Q. The capacities of two conductors are $C_{1}$ and $C_{2}$ and their respective potentials are $V_{1}$ and $V_{2}.$ If they are connected by a thin wire, then the loss of energy will be given by
A) $\frac{C_{1}C_{2}(V_{1} + V_{2})}{2(C_{1} + C_{2})}$
B) $\frac{C_{1}C_{2}(V_{1} - V_{2})}{2(C_{1} + C_{2})}$
C) $\frac{C_{1}C_{2}\left( V_{1} - V_{2} \right)^{2}}{2(C_{1} + C_{2})}$
D) $\frac{(C_{1} + C_{2})(V_{1} - V_{2})}{C_{1}C_{2}}$
Explanation
The correct answer is Option C: $\frac{C_{1}C_{2}\left( V_{1} - V_{2} \right)^{2}}{2(C_{1} + C_{2})}$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.