Question

The net capacitance of the system shown in the figure is -

• A. $\frac{25}{24}\frac{{ϵ}_0\text{A}}{d}$
• B. $\frac{24}{25}\frac{{ϵ}_0\text{A}}{d}$
• C. $\frac{21}{10}\frac{{ϵ}_0\text{A}}{d}$
• D. $\frac{10}{21}\frac{{ϵ}_0\text{A}}{d}$

Answer 1

The correct option is A $\frac{25}{24}\frac{{ϵ}_0\text{A}}{d}$


$C_1$ and $C_3$ are in series.

So, $C_{\text{net 1}}=\frac{C_1{C}_3}{C_1+C_3}$

$\Rightarrow C_{\text{net 1}}=\frac{1{ϵ}_{o}A/2d\times 3{ϵ}_{o}A/2d}{1{ϵ}_{o}A/2d+3{ϵ}_{o}A/2d}$

$\Rightarrow C_{\text{net 1}}=\frac{3{ϵ}_{o}A}{8d}$

Similarly, $C_2$ and $C_4$ are in series.

So, $C_{\text{net 2}}=\frac{C_2{C}_4}{C_2+C_4}$

$\Rightarrow C_{\text{net 2}}=\frac{2{ϵ}_{o}A/2d\times 4{ϵ}_{o}A/2d}{2{ϵ}_{o}A/2d+4{ϵ}_{o}A/2d}$

$\Rightarrow C_{\text{net 2}}=\frac{2{ϵ}_{o}A}{3d}$



Further, $C_{\text{net 1}}$ and $C_{\text{net 2}}$ are in parallel.

$\therefore C_{\text{net}}=C_{\text{net 1}}+C_{\text{net 2}}$

$\Rightarrow C_{\text{net}}=\frac{3{ϵ}_{o}A}{8d}+\frac{2{ϵ}_{o}A}{3d}=\frac{25}{24}\frac{{ϵ}_{o}A}{d}$