Two plates of a parallel plate capacitor after being charged from a constant voltage source are separated apart by means of insulated handles, then the

Solution (By Examveda Team)

When a parallel-plate capacitor is charged using a constant voltage source and then disconnected from the source, the charge Q on the capacitor remains constant because there is no conducting path for the charge to escape.

The capacitance of a parallel-plate capacitor is given by:

C = εA/d

where d is the separation between the plates.

When the plates are separated apart using insulated handles, the distance d increases. Therefore, the capacitance C decreases.

Using the capacitor relation:

Q = CV

Since Q remains constant and C decreases, the voltage V must increase to maintain the same charge:

V = Q/C

Thus, increasing the plate separation causes the voltage across the plates to increase.

Why the other options are incorrect:

Option B — Voltage decreases: Incorrect. Since capacitance decreases while charge remains constant, voltage increases.

Option C — Charge decreases: Incorrect. The plates are handled with insulated handles and the capacitor is disconnected from the source, so the charge remains essentially constant.

Option D — Charge increases: Incorrect. There is no external source connected to supply additional charge.

Key Concept: For an isolated charged capacitor, increasing plate separation causes capacitance to decrease and voltage to increase, while the charge remains constant.

Final Answer: Option A — Voltage across the plates increases.