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Electrostatic Potential and Capacitance

NCERT Class 12 Physics Chapter 2

Electrostatic Potential and Capacitance for Class 12 Physics, JEE and NEET: potential, equipotential surfaces, capacitors and dielectrics.

15 topics Updated 8 Nov 2026

For: Class 12, JEE and NEET

In the syllabus: JEE Class 12 syllabus, NEET Class 12 syllabus

With your textbook

NCERT Physics Part I (Class 12), Chapter 2 “Electrostatic Potential and Capacitance”, reprint 2026-27.

Open the official NCERT chapter (PDF, 3.9 MB, ncert.nic.in)

Checked on 19 Oct 2026.

Learn in this order

  1. Electrostatic potential energy

    Definition, why the Coulomb force is conservative, ΔU as external work, the zero at infinity.

    NCERT Section 2.1 10 min read 4 solved examples 6 practice questions

  2. Electrostatic potential

    Electrostatic potential as work per unit charge, potential difference, W = qΔV, volts and electron volts.

    NCERT Section 2.2 10 min read 4 solved examples 6 practice questions

  3. Potential due to a point charge

    The derivation of V = Q/4πεr, its sign, V against E with distance, and NCERT Example 2.1.

    NCERT Section 2.3 8 min read 4 solved examples 6 practice questions

  4. Potential due to an electric dipole

    Derivation of V = p cos θ/4πεr, its value at axial and equatorial points, and the 1/r fall-off.

    NCERT Section 2.4 9 min read 4 solved examples 6 practice questions

  5. Potential due to a system of charges

    Superposition, a charged shell inside and outside, and NCERT Examples 2.2 and 2.3.

    NCERT Section 2.5 10 min read 4 solved examples 7 practice questions

  6. Equipotential surfaces

    Properties, shapes for a point charge and a uniform field, and the relation between field and potential.

    NCERT Section 2.6 10 min read 4 solved examples 6 practice questions

  7. Potential energy of a system of charges

    Two charges, three or more charges, signs of U, and NCERT Example 2.4 on a square.

    NCERT Section 2.7 8 min read 4 solved examples 7 practice questions

  8. Potential energy in an external field

    U = −pE cos θ, a single charge’s U = qV, two charges, and NCERT Example 2.6.

    NCERT Section 2.8 11 min read 4 solved examples 6 practice questions

  9. Electrostatics of conductors

    Why E = 0 inside, charge on the surface, E = σ/ε just outside, and electrostatic shielding.

    NCERT Section 2.9 12 min read 4 solved examples 6 practice questions

  10. Dielectrics and polarisation

    Polar and non-polar molecules, induced dipoles, P = εχE, and bound surface charge ±σp.

    NCERT Section 2.10 11 min read 4 solved examples 6 practice questions

  11. Capacitors and capacitance

    Two conductors and an insulator, C = Q/V, the farad, µF and pF, and dielectric strength of air.

    NCERT Section 2.11 12 min read 4 solved examples 6 practice questions

  12. The parallel plate capacitor

    Field σ/ε between the plates, zero outside, C = εA/d derivation, and the force between the plates.

    NCERT Section 2.12 9 min read 4 solved examples 6 practice questions

  13. Effect of a dielectric on capacitance

    C = KC, the dielectric constant K = ε/ε, partial slabs, battery connected or not.

    NCERT Section 2.13 13 min read 4 solved examples 6 practice questions

  14. Combination of capacitors

    Same charge or same voltage, 1/C = 1/C + 1/C, C = C + C, and NCERT Example 2.9.

    NCERT Section 2.14 10 min read 4 solved examples 6 practice questions

  15. Energy stored in a capacitor

    Derivation of U = ½CV, energy density ½εE, energy lost in sharing charge, solved examples.

    NCERT Section 2.15 12 min read 4 solved examples 6 practice questions

Key formulas in this chapter

Electrostatic potential energy

  • ΔU = UP − UR = WRP
  • work by the field = −WRP
  • UP = W∞P (U = 0 at infinity)
  • (UP + α) − (UR + α) = UP − UR
  • work round a closed path = 0

Electrostatic potential

  • VP − VR = (UP − UR)/q
  • Wext = q(VP − VR)
  • U = qV
  • 1 V = 1 J C
  • 1 eV = 1.6 × 10 J

Potential due to a point charge

  • V = Q/(4πεr)
  • Er = Q/(4πεr) (signed radial component)
  • |V| = |E|r (one point charge, zero at infinity)
  • W = qV (from infinity)
  • W = −∫ Q dr′/(4πεr′), from ∞ to r

Potential due to a dipole

  • V = (q/4πε)(1/r − 1/r)
  • V = p cos θ/(4πεr), r ≫ a
  • far-field axis: V = ±p/(4πεr)
  • equatorial plane: V = 0
  • dipole V ∝ 1/r, point charge V ∝ 1/r

Potential due to a system of charges

  • V = V + V + … + V
  • V = (1/4πε) Σ q/rP
  • shell, r ≥ R: V = q/(4πεr)
  • shell, r ≤ R: V = q/(4πεR)
  • zero potential: q/r + q/r = 0

Equipotential surfaces

  • equipotential: V constant on the surface
  • E normal to the equipotential, towards lower V
  • |E| = −dV/dl along the field; |ΔV|/Δl for close surfaces
  • uniform field: E = V/d
  • 1 V/m = 1 N/C

Potential energy of a system of charges

  • U = qq/(4πεr)
  • U = U + U + U, U = kqq/r
  • n charges: one term per pair, n(n − 1)/2 terms
  • like charges: U > 0; unlike charges: U < 0
  • work to separate to infinity = −U

Potential energy in an external field

  • U = qV(r)
  • U = qV(r) + qV(r) + qq/(4πεr)
  • U(θ) = −pE cos θ = −p·E
  • W = pE(cos θ − cos θ)
  • 1 eV = 1.6 × 10 J

Electrostatics of conductors

  • inside a conductor: E = 0
  • surface field normal; charge only on the surface
  • V constant throughout the conductor
  • just outside: E = σ/ε
  • empty cavity: E = 0 (shielding)

Dielectrics and polarisation

  • P = dipole moment per unit volume, in C m
  • P = εχE (linear isotropic dielectric)
  • χ: a pure number, fixed by the dielectric
  • slab, P normal to two faces: σp = P
  • field inside a dielectric: reduced, not zero

Capacitors and capacitance

  • C = Q/V
  • 1 F = 1 C V = 1 C N m
  • 1 µF = 10 F, 1 nF = 10 F, 1 pF = 10 F
  • air breaks down at about 3 × 10 V m
  • isolated sphere: C = 4πεR

The parallel plate capacitor

  • σ = Q/A
  • between the plates: E = Q/(εA) = σ/ε
  • outside: E = 0 in the ideal infinite-sheet approximation
  • V = Ed = Qd/(εA)
  • C = εA/d

Effect of a dielectric on capacitance

  • C = εKA/d = KC
  • ε = εK
  • K = ε/ε = C/C (> 1)
  • Q fixed: V → V/K, E → E/K
  • slab of thickness t: C = εA/(d − t + t/K)

Combination of capacitors

  • series: 1/C = 1/C + 1/C + … + 1/C
  • series: same Q, V = V + V + …
  • two in series: C = CC/(C + C)
  • parallel: C = C + C + … + C
  • parallel: same V, Q = Q + Q + …

Energy stored in a capacitor

  • U = Q/(2C) = ½CV = ½QV
  • δW = V′δQ′ = (Q′/C)δQ′
  • u = ½εE
  • parallel plates: U = ½εE × Ad
  • loss on joining: CC(V − V)/[2(C + C)]

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