Question

The capacity of a parallel plate capacitor is $10\mu \text{F}$. This capacitor is divided into two equal parts and these parts are filled by media of dielectric constant $2$ and $4$. The new capacitance will be

• A. $20\mu \text{F}$
• B. $30\mu \text{F}$
• C. $10\mu \text{F}$
• D. $40\mu \text{F}$

Answer 1

The correct option is B $30\mu \text{F}$

Given:
$k_1=2;k_2=4;C=10\mu \text{F}$

As we know that capacitance of parallel plate capacitor is

$C=\frac{{ϵ}_{0}A}{d}=10\mu \text{F}.....\left(1\right)$

From given diagram, we can say that the two parts can be treated as two capacitors in parallel, with separation $d$ and plate area $\frac{A}{2}$.

So, effective capacitance will be

$C_{eq}=C_1+C_2=\frac{k_1{ϵ}_0\left(A/2\right)}{d}+\frac{k_2{ϵ}_0\left(A/2\right)}{d}$

$\Rightarrow C_{eq}=\frac{{ϵ}_{0}A(k_1+k_2)}{2d}$

from equation $\left(1\right)$ and given data

$\Rightarrow C_{eq}=10\times \frac{(2+4)}{2}=30\mu \text{F}$

Hence, option $\left(b\right)$ is correct.

Key concept: $C_{eq}=\frac{{ϵ}_{0}A(k_1+k_2)}{2d}$