A parallel plate capacitor of capacitance C is connected to a battery and charged to a potential difference V. Another capacitor of capacitance 2C is charged to a potential difference 2V by another battery. The batteries are disconnected and the capacitors are connected in parallel so that the positive plate of each is joined to the negative plate of the other. What is the common potential difference across the capacitors?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( V \)
After the batteries are disconnected, the charges can move only between the joined plates. So, by conservation of charge, each pair of joined plates keeps the algebraic sum of the charges it had. In parallel the capacitances add, and the common potential difference is the net charge divided by the total capacitance.
The charges stored before joining are:
\( Q_1=CV \)
\( \begin{aligned} Q_2 &= 2C\times2V \\ &= 4CV \end{aligned} \)
The positive plate of each capacitor is joined to the negative plate of the other. So on each pair of joined plates, the charges have opposite signs:
\( \begin{aligned} Q_{\mathrm{net}} &= Q_2-Q_1 \\ &= 4CV-CV \\ &= 3CV \end{aligned} \)
\( \begin{aligned} C_{\mathrm{total}} &= C+2C \\ &= 3C \end{aligned} \)
The common potential difference is:
\( \begin{aligned} V' &= \frac{Q_{\mathrm{net}}}{C_{\mathrm{total}}} \\ &= \frac{3CV}{3C} \\ &= V \end{aligned} \)