Question

The distance between the plates of a parallel plate capacitor is $0.05\text{m}$. An electric field of magnitude $3\times {10}^4\text{V/m}$ is established between the plates by connecting it to a battery. It is disconnected from the battery and an uncharged plate of thickness $0.01\text{m}$ and dielectric constant $k=2$ is inserted then, what will be the magnitude of difference in potential difference (in $\text{kV}$) between plates?

Answer 1

In case of a parallel plate capacitor the $\overrightarrow{E}$between plates is constant.

The potential difference between parallel plates before introduction of uncharged plate is given by $V=Ed$

$\Rightarrow V=(3\times {10}^4)\times (5\times {10}^{-2})$

$\Rightarrow V=1500\text{V}=1.5\text{kV}$

The capacitance, $C=\frac{{\epsilon }_{0}A}{d}.....\left(1\right)$

After the battery is removed, the capacitor becomes isolated and $\text{Charge=constant}$.

Let, the capacitance and potential difference be $C_1$ and $V_1$, when dielectric slab is inserted.Then $$C_1=\frac{A{\epsilon }_0}{d-t+\frac{t}{k}}$$$\Rightarrow C_1=\frac{A{\epsilon }_0}{(0.05-0.01)+\left(\frac{0.01}{2}\right)}=\frac{A{\epsilon }_0}{0.045}....\left(2\right)$

As, charge is constant, we can write $C_1{V}_1=CV$
Using equation $\left(1\right)$ and $\left(2\right)$, we have

$\frac{A{\epsilon }_0}{0.045}\times V_1=\frac{A{\epsilon }_0}{0.05}\times 1.5$

$\Rightarrow V_1=\frac{0.045}{0.05}\times 1.5=1.35\text{kV}$

Thus the magnitude of difference in potential difference for the two senario decribed is,
$\Delta V=|V_1-V|$
$\Rightarrow \Delta V=|1.35-1.5|$
$\therefore \Delta V=0.15\text{kV}$

Accepted answer : $0.15$