- #1
jablonsky27
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A friend asked me this question:
There is a capacitor with capacitance C, charge Q on either plate and an electric field E between the plates. The energy stored in the capacitor is then given by (C.V^2)/2, where V is E.(distance between plates).
Now, take another similar capacitor, and connect it across the original capacitor like shown.
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Then, due to redistribution of the chrges, the electric field between plates of each capacitor becomes E/2. The equivalent capacitance is 2C and the total energy stored in the 2 capacitors is now (C.V^2)/4.
What happens to the other (C.V^2)/4? I think so much energy is lost during redistribution of charges. But am not sure.. Any help?
There is a capacitor with capacitance C, charge Q on either plate and an electric field E between the plates. The energy stored in the capacitor is then given by (C.V^2)/2, where V is E.(distance between plates).
Now, take another similar capacitor, and connect it across the original capacitor like shown.
----||----
---------| |--------
----||----
Then, due to redistribution of the chrges, the electric field between plates of each capacitor becomes E/2. The equivalent capacitance is 2C and the total energy stored in the 2 capacitors is now (C.V^2)/4.
What happens to the other (C.V^2)/4? I think so much energy is lost during redistribution of charges. But am not sure.. Any help?
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