InterviewSolution
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InterviewSolution
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A capacitor of capacitance `C` is charged to a potential differece `V_(0)`. The terminals of the charged capacitor are then connected to those of an uncharged capacitor of capacitance `C//2`. a. Compute the original charge of the system. b. Find the final potential differce across each capacitor. c. Find the final energy of the system. d. Calculate the decrease in energy when the capacitors are connected. e. Where did the 'lost' energy go? |
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Answer» `Q=CV_(0)` They have equal potential difference, and thir combined charge must add up to the original charge Therefore, `V=Q_(1)/C_(1)=Q_(2)/C_(2)` and also `Q_(1)+Q_(2)=CV_(0)` `C_(1)=C` and `C_(2)=C/2`, So `Q_(1)/C=Q_(2)/((C//2))` or `Q_(2)=Q_(1)/Q_(2)` or `Q=3/2Q_(1) or Q_(1)=2/3Q` So `V=Q_(1)/C_(1)=2/3Q/C=2/3V_(0)` c. `U=1/2(Q_(1)^(2)/C_(1)+Q_(2)^(2)/C_(2))=1/2[(2/3Q)^(2)/C+(2(1/3Q)^(2))/C] ` `=1/2Q_(2)/C=1/2CV_(0)^(2)` d. The original `U` was `U=1/2CV_(0)^(2)` or `DeltaU=(-1)/6CV_(0)^(2)`. Thermal energy of capacitor, wires, etc., and electromagnetic radiation. |
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