91Ó°ÊÓ

Given :

Battery has potential difference : $∆V_{bat}$.

The plate separation is increased to : $2d$.

Formula used :

1) $C=\frac{{\mathrm{Î\mu }}_{0}A}{d}$

2)$C=\frac{Q}{∆V_C}$

(a) No, $∆V_c$does not change since the upper plate is still connected to the positive electrode of the battery by a conducting wire, therefore it is still at the potential of the positive electrode.

Similarly, because the bottom plate is connected to the negative electrode, its potential is unaffected.

As a result, there is no change in the potential difference between the top and bottom plates.

(b) The answer is yes. Because $C=\frac{{\mathrm{Î\mu }}_{0}A}{d}$, the capacitance reduces by a factor of two when the plate spacing $d$is twice.

(c) Yes.

In addition, the charge is reduced by a factor of two. $Q$reduces by the same factor as the capacitance since $C=\frac{Q}{∆V_C}$drops while$∆V_c$ remains constant.