The parallel-plate capacitor in Figure Q26.11 is connected to a battery having potential difference $∆V_{bat}$. Without breaking any of the connections, insulating handles are used to increase the plate separation to $2d$.

a. Does the potential difference $∆V_C$change as the separation increases? If so, by what factor? If not, why not?

b. Does the capacitance change? If so, by what factor? If not, why not?

c. Does the capacitor charge $Q$change? If so, by what factor? If not, why not?

Given :

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

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

Formula used :

1) $C=\frac{{\epsilon }_{0}A}{d}$

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 $C=\frac{{\epsilon }_{0}A}{d}$, 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.