Question

A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential and capacitance respectively are:

• A. constant, decreases, decreases
• B. increases, decreases, decreases
• C. constant, decreases, increases
• D. constant, increases, decreases

Answer 1

The correct option is C constant, decreases, increases

Since, the capacitor is isolated, charge will remain constant.

Capacitance of parallel plate capacitor is given by:
$C=\frac{{\epsilon }_{o}A}{d}$

As $d$ is increased, capacitance will decrease.

By definition of capacitance,

$Q=CV$

Since $Q$ is constant and $C$ is decreased, $V$ will increase.