The capacitance of arrangement of 4 plates of area $A$ at distance $d$ as shown in figure is

Question

Aicharya · Accepted Answer

$\begin{aligned} c^{\prime}=& c+c+c=3 c \ &=\frac{3 \times \varepsilon_{0} A}{d} \end{aligned}$ Plate 1 and 2 make one set of capacitors. Similarly plate 2 and 3 make one set of capacitor and plate 3 and 4 make one set of capacitor. Hence there are three capacitors joined in parallel. Hence equivalent capacitance is: $\mathrm{C}=\frac{3 \mathrm{~A} \varepsilon_{0}}{\mathrm{~d}}$

The capacitance of arrangement of 4 plates of area \(A\) at distance \(d\) as shown in figure is

Solution