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A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance to 2C is similarly charged to a potential difference 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of other. What is the final energy of the combination?
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A parallel plate capacitor of capacitance C is connected to a battery ...
Answer


Initial Energy


  • The energy of a capacitor is given by U = (1/2)CV^2.

  • The energy of the first capacitor is U1 = (1/2)CV^2.

  • The energy of the second capacitor is U2 = (1/2)(2C)(2V)^2 = 8CV^2.

  • The total energy of the system before they are connected in parallel is U = U1 + U2 = (9/2)CV^2.



Final Energy


  • When the capacitors are connected in parallel, the equivalent capacitance is Ceq = C + 2C = 3C.

  • The equivalent voltage across the capacitors is Veq = (V + (-2V))/2 = -V/2.

  • The energy of the combined capacitors is Ueq = (1/2)CeqVeq^2 = (1/2)(3C)(-V/2)^2 = (3/8)CV^2.



Explanation

When the capacitors are connected in parallel, the equivalent capacitance increases while the equivalent voltage decreases. The total energy of the system is conserved, but it is redistributed between the two capacitors. The first capacitor loses some energy while the second capacitor gains some energy. The final energy of the combined capacitors is less than the initial energy of the system due to the redistribution of energy.
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A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance to 2C is similarly charged to a potential difference 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of other. What is the final energy of the combination?
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