Question

Capacitance of a parallel plate capacitor varies directly as nth power of the distance between the plates. The value of n is

a-1
b-2
c-(-1)
d-(-2)

Answer 1

Given:

Capacitance of a parallel plate capacitor varies directly as nth power of the distance between the plates.

To find:

Value of 'n' ?

Calculation:

In a parallel plate capacitor,

The Electrostatic Field Intensity will be:

$E = \dfrac{ \sigma}{ \epsilon_{0}}$

Now, $\sigma$ (surface charge density) can be represented as:

$\implies E = \dfrac{ q}{ A\epsilon_{0}}$

Now, potential difference will be:

$\implies V = Ed = \dfrac{ qd}{ A\epsilon_{0}}$

Now, capacitance be C :

$C = \dfrac{q}{V}$

$\implies C = \dfrac{q}{(\dfrac{ qd}{ A\epsilon_{0}})}$

$\implies C = \dfrac{ A\epsilon_{0}}{d}$

$\implies C \propto \: {d}^{ - 1}$

So , value of 'n' is -1.