Formulas/physics/Electrostatic Potential/Capacitor with Partial Dielectric Slab

Capacitor with Partial Dielectric Slab

Capacitance when a dielectric slab of thickness t and constant K fills part of the gap d. Treated as two capacitors in series: one vacuum (d − t), one dielectric (t).

Class 12

Derivation

Setup

Parallel plate capacitor: plate area $A$ , total gap $d$ . A dielectric slab of thickness $t$ and constant $K$ fills part of the gap, leaving a vacuum gap of $(d - t)$ .

Series capacitor model

The two regions — vacuum and dielectric — are traversed by the same field lines. They see the same charge $Q$ on the plates. This is the condition for series connection.

Treat as two capacitors in series:

Vacuum layer: $C_1 = \dfrac{\varepsilon_0 A}{d - t}$
Dielectric layer: $C_2 = \dfrac{K\varepsilon_0 A}{t}$

Equivalent capacitance

\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d-t}{\varepsilon_0 A} + \frac{t}{K\varepsilon_0 A} = \frac{1}{\varepsilon_0 A}\left(d - t + \frac{t}{K}\right)

\boxed{C = \frac{\varepsilon_0 A}{d - t + t/K}}

Limiting cases

$t = 0$ (no slab): $C = \varepsilon_0 A/d = C_0$ ✓

$t = d$ (full slab): $C = K\varepsilon_0 A/d = KC_0$ ✓

$K = 1$ (vacuum slab): $C = \varepsilon_0 A/d = C_0$ ✓

Conducting slab ( $K \to \infty$ )

C = \frac{\varepsilon_0 A}{d - t}

A conducting slab of thickness $t$ effectively reduces the gap to $(d - t)$ .

Note

If the slab is placed parallel to the plates (as assumed here), the analysis is series. If multiple dielectric slabs are placed side by side (vertically), each covers a fraction of the area at the full gap — that is a parallel combination.