Electrostatic Potential and Capacitance

Physics · Class 12

Lesson 9 of 13 · 7 min

Capacitors and the parallel plate capacitor

NCERT §2.11 and §2.12

Meera's demo capacitor is two 10 cm × 10 cm metal plates, 2 mm apart, joined to a 100 V supply. How much charge does it hold, and why does the field stay between the plates?

Loading the full lesson

The lesson in notes

In short

A capacitor is two conductors separated by an insulator, usually carrying charges +Q and −Q with a potential difference V between them. Q is the charge on one plate; the total is zero.

Doubling Q doubles the field and so doubles V, so Q/V is fixed: C = Q/V, the capacitance. It depends on the shape, size and spacing of the conductors and on the insulator between them, not on Q or V.

Unit: 1 farad = 1 C V⁻¹ = 1 C² N⁻¹ m⁻¹. The farad is huge; practical capacitors are in μF (10⁻⁶ F), nF (10⁻⁹ F) and pF (10⁻¹² F).

A large C stores a large charge at a small V. This matters because a high V means a strong field that can ionise the air and let the charge leak away.

Dielectric strength is the largest field an insulator withstands before it breaks down; for air it is about 3 × 10⁶ V m⁻¹, which is 3 × 10⁴ V across a 1 cm gap.

Parallel plate capacitor (plate area A, gap d, d² ≪ A): outside the plates the fields of the two sheets cancel, and between them they add to E = σ/ε₀ = Q/(ε₀A), uniform and pointing from + to −.

V = Ed = Qd/(ε₀A), so C = ε₀A/d. Near the plate edges the field lines bulge outward (fringing), which this formula ignores.

Worked values: A = 1 m² and d = 1 mm give C = 8.85 × 10⁻⁹ F. For 1 F with d = 1 cm the plates would need an area of about 10⁹ m², roughly 30 km on each side.

Capacitors and the parallel plate capacitor | Electrostatic Potential and Capacitance | Lumi Learn