Answer & Solution
Answer: Option
A
Solution:
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.
