Question

A capacitor is half filled with a dielectric of dielectric constant $K=2$ as shown in figure -A. If the same capacitor has to be filled with same dielectric as shown in figure B, What would be the thickness of dielectric such that capacitor still has the same value of capacitance.

• A. $\frac{2d}{3}$
• B. $\frac{3d}{2}$
• C. $\frac{3d}{4}$
• D. $\frac{4d}{3}$

Answer 1

The correct option is A $\frac{2d}{3}$

The figure A in the given question can be visualised as combination of two capacitors in parallel with each having area of plate $\frac{A}{2}$ and separation between plates being ${}^{\prime }{d}^{\prime }$.


Thus , equivalent capacitance, $C_{eq}=C_1+C_2$

$\Rightarrow C_{eq}=\frac{K\left(\frac{A}{2}{\epsilon }_0\right)}{d}+\frac{\left(\frac{A}{2}\right){\epsilon }_0}{d}$

$\Rightarrow C_{eq}=\frac{(K+1)A{\epsilon }_0}{2d}$

$\Rightarrow C_{eq}=\frac{3A{\epsilon }_0}{2d}......\left(1\right)[\because K=2]$

The arrangement shown in figure B, represents a capacitor with area $A$ and separation $d$ which is partially filled with dielectric $K$.

Thus, capacitance of the given arrangement is given by

$C_{eq}^{\prime }=\frac{A{\epsilon }_0}{(d-t)+\frac{t}{K}}$

$\Rightarrow C^{\prime }eq=\frac{A{\epsilon }_0}{(d-t)+\frac{t}{2}}$

$\Rightarrow C_{eq}^{\prime }=\frac{A{\epsilon }_0}{d-\frac{t}{2}}........\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get,

$\frac{A{\epsilon }_0}{d-\frac{t}{2}}=\frac{3A{\epsilon }_0}{2d}$

$\Rightarrow 2d=3(d-\frac{t}{2})$

or, $2d=3d-\frac{3t}{2}$

or , $\frac{3t}{2}=d\therefore t=\frac{2d}{3}$

Thus, required thickness is $\frac{2d}{3}$

Hence, option (a) is the correct answer.