A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charged to a potential difference 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal to one is connected to the negative terminal of the other. The final energy of the configuration is :
Answer & explanation
Correct answer: option 2
The correct answer is Option 2:$\frac{3}{2} C V^2$
The diagrammatic representation of given problem is shown in figure.
The net charge shared between the two capacitors is
$Q'=Q_2-Q_1=4 C V-C V=3 C V$
The two capacitors will have the same potential, say V'. The net capacitance of the parallel combination of the two capacitors will be
$C'=C_1+C_2=C+2 C=3 C$
The potential of the capacitors will be
$V'=\frac{Q'}{C'}=\frac{3 C V}{3 C}=V$
The electrostatic energy of the capacitors will be
$E'=1 / 2 C' V^2=\frac{1}{2}(3 C) V^2$
$=\frac{3}{2} C V^2$