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A parallel plate capacitor of capacitanc...

A parallel plate capacitor of capacitance C has been charged, so that the potential difference between its plates is V. Now, the plates of this capacitor are connected to another uncharged capacitor of capacitance 2C. Find the common potential acquired by the system and loss of energy.

Text Solution

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Here `C_(1)=C,C_(2)=2C, V_(1)` and `V_(2)=0`
`implies ` Common potential `=V =(C_(1)V_(1)+C_(2)V_(2))/(C_(1)+C_(2))=(CV+0)/(3C)=(v)/(3)`
And `DeltaU=(1)/(2) (Cxx2C)/(3C)xx(V-0)^(2)=(1)/(3)CV^(2)`
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