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A parallel plate capacitor of capacitanc...

A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.

Text Solution

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Let the charge on the charged capacitor be Q.
Energy stored, `U_1= (Q^2)/(2C)`
when another uncharged simillar capacitor is connected with the first capacitor, the charge on the system remains constant and the capacitance becomes `C_1 = 2C`,
The energy stored in the system, `U_2 = (Q^2)/(4C)`
`:. U_2 : U_1 = (Q^2)/(4C) : (Q^2)/(2C) = 1 :2`
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