Question

A dielectric slab of thickness 't’ is kept between the plates of a parallel plate capacitor with plate separation 'd' (t < d). Derive the expression for the capacitance of the capacitor.

Answer 1

P and Q are two parallel plates. The distance between them is d.

A dielectric material of thickness t(t < d) and dielectric constant K is inserted in between the plates.

In the remaining space (d - t) there is air.

The electric field in the dielectric-filled portion = `E_1 = sigma/(epsi_0K)`

The electric field in the air-filled portion = `E_2 = sigma/(epsi_0)`

The potential difference between the plates = V

V = E₂ × (d - t) + E₁ × t

Or, V = `sigma/(2epsi_0) xx (d - t) + sigma/(epsi_0K) xx t`

Or, V = `q/(Aepsi_0) xx [(d - t) + t/K]`

Now, capacitance = C = `q/V`

Or, C = `(epsi_0A)/((d - t) + t/K)`