Q. Two identical capacitors, have the same capacitance $C.$ One of them is charged to potential $V_{1}$ and the other to $V_{2}.$ The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is
A) $\frac{1}{4}C(V_{1}^{2} - V_{2}^{2})$
B) $\frac{1}{4}C(V_{1}^{2} + V_{2}^{2})$
C) $\frac{1}{4}C\left( V_{1} - V_{2} \right)^{2}$
D) $\frac{1}{4}C\left( V_{1} + V_{2} \right)^{2}$
Explanation
(c) Initial energy of the system
$U_{i} = \frac{1}{2}CV_{1}^{2} + \frac{1}{2}CV_{2}^{2}$
When the capacitors are joined, common potential
$V = \frac{CV_{1} + CV_{2}}{2C} = \frac{V_{1} + V_{2}}{2}$
Final energy of the system
$U_{f} = \frac{1}{2}(2C)V^{2} = \frac{1}{2}2C\left( \frac{V_{1} + V_{2}}{2} \right)^{2} = \frac{1}{4}C\left( V_{1} + V_{2} \right)^{2}$
Decrease in energy $= U_{i} - U_{f} = \frac{1}{4}C\left( V_{1} - V_{2} \right)^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.