Question

A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates:

• A. decreases
• B. does not change
• C. becomes zero
• D. increases

Answer 1

Answer: (D)

increases

Answer 2

Area of the capacitor is represented by A, and the distance between the plates is represented by d.
$Q = CV$
Capacitance of a parallel plate capacitor
$C= \frac {ε_0A}{d}$
$C ∝ \frac 1d$
The capacitance C of the capacitor diminishes as the distance between its plates increases.
The potential difference between the plates V increases as capacitance decreases.

So, the correct option is (D): increases