NCERT Solutions for Class 12 Physics Chapter 1: Electric Charges and Fields (2026-27 Board Exam)

Electric Charges and Fields is the opening chapter of Class 12 Physics, and it carries weight far beyond the boards — it forms the base for Electrostatics questions in NEET, JEE Main/Advanced, and NDA as well. A weak grip here means struggling with Capacitance, Current Electricity, and even Magnetism later. This is why at Convex Classes, Jaipur, we make sure every student builds rock-solid clarity on this chapter before moving ahead.

Below are complete, step-by-step NCERT solutions for all in-chapter exercise questions of Chapter 1, solved the way we teach them in our Board + NEET + JEE + NDA batches.

Quick Chapter Overview

This chapter covers electric charge and its properties (quantisation, conservation), Coulomb’s Law, the concept of an electric field and field lines, electric dipoles and dipole moment, electric flux, and Gauss’s Law with its applications to simple charge distributions like spheres, wires, and charged plates.

Key formulas you must memorise:

• Coulomb’s Law: F = k·q₁q₂/r², where k = 9 × 10⁹ N·m²/C²
• Electric field: E = F/q = k·q/r²
• Dipole moment: p = q × d
• Torque on a dipole: τ = pE sinθ
• Electric flux: Φ = E·A·cosθ
• Gauss’s Law: Φ = q_enclosed/ε₀, where ε₀ = 8.854 × 10⁻¹² C²/N·m²

Solved Exercise Questions

Q1. Force between two charges of 2 × 10⁻⁷ C and 3 × 10⁻⁷ C, placed 30 cm apart in air.

Using F = kq₁q₂/r²: F = (9 × 10⁹ × 2 × 10⁻⁷ × 3 × 10⁻⁷) / (0.3)² F = 6 × 10⁻³ N (repulsive, since both charges are positive)

Q2. Electrostatic force of 0.2 N between spheres of charge 0.4 μC and −0.8 μC.

(a) Rearranging Coulomb’s Law for r: r² = k·q₁q₂/F = (9 × 10⁹ × 0.4 × 10⁻⁶ × 0.8 × 10⁻⁶) / 0.2 = 0.0144 r = 0.12 m

(b) By Newton’s third law, the force on the second sphere is equal and opposite — still 0.2 N, but attractive since the charges are of opposite sign.

Q3. Show that ke²/(Gmₑmₚ) is dimensionless and find its value.

Both the numerator (electrostatic force constant) and denominator (gravitational force constant) reduce to units of Newtons when multiplied by 1/r², so the ratio is a pure number.

Substituting e = 1.6 × 10⁻¹⁹ C, G = 6.67 × 10⁻¹¹ N·m²/kg², mₑ = 9.1 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg:

Ratio ≈ 2.3 × 10³⁹

This number represents how many times stronger the electrostatic force is compared to the gravitational force between a proton and an electron — a good way to appreciate why gravity is ignored in atomic-scale physics.

Q4. What does “charge is quantised” mean, and why is it ignored for macroscopic charges?

(a) Charge on any object always exists as an integer multiple of the elementary charge e (1.6 × 10⁻¹⁹ C). You cannot transfer half an electron — charge transfer happens in whole-number packets.

(b) At everyday (macroscopic) scales, the charges involved are billions of times larger than e, so treating charge as continuous introduces negligible error and greatly simplifies calculations.

Q5. How does charging by rubbing (like glass rod with silk) support charge conservation?

Rubbing does not create new charge — it transfers electrons from one surface to the other. One object gains electrons (becomes negative) and the other loses an equal number (becomes positive). The total charge of the two-body system before and after remains zero, exactly as the law of conservation of charge demands.

Q6. Net force on a 1 μC charge at the centre of square ABCD with corner charges qA=2μC, qB=−5μC, qC=2μC, qD=−5μC.

By symmetry, A and C (equal charges, equal distance from centre) exert equal and opposite forces on the central charge — they cancel. Similarly, B and D (equal charges, equal distance) cancel each other.

Net force = 0

Q7. Why are electrostatic field lines continuous, and why don’t two field lines ever cross?

(a) A field line traces the path a positive test charge would follow, and force on that charge acts continuously at every point in space — so the line cannot have gaps or sudden breaks.

(b) At any single point, the electric field has one unique direction and magnitude. If two lines crossed, that point would have two different field directions simultaneously, which is physically impossible.

Q8. Field and force at the midpoint of two charges qA=3μC and qB=−3μC, 20 cm apart.

(a) Distance from midpoint to each charge = 10 cm = 0.1 m E from +3μC = kq/r² = (9×10⁹ × 3×10⁻⁶)/(0.1)² = 2.7 × 10⁶ N/C, directed away from A (toward B) E from −3μC = same magnitude, 2.7 × 10⁶ N/C, also directed toward B (field lines point into negative charges)

Since both fields point the same way, they add up: E_net = 5.4 × 10⁶ N/C, directed from A to B

(b) F = qE = 1.5 × 10⁻⁹ × 5.4 × 10⁶ = 8.1 × 10⁻³ N, directed from B toward A (opposite to E, since the test charge is negative).

Q9. Total charge and dipole moment for qA=2.5×10⁻⁷ C at (0,0,−15cm) and qB=−2.5×10⁻⁷ C at (0,0,+15cm).

Total charge = qA + qB = 0 (this is exactly why it’s called a dipole — equal and opposite charges)

Separation, d = 30 cm = 0.3 m p = q × d = 2.5 × 10⁻⁷ × 0.3 = 7.5 × 10⁻⁸ C·m, directed from the negative charge toward the positive charge (i.e., along the negative z-axis, from B toward A).

Q10. Torque on a dipole of moment 4 × 10⁻⁹ C·m at 30° to a field of 5 × 10⁴ N/C.

τ = pE sinθ = 4 × 10⁻⁹ × 5 × 10⁴ × sin30° = 4 × 10⁻⁹ × 5 × 10⁴ × 0.5

τ = 1 × 10⁻⁴ N·m

Q11. Polythene rubbed with wool carries a charge of −3 × 10⁻⁷ C.

(a) Number of electrons transferred: n = q/e = (3 × 10⁻⁷)/(1.6 × 10⁻¹⁹) ≈ 1.87 × 10¹²

Electrons move from wool to polythene (wool ends up positively charged, polythene negatively charged).

(b) Yes — mass does transfer, since electrons have mass. m = n × mₑ = 1.87 × 10¹² × 9.1 × 10⁻³¹ ≈ 1.7 × 10⁻¹⁸ kg

This is an extremely tiny, practically undetectable amount of mass.

Q12. Force between identical spheres A and B, each with charge 6.5 × 10⁻⁷ C, 50 cm apart.

(a) F = kq²/r² = (9×10⁹ × (6.5×10⁻⁷)²)/(0.5)²

F = 1.52 × 10⁻² N (repulsive)

(b) When charge on each sphere doubles and distance halves: F’ = k(2q)²/(r/2)² = 16 × [kq²/r²] = 16 × 1.52 × 10⁻²

F’ = 0.243 N

Q13. A third identical uncharged sphere touches A, then B, then is removed. New force between A and B?

Let initial charge on A = B = q.

Step 1 — sphere 3 touches sphere A: charge splits equally → A = q/2, sphere 3 = q/2 Step 2 — sphere 3 (now carrying q/2) touches sphere B (still q): total = 3q/2, split equally → B = 3q/4, sphere 3 = 3q/4

Final: A = q/2, B = 3q/4

New force = k(q/2)(3q/4)/r² = (3/8) × [kq²/r²] = (3/8) × 1.52 × 10⁻²

New force ≈ 5.7 × 10⁻³ N (repulsive)

Q14. Signs of three charged particles from their tracks in a field, and which has the highest charge-to-mass ratio.

Particles moving toward the positive plate and away from the negative plate are negatively charged; a particle moving the opposite way is positively charged. The amount of deflection for a given field and velocity is proportional to the charge-to-mass ratio — the particle showing the largest deflection has the highest q/m ratio.

Q15. Flux of a uniform field E = 3 × 10³ N/C î through a 10 cm square.

(a) When the square’s plane is parallel to the yz-plane, its normal is along x, same direction as E, so θ = 0°: Φ = EA cosθ = 3×10³ × (0.1)² × cos0° = 30 N·m²/C

(b) When the normal makes 60° with the x-axis: Φ = 3×10³ × 0.01 × cos60° = 15 N·m²/C

Q16. Net flux of the same field through a cube of side 20 cm.

For a uniform field, the flux entering the cube through one set of faces exactly equals the flux leaving through the opposite faces. With no charge enclosed:

Net flux = 0

Q17. Net outward flux through a box’s surface is 8.0 × 10³ N·m²/C.

(a) By Gauss’s Law, q = Φε₀ = 8.0×10³ × 8.85×10⁻¹² ≈ 7.08 × 10⁻⁸ C

(b) No. Zero net flux only tells us the net enclosed charge is zero — the box could still contain equal amounts of positive and negative charge that cancel out.

Q18. Flux through a 10 cm square, with a +10 μC point charge 5 cm directly above its centre.

Since 5 cm is exactly half of 10 cm, the charge sits at the centre of an imaginary cube with the square as one face.

Total flux through the cube = q/ε₀ = (10×10⁻⁶)/(8.85×10⁻¹²) ≈ 1.13 × 10⁶ N·m²/C

By symmetry, this splits equally across all 6 faces:

Flux through the square = 1.13×10⁶ / 6 ≈ 1.88 × 10⁵ N·m²/C

Q19. Net flux through a cubic Gaussian surface (9 cm edge) enclosing a 2.0 μC point charge.

By Gauss’s Law, the shape and size of the surface don’t matter — only the enclosed charge does:

Φ = q/ε₀ = (2×10⁻⁶)/(8.85×10⁻¹²) ≈ 2.26 × 10⁵ N·m²/C

Q20. A spherical Gaussian surface (10 cm radius) shows flux of −1.0 × 10³ N·m²/C.

(a) Doubling the radius does not change the flux, since flux depends only on enclosed charge, not surface geometry:

Flux remains −1.0 × 10³ N·m²/C

(b) q = Φε₀ = −1.0×10³ × 8.85×10⁻¹² ≈ −8.85 × 10⁻⁹ C

Q21. Net charge on a conducting sphere (radius 10 cm) if the field 20 cm from its centre is 1.5 × 10³ N/C, pointing inward.

q = E·r²/k = (1.5×10³ × 0.04)/(9×10⁹) ≈ 6.67 × 10⁻⁹ C

Since the field points radially inward, the charge must be negative: q ≈ −6.67 × 10⁻⁹ C

Q22. Conducting sphere, diameter 2.4 m, surface charge density 80.0 μC/m².

(a) Q = σ × 4πr² = 80×10⁻⁶ × 4π(1.2)² ≈ 1.45 × 10⁻³ C

(b) Φ = Q/ε₀ = 1.45×10⁻³/8.85×10⁻¹² ≈ 1.64 × 10⁸ N·m²/C

Q23. Linear charge density of an infinite line charge producing 9 × 10⁴ N/C at 2 cm.

Using E = λ/(2πε₀r): λ = E × 2πε₀ × r = 9×10⁴ × 2π × 8.85×10⁻¹² × 0.02

λ ≈ 1.11 × 10⁻⁷ C/m

Q24. Two parallel plates with opposite surface charge densities of magnitude 17.0 × 10⁻²² C/m². Find E in three regions.

In the regions outside the plates, the fields due to the two plates cancel exactly, so:

E (outside, both sides) = 0

Between the plates, the fields add up: E = σ/ε₀ = (17×10⁻²²)/(8.85×10⁻¹²) ≈ 1.92 × 10⁻¹⁰ N/C

ADDITIONAL EXERCISES

Q25. Radius of an oil drop (12 excess electrons) held stationary in a field of 2.55 × 10⁴ N/C (Millikan’s experiment). Oil density = 1.26 g/cm³.

For the drop to be stationary, electric force balances gravity: qE = (4/3)πr³ρg

q = 12 × 1.6×10⁻¹⁹ = 1.92×10⁻¹⁸ C qE = 1.92×10⁻¹⁸ × 2.55×10⁴ = 4.896×10⁻¹⁴ N

Solving for r: r³ = 3qE / (4πρg) = 3(4.896×10⁻¹⁴) / (4π × 1260 × 9.81) ≈ 9.46×10⁻¹⁹

r ≈ 9.8 × 10⁻⁷ m

Q26. Which field-line patterns cannot represent real electrostatic fields?

A valid electrostatic field pattern must satisfy: field lines meet a conductor’s surface at right angles, lines start on positive charges and end on negative ones (never the reverse), lines never cross, and they never form closed loops in charge-free regions. Any pattern violating one of these rules is not physically possible.

Q27. Force and torque on a dipole (p = 10⁻⁷ C·m, along −z) in a field increasing along +z at 10⁵ N/C per metre.

Since the dipole axis is aligned with the direction along which the field is changing, the two ends of the dipole experience slightly different field strengths, producing a net force:

F = p × (dE/dz) = 10⁻⁷ × 10⁵ = 1 × 10⁻² N, directed along the negative z-axis (same direction as p, since the negative end sits in the stronger field).

Since the dipole is aligned (not tilted) with the field, θ = 0°, so:

Torque = pE sinθ = 0

Q28. Charge distribution on a conductor with a cavity, and electrostatic shielding.

(a) Inside a conductor in electrostatic equilibrium, the field is always zero. Applying Gauss’s Law to any surface drawn just inside the conductor (enclosing the cavity) gives zero flux, meaning zero enclosed charge. Since the cavity itself is empty, the entire charge Q must reside on the conductor’s outer surface.

(b) When an insulated charge q is placed inside the cavity, a Gaussian surface around the cavity must still enclose zero net charge (since the field inside the conductor material is zero). This forces an induced charge of −q on the cavity’s inner wall. By conservation of charge on the conductor (which originally held Q), a charge of +q gets pushed to the outer surface, making the total outer charge Q + q.

(c) This principle is used for electrostatic shielding — placing a sensitive instrument inside a hollow conducting enclosure protects it from external electric fields, since the field inside any cavity within a conductor stays zero regardless of the fields outside.

Q30. Deriving the field of an infinite line charge using Coulomb’s Law (without Gauss’s Law).

Treat the wire as a series of infinitesimal charge elements dq = λdl. Using Coulomb’s Law, compute the field contribution dE from each element at a perpendicular distance r from the wire, then integrate over the entire length. By symmetry, the components parallel to the wire cancel, and only the perpendicular components survive. Carrying out this integration over an infinite wire yields:

E = λ / (2πε₀r)

— the same result Gauss’s Law gives far more efficiently, which is exactly why Gauss’s Law is preferred for symmetric charge distributions.

Q31. Quark composition of a proton and a neutron.

Up quark charge = +2/3 e, down quark charge = −1/3 e.

Proton (uud): (2/3 + 2/3 − 1/3)e = +e ✓ Neutron (udd): (2/3 − 1/3 − 1/3)e = 0 ✓

Q32. Why is equilibrium at a null point (E = 0) always unstable?

(a) If equilibrium were stable, a small displacement in any direction would need to produce a restoring force pointing back toward the null point — meaning field lines would point inward from all directions around that point. But Gauss’s Law says the flux through a small closed surface around a point with no enclosed charge must be zero, which is impossible if every field line points inward. So stable equilibrium at a null point cannot exist.

(b) For two equal, like charges, the midpoint is the null point. Along the line joining them, a small displacement does produce a restoring force (stable in that direction). But perpendicular to that line, the resultant force pushes the test charge further away (unstable). Since true stability requires restoring forces in every direction, the equilibrium is overall unstable.

Q33. Show that vertical deflection of a charged particle entering a field region is y = qEL²/(2mvₓ²).

Time spent between the plates: t = L/vₓ Vertical acceleration: a = qE/m Vertical deflection: y = ½at² = ½(qE/m)(L/vₓ)²

y = qEL² / (2mvₓ²)

This is mathematically identical to projectile motion under gravity, with qE/m playing the role of g, and the horizontal velocity vₓ playing the role of the projectile’s launch speed.

Q34. Where does the electron (vₓ = 2.0×10⁶ m/s) strike the upper plate? Plate separation 0.5 cm, E = 9.1 × 10² N/C.

The electron must deflect by half the plate separation (0.25 cm = 2.5×10⁻³ m) to strike a plate. Using y = eEL²/(2mₑvₓ²), solve for L:

L² = 2mₑvₓ²y / (eE) = [2 × 9.1×10⁻³¹ × (2×10⁶)² × 2.5×10⁻³] / (1.6×10⁻¹⁹ × 9.1×10²)

L ≈ 1.12 × 10⁻² m ≈ 1.12 cm

Chapter 1 Summary — Electric Charges and Fields

• Charge exists in two types (positive, negative), is always quantised in units of e, and is always conserved in any isolated system.
• Coulomb’s Law governs the force between point charges: F = kq₁q₂/r².
• The electric field at a point is force per unit positive test charge: E = F/q.
• Field lines start on positive charges, end on negative ones, never cross, and are always perpendicular to a conductor’s surface.
• An electric dipole (equal, opposite charges separated by a distance) has dipole moment p = qd, and experiences a torque τ = pE sinθ in a uniform field.
• Electric flux through a surface is Φ = E·A cosθ.
• Gauss’s Law (Φ = q_enclosed/ε₀) is the fastest route to finding fields for symmetric charge distributions — spheres, infinite wires, and infinite charged sheets.

Frequently Asked Questions

Q1. Is Chapter 1 (Electric Charges and Fields) important for NEET and JEE, or only for boards?

It’s important for all three. Electrostatics questions in NEET and JEE frequently draw directly from these NCERT exercise problems — Gauss’s Law applications and dipole questions especially repeat in slightly modified forms almost every year.

Q2. How many questions come from this chapter in CBSE Class 12 board exams?

Typically 1–2 questions directly, but the concepts (Coulomb’s Law, Gauss’s Law, dipole behaviour) also support numericals in later chapters like Capacitance and Current Electricity, so the actual weightage is higher than it looks.

Q3. What’s the best way to approach Gauss’s Law numericals?

Always ask first: what is the enclosed charge? Flux depends only on that — not on the shape, size, or distance of the Gaussian surface. Most mistakes happen when students try to overcomplicate symmetric problems.

Q4. Should Class 9th/10th students worry about this chapter already?

Not the numericals, but building comfort with basic charge behaviour and force concepts early (as we guide students in our Foundation batches) makes Class 11-12 Electrostatics far less intimidating when it actually arrives.

Get These Concepts Crystal Clear with Convex Classes, Jaipur

Numericals like these are exactly where students lose easy marks — not because the concept is hard, but because the approach isn’t practiced enough. At Convex Classes, Jaipur, our Board + NEET + JEE + NDA batches are built to fix exactly this, with chapter-wise problem practice, doubt-clearing sessions, and regular tests.

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