and these are as follows. 

Introduction

Electric Charge

Conductors and Insulators

Charging by Induction

Basic properties of Electric Charge

Forces between multiple charges

Electric fields

Electric field lines

Electric flux

Electric Dipole

Dipole in a uniform external field

Continuous charge distribution

Gauss’s Law

Application of Gauss’s Law

Students who want to get an in-depth knowledge of these topics can consult the class 12 Physics chapter 1 NCERT solutions provided by Vedantu.

4. Are the NCERT Solutions for Class 12 Physics Chapter 1 sufficient for the exam preparation?

Yes, the NCERT Solutions for Class 12 Physics Chapter 1 is sufficient for the exam preparation. The solutions are well detailed and include every important point that will help students to score high. Each numerical problem is explained in a simple manner so that students can clear their basic concepts and learn the approach to solve questions in Physics. They can refer to Vedantu’s revision notes that contain all the important formulas and key points. The solution PDFs and other study materials such as important questions and revision notes can also be downloaded from the Vedantu  app as well for free of cost.

5. How do you solve electric charges and fields?

Electric charges and fields are the first chapters of Class 12 Physics. Electric charges and fields are the basics to understand the concepts in further chapters. Solve every problem from your NCERT Physics book. Refer to Vedantu’s Solutions for Class 12 Physics to understand every concept. Every minute detail is included in this material, with an emphasis on strengthening your concept. You can download the Solutions for free from Vedantu’s website. 

6. What is the first chapter of Class 12 Physics?

The first chapter of Class 12 Physics is Electric charges and Fields. It introduces the concepts of charges to the student and how it affects other materials. Students will learn about insulators, conductors, their working procedures, etc. Coulomb’s law and its applications are covered in detail. Vedantu’s NCERT Solution PDF for Class 12 Physics Chapter 1 consists of 34 questions to help students gain command over this topic. It is the best guide to understand the concepts. 

7. What is electric charge and fields?

Electric Charge is the property of a material due to which it experiences a force in an electric, magnetic, or electromagnetic field. There are two different types of charges, positive charge and negative charge. The region around an electrically charged particle up to which it can exert an electrical force on another electrically charged particle is termed an electric field. It can cause either the force of attraction or repulsion, depending on the nature of the charges. 

8. What are some tips to utilise NCERT books effectively for Class 12 exams?

Following are the basic and helpful tips to utilise NCERT books effectively for Class 12 exams:

Make a timetable and stick to it. Your timetable should be unique to you and should be according to your needs and capabilities.

Do not skip topics. Cover the entire syllabus. If you get stuck at any point, you should immediately get your doubts cleared. Refer to Vedantu’s Solutions for Class 12 to clear all your doubts and strengthen your concepts.

Analyse the solutions after you have successfully solved a question. It will help to develop problem-solving capabilities. 

NCERT Solutions Class 12 Physics

CBSE Class 12 Physics Chapter 1 Revision Notes – Electric Charges and Fields

Electric charges and fields  class 12 cbse revision notes.

Electric Charges and Fileds is the first chapter of CBSE Class 12 Physics. Everyone has the experience of watching a spark or hearing a crackle from a sweater during the dry weather. Moreover, this certainly gives rise to an important question that what can be a possible explanation for this particular phenomenon. Further, students must consider an example of an electric discharge that is lightning. In addition, this lightning occurs in the sky at the time of thunderstorms.

Most noteworthy, the reason for these experiences is the discharge of electric charges through the body of the individual. Further, the accumulation of these electric charges had taken place because of the rubbing of insulating surfaces. Moreover, this takes place due to the creation of static electricity. Thus, it happens to be the topic whose discussion will take place in ray optics class 12 notes.

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Sub-topics covered under Electric Charges and Fields

• Conductors and Insulators
• Electric Charge
• Basic Properties of Electric Charge
• Coulomb’s Law
• Electric Field
• Electric Field Lines
• Gauss’s Law
• Applications of Gauss’s Law
• Electric Flux
• Electric Dipole
• Dipole in a Uniform External Field

You can download CBSE Class 12 Physics Chapter 1 Revision Notes by clicking on the download button below

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CBSE Class 12 Physics Revision Notes

• CBSE Class 12 Physics Chapter 14 – Semiconductor Electronic Class 12 Notes
• CBSE Class 12 Physics Chapter 13 – Nuclei Class 12 Notes
• CBSE Class 12 Physics Chapter 12 – Atoms Revision Notes
• CBSE Class 12 Physics Chapter 11 Notes – Dual Nature of Radiation and Matter
• CBSE Class 12 Physics Chapter 10 Notes – Wave Optics
• CBSE Class 12 Physics Chapter 9 Notes – Ray Optics and Optical Instruments
• CBSE Class 12 Physics Chapter 8 Revision Notes – Electromagnetic Waves
• CBSE Class 12 Physics Chapter 7 – Alternating Current Class 12 Notes
• CBSE Class 12 Physics Chapter 6 – Electromagnetic Induction Class 12 Notes
• CBSE Class 12 Physics Chapter 5 Notes – Magnetism and Matter

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1569+ Class 12 Physics Case Study Questions PDF Download

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So, you’re in Class 12 and the concept of Physics case study questions is beginning to loom large. Are you feeling a little lost? Fear not! This article will guide you through everything you need to know to conquer these challenging yet rewarding Class 12 Physics Case Study Questions.

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We have provided here Case Study questions for the Class 12 Physics for board exams. You can read these chapter-wise Case Study questions. These questions are prepared by subject experts and experienced teachers. The answer and solutions are also provided so that you can check the correct answer for each question. Practice these questions to score excellent marks in your final exams.

Physics is more than just formulas and equations, right? It’s a way to interpret the natural world, and case studies provide a perfect opportunity for students to apply theoretical knowledge to real-world situations. So, what’s the importance of case studies in class 12 Physics? Well, they help you to solidify your understanding and enhance your analytical skills, which are invaluable for exams and beyond.

Table of Contents

Case Study-Based Questions for Class 12 Physics

There are total of 14 chapters Electric Charges And Fields, Electrostatic Potential And Capacitance, Current Electricity, Moving Charges And Magnetism, Magnetism And Matter, Electromagnetic Induction, Alternating Current, Electromagnetic Waves, Ray Optics and Optical Instruments, Wave Optics, Dual Nature Of Radiation And Matter, Atoms, Nuclei, Semiconductor Electronics Materials Devices, And Simple Circuits.

• Case Study Based Questions on class 12 Physics Chapter 1 Electric Charges And Fields
• Case Study Based Questions on Class 12 Physics Chapter 2 Electrostatic Potential And Capacitance
• Case Study Based Questions on Class 12 Physics Chapter 3 Current Electricity
• Case Study Based Questions on Class 12 Physics Chapter 4 Moving Charges And Magnetism
• Case Study Based Questions on Class 12 Physics Chapter 5 Magnetism And Matter
• Case Study Based Questions on Class 12 Physics Chapter 6 Electromagnetic Induction
• Case Study Based Questions on Class 12 Physics Chapter 7 Alternating Current
• Case Study Based Questions on Class 12 Physics Chapter 8 Electromagnetic waves
• Case Study Based Questions on Class 12 Physics Chapter 9 Ray Optics and Optical Instruments
• Case Study Based Questions on class 12 Physics Chapter 10 Wave Optics
• Case Study Based Questions on class 12 Physics Chapter 11 Dual Nature of Matter and Radiation
• Case Study Based Questions on class 12 Physics Chapter 12 Atoms
• Case Study Based Questions on class 12 Physics Chapter 13 Nuclei
• Case Study Based Questions on class 12 Physics Chapter 14 Semiconductor Electronics
• Class 12 Chemistry Case Study Questions
• Class 12 Biology Case Study Questions
• Class 12 Maths Case Study Questions

Class 12 students should go through important Case Study problems for Physics before the exams. This will help them to understand the type of Case Study questions that can be asked in Grade 12 Physics examinations. Our expert faculty for standard 12 Physics have designed these questions based on the trend of questions that have been asked in last year’s exams. The solutions have been designed in a manner to help the grade 12 students understand the concepts and also easy-to-learn solutions.

How to Approach Physics Case Study Questions

Ah, the million-dollar question! How do you tackle these daunting case study questions? Let’s dive in.

Effective Reading Techniques

A vital part of cracking case study questions is understanding the problem correctly. Don’t rush through the question, take your time to fully grasp the scenario, and don’t skip over the details – they can be critical!

Conceptual Analysis

Once you’ve understood the problem, it’s time to figure out which physics principles apply. Remember, your foundational knowledge is your greatest weapon here.

Solving Strategies

Next, put pen to paper and start solving! Follow a systematic approach, check your calculations, and make sure your answer makes sense in the context of the problem.

Best Books for Class 12 Physics Exams

Class 12 Physics Syllabus 2024

Chapter–1: Electric Charges and Fields

Electric charges, Conservation of charge, Coulomb’s law-force between two point charges, forces between multiple charges; superposition principle, and continuous charge distribution.

Electric field, electric field due to a point charge, electric field lines, electric dipole, electric field due to a dipole, torque on a dipole in a uniform electric field.

Electric flux, statement of Gauss’s theorem and its applications to find field due to infinitely long straight wire, uniformly charged infinite plane sheet, and uniformly charged thin spherical shell (field inside and outside).

Chapter–2: Electrostatic Potential and Capacitance

Electric potential, potential difference, electric potential due to a point charge, a dipole and system of charges; equipotential surfaces, electrical potential energy of a system of two-point charges and of electric dipole in an electrostatic field.

Conductors and insulators, free charges and bound charges inside a conductor. Dielectrics and electric polarization, capacitors and capacitance, combination of capacitors in series and in parallel, capacitance of a parallel plate capacitor with and without dielectric medium between the plates, energy stored in a capacitor (no derivation, formulae only).

Chapter–3: Current Electricity

Electric current, flow of electric charges in a metallic conductor, drift velocity, mobility and their relation with electric current; Ohm’s law, V-I characteristics (linear and non-linear), electrical energy and power, electrical resistivity and conductivity, temperature dependence of resistance, Internal resistance of a cell, potential difference and emf of a cell, combination of cells in series and in parallel, Kirchhoff’s rules, Wheatstone bridge.

Chapter–4: Moving Charges and Magnetism

Concept of magnetic field, Oersted’s experiment.

Biot – Savart law and its application to current carrying circular loop.

Ampere’s law and its applications to infinitely long straight wire. Straight solenoid (only qualitative treatment), force on a moving charge in uniform magnetic and electric fields.

Force on a current-carrying conductor in a uniform magnetic field, force between two parallel current-carrying conductors-definition of ampere, torque experienced by a current loop in uniform magnetic field; Current loop as a magnetic dipole and its magnetic dipole moment, moving coil galvanometerits current sensitivity and conversion to ammeter and voltmeter.

Chapter–5: Magnetism and Matter

Bar magnet, bar magnet as an equivalent solenoid (qualitative treatment only), magnetic field intensity due to a magnetic dipole (bar magnet) along its axis and perpendicular to its axis (qualitative treatment only), torque on a magnetic dipole (bar magnet) in a uniform magnetic field (qualitative treatment only), magnetic field lines.

Magnetic properties of materials- Para-, dia- and ferro – magnetic substances with examples, Magnetization of materials, effect of temperature on magnetic properties.

Chapter–6: Electromagnetic Induction

Electromagnetic induction; Faraday’s laws, induced EMF and current; Lenz’s Law, Self and mutual induction.

Chapter–7: Alternating Current

Alternating currents, peak and RMS value of alternating current/voltage; reactance and impedance; LCR series circuit (phasors only), resonance, power in AC circuits, power factor, wattless current. AC generator, Transformer.

Chapter–8: Electromagnetic Waves

Basic idea of displacement current, Electromagnetic waves, their characteristics, their transverse nature (qualitative idea only). Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays) including elementary facts about their uses.

Chapter–9: Ray Optics and Optical Instruments

Ray Optics:  Reflection of light, spherical mirrors, mirror formula, refraction of light, total internal reflection and optical fibers, refraction at spherical surfaces, lenses, thin lens formula, lens maker’s formula, magnification, power of a lens, combination of thin lenses in contact, refraction of light through a prism.

Optical instruments:  Microscopes and astronomical telescopes (reflecting and refracting) and their magnifying powers.

Chapter–10: Wave Optics

Wave optics: Wave front and Huygen’s principle, reflection and refraction of plane wave at a plane surface using wave fronts. Proof of laws of reflection and refraction using Huygen’s principle. Interference, Young’s double slit experiment and expression for fringe width (No derivation final expression only), coherent sources and sustained interference of light, diffraction due to a single slit, width of central maxima (qualitative treatment only).

Chapter–11: Dual Nature of Radiation and Matter

Dual nature of radiation, Photoelectric effect, Hertz and Lenard’s observations; Einstein’s photoelectric equation-particle nature of light.

Experimental study of photoelectric effect

Matter waves-wave nature of particles, de-Broglie relation.

Chapter–12: Atoms

Alpha-particle scattering experiment; Rutherford’s model of atom; Bohr model of hydrogen atom, Expression for radius of nth possible orbit, velocity and energy of electron in his orbit, of hydrogen line spectra (qualitative treatment only).

Chapter–13: Nuclei

Composition and size of nucleus, nuclear force

Mass-energy relation, mass defect; binding energy per nucleon and its variation with mass number; nuclear fission, nuclear fusion.

Chapter–14: Semiconductor Electronics: Materials, Devices and Simple Circuits

Energy bands in conductors, semiconductors and insulators (qualitative ideas only) Intrinsic and extrinsic semiconductors- p and n type, p-n junction

Semiconductor diode – I-V characteristics in forward and reverse bias, application of junction diode -diode as a rectifier.

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Physics Case Study for Class 12

While preparing for the board exams, students are being judged on different levels of skills, such as writing, reading, etc. Physics Case Study for Class 12 questions are one of them that helps in assessing critical thinking.

The Central Board of Secondary Education will be asking the case study questions in the Class 12 board examination. Therefore, here on this page, we have provided the Physics Case Study for Class 12 at free of cost. Our subject matter experts have prepared Case Study questions so that Apart from the basic standard questions, students can have a variety of problems to solve. 

Just like MCQs, and other written types questions Physics Case Study for Class 12 questions will impact the overall performance of a student. Therefore for the convenience of the students we have provided the download links here, so that they can easily access the Class 12 Case Study.

Download Chapter Wise Physics Case Study for Class 12 Question and Answers PDF 

There are lots of chapters in class 12 Physics from which Class 12 Case Study Questions can be framed. Going through such types of questions help the students to assess their understanding level in the topics discussed in NCERT Class 12 Physics Books. By practicing the Class 12 Case Study Questions for Physics students will be very confident to ace the board exam. Also the Physics case study will be very useful for the NEET exam preparation.

Doing a regular practice of Class 12 Physics Case Study questions is a great way to score higher marks in the board exams as it will help students to develop a grip on the concepts. 

Here our subject experts have crafted the Chapter Wise Case Study For Class 12. Download Chapter Wise CBSE Case Study Class 12 Question and Answers PDF from the below given links.

Case Study Questions Class 12 Chapter 1 Electric Charges and Fields

Case Study Questions Class 12 Chapter 2 Electrostatic Potential And Capacitance

Case Study Questions Class 12 Chapter 3 Current Electricity

Case Study Questions Class 12 Chapter 4 Moving Charges And Magnetism

Case Study Questions Class 12 Chapter 5 Magnetism And Matter

Case Study Questions Class 12 Chapter 6 Electromagnetic Induction

Case Study Questions Class 12 Chapter 7 Alternating Current

Case Study Questions Class 12 Chapter 8 Electromagnetic Waves

Case Study Questions Class 12 Chapter 9 Ray Optics & Optical Instruments

Case Study Questions Class 12 Chapter 10 Wave Optics

Case Study Questions Class 12 Chapter 11 Dual Nature Radiation & Matter

Case Study Questions Class 12 Chapter 12 Atoms

Case Study Questions Class 12 Chapter 13 Nuclei

Case Study Questions Class 12 Chapter 14 Semiconductor Electronics - Materials, Devices & Simple Circuits

Case study types of questions are generally descriptive that helps to gather more information easily so, it is kinda easy to answer. However, our subject matter experts have given the solutions of all the Physics Case Study for Class 12 Physics questions.

Passage Based Class 12 Physics Case Study Questions in PDF

CBSE Class 12 Case studies are known as Passage Based Questions. These types of problems usually contain a short/long paragraph with 4 to 5 questions. 

Students can easily solve Passage Based Class 12 Case Study Questions by reading those passages. By reading the passage students will get the exact idea of what should be the answers. Because the passage already contains some vital information or data. However a better understanding of the basic concepts that can be learned from the NCERT Class 12 Textbooks will aid in solving the Case based questions or passage based questions.

How to Download CBSE Case Study of Class 12 Physics ?

Follow the below given simple steps to know how to download CBSE Case Study of Class 12 Physics:-

• Open Selfstudys website in your browser
• Go to the navigation menu that look like this
• Now, click on CBSE and then Case Study respectively
• A new page will open, where you have to click on “Class 12”
• Now, you are ready to select the subject for which you want to download the case study questions.

How to Solve Physics Case Study Based Questions of Class 12?

There are very simple methods that a student should keep in mind while solving Physics Case Study for Class 12:

• Read each line of paragraph carefully and pay attention to the given data/numbers. Often questions are framed according to the highlighted data of the passage.
• Since case study questions are often framed in Multiple choice questions, students should have the knowledge of elimination methods in MCQs.
• Having a good understanding of the topics that are discussed in CBSE Class 12 Books are ideal to Solve Case Study Based Questions of Class 12.

Features Of Class 12 Physics Case Study Questions And Answers PDF

The three most noticeable features of Class 12 Physics Case Study Questions And Answers Pdf are -

• It Is Free To Use:  Keeping in mind the need of students and to help them in doing Self Study, our team has made all the PDF of Physics Case Study for Class 12 free of cost.
• Answers Are Given:  Not only the PDFs are free provided but answers are given for all the questions of Physics Case Study for Class 12. 
• PDF Can Be Downloaded Or Viewed Online:  Many students don’t like to download the PDFs on their device due to the shortage of storage. Therefore, CBSE Class 12 Case Study questions are made available here in online format, so that students can view them online. However, through the Selfstudys app, a learner can download the PDFs too.

Benefits of Using CBSE Class 12 Physics Case Study Questions and Answers

The CBSE Class 12 Physics Case Study Questions and Answers can help a student in several ways:

• In Exam Preparation:  Those who go through the Physics Case Study for Class 12 will find support during the exam preparation as case based questions are also asked in the CBSE Class 12 Board examination. 
• Help in brushing up the previous learnings:  No matter how brilliant you are, you have to revise the studied topics time and again to keep them refreshed. And in this task, the CBSE Class 12 Case Study Questions and Answers can help a lot.
• To develop the critical thinking:  Able to analyse information and make an objective judgement is a skill that is known as critical thinking. A student of class 12 can use Physics Case Study for Class 12 with answers to develop critical thinking so that they can make better decisions in their life and in the board examination.

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Class 12 Physics Case Study Questions

Please refer to the chapter-wise Class 12 Physics Case Study Questions given below. These questions are expected to come in your class 12 Physics board exams. Our expert teachers have developed these Case Study questions and answers based on the latest examination guidelines and updated syllabus issued by CBSE, NCERT, and KVS. These will help you to understand all topics too and bet way to write answers in your exams. Revise these solved problems prior to your exams to score better marks in Class 12 exams.

Case Study Questions Class 12 Physics

Please click on the links below to access free important Case Study questions and answers for Grade 12 Physics. All solved questions have been designed based on the latest pattern of questions asked in recent examinations.

Chapter-wise Solved Case Study Questions for Class 12 Physics

Class 12 students should go through important Case Study problems for Physics before the exams. This will help them to understand the type of Case Study questions that can be asked in Grade 12 Physics examinations. Our expert faculty for standard 12 Physics have designed these questions based on the trend of questions that have been asked in last year’s exams. The solutions have been designed in a manner to help the grade 12 students understand the concepts and also easy to learn solutions.

You can click on the chapter-wise Case Study links given above and read the Class 12th Physics Case Study questions and answers provided by us. If you face any issues or have any questions please put your questions in the comments section below. Our teachers will be happy to provide you answers.

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Case Study Chapter 6 Electromagnetic Induction Class 12 Physics

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• Chapter 1: Electric Charges And Fields

CBSE Class 12 Physics Notes Chapter 1 Electric Charges and Fields

Have you ever heard a crackle or seen a spark while taking off your synthetic clothes or sweater, especially in dry weather? How about some explanation about this phenomenon? Another example of electric discharge is the lightning that we see in the sky during thunderstorms. All these experiences are the result of the discharge of electric charges through your body that accumulated due to the rubbing of insulating surfaces. It is also caused due to the generation of static electricity. CBSE Class 12 Physics Notes Chapter 1 Electric Charges and Fields discusses these facts in detail. Electrostatics deals with the study of forces, fields and potentials arising from static charges.

For more information on Charging by Induction, watch the below video

Electric Charge

The term ‘electricity’ is derived from Elektron, a Greek word meaning amber. The properties of matter, atoms and molecules are determined by the magnetic and electric forces present in them. There are also only 2 kinds of an entity called the electric charge .

An experiment conducted also suggested that there are two kinds of electrification wherein (i) like charges repel and (ii) unlike charges attract each other. The property that differentiates these 2 kinds of charges is called the polarity of charge.

Conductors and Insulators

When an experiment was conducted on electric charges due to frictional electricity, it was found that conductors assist in the movement of electric charge, but insulators do not behave in the same manner. Metal, Earth, and human bodies are all examples of conductors, while porcelain, nylon, and wood all offer high resistance to the passage of electricity through them as they are insulators.

What Are the Properties of Electric Charge?

An electric charge has three fundamental properties:

• Quantization- This property states that the total charge of a body represents the integral multiple of a basic quantum of charge.
• Additive- This property of electric charges represents the total charge of a body as the algebraic sum of all the singular charges acting on the system.
• Conservation- This property states that the total charge of a system remains unaffected by time. In other words, when objects get charged due to friction, a transfer of charge from one object to another occurs. Charges can neither be created nor destroyed.

Coulomb’s Law

Coulomb’s law states that the mutual electrostatic force existing between two point charges, A and B, is proportional to their product which is AB and inversely proportional to the square of the distance between them (

Mathematically,

This law consists of constant terms, which are also called a constant of proportionality and are represented by ‘k’, and its values are

For more information on Coulomb’s Law, watch the below video

Forces between Multiple Charges

Even if the mutual electric force between two charges is given by Columb’s law , it does not help to calculate the force on a charge where there is not one but several charges around.  It has been proved via an experiment that force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time.

Superposition Principles

According to the superposition principle, the property of two charges to repel and attract each other remains unaffected even though there is the presence of a third additional charge.

Consider q 1,  q 2 and  q 3 as three charges of a system. Here, if the force on q 1 due to q 2  is denoted by F 12.  

Then, \[F_{12}=\frac{1}{4\pi \varepsilon}\frac{q_{1}q_{2}}{r^{2}_{12}}r_{12}\]

Likewise, the force on q 1  due to q 3,   denoted by F 13  is \[F_{13}=\frac{1}{4\pi \varepsilon}\frac{q_{1}q_{3}}{r^{2}_{13}}r_{13}\]

Thus, the total force F 1  on q 1  due to 2 charges q 2 and  q 3  is  \[F_{1}=F_{12}+F_{13}\]

\[F_{1}=\frac{1}{4\pi \varepsilon}\frac{q_{1}q_{2}}{r^{2}_{12}}r_{12}+\frac{1}{4\pi \varepsilon}\frac{q_{1}q_{3}}{r^{2}_{13}}r_{13}\]

Properties of Electric Field Lines

Some of the general properties of field lines are:

• Field lines show a continuous curve without having any breakage in a charge-free region.
• Two-line never cross each other.
• These electric field lines start on the positive charge and end on the negative charge.
• Electrostatic field lines do not form any closed loops.

Electric Flux

The total number of electric field lines passing a given area in a unit of time is defined as the electric flux. However, we note that there is no flow of a physically observable quantity like in the case of liquid flow. Coming to the definition,  Electric flux  Δθ through an area element ΔS is defined by

Δθ= E.ΔS= E ΔS  cosθ

This is proportional to the number of field lines cutting the area element. The angle θ here is the angle between E and ∆S. In a closed surface, where the convention is already stated, θ is the angle between E and the outward normal to the area element. To calculate the total flux through any given surface, divide the surface into small area elements, calculate the flux at each element and add them up. Thus, the total flux θ through a surface S is θ ~ Σ E. ∆S. The approximation symbol is used because the electric field E is taken to be constant over the small area element.

For more information on Electric Flux and Electric Force, watch the below videos

Electric Dipole

It is a pair of equal or opposite charges, A and -B, which are separated by distance 2x. The dipole moment vector (let’s assume it is p ) has a magnitude 2Ax and is in the direction of the dipole axis from -B to A

For more information on Flux through a Sphere, watch the below video

To know more about Electric Charge and Electric Field, keep visiting BYJU’S website.

Important Question

• (a) The Statement needs an explanation ‘electric charge of a body is quantized’.

(b) Why can one ignore the quantization of electric charge when dealing with macroscopic, i.e., large-scale charges?

2. Explain how the law of conservation of charge works on this phenomenon when charges appear on both the silk cloth and the glass rod when they are rubbed together.?

(a)Why electrostatic field line is not a continuous curve and cannot have sudden breaks? (b) Why do two field lines never cross each other at any point?

Related Topics:

NCERT Solutions for Electric Charges and Fields

NCERT Exemplars for Electric Charges and Fields

Electric Charge and Field Important Questions

CBSE Notes Class 12 Physics

Frequently Asked Questions on CBSE Class 12 Physics Notes Chapter 1 Electric Charges and Fields

What is meant by induction.

Induction is the property of a conductor to become electrified when there is the presence of a charged body present nearby.

What does Coulomb’s Law state?

Coulomb’s Law: The force of attraction or repulsion acting along a straight line between two electric charges is directly proportional to the product of the charges and inversely to the square of the distance between them.

What is electric flux?

The measure of a count of the number of electric field lines crossing an area is known as electric flux.

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NCERT Solutions For Class 12 Physics Chapter 1 Electric Charges and Fields

NCERT Solutions for class 12 Physics Chapter 1 Electric Charges and Fields is prepared by our senior and renowned teachers of Physics Wallah. Read now.

February 9, 2024

Table of Contents

NCERT Solutions for class-12 Physics  Chapter 1 Electric Charges and Fields is prepared by our senior and renowned teachers of Physics Wallah primary focus while solving these questions of class-12 in NCERT textbook, also do read theory of this Chapter 1 Electric Charges and Fields while going before solving the NCERT questions. You can download and share  NCERT Solutions  of Class 12 Physics from Physics Wallah.

NCERT Solutions Class 12 Science Chapter 1 Overview

Developing a solid understanding of each chapter is crucial for students. Chapter 1 of NCERT Solutions for Class 12 Science covers significant topics. To fully comprehend the concepts presented in this chapter and effectively utilize the provided solutions, it’s recommended that students meticulously study each topic.

Physics Wallah’s instructors have specifically created these solutions to aid in understanding the concepts within this chapter. The goal is to equip students to confidently tackle tests after reviewing and practicing these solutions.

UP Board 12 result 2024

NCERT Solutions For Class 12 Physics Chapter 1

Answer The Following Question Answers of class 12 physics Chapter 1 – Electric Charges and Fields:

Question 1. What is the force between two small charged spheres having charges of 2 × 10−7 C and 3 × 10−7 C placed 30 cm apart in air?

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Question 2. The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge − 0.8 μC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

Solution : (a) Electrostatic force on the first sphere, F = 0.2 N

Charge on this sphere, q1 = 0.4 μC = 0.4 × 10−6 C

Charge on the second sphere, q2 = − 0.8 μC = − 0.8 × 10−6 C

Electrostatic force between the spheres is given by the relation,

Where, ∈0 = Permittivity of free space

The distance between the two spheres is 0.12 m.

(b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N.

Question 3. Check that the ratio ke2/G memp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

G = Gravitational constant

Its unit is N m2 kg−2.

me and mp = Masses of electron and proton.

Their unit is kg.

e = Electric charge.

Its unit is C.

∈0 = Permittivity of free space

Its unit is N m2 C−2.

Hence, the given ratio is dimensionless.

e = 1.6 × 10−19 C

G = 6.67 × 10−11 N m2 kg-2

me= 9.1 × 10−31 kg

mp = 1.66 × 10−27 kg

Hence, the numerical value of the given ratio is

This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.

Question 4. (a) Explain the meaning of the statement ‘electric charge of a body is quantised’.

(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

Solution : (a) Electric charge of a body is quantized. This means that only integral (1, 2, …., n) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge.

(b) In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous.

Question 5. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Solution : Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are created in pairs. This phenomenon of charging is called charging by friction. The net charge on the system of two rubbed bodies is zero. This is because equal amount of opposite charges annihilate each other. When a glass rod is rubbed with a silk cloth, opposite natured charges appear on both the bodies. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies.

Question 6. Four point charges qA = 2 μC, qB = −5 μC, qC = 2 μC, and qD = −5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?

Solution : The given figure shows a square of side 10 cm with four charges placed at its corners. O is the centre of the square.

(Sides) AB = BC = CD = AD = 10 cm

A charge of amount 1μC is placed at point O.

Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on 1 μC charge at centre O is zero.

Question 7. (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?

(b) Explain why two field lines never cross each other at any point?

Solution : (a) An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.

(b) If two field lines cross each other at a point, then electric field intensity will show two directions at that point. This is not possible. Hence, two field lines never cross each other.

Question 8. Two point charges qA = 3 μC and qB = −3 μC are located 20 cm apart in vacuum.

(a) What is the electric field at the midpoint O of the line AB joining the two charges?

(b) If a negative test charge of magnitude 1.5 × 10−9 C is placed at this point, what is the force experienced by the test charge?

Solution : (a) The situation is represented in the given figure. O is the mid-point of line AB.

Distance between the two charges, AB = 20 cm

∴AO = OB = 10 cm

Net electric field at point O = E

Electric field at point O caused by +3μC charge,

Magnitude of electric field at point O caused by −3μC charge,

= 5.4 × 106 N/C along OB

Therefore, the electric field at mid-point O is 5.4 × 106 N C−1 along OB.

(b) A test charge of amount 1.5 × 10−9 C is placed at mid-point O.

q = 1.5 × 10−9 C

Force experienced by the test charge = F

= 1.5 × 10−9 × 5.4 × 106

= 8.1 × 10−3 N

The force is directed along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A.

Therefore, the force experienced by the test charge is 8.1 × 10−3 N along OA.

Question 9. A system has two charges qA = 2.5 × 10−7 C and qB = −2.5 × 10−7 C located at points A: (0, 0, − 15 cm) and B: (0, 0, + 15 cm), respectively. What are the total charge and electric dipole moment of the system?

Solution : Both the charges can be located in a coordinate frame of reference as shown in the given figure.

At A, amount of charge, qA = 2.5 × 10−7C

At B, amount of charge, qB = −2.5 × 10−7 C

Total charge of the system,

q = qA + qB

= 2.5 × 10−7 C − 2.5 × 10−7 C

Distance between two charges at points A and B,

d = 15 + 15 = 30 cm = 0.3 m

Electric dipole moment of the system is given by,

p = qA × d = qB × d

= 2.5 × 10−7 × 0.3

= 7.5 × 10−8 C m along positive z-axis

Therefore, the electric dipole moment of the system is 7.5 × 10−8 C m along positive z−axis.

Question 10. An electric dipole with dipole moment 4 × 10−9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 104 N C−1. Calculate the magnitude of the torque acting on the dipole.

Solution : Electric dipole moment, p = 4 × 10−9 C m

Angle made by p with a uniform electric field, θ = 30°

Electric field, E = 5 × 104 N C−1

Torque acting on the dipole is given by the relation,

τ = pE sinθ

Therefore, the magnitude of the torque acting on the dipole is 10−4 N m.

Question 11. A polythene piece rubbed with wool is found to have a negative charge of 3 × 10−7 C.

(a) Estimate the number of electrons transferred (from which to which?)

(b) Is there a transfer of mass from wool to polythene?

Solution : (a) When polythene is rubbed against wool, a number of electrons get transferred from wool to polythene. Hence, wool becomes positively charged and polythene becomes negatively charged.

Amount of charge on the polythene piece, q = −3 × 10−7 C

Amount of charge on an electron, e = −1.6 × 10−19 C

Number of electrons transferred from wool to polythene = n

n can be calculated using the relation,

= 1.87 × 1012

Therefore, the number of electrons transferred from wool to polythene is 1.87 × 1012.

There is a transfer of mass taking place. This is because an electron has mass,

me = 9.1 × 10−3 kg

Total mass transferred to polythene from wool,

= 9.1 × 10−31 × 1.85 × 1012

= 1.706 × 10−18 kg

Hence, a negligible amount of mass is transferred from wool to polythene.

Question 12. (a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10−7 C? The radii of A and B are negligible compared to the distance of separation.

(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Solution : (a) Charge on sphere A, qA = Charge on sphere B, qB = 6.5 × 10−7 C

Distance between the spheres, r = 50 cm = 0.5 m

Force of repulsion between the two spheres,

∈0 = Free space permittivity

= 1.52 × 10−2 N

Therefore, the force between the two spheres is 1.52 × 10−2 N.

(b) After doubling the charge, charge on sphere A, qA = Charge on sphere B, qB = 2 × 6.5 × 10−7 C = 1.3 × 10−6 C

The distance between the spheres is halved.

= 16 × 1.52 × 10−2

Therefore, the force between the two spheres is 0.243 N.

Question 13. Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

Note: If a conducting uncharged material is brought in contact with a charged surface then the charges are shared uniformly between the two bodies.

Given: Charge on sphere 1, q 1  = 6.5× 10 -7  C

Charge on sphere 2, q 2  = 6.5× 10 -7  C

Charge on sphere 3, q 3  = 0

Step 1: The uncharged sphere is brought in contact with sphere 1. Since sphere 1 has charge ‘q’, it gets distributed among sphere 1 and sphere 3.

Now, charge on sphere 1 = q/2

Charge on sphere 2 = q/2

At this point the sphere 3 which was initially uncharged has a charge “q/2”.

Step 2: Now sphere 3 is brought in contact with sphere 2 due to which 1/4 × q will flow from sphere 2 to sphere 3. Now sphere 2 and sphere 3 have “3/4 × q” charge.

Now, q 1  = ½ × 6.5 × 10 -7  C

⇒ 3.25× 10 -7  C

q 2  = ¾ × 6.5× 10 -7  C

⇒ 4.87 × 10 -7  C

q 3  = ¾ × 6.5× 10 -7  C

Question 14. Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Opposite charges attract each other and same charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged.

The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.

Question 15. Consider a uniform electric field E = 3 × 103 îN/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?

Solution : Given:

Electric field E = 3× 10 3   N/C

Side of square, s = 10 cm

a)  Flux of field through square whose plane is parallel to yz plane.

We understand that the normal to the plane is parallel to the direction of field.

Ø = E × A× cos(θ)            …(1)

Where, E = Electric field

A = Area through which we have to calculate flux

θ = Angle between normal to surface and the Electric field

A = .01 m 2

Plugging values, of E, A and θ in equation (1)

Ø = 3× 10 3  NC -1 × 0.01m 2 × cos0°

Ø = 30 Nm 2 C -1

b)  If normal to its (square’s) plane makes 60° with the X axis.

Ø = E × A× cos(θ)

Ø = 3× 10 3  NC -1 × 0.01m 2  × cos60°

Ø = 15 Nm 2 C -1

Question 16. What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

Solution :  All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero.

Question 17. Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 103 N m2/C. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?

a)           Ø = 8.0× 10 3  Nm 2 C -1

Let net charge inside the box = q

We know that,

Flux, Ø = q/ε 0                 ..(1)

Where, q = net charged enclosed

ε 0  = permittivity of free space

ε 0  = 8.85× 10 -12  N -1  m -2 C 2

Plugging values of Ø and ε 0  in equation (1) we get,

q = Ø × ε 0

⇒ q = 8.0 × 10 3  Nm 2 C -1 × 8.85× 10 -12 N -1 C 2 m -2

⇒ q = 7.08 × 10 -8  C

Hence, the net charge inside the box is 0.07 µC.

(b) No, we cannot conclude that the body doesn’t have any charge. The flux is due to the Net charge of the body. There may still be equal amount of positive and negative charges. So, it is not necessary that if flux is zero then there will be no charges.

Question 18. A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)

Assume the charge to be enclosed by a cube, where the square is one of its sides.

Now, let us find the total flux through the imaginary cube.

Flux, Ø = q/ε            …(1)

Now plugging the values of q and ε 0  in equation (2)

⇒ Ø = 11.28 × 10 5  Nm -2 C -1

We understand that flux through all the faces of cube will be equal;

Let flux through the square = Ø a

Explanation: The net flux will be distributed equally among all 6 faces of the cube. Hence, the square will have one sixth of the total flux.

Ø a  = 1.88 Nm -2 C -1

Question 19. A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

Total charge inside the cube, q = 2.0 µC

Edge length of Cube, a = 9.0 cm

Question 20. A point charge causes an electric flux of −1.0 × 103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centered on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?

Ø = -1.0× 10 3  Nm 2 C -1

r­ 1  = 10.0 cm.

a)  Flux if the radius of the Gaussian surface is doubled.

If the radius is doubled then the flux would remain same i.e. -1.0× 10 3  Nm 2 C -1 .

The geometry of the Gaussian surface doesn’t affect the total flux through it. The net charge enclosed by Gaussian surface determines the net flux.

b)  Let value of point charge enclosed by Gaussian surface = q

Flux, Ø = q/ε 0            …(1)

Now plugging, the values Ø and ε 0  in equation (1).

⇒ q = -1.0× 10 3 Nm 2 C -1  × 8.85× 10 -12 N -1 m -2 C 2

⇒ q = -8.85 × 10 -9  C

The charge enclosed by the surface is -8.85 × 10 -9  C.

Question 21. A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C and points radially inward, what is the net charge on the sphere?

Radius of charged sphere, r = 10 cm

Electric field, 20 cm away from centre of sphere, E = 1.5× 10 3  NC -1

We know that, electric field intensity at a point P, located at a distance R, due to net charge q is given by,

Now plugging the values of q and R in equation (1)

q = 4× π× ε 0 × R 2 × E

⇒ q = 4× 3.14× 8.85× 10 -12 N -1 m -2 C -2 × (0.2m) 2  × 1.5× 10 3 C

⇒ q = 6.67× 10 -9 C

The net charge on the sphere is -6.67× 10 -9 C. Since the electric field points radially inwards, we can infer that charges on sphere are negative.

Question 22. A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?

Solution : Given: radius of sphere, r = 1.2 m

Surface charge density, σ = 80.0 μC/m 2

a)  Let charge on sphere = q

We understand that,

Total charge, Q = Surface charge density× surface area.

Surface area of sphere, S = 4πr 2

S = 18.08 m 2

Q = 80× 10 -6 Cm -2 × 18.08m -2

⇒ Q = 1.447 × 10 -3 C

b)  Let total electric flux leaving the surface of sphere =

Where, Q = net charged.

By, plugging the values of q and ε 0  in equation (1), we get,(b)

Question 23. An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density.

E = 9× 10 4  NC -1

Let the linear charge density = λ Coulomb/metre

Where, λ = linear charge density.

d = distance

From equation (1) we have,

⇒ λ = E× 2× π× ε 0 × d

⇒ λ = 9× 10 4  NC -1 × 2× π×8.85× 10 -12  N -1  m -2 C 2 × 0.02m

⇒ λ = 10 µ Cm -1

The linear charge density is 10 µ Cm -1 .

Question 24. Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10−22 C/m2. What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?

Surface charge density on plate A, σ A  = -17.0 × 10 -22  Cm -2

Surface charge density on plate B, σ B  = 17.0 × 10 -22  Cm -2

The arrangement of plates are as shonw:

Let electric field in region 1 = E 1

Region 2 = E 2

Region 3 = E 3

The electric field in region I and region III is zero because no charge is present in these regions.

E 3  = σ/ε 0    …(1)

Where, σ Surface charge density

Now, plugging the Values in equation (1).

⇒ E 3  = 1.92 × 10 -10  NC -1

The electric field in the region enclosed by the plates is found to be 1.92 × 10 -10  NC -1 .

The electric field in region III is 1.92 × 10 -10  NC -1 .

ADDITIONAL EXERCISES

Question 25. An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 N C−1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm−3. Estimate the radius of the drop. (g = 9.81 m s−2; e = 1.60 × 10−19 C).

Solution : If the oil drop is stationary, the net force on it must be Zero or the resultant of all the forces on oil drop is zero. There are two forces acting on the oil drop, its weight or force due to Earth’s gravity which is pulling it vertically downwards, and Electrostatic force which is acting in vertically upward direction.Both forces should be equal in magnitude and opposite direction so that they cancel each other.

The arrangement is shown in the figure

Note: Electric Field is in vertically downward direction, because force on an negatively charged body in opposite direction of the field, so force on the drop is in vertically upward direction, which balances weight of the drop acting in vertically downward direction.

There are nine excess electrons which make the drop negatively charged because the electron is negatively charged, the net magnitude charge on a body is given by

n is the excess of electrons or protons on the body,

the charge of an electron is denoted by e

e = 1.60 × 10 –19  C

here since there are twelve excess electrons so

i.e. q = 12 × 1.60 × 10 –19  C

= 1.92 × 10 -18  C

Now the magnitude electrostatic force on a charged particle held in an electric field is given by

where, F force is acting on a particle having charge q held in an electric field of magnitude E

here, charge on the oil drop is

q = 1.92 × 10 -18  C

magnitude of Electric field is

E = 2.55 × 10 4  NC –1

So Electrostatic force on the oil drop is

F = 1.92 × 10 -18  C × 2.55 × 10 4  NC –1

= 4.896 ×10 -14  N

This force is acting in vertically upward direction, so Weight should have same magnitude of Force and is acting in vertically downward direction.

Let us assume oil drop to be Spherical in shape, so the volume of drop will be

Both the forces should be equal in magnitude so equating them

i.e. putting F = W

4.896 ×10 -14  N = (4/3 πr 3 ) ×1.26 × 10 3  Kgm -3  × 9.81ms -2

Question 26. Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?

(a) The field lines showed in (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor.

(b) The field lines showed in (b) do not represent electrostatic field lines because the field lines cannot emerge from a negative charge and cannot terminate at a positive charge.

(c) The field lines showed in (c) represent electrostatic field lines. This is because the field lines emerge from the positive charges and repel each other.

(d) The field lines showed in (d) do not represent electrostatic field lines because the field lines should not intersect each other.

(e) The field lines showed in (e) do not represent electrostatic field lines because closed loops are not formed in the area between the field lines.

Question 27. In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 105 NC−1 per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10−7 Cm in the negative z-direction?

Solution : A dipole is a system consisting of two charges equal in magnitude and opposite in nature i.e. a positive charge + q and a negative charge –q separated by some distance d

Electric dipole moment of a dipole is given by,

Where, ‘q’ is magnitude of either of charge (in column)

And ‘d’ is separation between the pair of charges (in metres)

Dipole moment is a vector quantity and its direction is from negative charge to positive charge

We are given dipole moment of the system,

P = 10 -7  Cm , along negative Z axis

Here Electric Field is varying at the rate of 10 5  NC –1  per metre in positive Z direction

i.e. dE/dl = 10 5  NC –1 m -1  along Z direction

so the electric field at any point on z axis at distance of a metres

Force on a charged particle in an electric field is given by

Where q is the magnitude of charge and E is the magnitude of Electric Field and the force is same as the direction of electric field in case of a positively charged particle and opposite to the direction of field in case of negatively charged particle

so Force on the + q charge located at a distance l from origin will be

So whatever be the magnitude of  P  and  E  putting  in the equation  = PESin

We get the torque  τ  = 0 Nm

So the torque of the system is zero

Question 28. (a) A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor. (b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [Fig. 1.36(b)]. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

In Electrostatic we deal with charges at rest so there is no current inside or on the surface of the conductor i.e. in the static situation, the electric field is zero everywhere inside the conductor. This is the basic property of a conductor. A conductor has free electrons. As long as electric field is not zero, the free charge carriers would experience force and drift and re distribute themselves in such a way so that net electric field inside a conductor is zero.

We will use the above stated property to solve the problems

(a)Here in this situation a conductor with a cavity having no charge inside is given a charge Q on outer surface, we have to show that all charge resides on the external surface only and there is no charge on the inner surface.

From Gauss Theorem we know, net electric flux through an Gaussian surface enclosing a charge is equal to net charge enclosed by surface divided by permittivity of free space

we know that net electric field everywhere inside a conductor is zero

so the electric flux through this Gaussian surface would also be zero, this means no charge is enclosed by the Gaussian surface, since cavity does not contain any charge so no charge should be induced on the cavity’s metal surface as well so whole of the charge Q is distributed on the outer surface only.

(b) Again using Gauss Theorem we know,

If electric field at all the points on a Gaussian surface is zero then the net charge enclosed by the Gaussian surface is also zero, or there is no charge inside the Gaussian surface

Now if we consider a Gaussian surface enclosing the cavity inside the conductor  as shown in the figure

so the electric flux through this Gaussian surface would also be zero, this means no charge is enclosed by the Gaussian surface, but the cavity has a charge q , this means a charge –q must be induced on the inner metal surface because net charge enclosed by Gaussian surface must be zero (q + (-q) = 0) , but the metal surface has only been given charge Q so net charge of system must only be q + Q , i.e. a charge q is induced on outer surface of the conductor as which makes the total charge on the conductor equal to Q only Q + (-q) + q = Q

and total charge of the system as

Q + (-q) + q + q = Q + q

This arrangement is also in accordance with the law of conservation of charge

note: Only Q charge has been given on the conductor’s outer surface , the charge appearing (Q + q) on it is only due to induction due to charge inside the cavity.

(c) To shield the sensitive instrument from strong electrostatic fields in its environment. We need to enclose it inside a metal piece’s Cavity, this process is also known as electrostatic shielding.

we consider a metal Block with a cavity having no charge inside and Instrument inside it  as shown in the figure

any cavity inside a metal surface the electric field is zero no matter in which environment it is placed. So in order to save the Instrument from External Electrostatic fiels we need to keep it inside a metal Cavity.

Question 30. Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

Solution : Let AB be a long thin wire of uniform linear charge density λ.Let us consider the electric field intensity due to AB at point P at a distance h from it as shown in the figure.

Question 31. It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge (+2/3) e, and the ‘down’ quark (denoted by d) of charge (−1/3) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.

Question 32. (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.

(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

Solution : (a) Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and displaced from its position in any direction, then it experiences a restoring force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards towards the null point. There is a net inward flux of electric field through a closed surface around the null point. According to Gauss’s law, the flux of electric field through a surface, which is not enclosing any charge, is zero. Hence, the equilibrium of the test charge can be stable.

(b) Two charges of same magnitude and same sign are placed at a certain distance. The mid-point of the joining line of the charges is the null point. When a test charged is displaced along the line, it experiences a restoring force. If it is displaced normal to the joining line, then the net force takes it away from the null point. Hence, the charge is unstable because stability of equilibrium requires restoring force in all directions.

Question 33. A particle of mass m and charge (−q) enters the region between the two charged plates initially moving along x-axis with speed vx (like particle 1 in Fig. 1.33). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL2/ (2mV x 2 ).

Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.

Solution : Given,

Mass of the particle = m

Charge of the particle = -q

Velocity of the particle = v x

Length of the plate = L

Electric field between the plates = E

From Newton’s second law of motion,

This is the vertical displacement of the particle at the far edge of the plate.

This motion is very similar to the motion of a projectile in a gravitational field. In a gravitational field, the force acting on the particle is mg and in the given case it is qE. The trajectory followed by the object will be similar in both the cases.

NOTE: A projectile is any object thrown into space by the exertion of a force. The path followed by a projectile is known as its trajectory.

Question 34. Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vx= 2.0 × 106 m s−1. If E between the plates separated by 0.5 cm is 9.1 × 102 N/C, where will the electron strike the upper plate? (| e | =1.6 × 10−19 C, me = 9.1 × 10−31 kg.)

Velocity of the electron, v x  = 2.0 × 10 6  m s –1

Separation of plates, l = 0.5cm = 0.5 × 10 -2  m

Electric field between the plates, E = 9.1 × 10 2  N C -1

Charge on the electron, q = |e| = 1.6 × 10 -19  C

Mass of the electron, m e  = 9.1 × 10 -31  kg

The vertical deflection of the particle (s) is given by

NCERT Solutions For Class 12 Physics FAQs

Electric charge is a fundamental property of matter that determines its interactions with electric fields. It can be positive or negative and is quantized in nature, meaning it exists as discrete units.

An electric field is a region around a charged object where another charged object experiences a force. It is a vector quantity and is defined as the force experienced per unit positive charge placed at a point in space. Electric field lines represent the direction and strength of the electric field.

The principle of superposition states that the net electric force on a charged particle due to multiple charges is the vector sum of the forces exerted by individual charges acting independently. In other words, the total force on a charge in the presence of multiple charges is the sum of forces exerted by each individual charge.

An electric dipole consists of two equal and opposite charges separated by a small distance. It is characterized by a dipole moment, which is the product of one of the charges and the separation between the charges. An electric dipole experiences a torque in an electric field.

The SI unit of electric charge is the Coulomb (C).

NCERT Solutions For Class 12 Physics Chapter 8 Electromagnetic Waves

NCERT Solutions For Class 12 Physics Chapter 3 Current Electricity

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NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields in Hindi and English Medium PDF study online through the solutions updated for new session 2024-25. Question answers of chapter 1 are modified as per rationalised NCERT textbooks published for academic year 2024-25. The Additional Exercises questions are now deleted from the syllabus.

Chapter 1 Electric Charges and Fields Solutions

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1. Net capacitance of three identical capacitors connected in parallel is 12 microfarad. What will be the net capacitance when two of them are connected in (i) parallel (ii) series? 2. A charge Q is distributed over a metal sphere of radius R. What is the electric field and electric potential at the centre? Answer: E = 0, V = kQ/R.

3. Write the ratio of electric field intensity due to a dipole at a point on the equatorial line to the field at a point at a point on the axial line, when the points are at the same distance from the centre of dipole.

4. An uncharged conductor A placed on an insulating stand is brought near a charged insulated conductor B. What happens to the charge and potential of B? 5. Why is the potentiometer preferred to a voltmeter for measuring emf of a cell?

1. In an uniform electric field of strength E, a charged particle Q moves point A to point B in the direction of the field and back from B to A. Calculate the ratio of the work done by the electric field in taking the charge particle from A to B and from B to A.

2. How does the relaxation time of electron in the conductor change when temperature of the conductor decreases.

3. A copper wire of resistance R is uniformly stretched till its length is increased to n times its original length. What will be its new resistance?

4. A car battery is of 12V. Eight dry cells of 1.5 V connected in series also give 12V, but such a combination is not used to start a car. Why?

5. An oil drop of mass m carrying charge –Q is to be held stationary in the gravitational field of the earth. What is the magnitude and direction of the electrostatic field required for this purpose? Answer: E = mg/Q, downward.

Important Questions on 12th Physics Chapter 1

Explain the meaning of the statement ‘electric charge of a body is quantised’..

Electric charge of a body is quantized. This means that only integral (1, 2, …., n) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge.

Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, quantisation of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous.

When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are created in pairs. This phenomenon of charging is called charging by friction. The net charge on the system of two rubbed bodies is zero. This is because equal amount of opposite charges annihilate each other. When a glass rod is rubbed with a silk cloth, opposite nature charges appear on both the bodies. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies.

An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?

Electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.

Explain why two field lines never cross each other at any point?

If two field lines cross each other at a point, then electric field intensity will show two directions at that point. This is not possible. Hence, two field lines never cross each other.

What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero.

Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.

Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and displaced from its position in any direction, then it experiences a restoring force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards towards the null point. There is a net inward flux of electric field through a closed surface around the null point. According to Gauss’s law, the flux of electric field through a surface, which is not enclosing any charge, is zero. Hence, the equilibrium of the test charge can be stable.

Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

Two charges of same magnitude and same sign are placed at a certain distance. The mid-point of the joining line of the charges is the null point. When a test charged is displaced along the line, it experiences a restoring force. If it is displaced normal to the joining line, then the net force takes it away from the null point. Hence, the charge is unstable because stability of equilibrium requires restoring force in all directions.

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