NCERT Solutions for Class 12 Physics Chapter 1: Electric Charges and Fields

These Class 12 Physics Chapter 1 solutions cover Electric Charges and Fields, the opening chapter of electrostatics in the NCERT textbook (session 2026–27). Every one of the 23 end-of-chapter Exercises is reproduced verbatim and solved step by step — each numerical worked out with full units and cross-checked against the NCERT answer key — alongside key formulas, extra practice, MCQs, Assertion–Reason questions and FAQs.

Class: 12 Subject: Physics Chapter: 1 Name: Electric Charges and Fields Exercises: 23 questions Session: 2026–27
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Class 12 Physics Chapter 1 – Overview

Chapter 1, Electric Charges and Fields, introduces electrostatics — the study of forces, fields and potentials arising from charges at rest. It begins with the two kinds of electric charge (positive and negative), the rules that like charges repel and unlike charges attract, and the three basic properties of charge: additivity, conservation and quantisation (q = ne). It then builds up Coulomb’s law for the force between point charges, the superposition principle for many charges, and the idea of the electric field E as the force per unit positive test charge. The chapter develops electric field lines, the electric dipole and its dipole moment, the torque on a dipole in a uniform field, electric flux, and finally Gauss’s law with its applications to an infinite line charge, an infinite charged sheet and a uniformly charged thin shell.

Key Concepts & Definitions

Electric charge: a scalar property of matter that produces and responds to electric forces; it is of two kinds, positive and negative, and is measured in coulomb (C).

Quantisation of charge: any free charge is an integral multiple of the elementary charge, q = ne, where e = 1.6 × 10−19 C.

Conservation of charge: the total charge of an isolated system stays constant; charge is only transferred, never created or destroyed.

Coulomb’s law: the electrostatic force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.

Electric field (E): the force experienced by a unit positive test charge placed at a point, without disturbing the source charges; a vector measured in N C−1.

Electric dipole: a pair of equal and opposite charges +q and −q separated by a small distance 2a; its dipole moment p = q × 2a points from −q to +q.

Electric flux (Φ): a measure of the number of field lines crossing a surface, Φ = E·S = E S cosθ; SI unit N C−1 m2.

Gauss’s law: the net electric flux through any closed surface equals the net charge enclosed divided by ε0.

Important Formulas

Coulomb’s law: F = (1/4πε0) · q1q2/r2,   where 1/4πε0 = 9 × 109 N m2 C−2 and ε0 = 8.854 × 10−12 C2 N−1 m−2.

Quantisation: q = ne,   e = 1.6 × 10−19 C.

Electric field of a point charge: E = (1/4πε0) · q/r2;   force on a charge F = qE.

Dipole moment: p = q × 2a (vector from −q to +q).

Field of a dipole (r >> a): on axis E = (1/4πε0)·2p/r3; on equatorial line E = (1/4πε0)·p/r3.

Torque on a dipole: τ = pE sinθ (&vec;τ = &vec;p × &vec;E).

Electric flux: Φ = E·S = E S cosθ.

Gauss’s law: Φ = ∮ E·dS = qenclosed/ε0.

Infinite line charge: E = λ/(2πε0r). Infinite charged sheet: E = σ/(2ε0). Between two oppositely charged plates: E = σ/ε0 (zero outside).

NCERT Exercises – Solutions (1.1 to 1.23)

Questions are reproduced verbatim from the NCERT textbook. Charges written as µC below appear as “micro coulomb” in the book; all answers are cross-checked against the NCERT answer key.

1.1 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?

SOLUTION Given: q1 = 2 × 10−7 C, q2 = 3 × 10−7 C, r = 30 cm = 0.30 m. F = (1/4πε0) · q1q2/r2 = (9 × 109) × (2 × 10−7)(3 × 10−7) / (0.30)2 = (9 × 109) × (6 × 10−14) / (0.09) = (5.4 × 10−4) / 0.09 F = 6 × 10−3 N. Since both charges are positive, the force is repulsive.

1.2 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 Given: q1 = 0.4 µC = 4 × 10−7 C, q2 = −0.8 µC = −8 × 10−7 C, F = 0.2 N. (a) From F = (1/4πε0) · |q1q2|/r2,   r2 = (9 × 109) × (4 × 10−7)(8 × 10−7) / 0.2 = (9 × 109) × (3.2 × 10−13) / 0.2 = (2.88 × 10−3) / 0.2 = 1.44 × 10−2 m2. r = √(1.44 × 10−2) = 0.12 m = 12 cm. (b) By Newton’s third law, the force on the second sphere is equal and opposite: 0.2 N, attractive (the charges are of opposite sign).

1.3 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?

SOLUTION Here k = 1/4πε0 = 9 × 109 N m2 C−2, e = 1.6 × 10−19 C, G = 6.67 × 10−11 N m2 kg−2, me = 9.11 × 10−31 kg, mp = 1.67 × 10−27 kg. Dimensions: ke2 has units (N m2 C−2)(C2) = N m2;   G memp has units (N m2 kg−2)(kg2) = N m2. The two are identical, so the ratio is dimensionless. Value: ke2/(G memp) = (9 × 109)(1.6 × 10−19)2 / [(6.67 × 10−11)(9.11 × 10−31)(1.67 × 10−27)] Numerator = 9 × 109 × 2.56 × 10−38 = 2.30 × 10−28. Denominator = 6.67 × 10−11 × 1.52 × 10−57 = 1.01 × 10−67. Ratio = (2.30 × 10−28)/(1.01 × 10−67) = 2.4 × 1039. This is the ratio of the electric force to the gravitational force (at the same separation) between an electron and a proton, showing how enormously stronger the electrostatic force is.

1.4 (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) ‘Charge is quantised’ means the charge on any body can only be an integral multiple of the elementary charge e, i.e. q = ne where n = 0, ±1, ±2, … A body cannot carry a charge such as 1.5e or 0.3e; charge always comes in whole packets of e = 1.6 × 10−19 C. (b) The elementary charge e is extremely small. A macroscopic charge of even 1 µC contains about 6 × 1012 electronic charges, so adding or removing one e changes it by a negligible, unmeasurable fraction. The charge therefore appears continuous, and quantisation can be safely ignored at the large scale.

1.5 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 When the rod and silk are rubbed, no new charge is created; electrons are simply transferred from the glass rod to the silk. The rod loses electrons and becomes positively charged, while the silk gains exactly the same number of electrons and becomes equally negatively charged. The two charges are equal in magnitude and opposite in sign, so the total charge of the (rod + silk) system stays zero, exactly as it was before rubbing. Charge is merely transferred from one body to another — this is fully consistent with the law of conservation of charge.

1.6 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 centre O is equidistant from all four corners. The diagonally opposite charges A and C are equal (both +2 µC) and lie on the same diagonal on opposite sides of O, so the forces they exert on the central 1 µC charge are equal in magnitude but opposite in direction — they cancel. Similarly, B and D (both −5 µC) lie on the other diagonal at equal distances on opposite sides, so their forces on the central charge also cancel. Hence the net force on the 1 µC charge at the centre is zero (0 N).

1.7 (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) A field line shows the path a unit positive test charge would follow; the moving charge experiences a continuous force and cannot jump from one point to another. A sudden break would mean the charge disappears for an instant or the field is undefined in between, which is impossible. So a field line is a continuous, unbroken curve (in a charge-free region). (b) If two field lines crossed, then at the point of intersection there would be two tangents, implying two different directions of the electric field at the same point. But the electric field at any point has a single, unique direction. Hence two field lines can never cross.

1.8 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 Given: qA = 3 µC = 3 × 10−6 C, qB = −3 µC, AB = 20 cm, so each charge is r = 10 cm = 0.10 m from the midpoint O. (a) Field at O due to +qA points from A towards O (away from A), i.e. towards B. Field due to −qB points from O towards B (towards the negative charge). Both fields point in the same direction (from A to B, i.e. along OB) and have equal magnitude: EA = (9 × 109)(3 × 10−6)/(0.10)2 = (2.7 × 104)/0.01 = 2.7 × 106 N C−1. Total E = EA + EB = 2 × 2.7 × 106 = 5.4 × 106 N C−1, directed from A to B (along OB). (b) Force on the test charge q = 1.5 × 10−9 C: F = qE = (1.5 × 10−9)(5.4 × 106) = 8.1 × 10−3 N. Since the test charge is negative, the force is opposite to E, i.e. directed from B to A (along OA).

1.9 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 Total charge = qA + qB = (2.5 × 10−7) + (−2.5 × 10−7) = 0 C. The system is an electric dipole. Separation between the charges, 2a = distance from z = −15 cm to z = +15 cm = 30 cm = 0.30 m. Dipole moment p = q × 2a = (2.5 × 10−7) × (0.30) = 7.5 × 10−8 C m. Its direction is from the negative charge (at +15 cm) to the positive charge (at −15 cm), i.e. along the negative z-axis (magnitude 7.5 × 10−8 C m along the z-axis).

1.10 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 Given: p = 4 × 10−9 C m, E = 5 × 104 N C−1, θ = 30°. Torque τ = pE sinθ = (4 × 10−9)(5 × 104)(sin 30°) = (4 × 10−9)(5 × 104)(0.5) = (2 × 10−4)(0.5) = 1 × 10−4 N m.

1.11 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) Since the polythene becomes negative, electrons are transferred from wool to polythene. Number of electrons n = q/e = (3 × 10−7)/(1.6 × 10−19) = 1.875 × 1012 ≈ 2 × 1012 electrons. (b) Yes. Because electrons carry mass, mass is transferred along with charge — from wool to polythene. Mass transferred = n × me = (1.875 × 1012)(9.11 × 10−31 kg) ≈ 1.7 × 10−18 kg (about 2 × 10−18 kg) — a completely negligible amount.

1.12 (a) Two insulated charged copper spheres A and B have their centres 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) qA = qB = 6.5 × 10−7 C, r = 50 cm = 0.50 m. F = (9 × 109)(6.5 × 10−7)2/(0.50)2 = (9 × 109)(4.225 × 10−13)/0.25 = (3.80 × 10−3)/0.25 = 1.5 × 10−2 N (repulsive). (b) Each charge is doubled (×2) and the distance is halved (×½). Since F ∝ q1q2/r2, the factor is (2)(2)/(½)2 = 4/0.25 = 16. New force F′ = 16 × (1.52 × 10−2) = 0.24 N (repulsive).

1.13 Figure 1.30 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?

SOLUTION A charged particle is deflected by the field force qE; particles attracted towards the negative plate are positive and those attracted towards the positive plate are negative. From the tracks in Fig. 1.30, particles 1 and 2 deflect towards the positive plate, so charges 1 and 2 are negative, while particle 3 deflects towards the negative plate, so charge 3 is positive. The deflection of a particle is greater when its charge-to-mass ratio (q/m) is larger, because the deflection ∝ force/mass = qE/m. Particle 3 shows the greatest deflection, so particle 3 has the highest charge-to-mass ratio.

1.14 Consider a uniform electric field E = 3 × 103 î N/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 Side of square = 10 cm = 0.10 m, so area S = (0.10)2 = 0.01 m2. E = 3 × 103 N/C along the x-direction. (a) If the plane is parallel to the yz plane, its normal is along the x-axis, i.e. parallel to E (θ = 0°). Flux Φ = E S cos 0° = (3 × 103)(0.01)(1) = 30 N m2 C−1. (b) Now the normal makes θ = 60° with the x-axis. Flux Φ = E S cos 60° = (3 × 103)(0.01)(0.5) = 15 N m2 C−1.

1.15 What is the net flux of the uniform electric field of Exercise 1.14 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

SOLUTION The field is uniform, so the number of field lines entering the cube equals the number leaving it. Equivalently, the flux into the two faces perpendicular to E (one face +E S, the opposite face −E S) cancels, and the four faces parallel to E contribute zero flux. Therefore the net flux through the closed cube is zero. (This is consistent with Gauss’s law: there is no charge enclosed inside the cube.)

1.16 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?

SOLUTION (a) By Gauss’s law, Φ = q/ε0, so q = ε0Φ = (8.854 × 10−12)(8.0 × 103) = 7.08 × 10−8 C ≈ 0.07 µC. (b) No. Zero net flux only tells us that the net charge inside is zero. The box could still contain equal amounts of positive and negative charge (e.g. +q and −q) whose contributions cancel, giving zero net flux.

1.17 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. 1.31. 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.)

SOLUTION If the charge is 5 cm above the centre of a 10 cm square, the square can be regarded as one of the six faces of a cube of edge 10 cm with the charge at the cube’s centre. By Gauss’s law the total flux through the whole cube is q/ε0. By symmetry, the flux through one face is one-sixth of this: Φface = q/(6ε0) = (10 × 10−6)/(6 × 8.854 × 10−12) = (10−5)/(5.31 × 10−11) = 1.9 × 105 N m2 C−1.

1.18 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?

SOLUTION By Gauss’s law the flux through a closed surface depends only on the enclosed charge, not on the size or shape of the surface. Φ = q/ε0 = (2.0 × 10−6)/(8.854 × 10−12) = 2.26 × 105 N m2 C−1 ≈ 2.2 × 105 N m2 C−1. (The 9.0 cm edge is irrelevant.)

1.19 A point charge causes an electric flux of −1.0 × 103 N m2/C to pass through a spherical Gaussian surface of 10.0 cm radius centred 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?

SOLUTION (a) By Gauss’s law the flux depends only on the enclosed charge, not on the radius. Doubling the radius does not change the enclosed charge, so the flux is unchanged: −1.0 × 103 N m2/C. (b) q = ε0Φ = (8.854 × 10−12)(−1.0 × 103) = −8.85 × 10−9 C ≈ −8.8 nC. (The negative flux indicates a negative charge.)

1.20 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?

SOLUTION Outside a conducting sphere the field is the same as that of a point charge at the centre: E = (1/4πε0) · q/r2, with r = 20 cm = 0.20 m. q = E r2/(1/4πε0) = (1.5 × 103)(0.20)2/(9 × 109) = (1.5 × 103)(0.04)/(9 × 109) = (60)/(9 × 109) = 6.67 × 10−9 C. Since the field points radially inward, the charge is negative: q = −6.67 nC.

1.21 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 Diameter = 2.4 m, so radius R = 1.2 m. Surface charge density σ = 80.0 µC/m2 = 80.0 × 10−6 C/m2. (a) Charge q = σ × (surface area) = σ × 4πR2 = (80.0 × 10−6) × 4π(1.2)2 = (80.0 × 10−6) × (18.10) = 1.45 × 10−3 C. (b) Total flux Φ = q/ε0 = (1.45 × 10−3)/(8.854 × 10−12) = 1.6 × 108 N m2 C−1.

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

SOLUTION For an infinite line charge, E = λ/(2πε0r). So λ = E × 2πε0r = E r/(2 × 9 × 109). Here E = 9 × 104 N/C, r = 2 cm = 0.02 m. λ = (9 × 104)(0.02)/(1.8 × 1010) = (1.8 × 103)/(1.8 × 1010) = 1.0 × 10−7 C/m (= 0.1 µC/m).

1.23 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?

SOLUTION For two parallel plates with equal and opposite surface charge densities, the fields outside cancel and the field between them adds up. (a) In the outer region of the first plate, the fields of the two sheets are equal and opposite, so E = 0. (b) In the outer region of the second plate, similarly E = 0. (c) Between the plates, E = σ/ε0 = (17.0 × 10−22)/(8.854 × 10−12) = 1.9 × 10−10 N/C, directed from the positive plate to the negative plate. (The NCERT key rounds this to “1.9 N/C”; the magnitude computed from the given σ is 1.9 × 10−10 N/C.)

Extra Practice Questions

Short Answer Type Questions

Q1. State Coulomb’s law and write its vector form.

ANSWERCoulomb’s law states that the electrostatic force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them, acting along the line joining them. In vector form, the force on q1 due to q2 is &vec;F12 = (1/4πε0)(q1q2/r2) r̂12.

Q2. Define electric field intensity and give its SI unit.

ANSWERThe electric field intensity at a point is the force experienced per unit positive test charge placed at that point, E = F/q (in the limit q → 0). It is a vector. Its SI unit is newton per coulomb (N C−1), equivalently volt per metre (V m−1).

Q3. Why must a test charge be vanishingly small?

ANSWERA finite test charge would exert its own force on the source charges and displace them, changing the very field we want to measure. By making the test charge negligibly small (q → 0), the source distribution is left undisturbed, while the ratio F/q remains finite and gives the true field.

Q4. State Gauss’s law and name one quantity it does not depend on.

ANSWERGauss’s law states that the net electric flux through a closed surface equals the net charge enclosed divided by ε0: Φ = qenclosed/ε0. The flux does not depend on the size or shape of the Gaussian surface, nor on the location of the charge inside it.

Q5. Two charges +4q and +q are separated by a distance d. Where on the line joining them is the net field zero?

ANSWERThe null point lies between the charges, nearer the smaller charge. Setting the magnitudes equal at distance x from +4q: 4q/x2 = q/(d − x)2 gives 2(d − x) = x, so x = 2d/3 from the +4q charge (i.e. d/3 from the +q charge).

Long Answer Type Questions

Q1. Derive an expression for the torque on an electric dipole placed in a uniform electric field, and state when it is maximum and zero.

ANSWERConsider a dipole of charges +q and −q separated by 2a placed in a uniform field E at angle θ to E. Each charge feels a force qE, equal in magnitude but opposite in direction, so the net force is zero. These two forces form a couple. The perpendicular distance between their lines of action is 2a sinθ. Hence the torque τ = (force) × (arm) = qE × 2a sinθ = (q × 2a)E sinθ = pE sinθ, where p = q × 2a is the dipole moment. In vector form &vec;τ = &vec;p × &vec;E. The torque is maximum (τ = pE) when θ = 90° (dipole perpendicular to the field) and zero when θ = 0° or 180° (dipole parallel or antiparallel to the field). The torque always tends to align the dipole with the field.

Q2. Using Gauss’s law, derive the electric field due to an infinitely long uniformly charged straight wire.

ANSWERLet the wire have linear charge density λ. By symmetry the field is radial and has the same magnitude at all points equidistant from the wire. Choose a coaxial cylindrical Gaussian surface of radius r and length l. The flux through the two flat ends is zero (field is parallel to them); through the curved surface it is E × (2πrl). The charge enclosed is λl. By Gauss’s law, E(2πrl) = λl/ε0. Therefore E = λ/(2πε0r), directed radially outward for positive λ. The field falls off as 1/r (not 1/r2), a direct consequence of the line geometry.

Q3. State the superposition principle and explain how it is used to find the field due to a system of charges.

ANSWERThe superposition principle states that the force (or field) on a charge due to several charges is the vector sum of the forces (or fields) due to each charge taken one at a time, and each individual interaction is unaffected by the presence of the others. To find the resultant field at a point P due to charges q1, q2, …, qn, we first compute the field of each charge separately using Ei = (1/4πε0)(qi/riP2) r̂iP, then add these vectors (using components or the parallelogram law) to get E = Σ Ei. All of electrostatics follows from Coulomb’s law combined with this principle.

MCQs & Assertion–Reason

1. The SI unit of electric charge is the:

(a) volt    (b) coulomb    (c) newton    (d) ampere

2. The statement “q = ne” expresses the property of:

(a) conservation of charge    (b) additivity of charge    (c) quantisation of charge    (d) polarity of charge

3. The force between two point charges in vacuum is F. If the distance between them is doubled, the force becomes:

(a) 2F    (b) F/2    (c) F/4    (d) 4F

4. The value of 1/4πε0 in SI units is approximately:

(a) 9 × 109 N m2 C−2    (b) 8.85 × 10−12    (c) 1.6 × 10−19    (d) 6.25 × 1018

5. The SI unit of electric field is:

(a) N C    (b) N C−1    (c) C N−1    (d) N m2 C−1

6. Electric field lines:

(a) can cross each other    (b) form closed loops in electrostatics    (c) never cross each other    (d) start at negative charges

7. The torque on an electric dipole in a uniform field is maximum when the angle between p and E is:

(a) 0°    (b) 45°    (c) 90°    (d) 180°

8. The net flux through a closed surface depends on:

(a) the shape of the surface    (b) the net charge enclosed    (c) the size of the surface    (d) charges outside the surface

9. The electric field due to an infinite charged plane sheet of surface density σ is:

(a) σ/ε0    (b) σ/2ε0    (c) σ/4πε0    (d) 2σ/ε0

10. The number of electrons in a charge of −1 C is about:

(a) 1.6 × 10−19    (b) 6.25 × 1018    (c) 9 × 109    (d) 6.02 × 1023

Answer key: 1-(b), 2-(c), 3-(c), 4-(a), 5-(b), 6-(c), 7-(c), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: The net charge of an isolated system always remains constant.

Reason: Charge can only be transferred from one body to another, not created or destroyed.

A-R 2. Assertion: Two electric field lines never intersect each other.

Reason: At the point of intersection the field would have two directions, which is impossible.

A-R 3. Assertion: The electric flux through a closed surface enclosing an electric dipole is zero.

Reason: The net charge enclosed by the surface is zero.

A-R 4. Assertion: The electric field inside a uniformly charged spherical shell is zero.

Reason: A Gaussian surface inside the shell encloses no charge.

A-R 5. Assertion: A charge of 2.5e can exist on a body.

Reason: Electric charge is quantised and exists only as an integral multiple of e.

Answer key: 1-(A), 2-(A), 3-(A), 4-(A), 5-(D).

Common Mistakes & Exam Tips

Common mistakes to avoid

  • Forgetting to convert centimetres to metres before squaring the distance in Coulomb’s law — always work in SI units.
  • Confusing µC (10−6 C) with mC (10−3 C); read the prefix carefully.
  • Adding electric fields as scalars when they point in different directions — E is a vector, so add by components or geometry.
  • Using E = σ/2ε0 (single sheet) when the question asks for the field between two oppositely charged plates, where E = σ/ε0.
  • Thinking Gauss-law flux depends on the size/shape of the surface or on outside charges — it depends only on the net charge enclosed.
  • Mixing up the directions of field lines: they go outward from positive and inward to negative charges.

How to score full marks in this chapter

Write the formula, substitute the values with units, then give the boxed final answer with its direction (for vectors). For symmetry questions like Exercise 1.6, explain in one line why the forces cancel rather than just writing “zero”. For Gauss’s law numericals, always start with Φ = q/ε0 and use ε0 = 8.854 × 10−12. Quote the value 1/4πε0 = 9 × 109 N m2 C−2 while solving Coulomb-law problems, and remember e = 1.6 × 10−19 C for counting electrons.

Frequently Asked Questions

What is Class 12 Physics Chapter 1 Electric Charges and Fields about?

Chapter 1 introduces electrostatics — the properties of electric charge (additivity, conservation, quantisation), Coulomb’s law and the superposition principle, the electric field and field lines, the electric dipole and torque on it, electric flux and Gauss’s law with its applications to a line charge, a charged sheet and a charged shell.

How many exercise questions are there in Class 12 Physics Chapter 1?

The NCERT textbook has 23 end-of-chapter Exercises (1.1 to 1.23). All of them are reproduced verbatim and solved step by step on this page, with every numerical worked out in SI units and cross-checked against the NCERT answer key.

State Coulomb’s law in one line.

The electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them: F = (1/4πε0)·q1q2/r2.

Are these Class 12 Physics Chapter 1 solutions free?

Yes. All solutions are free and follow the official NCERT Physics textbook for session 2026–27.

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