Electrostatics Question 546

Question: 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 [MP PMT 1986]

Options:

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}{{(V _{1}-V _{2})}^{2}}}{2(C _{1}+C _{2})} $

D) $ \frac{(C _{1}+C _{2})(V _{1}-V _{2})}{C _{1}C _{2}} $

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Answer:

Correct Answer: C

Solution:

Initial energy

$ U _{i}=\frac{1}{2}C _{1}V _{1}^{2}+\frac{1}{2}C _{2}V _{2}^{2} $ ,

Final energy $ U _{f}=\frac{1}{2}(C _{1}+C _{2})V^{2} $ (where $ V=\frac{C _{1}V _{1}+C _{2}V _{2}}{C _{1}C _{2}}) $

Hence energy loss $ \Delta U=U _{i}-U _{f}=\frac{C _{1}C _{2}}{2(C _{1}+C _{2})}{{(V _{1}-V _{2})}^{2}} $



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