Question

Four identical metal plates are located in air at equal distances $d$ from one another. The area of each plate is equal to $A$. Find the capacitance of the system between points $A$ and $B$ if the plates are interconnected as shown (a) in Fig. (a) (b) in Fig. (b).

Answer 1

(a) In given arrangement, we have 3 capacitor of equal capacitance C and the first and third plates are at same potential Hence we can resolve the network into simple form.

Ref. image

$C_{eq}=\frac{(C+C)C}{(C+C)+C}=\frac{2}{3}C$

(b) Let us impart the charges +q and -q to the plates 1 and 2 and then distribute them to other plates using charge conservation and electric induction

As potential difference between plate 1 and 2 is zero

$\frac{-q_1}{C}+\frac{q_2}{C}=\frac{q_1}{C}=0\Rightarrow q_2=2q_1$

The potential difference between point A and B

$V=V_A-V_B=q_2/C$

Hence the sought capacitance

$C_{eq}=\frac{q}{V}=\frac{q_1+q_2}{q_2/C}=\frac{3q_1}{2q_1/C}=\frac{3}{2}C$