Question

A parallel plate capacitor has the space between its plates filled by two slabs of thickness $d/2$ each with dielectric constants $k_1$ and $k_2$. Given that $d$ is the plate separation of the capacitor, the effective capacitance is

• A. $\frac{{ϵ}_{0}A(k_1+k_2)}{2d}$
• B. $\frac{2{ϵ}_{0}A}{d}\frac{k_1{k}_2}{k_1+k_2}$
• C. $\frac{{ϵ}_{0}A}{d}\frac{k_1{k}_2}{k_1-k_2}$
• D. $\frac{{ϵ}_{0}A(k_1-k_2)}{d}$

Answer 1

The correct option is B $\frac{2{ϵ}_{0}A}{d}\frac{k_1{k}_2}{k_1+k_2}$

The parts behave like 2 capacitors connected in series

This implies,
$C_1=\frac{k_1{ϵ}_{0}A}{d/2}=\frac{2k_1{ϵ}_{0}A}{d}$ and

$C_2=\frac{k_2{ϵ}_{0}A}{d/2}=\frac{2k_2{ϵ}_{0}A}{d}$

Now the effective capacitance for series combination,

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$

$\Rightarrow C_{eq}=\frac{C_1{C}_2}{C_1+C_2}=\frac{2{ϵ}_{0}A}{d}\frac{k_1{k}_2}{(k_1+k_2)}$

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

Key concept- Two or more dielectrics placed parallel to the capacitor plates can be assumed to be in series.