Question

Addition of a metal slab of thickness "a" between the plates of a parallel plate capacitor of plate separation "d" is equivalent to introducing a dielectric with dielectric constant "K" between the plates. Find an expression for "K" in terms of "a" and "d"?

Answer 1

Answer:

$K = \frac{d}{d+a}$

Explanation:

The capacitance of a capacitor in terms of the dielectric constant, area of the plate and the distance separating the plate is given by:

$C = \frac{\epsilon A}{d}$

Where A = Area of the plate

d = distance between the plates

$\epsilon =$ dielectric constant

Case 1:

When a meta slab of thickness, a, is added between the plates of the parallel plate capacitor , the effective separation between the plates becomes d+a

Therefore the capacitance of the capacitor becomes:

  $C = \frac{\epsilon A}{d + a}$ .......................(1)

Case 2:

Introducing a dielectric with dielectric constant K between the plates, the capacitance of the capacitor becomes:

$C = \frac{K\epsilon A}{d}$.........................(2)

Equating (1) and (2)

$\frac{K\epsilon A}{d} = \frac{\epsilon A}{d+a}\\\frac{K}{d} = \frac{1}{d+a} \\K = \frac{d}{d+a}$