Class 12 Physics NCERT Solutions

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Electric Charges and Fields Class 12 Physics NCERT Solutions

The complete NCERT exercise solutions for Chapter 1, Electric Charges and Fields — 23 questions from 1.1 to 1.23, each worked through step by step in the CBSE marking pattern. Coulomb's law, electric field and field lines, the electric dipole, and Gauss's law with its applications.

Class:12Subject:PhysicsChapter:1

3 Key Formulas25 Practice MCQs

DWritten byDeep Narayan

UpdatedSeptember 2026

Key Concept Summary

How many questions are in NCERT Class 12 Physics Chapter 1?

Chapter 1 carries 23 exercise questions, numbered 1.1 to 1.23. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

Table of Contents24 sections

Revision & Practice

Chapter Overview

Electric Charges and Fields opens electrostatics with Coulomb's law, the electric field of point and continuous charge distributions, dipoles, and Gauss's law. Boards test the standard forms hard: the inverse-square force, field of a dipole on axis and equatorial line, flux through closed surfaces, and the three classic Gauss applications. Every question below is from the NCERT Class 12 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

Start with F = kq₁q₂/r² with k = 9 × 10⁹ N m² C⁻² and keep charges in coulombs and distances in metres. For fields, E = F/q₀ and E = kq/r². For flux, link the net outward flux to the enclosed charge with Φ = qₑₙ𝒄/ε₀ and remember a charge inside a closed surface contributes the same flux whatever the size or shape of the surface. Direction words — repulsive, attractive, radially inward, left-to-right — carry half the marks.

$Ex 1.1 — Coulomb force between two point charges in air$
$Ex 1.2 — Force-distance relation and Newton’s third law$
$Ex 1.3 — Ratio ke²/Gmₑmₚ: dimensionless, and its value$
$Ex 1.4 — Meaning of quantisation of charge$
$Ex 1.5 — Rubbing glass with silk and conservation of charge$
$Ex 1.6 — Net force on a charge at the centre of a square$
$Ex 1.7 — Why field lines are continuous and never cross$
$Ex 1.8 — Field of a dipole-like pair at the midpoint$
$Ex 1.9 — Total charge and dipole moment of a two-charge system$
$Ex 1.10 — Torque on a dipole in a uniform field$
$Ex 1.11 — Electrons transferred to a rubbed polythene piece$
$Ex 1.12 — Coulomb repulsion of two spheres, then doubled and halved$
$Ex 1.13 — Reading charge signs from tracks in a uniform field$
$Ex 1.14 — Flux of a uniform field through a square$
$Ex 1.15 — Net flux through a cube in a uniform field$
$Ex 1.16 — Net charge inside a box from its outward flux$
$Ex 1.17 — Flux through one face of a cube: the 1/6 trick$
$Ex 1.18 — Flux through a cubic Gaussian surface$
$Ex 1.19 — Flux on doubling the Gaussian radius; point charge value$
$Ex 1.20 — Charge on a conducting sphere from its external field$
$Ex 1.21 — Charge and flux from a known surface charge density$
$Ex 1.22 — Linear charge density of an infinite line charge$
$Ex 1.23 — Field of two large oppositely charged plates$

NCERT Exercise 1.1 — Coulomb Force Between Two Point Charges in Air

1Exercise question

Step-by-step solution

1Use Coulomb's law: F = kq₁q₂/r² with k = 9 × 10⁹ N m² C⁻².
2Convert r = 30 cm = 0.30 m. Then F = (9 × 10⁹)(2 × 10⁻⁷)(3 × 10⁻⁷)/(0.30)².
3F = (9 × 10⁹ × 6 × 10⁻¹⁴)/(0.09) = (5.4 × 10⁻⁴)/(0.09) = 6 × 10⁻³ N.
4Both charges are positive, so the force is repulsive.

Final answer

F = 6 × 10⁻³ N, repulsive (the two like charges push each other apart).

NCERT Exercise 1.2 — Force–Distance Relation and Newton's Third Law

1Exercise question

Step-by-step solution

1(a) F = k|q₁q₂|/r², so r² = k|q₁q₂|/F.
2r² = (9 × 10⁹ × 0.4 × 10⁻⁶ × 0.8 × 10⁻⁶)/0.2 = (9 × 10⁹ × 0.32 × 10⁻¹²)/0.2.
3r² = (2.88 × 10⁻³)/0.2 = 1.44 × 10⁻² m², hence r = 0.12 m = 12 cm.
4(b) By Newton's third law the force on the second sphere has the same magnitude, 0.2 N.
5The charges are of opposite sign, so the force between them is attractive.

Final answer

(a) r = 12 cm. (b) Force on the second sphere = 0.2 N, directed towards the first sphere.

NCERT Exercise 1.3 — Ratio ke²/Gmₑmₚ: Dimensionless, and Its Value

1Exercise question

Step-by-step solution

1Dimension of k: [M L³ T⁻⁴ A⁻²] (from F = kq²/r²). e² carries [A² T²], so ke² has [M L³ T⁻²] = N m².
2G has [M⁻¹ L³ T⁻²] and mₑmₚ has [M²], so Gmₑmₚ has [M L³ T⁻²] = N m².
3Both numerator and denominator have the same dimensions, so the ratio is dimensionless.
4Numerically, ke² = (9 × 10⁹)(1.6 × 10⁻¹⁹)² = 2.3 × 10⁻²⁸ N m².
5Gmₑmₚ = (6.67 × 10⁻¹¹)(9.1 × 10⁻³¹)(1.67 × 10⁻²⁷) ≈ 1.0 × 10⁻⁶⁷ N m².
6Ratio = 2.3 × 10⁻²⁸/1.0 × 10⁻⁶⁷ ≈ 2.3 × 10³⁹.

Final answer

The ratio ke²/Gmₑmₚ ≈ 2.3 × 10³⁹ — it measures how enormously stronger the electric force is than the gravitational force between an electron and a proton.

NCERT Exercise 1.4 — Meaning of Quantisation of Charge

1Exercise question

Step-by-step solution

1(a) Quantisation means charge on a body is always an integral multiple of e, the elementary charge: q = ±ne, n = 1, 2, 3, …
2Charge can be added or removed only in whole multiples of e, never in fractions of e.
3(b) Because e = 1.6 × 10⁻¹⁹ C is extremely small — macroscopic charges are huge multiples of e, so the discrete steps are too fine to detect and the charge behaves as if continuous.

Final answer

Quantisation means q = ±ne; it is ignored at macroscopic scale because e is so small that the charge looks continuous.

NCERT Exercise 1.5 — Rubbing Glass with Silk and Conservation of Charge

1Exercise question

Step-by-step solution

1Rubbing transfers electrons, not charge creation: electrons move from the glass rod to the silk cloth.
2The glass rod loses electrons and becomes positively charged; the silk gains the same number of electrons and becomes negatively charged.
3The positive charge on one equals the negative charge on the other, so the total charge of the system remains zero before and after rubbing.
4This holds for every pair of bodies — the algebraic sum of the charges produced is always zero, which is the law of conservation of charge.

Final answer

The two charges always appear in equal and opposite amounts — total charge stays constant, conserving the initial (zero) charge of the system.

NCERT Exercise 1.6 — Net Force on a Charge at the Centre of a Square

1Exercise question

Step-by-step solution

1The centre is equidistant (r = 5√2 cm = 7.07 × 10⁻² m) from all four corners.
2qA and qC are equal and diagonally opposite; the forces they exert on the central +1 µC charge are equal in magnitude and opposite in direction, so they cancel.
3qB and qD are equal and diagonally opposite; their forces on the central charge again cancel.
4Hence the vector sum of all four forces is zero.

Final answer

The net force on the charge at the centre is zero (symmetry of the diagonally opposite pairs).

NCERT Exercise 1.7 — Why Field Lines Are Continuous and Never Cross

1Exercise question

Step-by-step solution

1(a) A field line shows the direction of the force on a (conceptual) test charge placed on it; at every point the force has a definite direction, so the line traced through the field cannot jump discontinuously.
2The field direction varies smoothly, which forces the field line to be a smooth, continuous curve without breaks.
3(b) At the crossing point of two lines the field would have to point in two different directions at once, which is impossible.
4At a crossing, the tangent to each line would give a different field direction — a contradiction, so field lines never intersect (they also never enter regions where E = 0 except at charges).

Final answer

A break would mean an undefined field direction; a crossing would mean two directions at one point — both impossible for a well-defined field.

NCERT Exercise 1.8 — Field of a Dipole-Like Pair at the Midpoint

1Exercise question

Step-by-step solution

1(a) At O the distance to each charge is r = 10 cm = 0.10 m.
2E due to each charge has magnitude kq/r² = (9 × 10⁹ × 3 × 10⁻⁶)/(0.10)² = 2.7 × 10⁶ N/C.
3Field lines leave the +qA charge and enter the –qB charge, so both individual fields at O point from A towards B (rightward).
4Resultant E = 2.7 × 10⁶ + 2.7 × 10⁶ = 5.4 × 10⁶ N/C, directed along AB from A to B.
5(b) F = qE = (1.5 × 10⁻⁹)(5.4 × 10⁶) = 8.1 × 10⁻³ N.
6The test charge is negative, so it is pulled opposite to E — back towards A.

Final answer

(a) E = 5.4 × 10⁶ N/C along AB from A to B. (b) F = 8.1 × 10⁻³ N, directed from B towards A.

NCERT Exercise 1.9 — Total Charge and Dipole Moment of a Two-Charge System

1Exercise question

Step-by-step solution

1Total charge = qA + qB = 2.5 × 10⁻⁷ – 2.5 × 10⁻⁷ = 0.
2Dipole moment p = q × 2a, where 2a is the separation = 30 cm = 0.30 m.
3p = 2.5 × 10⁻⁷ × 0.30 = 7.5 × 10⁻⁸ C m.
4Dipole moment points from the negative to the positive charge: −q sits at B (+15 cm) while +q sits at A (−15 cm), so p points from +z toward −z, along the −z axis.

Final answer

Total charge = 0; dipole moment = 7.5 × 10⁻⁸ C m directed from –q to +q (along −z).

NCERT Exercise 1.10 — Torque on a Dipole in a Uniform Field

1Exercise question

Step-by-step solution

1Torque on a dipole: τ = pE sinθ.
2τ = (4 × 10⁻⁹)(5 × 10⁴) sin 30°.
3τ = (2 × 10⁻⁴)(0.5) = 1 × 10⁻⁴ N m.

Final answer

τ = 1 × 10⁻⁴ N m (the torque tries to align the dipole with the field).

NCERT Exercise 1.11 — Electrons Transferred to a Rubbed Polythene Piece

1Exercise question

Step-by-step solution

1(a) q = ne, so n = q/e = (3 × 10⁻⁷)/(1.6 × 10⁻¹⁹) ≈ 1.9 × 10¹² electrons.
2The polythene is negatively charged, so it has gained electrons — they are transferred from the wool to the polythene.
3(b) Yes. Each transferred electron carries mass mₑ ≈ 9.1 × 10⁻³¹ kg.
4Mass transferred ≈ 1.9 × 10¹² × 9.1 × 10⁻³¹ ≈ 1.7 × 10⁻¹⁸ kg — present but far too small to measure.

Final answer

(a) ≈ 1.9 × 10¹² electrons, transferred from wool to polythene. (b) Yes — ≈ 1.7 × 10⁻¹⁸ kg of mass moves with the electrons.

NCERT Exercise 1.12 — Coulomb Repulsion of Two Spheres, Then Doubled and Halved

1Exercise question

Step-by-step solution

1(a) F = kq₁q₂/r² with q₁ = q₂ = 6.5 × 10⁻⁷ C and r = 0.50 m.
2F = (9 × 10⁹)(6.5 × 10⁻⁷)²/(0.50)² = (9 × 10⁹ × 4.225 × 10⁻¹³)/0.25.
3F = (3.8 × 10⁻³)/0.25 = 1.5 × 10⁻² N.
4(b) Each charge doubled means q₁q₂ becomes 4×; distance halved means 1/r² becomes 4×.
5New F = F₀ × 4 × 4 = 1.5 × 10⁻² × 16 = 2.4 × 10⁻¹ N.

Final answer

(a) 1.5 × 10⁻² N. (b) 2.4 × 10⁻¹ N — doubling charges and halving distance multiplies the force by 16.

NCERT Exercise 1.13 — Reading Charge Signs from Tracks in a Uniform Field

1Exercise question

Step-by-step solution

1In a uniform field the force is F = qE, and the deflection of a track tells the sign of q for a given field direction.
2Particles A and C curve to one side of the field direction while B curves to the other, so A and C carry a charge opposite to B.
3Taking the field directed away from the positive plate, A and C are drawn toward the negative plate — they are negatively charged, while B is positively charged.
4The deflection is proportional to |q|/m. The most deflected track belongs to B, so B has the highest |q|/m.

Final answer

A and C are negatively charged and B is positively charged; particle B has the highest charge-to-mass ratio.

NCERT Exercise 1.14 — Flux of a Uniform Field Through a Square

1Exercise question

Step-by-step solution

1(a) The square lies in the yz plane, so its area vector points along +x (parallel to E).
2A = (0.10)² = 0.01 m². Φ = E·A = E A cos 0° = (3 × 10³)(0.01) = 30 N m²/C.
3(b) With the normal at 60° to the x-axis: Φ = E A cos 60° = 30 × 0.5 = 15 N m²/C.

Final answer

(a) 30 N m²/C. (b) 15 N m²/C.

NCERT Exercise 1.15 — Net Flux Through a Cube in a Uniform Field

1Exercise question

Step-by-step solution

1The field is uniform and directed along +x.
2The flux entering the left (x = 0) face equals the flux leaving the right face, because E and the face areas are identical.
3All other faces have their normals perpendicular to E, so their flux is zero.
4Hence the net flux through the closed cube is zero.

Final answer

Net flux = 0 — the uniform field has no sources inside the cube.

NCERT Exercise 1.16 — Net Charge Inside a Box from Its Outward Flux

1Exercise question

Step-by-step solution

1(a) Gauss's law: Φ = q/ε₀, so q = Φ ε₀.
2q = (8.0 × 10³)(8.85 × 10⁻¹²) = 7.1 × 10⁻⁸ C.
3(b) No. A zero net flux means only that the net charge enclosed is zero.
4The box could still contain equal amounts of positive and negative charge (or a balanced distribution) whose fields cancel on the boundary.

Final answer

(a) q = 7.1 × 10⁻⁸ C. (b) No — zero net flux implies zero net charge, but equal positive and negative charges inside are still possible.

NCERT Exercise 1.17 — Flux Through One Face of a Cube: The 1/6 Trick

1Exercise question

Step-by-step solution

1Imagine a cube of edge 10 cm with the square as one face; the charge sits at the centre of the cube, 5 cm above the square's centre.
2By symmetry the charge sends equal flux through all six faces.
3Total flux = q/ε₀ = (10 × 10⁻⁶)/(8.85 × 10⁻¹²) = 1.13 × 10⁶ N m²/C.
4Flux through the square = one sixth of the total = 1.13 × 10⁶/6 ≈ 1.9 × 10⁵ N m²/C.

Final answer

Φ = q/6ε₀ ≈ 1.9 × 10⁵ N m²/C through the square.

NCERT Exercise 1.18 — Flux Through a Cubic Gaussian Surface

1Exercise question

Step-by-step solution

1The charge is enclosed, so the net flux is fixed by Gauss's law regardless of the cube's size or orientation.
2Φ = q/ε₀ = (2.0 × 10⁻⁶)/(8.85 × 10⁻¹²) = 2.26 × 10⁵ N m²/C.
3The 9.0 cm edge length does not enter the answer.

Final answer

Φ ≈ 2.3 × 10⁵ N m²/C (independent of the cube's edge length).

NCERT Exercise 1.19 — Flux on Doubling the Gaussian Radius; Point Charge Value

1Exercise question

Step-by-step solution

1(a) The flux depends only on the enclosed charge, not on the surface radius.
2Doubling the radius (still enclosing the point charge) leaves the flux unchanged: –1.0 × 10³ N m²/C.
3(b) Gauss's law: q = Φ ε₀ = (–1.0 × 10³)(8.85 × 10⁻¹²) = –8.85 × 10⁻⁹ C ≈ –8.9 × 10⁻⁹ C.
4The negative flux (inward) confirms the charge is negative.

Final answer

(a) Flux stays –1.0 × 10³ N m²/C. (b) q ≈ –8.9 × 10⁻⁹ C.

NCERT Exercise 1.20 — Charge on a Conducting Sphere from Its External Field

1Exercise question

Step-by-step solution

1Outside the sphere the field is that of a point charge: E = kq/r².
2q = Er²/k = (1.5 × 10³)(0.20)²/(9 × 10⁹) = (1.5 × 10³ × 0.04)/(9 × 10⁹).
3q = 60/(9 × 10⁹) = 6.7 × 10⁻⁹ C.
4The field points inward (toward the sphere), so the charge is negative: q ≈ –6.7 × 10⁻⁹ C.

Final answer

q ≈ –6.7 × 10⁻⁹ C (6.7 × 10⁻⁹ C of negative charge).

NCERT Exercise 1.21 — Charge and Flux from a Known Surface Charge Density

1Exercise question

Step-by-step solution

1(a) Radius R = 1.2 m. Surface area = 4πR² = 4π(1.2)² = 18.1 m².
2q = σ × area = (80.0 × 10⁻⁶)(18.1) = 1.45 × 10⁻³ C.
3(b) Total flux = q/ε₀ = (1.45 × 10⁻³)/(8.85 × 10⁻¹²) = 1.6 × 10⁸ N m²/C.

Final answer

(a) q ≈ 1.45 × 10⁻³ C. (b) Φ ≈ 1.6 × 10⁸ N m²/C.

NCERT Exercise 1.22 — Linear Charge Density of an Infinite Line Charge

1Exercise question

Step-by-step solution

1Field of an infinite line charge: E = λ/(2πε₀r).
2λ = 2πε₀ r E.
3λ = (2π × 8.85 × 10⁻¹²)(0.02)(9 × 10⁴).
4λ = (5.56 × 10⁻¹¹)(0.02)(9 × 10⁴) ≈ 1.0 × 10⁻⁷ C/m.

Final answer

λ ≈ 1.0 × 10⁻⁷ C/m.

NCERT Exercise 1.23 — Field of Two Large Oppositely Charged Plates

1Exercise question

Step-by-step solution

1Each plate alone produces E = σ/2ε₀ on either side, directed away from a positive plate and toward a negative one.
2(a),(b) In the outer regions the two plates' fields are equal and opposite, so they cancel: E = 0.
3(c) Between the plates the fields add: E = σ/ε₀.
4E = (17.0 × 10⁻²²)/(8.85 × 10⁻¹²) ≈ 1.9 × 10⁻¹⁰ N/C, directed from the positive to the negative plate.

Final answer

(a) E = 0. (b) E = 0. (c) E ≈ 1.9 × 10⁻¹⁰ N/C between the plates, from positive to negative.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Coulomb's law

F = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r ^{2}}

Electric field

E = \frac{F}{q} = \frac{1}{4 π ε _{0}} \frac{q}{r ^{2}}

Gauss's law

\oint E \cdot d A = \frac{q _{e n c}}{ε _{0}}

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

Gauss's law only gives the field directly when the charge distribution has enough symmetry — a sphere, a cylinder or an infinite sheet, and nothing else.
Field lines never cross and their density shows field strength, but a plain diagram of lines is never a substitute for computing E from Coulomb's law.

FAQ

Frequently asked questions

How many questions are in NCERT Class 12 Physics Chapter 1 (Electric Charges and Fields)?

There are 23 exercise questions in this chapter, numbered 1.1 to 1.23. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Electric Charges and Fields Class 12 Physics?

The formulas this chapter's questions actually turn on are: Coulomb's law, Electric field, Gauss's law. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Electric Charges and Fields important for JEE Main and NEET?

Very important — Coulomb's law, field and Gauss's law form the electrostatic core and are asked directly in both JEE Main and NEET, with the shell and sheet applications near-guaranteed.

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