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

Answer 1

The correct option is A increases

Capacitance $C=\epsilon \frac{A}{d}$ , $A-areaofthecapacitor;d-distancebetweentheplates$

For a charged capacitor $Q=CV$

As the distance between the plates of the capacitor is increased the capacitance $C$ decreases $(\because C\propto \frac{1}{d})$

As capacitance decreases the potential difference between the plates $Vincreases$