Question

A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved farther apart by pulling at them by means of insulating handles, then:

• A. the capacitance of the capacitor increases
• B. the energy stored in the capacitor decreases
• C. the charge on the capacitor decreases
• D. the voltage across the capacitor increases

Answer 1

Answer: (D)

Let the battery of potential difference $V$ be used to charge the capacitor of capacitance $C$.

$\therefore$ the charge stored in the capacitor $q=CV$

Now the battery is disconnected, so the the charge of the capacitor becomes constant

i.e $q=constant$ OR $CV=constant$ .............(1)

Capacitance of parallel plate capacitor $C=\frac{A{ϵ}_o}{d}$

So if the distance between the plates is increased, then the capacitance will decrease which is compensated by the increase in voltage across the capacitor according to equation (1).

Also the energy stored in the capacitor $E=\frac{q^2}{2C}$

$⟹E\propto \frac{1}{C}$ $(\because q=constant)$

Thus energy will increase due to the decrease in capacitance.