Q. 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 connected to another battery and is charged to potential difference $2V.$ The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is
A) Zero
B) $\frac{25CV^{2}}{6}$
C) $\frac{3CV^{2}}{2}$
D) $\frac{9CV^{2}}{2}$
Explanation
(c) Total charge $= (2C)(2V) + (C)( - V) = 3CV$
$\therefore$ Common potential $= \frac{3CV}{3C} = V$
$\therefore$ Energy $= \frac{1}{2}(3C)(V)^{2} = \frac{3}{2}CV^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.