Question

A parallel-plate capacitor has a capacitance of $4\mu F$. Initial charge on the capacitor is $20\mu C$. Which of the following changes would increase potential difference between its plates to $10V$?

• A. Doubling the area of the capacitor plates
• B. Doubling the charge on each plate
• C. Decreasing the the distance between the plates by half
• D. Decreasing the charge on each plate by half
• E. Decreasing the time taken to charge the capacitor by half

Answer 1

The correct option is B Doubling the charge on each plate

Given : $C=4\mu F$ $Q_i=20\mu C$

Thus potential drop across the capacitor initially $V_i=\frac{Q_i}{C}=\frac{20\mu C}{4\mu F}=5$ volts

Let new charge on the plates of capacitor be $Q_f$

$\therefore$ $V_f=\frac{Q_f}{C}$

OR $10V=\frac{Q_f}{4\mu F}$ $⟹Q_f=40\mu C=2Q_i$

Hence option B is correct.