Question

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF =${10}^{-12}$ F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

Answer 1

$\text{Step 1: Initial capacitance}$

Let the initial distance between the plates be $d.$

Let $A$ be the area of each plate.

Capacitance of the parallel plate capacitor is given by:

$C_i=\frac{A{\in }_0}{d}$ $....\left(1\right)$

$\text{Step 2: Capacitance after inserting dielectric}$

Let $k$ be the dielectric constant.

$C_f=\frac{kA{\in }_0}{d^{\prime }}$

As the distance between plates is reduced to half, new distance is $d^{\prime }=\frac{d}{2}$

$\therefore C_f=\frac{2kA{\in }_0}{d}=2k{C}_i=2\times 6\times 8pF=96pF$