Question

A parallel plate capacitor of capacitance (C₀) has plates of area (A) with separation d between them. When it is connected to a battery of voltage (V₀), it has charge of magnitude (Q₀ ) on its plates. It is then disconnected from the battery and the plates are pulled apart to a separation (2d) without discharging them. After the plates are 2 d apart, the new capacitance and the potential difference between the plates are

a. 0.5C₀&2 V₀
​b. C₀&0.5 V₀
c. C₀& V₀
​d. C₀&2 V₀
​e. 2C₀&2 V₀​
f. All the above are not correct

Answer 1

After disconnecting the capacitor from the battery and doubling the separation between the plates, the new capacitance is half the original, and the potential difference is doubled.

Step-by-step explanation:

The student's question relates to the change in capacitance and the potential difference of a parallel plate capacitor when the separation between the plates is increased. Given that the capacitance (C₀) of a parallel plate capacitor is inversely proportional to the separation (d) between the plates, when the distance is doubled to 2d, the new capacitance becomes half, that is 0.5C₀. As the capacitor was disconnected from the battery before the plates were moved apart, the charge (Q₀) remains the same. This implies that since capacitance (C) is the charge (Q) divided by the potential difference (V), the potential difference now must be double, i.e., 2V₀. Hence, the correct answer is (a) 0.5C₀ & 2V₀.