Question

Five identical capacitor plates, each of area A, are arranged such that the adjacent plates are at a distance d apart. The plates are connected to a battery of e.m.f. E volt as shown. The charges on plates 1 and 4 respectively are :

• A. $\left(\frac{{ϵ}_{0}A}{d}\right)E$ and $\left(\frac{-2{ϵ}_{0}A}{d}\right)E$
• B. $\left(\frac{2{ϵ}_{0}A}{d}\right)E$ and $\left(\frac{-{ϵ}_{0}A}{d}\right)E$
• C. $\left(\frac{-{ϵ}_{0}A}{d}\right)E$ and $\left(\frac{2{ϵ}_{0}A}{d}\right)E$
• D. $\left(\frac{-2{ϵ}_{0}A}{d}\right)E$ and $\left(\frac{{ϵ}_{0}A}{d}\right)E$

Answer 1

Answer: (A)

The arrangement is equivalent to four capacitors each of capacity $C=\frac{{ϵ}_{0}A}{d}$.

$\therefore$ charge on each capacitor is

$q=CV=CE=\frac{{ϵ}_{o}AE}{d}$

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 $+\frac{{ϵ}_{0}AE}{d}$ 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

$-\frac{2{ϵ}_{0}AE}{d}$.