Question

Five identical plates, each of area A are arranged such that adjacent plates are at a distance d apart and are numbered as 1, 2, 3, 4 and 5 from left to right. The plates are connected to a battery of voltage V as shown in figure If $Q_1$ is the charge on the plate 1 and $Q_4$ on plate 4, then,

• A. $Q_1=\frac{{ϵ}_{0}AV}{d}$
• B. $Q_1=\frac{2{ϵ}_{0}AV}{d}$
• C. $Q_4=\frac{-2{ϵ}_{0}AV}{d}$
• D. $Q_4=-\frac{{ϵ}_{0}AV}{d}$

Answer 1

Answer: (A), (D)

The correct options are
A $Q_1=\frac{{ϵ}_{0}AV}{d}$
D $Q_4=-\frac{{ϵ}_{0}AV}{d}$

The arrangement is equivalent to four capacitors each of capacity $C=\frac{A{ϵ}_0}{d}$

∴ charge on each capacitor is $Q=CV=\frac{AV{ϵ}_0}{d}$

Now, the plate 1 is common only to one condensor and is connected to the positive terminal of the battery, therefore the charge on it is $+\frac{AV{ϵ}_0}{d}$

while the place 4 is common to two capacitors (formed between plate 3 and 4 and between 4 and 5, therefore it has a charge $-2\frac{AV{ϵ}_0}{d}$.