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Electrostatic Potential and Capacitance

Capacitors and dielectrics

Calculate capacitance with and without a dielectric.

A capacitor stores charge and energy. It is two conductors (plates) close together, separated by an insulator. When connected to a battery, one plate gets +Q and the other –Q.

The charge is proportional to the voltage. The ratio is the capacitance: C = Q / V. Unit: farad (F). One farad is huge, so we use µF (10⁻⁶ F) and pF (10⁻¹² F).

For parallel plates of area A, distance d apart, in air: C = ε₀A / d. Bigger plates hold more; a smaller gap holds more.

Example: plates of 1 m², 1 mm apart: C = 8.854 × 10⁻¹² / 10⁻³ ≈ 8.9 nF. Even a square metre of plate gives only nanofarads, which shows how big a farad is.

Dielectrics: fill the gap with an insulator such as glass or mica, of dielectric constant K. Its molecules line up (it is polarised), and their charges partly cancel the field. The field inside drops to E₀/K, so for the same charge V drops too, and:

C = Kε₀A / d, K times bigger. With K = 5, the 8.9 nF capacitor becomes about 44 nF.

Inserting a dielectric, in two cases:

Battery still connectedBattery removed first
Fixed quantityVQ
C× K× K
Q× Ksame
Vsame÷ K
E between platessame÷ K

If the slab fills only part of the gap (thickness t), C = ε₀A / (d – t + t/K).

C = Kε₀A/d.
A dielectric cuts the field to E₀/K and multiplies C by K.

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