Question

Consider the situation shown in the figure. The capacitor A has a charge $q$ on it, whereas B is uncharged. The charge appearing on the capacitor B, after a long time of closing the switch is:

• A. $\frac{q}{2}$
• B. zero
• C. $q$
• D. $2q$

Answer 1

Answer: (B)

The charge on positive plate of A remains same.

$\therefore$ on the negative plate, amount of charge must be same.
In above Gaussian surface, we know $E$ in conductor is zero.

$\therefore$ total charge enclosed in Gaussian surface must be 0.
$\therefore$ charge cannot decrease from negative plate.