NCERT Solutions for Class 12 Physics Chapter 2: Electrostatic Potential and Capacitance (NCERT 2026–27)

These Class 12 Physics Chapter 2 solutions cover Electrostatic Potential and Capacitance with every NCERT exercise (2.1–2.11) solved step by step, units shown and answers cross-checked against the official NCERT answer key. The chapter builds on Chapter 1 (Electric Charges and Fields) to define electric potential, equipotential surfaces, the energy of charge systems, and the working of capacitors and dielectrics — a high-weightage topic for the CBSE board exam and JEE/NEET.

Class: 12 Subject: Physics Chapter: 2 Title: Electrostatic Potential and Capacitance Exercises: 2.1 – 2.11 Session: 2026–27
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Class 12 Physics Chapter 2 Solutions – Overview

Chapter 2, Electrostatic Potential and Capacitance, treats the electrostatic field through energy rather than force. Because the Coulomb force is conservative, we can define an electrostatic potential energy and, per unit charge, an electric potential V. The chapter derives the potential of a point charge, an electric dipole and a system of charges, introduces equipotential surfaces and the field–potential relation E = −dV/dr, and computes the potential energy of charge configurations and of dipoles in external fields. It then studies conductors (field zero inside, charge on the surface, electrostatic shielding), dielectrics and polarisation, and finally capacitors — capacitance, the parallel-plate capacitor with and without a dielectric, series and parallel combinations, and the energy stored in a capacitor. The exercises are mostly numerical, so units and sign conventions must be handled carefully.

Key Concepts & Definitions

Electric potential (V): work done by an external force in bringing a unit positive charge, without acceleration, from infinity to a point. SI unit: volt (1 V = 1 J C−1). It is a scalar.

Potential difference (VP − VR): work per unit charge to move a charge from R to P; equals (UP − UR)/q and is path-independent.

Equipotential surface: a surface on which V is constant; no work is done moving a charge over it, and the electric field is everywhere normal to it.

Capacitance (C): C = Q/V, the charge stored per unit potential difference; depends only on geometry and the dielectric. SI unit: farad (1 F = 1 C V−1).

Dielectric constant (K): the factor (>1) by which capacitance rises when a dielectric fully fills the space between the plates; K = C/C0 = ε/ε0.

Electrostatic shielding: the field inside a charge-free cavity of a conductor is zero, whatever the external field — used to protect sensitive instruments.

Important Formulas (Chapter 2)

Potential of a point charge: V = Q / (4πε0r), with 1/(4πε0) = 9 × 109 N m2 C−2.

Potential of a dipole (r >> a): V = p·r̂ / (4πε0r2) = p cosθ / (4πε0r2); zero on the equatorial plane.

System of charges: V = (1/4πε0) Σ qi/riP (algebraic sum).

Field–potential relation: E = −dV/dr; E points toward decreasing potential.

P.E. of two charges: U = q1q2 / (4πε0r12). Dipole in field: U = −p·E = −pE cosθ.

Field at a conductor surface: E = σ/ε0 (normal to the surface).

Parallel-plate capacitor: C0 = ε0A/d (vacuum); C = Kε0A/d (with dielectric).

Combinations: series 1/C = 1/C1 + 1/C2 + …  •  parallel C = C1 + C2 + …

Energy stored: U = ½QV = ½CV2 = Q2/2C; energy density u = ½ε0E2.

NCERT Exercises (2.1–2.11) — Solutions

All questions are reproduced verbatim from the NCERT textbook; the worked answers are original and the numerical results have been checked against the NCERT answer key. Take 1/(4πε0) = 9 × 109 N m2 C−2 and ε0 = 8.85 × 10−12 C2 N−1 m−2.

2.1 Two charges 5 × 10−8 C and −3 × 10−8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

SOLUTION Put the positive charge q1 = 5 × 10−8 C at the origin O and the negative charge q2 = −3 × 10−8 C at A, 16 cm away on the x-axis. Let the potential be zero at a point distant x cm from O. Case 1 — point between the charges (0 < x < 16): the distance from q2 is (16 − x). Setting V = 0: k(5 × 10−8)/x − k(3 × 10−8)/(16 − x) = 0 ⇒ 5/x = 3/(16 − x). 5(16 − x) = 3x ⇒ 80 = 8x ⇒ x = 10 cm from the positive charge. Case 2 — point beyond A (x > 16): the distance from q2 is (x − 16): 5/x = 3/(x − 16) ⇒ 5x − 80 = 3x ⇒ 2x = 80 ⇒ x = 40 cm from the positive charge. Answer: the potential is zero at 10 cm and 40 cm from the positive charge, on the side of the negative charge.

2.2 A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the centre of the hexagon.

SOLUTION In a regular hexagon, the distance from the centre to each vertex equals the side, so r = 10 cm = 0.10 m for all six charges. Potential is a scalar, so V = 6 × kq/r = 6 × (9 × 109 × 5 × 10−6)/0.10. kq/r = (9 × 109 × 5 × 10−6)/0.10 = 4.5 × 104/0.10 = 4.5 × 105 V. V = 6 × 4.5 × 105 = 2.7 × 106 V.

2.3 Two charges 2 μC and −2 μC are placed at points A and B 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface?

SOLUTION (a) The two charges are equal and opposite (a dipole). Any point equidistant from A and B is at equal distance from a +2 μC and a −2 μC charge, so the potentials cancel and V = 0. Hence the equipotential surface of zero potential is the plane that is the perpendicular bisector of AB — the plane normal to AB passing through its mid-point. (b) The electric field is always normal to an equipotential surface, so on this plane the field is perpendicular to the plane, directed from the positive charge A toward the negative charge B (i.e. along AB).

2.4 A spherical conductor of radius 12 cm has a charge of 1.6 × 10−7 C distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the centre of the sphere?

SOLUTION Given R = 12 cm = 0.12 m, Q = 1.6 × 10−7 C. (a) Inside (r < R): the field inside a charged conductor is zero, because all the charge resides on the surface. (b) Just outside (r = R): E = kQ/R2 = (9 × 109 × 1.6 × 10−7)/(0.12)2 = 1440/0.0144 = 1 × 105 N C−1, directed radially outward. (c) At r = 18 cm = 0.18 m: the charge acts as if concentrated at the centre, so E = kQ/r2 = 1440/(0.18)2 = 1440/0.0324 = 4.4 × 104 N C−1, radially outward.

2.5 A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1 pF = 10−12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

SOLUTION For air, C0 = ε0A/d = 8 pF. Halving the separation (d → d/2) doubles the capacitance, and filling with a dielectric of K = 6 multiplies it by 6: C = K ε0A/(d/2) = 2K × C0. C = 2 × 6 × 8 pF = 96 pF.

2.6 Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

SOLUTION (a) In series, 1/C = 1/9 + 1/9 + 1/9 = 3/9 = 1/3 pF−1, so C = 3 pF. (b) The same charge flows through each: Q = CV = 3 × 10−12 × 120 = 3.6 × 10−10 C. Voltage across each capacitor = Q/C1 = 3.6 × 10−10/(9 × 10−12) = 40 V. (By symmetry, 120 V splits equally: 120/3 = 40 V each.)

2.7 Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

SOLUTION (a) In parallel, C = C1 + C2 + C3 = 2 + 3 + 4 = 9 pF. (b) The same voltage V = 100 V is across each, so Q = CV: On 2 pF: Q1 = 2 × 10−12 × 100 = 2 × 10−10 C. On 3 pF: Q2 = 3 × 10−12 × 100 = 3 × 10−10 C. On 4 pF: Q3 = 4 × 10−12 × 100 = 4 × 10−10 C.

2.8 In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

SOLUTION A = 6 × 10−3 m2, d = 3 mm = 3 × 10−3 m. C = ε0A/d = (8.85 × 10−12 × 6 × 10−3)/(3 × 10−3) = (5.31 × 10−14)/(3 × 10−3) = 1.77 × 10−11 F ≈ 18 pF. Charge on each plate: Q = CV = 1.77 × 10−11 × 100 = 1.8 × 10−9 C.

2.9 Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) while the voltage supply remained connected. (b) after the supply was disconnected.

SOLUTION The mica (K = 6) completely fills the 3 mm gap, so the capacitance increases six-fold: C = K C0 = 6 × 18 pF = 108 pF in both cases. (a) Supply connected: the voltage stays fixed at V = 100 V. The charge rises to Q = CV = 108 × 10−12 × 100 = 1.08 × 10−8 C (about six times the earlier charge). Extra charge flows from the battery. (b) Supply disconnected: the charge is isolated and stays at Q = 1.8 × 10−9 C. With C now 108 pF, the voltage drops to V = Q/C = (1.8 × 10−9)/(108 × 10−12) = 16.6 V (the potential difference falls by a factor of 6).

2.10 A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor?

SOLUTION U = ½CV2 = ½ × 12 × 10−12 × (50)2. = ½ × 12 × 10−12 × 2500 = 6 × 10−12 × 2500 = 1.5 × 10−8 J.

2.11 A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

SOLUTION Initial energy: Ui = ½CV2 = ½ × 600 × 10−12 × (200)2 = ½ × 600 × 10−12 × 40000 = 1.2 × 10−5 J. After connection: the initial charge Q = CV = 600 × 10−12 × 200 = 1.2 × 10−7 C is shared equally between two identical capacitors, so the common voltage is V′ = Q/(C + C) = (1.2 × 10−7)/(1200 × 10−12) = 100 V. Final energy: Uf = ½(C + C)V′2 = ½ × 1200 × 10−12 × (100)2 = 0.6 × 10−5 J. Energy lost = Ui − Uf = 1.2 × 10−5 − 0.6 × 10−5 = 6 × 10−6 J. (The lost energy is dissipated as heat and electromagnetic radiation during the transient current.)

Extra Practice Questions

Short Answer Type Questions

Q1. Why is electric potential a scalar quantity while electric field is a vector?

ANSWERPotential is defined as work done per unit charge (energy/charge), and work and energy are scalars; it has only magnitude and sign, so potentials simply add algebraically. The field is force per unit charge, and force has direction, so the field is a vector and must be added by vector addition.

Q2. Why is the work done in moving a charge over an equipotential surface zero?

ANSWERWork done = q(VB − VA). On an equipotential surface V is constant, so VB − VA = 0 and hence W = 0. Equivalently, the field is normal to the surface, so any displacement along the surface is perpendicular to the force.

Q3. The capacitance of a parallel-plate capacitor increases when a dielectric is inserted. Explain why.

ANSWERThe dielectric polarises and develops induced surface charges that set up a field opposing the applied field, reducing the net field and so the potential difference V (for a fixed charge). Since C = Q/V, a smaller V means a larger C; quantitatively C = KC0 with K > 1.

Q4. Two metal spheres of different radii are given equal charges. Which is at higher potential?

ANSWERFor an isolated sphere V = kQ/R. With Q equal, the smaller sphere (smaller R) is at the higher potential, since V is inversely proportional to R.

Q5. A charged capacitor is disconnected from the battery and its plate separation is then increased. How do the charge, capacitance, voltage and stored energy change?

ANSWERCharge Q is constant (isolated). C = ε0A/d decreases as d increases. V = Q/C increases. Energy U = Q2/2C increases — the extra energy comes from the work done by the external agent in pulling the plates apart against their attraction.

Long Answer Type Questions

Q1. Derive an expression for the capacitance of a parallel-plate capacitor with vacuum between its plates.

ANSWERLet each plate have area A and carry charge ±Q, with separation d (d2 << A so the field is uniform). The surface charge density is σ = Q/A. Between the plates the fields of the two sheets add, giving a uniform field E = σ/ε0 = Q/(ε0A), while the field outside is zero. The potential difference is V = Ed = Qd/(ε0A). By definition the capacitance is C = Q/V = Q / [Qd/(ε0A)] = ε0A/d. Thus C depends only on the geometry — directly on plate area and inversely on separation.

Q2. Obtain the expression for the energy stored in a capacitor and hence the energy density of an electric field.

ANSWERCharging a capacitor means transferring charge bit by bit. When charge Q′ is already present, the potential difference is V′ = Q′/C, so transferring a further δQ′ needs work δW = V′δQ′ = (Q′/C)δQ′. Integrating from 0 to Q: W = ∫0Q(Q′/C)dQ′ = Q2/2C. This is stored as potential energy: U = Q2/2C = ½CV2 = ½QV. For a parallel-plate capacitor, substituting C = ε0A/d and V = Ed gives U = ½ε0E2(Ad). Since Ad is the volume between the plates, the energy density is u = U/(Ad) = ½ε0E2, a result valid for any electric field.

Q3. State and prove the four key electrostatic properties of a conductor.

ANSWER(1) Field inside is zero: free electrons rearrange until any internal field vanishes; otherwise they would keep drifting, contradicting the static condition. (2) Excess charge resides on the surface: by Gauss’s law, for any closed surface drawn inside the conductor the enclosed field is zero, so the enclosed charge is zero — charge can only be on the outer surface. (3) Field just outside is normal to the surface, E = σ/ε0: any tangential component would push surface charges, contradicting equilibrium; a Gaussian pill-box at the surface gives E = σ/ε0. (4) Potential is constant throughout the conductor: since E = 0 inside and has no tangential component on the surface, no work is done moving a test charge within or on it, so V is the same everywhere — the whole conductor is an equipotential.

MCQs & Assertion–Reason

1. The SI unit of electric potential is:

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

2. The work done in moving a charge between two points on an equipotential surface is:

(a) maximum    (b) infinite    (c) zero    (d) negative

3. The electric potential due to a point charge varies with distance r as:

(a) 1/r    (b) 1/r2    (c) r    (d) r2

4. The capacitance of a parallel-plate capacitor depends on:

(a) the charge on the plates    (b) the voltage applied    (c) the plate area, separation and dielectric    (d) the current

5. Three 9 pF capacitors are joined in series. Their equivalent capacitance is:

(a) 27 pF    (b) 9 pF    (c) 3 pF    (d) 18 pF

6. Inserting a dielectric of constant K (plates fully filled) changes the capacitance to:

(a) C0/K    (b) KC0    (c) C0    (d) K2C0

7. The electric field inside a charged hollow conductor (no cavity charge) is:

(a) maximum at the centre    (b) zero    (c) σ/ε0    (d) infinite

8. The potential of an electric dipole on its equatorial plane is:

(a) maximum    (b) zero    (c) negative    (d) infinite

9. The energy stored in a capacitor of capacitance C charged to voltage V is:

(a) CV    (b) ½CV2    (c) CV2    (d) 2CV2

10. 1 electron volt (eV) equals:

(a) 1.6 × 10−19 J    (b) 1 J    (c) 9 × 109 J    (d) 1.6 × 1019 J

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

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: Electric field is always perpendicular to an equipotential surface.

Reason: If the field had a tangential component, work would be done in moving a charge along the surface, contradicting the constancy of potential.

A-R 2. Assertion: The electrostatic potential is constant throughout the volume of a conductor.

Reason: The electric field inside a conductor is zero in the static situation, so no work is done in moving a charge within it.

A-R 3. Assertion: When a dielectric is inserted into a charged isolated capacitor, its capacitance decreases.

Reason: The dielectric reduces the net electric field between the plates.

A-R 4. Assertion: The potential at the centre of a uniformly charged spherical shell is the same as at its surface.

Reason: The electric field inside the shell is zero, so no work is done moving a charge from the surface to the centre.

A-R 5. Assertion: When two charged capacitors are connected, electrostatic energy may be lost.

Reason: A transient current flows during charge redistribution and dissipates energy as heat and radiation.

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

Common Mistakes & Exam Tips

Watch out for these

  • Adding potentials as if they were vectors — potential is a scalar; add the signed values algebraically.
  • Dropping the sign of the charge when computing potential or potential energy (unlike field magnitude, sign matters here).
  • Confusing series and parallel formulae for capacitors (reciprocals add in series, capacitances add in parallel).
  • Forgetting that with the battery connected V is fixed (so Q changes), but with the battery disconnected Q is fixed (so V changes) — the key to Exercise 2.9.
  • Mixing units — convert cm to m and pF/μC to SI before substituting.
  • Writing energy as ½CV instead of ½CV2.

How to score full marks in this chapter

Show every step with units, and quote the constant you use (k = 9 × 109 or ε0 = 8.85 × 10−12). For capacitor word-problems, first decide whether the battery is connected (V constant) or disconnected (Q constant) before applying C = Q/V. State the formula, substitute, then simplify — examiners award method marks. For theory, link each property of a conductor to Gauss’s law or the static condition, and always justify why the field is normal to equipotential surfaces.

Frequently Asked Questions

What is Class 12 Physics Chapter 2 about?

Chapter 2, Electrostatic Potential and Capacitance, defines electric potential and potential energy, equipotential surfaces, the field–potential relation, the behaviour of conductors and dielectrics, and capacitors — including the parallel-plate capacitor, series and parallel combinations, and energy stored. It contains NCERT exercises 2.1 to 2.11.

How many exercises are there in Class 12 Physics Chapter 2?

There are 11 numbered exercise questions (2.1 to 2.11), most of them numerical. All are solved step by step on this page with units, and the results are matched to the official NCERT answer key.

What is the difference between connecting and disconnecting the battery when a dielectric is inserted?

With the battery connected the voltage stays fixed, so the charge increases six-fold (for K = 6). With the battery disconnected the charge stays fixed, so the voltage drops by a factor of K. This is exactly the contrast tested in Exercise 2.9.

Are these Class 12 Physics Chapter 2 solutions free?

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

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