Question

A parallel plate capacitor is charged from a cell and then isolated from it. The separation between the plates is now increased :

• A. The force of attraction between the plates will decrease
• B. The field in the region between the plates will not change
• C. The energy stored in the capacitor will increase
• D. The potential difference between the plates will decrease

Answer 1

Answer: (B), (C)

The correct options are
B The field in the region between the plates will not change
C The energy stored in the capacitor will increase

The force of attraction between the plates is $F=\frac{q^2}{2A{ϵ}_0}$ and the field between the plates is $E=\frac{q}{A{ϵ}_0}$
As the capacitor is isolated so charge remains unchanged.
As F and E is independent of separation d so they remain unchanged.
The parallel plate capacitance is $C=\frac{A{ϵ}_0}{d}$
Thus, if d is increased, C will decrease.
Potential energy is $U=\frac{q^2}{2C}$
As q is constant and C decreases so U will increase.
Potential difference between plates is $V=Ed$
As E unchanged and d increase so V will increase.