Question

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

• A. $2.5$
• B. $4$
• C. $5$
• D. $1.25$

Answer 1

The correct option is C $5$

As the battery is disconnected, charge $q$ will remain the same.

It is given that final potential $V$ is the same.

From $q=CV$ , we can say that the final capacitance should also be same as the initial capacitance i.e.,

$C_1=C_2....\left(1\right)$ ​

If the area of the plates is $A$ and the intial separation between them is $d$ then the capacitance will be

$C_1=\frac{{\epsilon }_{0}A}{d}....\left(2\right)$ ​

As we know that capacitance of partially filled dielectric slab of thickness $t$ and dielectric constant $K$ is given by

$C_2=\frac{{\epsilon }_{0}A}{d^{\prime }-t+\frac{t}{K}}....\left(3\right)$ ​

Given,
$t=2\text{mm}$ and $d^{\prime }=d+1.6$

From equation $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$, we have

$\frac{{\epsilon }_{0}A}{d}=\frac{{\epsilon }_{0}A}{(d+1.6)-2+\frac{2}{K}}$

On comaparing the both sides term we get,

$1.6-2+\frac{2}{K}=0$

$\Rightarrow \frac{2}{K}=0.4$

$\therefore K=5$

Hence, option (a) is the correct answer.

$\text{Alternate Solution:}$
Let, the initial voltage across the capacitor is $V_0$ and the initial Electric field be $E_0$

$\Rightarrow V_0=E_{0}d....\left(1\right)$

Let dielectric constant be $K$. So, the new voltage,

$V^{\prime }=E_0(d-t+\frac{t}{K})$

Given, $t=2\text{mm}$

To increase the voltage from $V^{\prime }$ to $V_0$, separation between the plates is increased by $1.6\text{mm}$.So,

$V_0=E_0(d-t+\frac{t}{K}+1.6)....\left(2\right)$

Comparing the equations $\left(1\right)$ and $\left(2\right)$, we get

$-t+\frac{t}{K}+1.6=0$

$\Rightarrow -2+\frac{2}{K}+1.6=0$

$\Rightarrow K=5$

Hence, option (a) is the correct answer.

Key concept: 1. For an isolated charged capacitor, the electric field inside the dielectric is reduced by the factor of dielectric constant of the dielectric. 2. If the distance between the plates of a parallel plate capacitor is small, the electric field can be assumed to be uniform.