A parallel plate capacitor is connected to a battery as shown in … - Vidyadip

Question

A parallel plate capacitor is connected to a battery as shown in Fig. Consider two situations:
A: Key K is kept closed and plates of capacitors are moved apart using insulating handle.
B: Key K is opened and plates of capacitors are moved apart using insulating handle. Choose the correct option(s).

✓

Answer

In A : V remains same and hence Q changes.
In B : Q remains same and hence V changes.

In case A, key K is kept closed and plates of capacitors are moved apart using insulating handle (i.e. distance between plates is increasing). As capacitance $\text{C}=\Big[\frac{(\text{K}\epsilon_0\text{A})}{\text{d}}\Big]=\text{C}\propto\text{d}$ (separation between plates) so, capacitance will decrease & amount of charge stored will also decrease (as Q = CV). Here, there will be no change in potential difference.
In case B, key K is opened and plates of capacitors are moved apart using insulating handle, by conservation of charge, charge stored by the capacitor remains same. Plates of capacitance are moving apart so capacitance will decrease and with the decreases of capacitance, potential difference V increases $\Big(\text{as V}=\frac{\text{Q}}{\text{C}}\Big)$.

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