The capacities of two conductors are C 1 and C 2 and their respective potentials are V 1…
Physics · JEE Main · NTA Exams — Electrostatics
The capacities of two conductors are C1 and C2 and their respective potentials are V1 and V2. If they are connected by a thin wire then the loss of energy will be
- \(\frac{C_{1} C_{2}\left(V_{1}+V_{2}\right)}{2\left(C_{1}+C_{2}\right)}\)
- \(\frac{\mathrm{C}_{1} \mathrm{C}_{2}\left(\mathrm{~V}_{1}-\mathrm{V}_{2}\right)}{2\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)}\)
- \(\frac{\mathrm{C}_{1} \mathrm{C}_{2}\left(\mathrm{~V}_{1}-\mathrm{V}_{2}\right)^{2}}{2\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)}\)
- \(\frac{\left(C_{1}+C_{2}\right)\left(V_{1}-V_{2}\right)}{C_{1} C_{2}}\)
Answer
(C) C _ 1 C _ 2 ( ~V _ 1 - V _ 2 )^ 2 2 ( C _ 1 + C _ 2 )
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