Physics - Capacitor Question with Solution | TestHub

(i) In case of a capacitor as $\mathrm{E}=(\mathrm{V}/\mathrm{d})$, the potential difference between the plates before the introduction of metal plate

$$\mathrm{V}=\mathrm{E}\times \mathrm{d}=3\times 10^4\times 0.05=1.5\mathrm{kV}$$

(ii) Now as after charging battery is removed, capacitor is isolated so $\mathrm{q}=$ constant. If ${\mathrm{C}}^{\prime }$ and ${\mathrm{V}}^{\prime }$ are the capacity and potential after the introduction of plate

$$\mathrm{q}=\mathrm{CV}={\mathrm{C}}^{\prime }{\mathrm{V}}^{\prime }\text{ i.e., }$$

${\mathrm{V}}^{\prime }=\frac{\mathrm{C}}{{\mathrm{C}}^{\prime }}\mathrm{V}$

And as $C=\frac{{\epsilon }_{0}A}{d}$ and $C^{\prime }=\frac{{\epsilon }_{0}A}{(d-t)+(t/K)},V^{\prime }=\frac{(d-t)+(t/K)}{d}\times V$

So in case of metal plate as $\mathrm{K}=\infty ,{\mathrm{V}}_{\mathrm{M}}=\left[\frac{\mathrm{d}-\mathrm{t}}{\mathrm{d}}\right]\times \mathrm{V}=\left[\frac{0.05-0.01}{0.05}\right]\times 1.5=1.2\mathrm{kV}$ And if instead of metal plate, dielectric with $\mathrm{K}=2$ is introduced

${\mathrm{V}}_{\mathrm{D}}=\left[\frac{(0.05-0.01)+(0.01/2)}{0.05}\right]\times 1.5=1.35\mathrm{kV}$