Question

The capacitance of a parallel plate capacitor having plate area $A$ and separation $2d$ with dielectrics inserted as shown in figure is

• A. $\frac{3k{ϵ}_{0}A}{2d}$
• B. $\frac{5k{ϵ}_{0}A}{12d}$
• C. $\frac{k{ϵ}_{0}A}{d}$
• D. $\frac{7k{ϵ}_{0}A}{12d}$

Answer 1

The correct option is D $\frac{7k{ϵ}_{0}A}{12d}$

From the given diagram, we can see that the top two dielectrics constitute two capacitors in series $(C_1$ and $C_2$) which are in parallel with the lower one $\left(C_3\right)$.

So, redraw the circuits in reduce form


Let $\frac{k{ϵ}_{0}A}{d}=C\left(\text{say}\right)$

Now, the capacitance of each term will be

$C_1=\frac{k{ϵ}_0\left(A/2\right)}{d}=\frac{k{ϵ}_{0}A}{2d}=\frac{C}{2}$

$C_2=\frac{2k{ϵ}_0\left(A/2\right)}{d}=\frac{k{ϵ}_{0}A}{d}=C$

$C_1$ and $C_2$ are in series. Their equivalent is given by

$C_{series}=\frac{C_1{C}_2}{C_1+C_2}=\frac{\frac{C}{2}\times C}{\frac{C}{2}+C}=\frac{C}{3}$

Now, $C_3=\frac{k{ϵ}_0\left(A/2\right)}{2d}=\frac{k{ϵ}_{0}A}{4d}=\frac{C}{4}$

So, the equivalent capacitance of the whole circuit is,

$C_{eq}=C_{series}+C_3=\frac{C}{3}+\frac{C}{4}=\frac{7C}{12}$

$\therefore C_{eq}=\frac{7k{ϵ}_{0}A}{12d}$

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