Question

The plates of a parallel plate capacitor are charged up to $100V$. Now, after removing the battery, a $2mm$ thick plate is inserted between the plates. Then, to maintain the same potential difference, the distance between the plates is increased by $1.6mm$. Dielectric constant of the plate is

• A. 5
• B. 1.25
• C. 4
• D. 2.5

Answer 1

The correct option is A 5

Initially,
and finally
and that can be redrawn as

As $Q$, $V$ are constant $C$ should also be constant.
$C_f=\frac{C_1\times C_2}{C_1+C_2}=\frac{\frac{k{\epsilon }_{0}A}{2}\times \frac{{\epsilon }_{0}A}{(d-0.4)}}{\frac{k{\epsilon }_{0}A}{2}+\frac{{\epsilon }_{0}A}{(d-0.4)}}$
Therfore,

$\frac{{\epsilon }_{0}A}{d}=\frac{\frac{k{\epsilon }_{0}A}{2}\times \frac{{\epsilon }_{0}A}{(d-0.4)}}{\frac{k{\epsilon }_{0}A}{2}+\frac{{\epsilon }_{0}A}{(d-0.4)}}$

On solving this, you get $k=5$.