Question

The separation between plates of charged parallel plate capacitor is increased (battery was removed after charging), then

• A. Field between plates decreases.
• B. Energy density between plates increases.
• C. Force between plates remains same.
• D. Potential difference between plates remains same.

Answer 1

The correct option is C Force between plates remains same.

$F_{\text{between plates}}=\frac{q^2}{2{\epsilon }_{0}A}$, remains constant as it is only dependent on charge which is not changing since battery is removed.

As field is dependent only on charge, so it remains same.

Potential difference is dependent on distance between plates, so it changes.

Energy density is inversely propotional to distance, so it decreases.