Question

Each of the three plates shown in the figure has area of $400{\text{cm}}^2$ on one side and the gap between the adjacent plates is $8.85\text{mm}$. The equivalent capacitance of the system is

• A. $70.8\text{pF}$
• B. $80\text{pF}$
• C. $160\text{pF}$
• D. $40\text{pF}$

Answer 1

The correct option is B $80\text{pF}$


Let potential at C be $V_1$ and
potential at D be $V_2$

$\Rightarrow$$V_{\text{A}}=V_{\text{B}}=V_{\text{D}}=V_2$

Two plates facing each other form a parallel plate capacitor. So there are two parallel plate capacitors between the terminals. Lets understand it better by redrawing the circuit


The two capacitors are in parallel.
Capacitance of each capacitor, $C=\frac{{\epsilon }_{o}A}{d}$

Equivalent capacitance , $C_{\text{eq}}=C+C=2C$

$\Rightarrow C_{\text{eq}}=\frac{2{\epsilon }_{o}A}{d}=\frac{2\times 8.85\times {10}^{-12}\times 400\times {10}^{-4}}{8.85\times {10}^{-3}}$

$C_{\text{eq}}=80\times {10}^{-12}\text{F}$
$C_{\text{eq}}=80\text{pF}$