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Question
A parallel plate capacitor (A) of capacitance C is charged by a battery to voltage V. The battery is disconnected and an uncharged capacitor (B) of capacitance 2C is connected across A. Find the ratio of total electrostatic energy stored in A and B finally and that stored in A initially.
Solution
The electric field in the region between the plates depends on the change given to the conducting plates,
For capacitor 'A' of capacitance C, having voltage V, the charge on it is given by Q = CV.
On connecting this capacitor A with capacitor B changes get distributed.
The total electrostatic energy stored in A is given by
`U_A = 1/2CV^2 = 1/2 Q^2/C` .........(i)
Since capacitor B is connected across capacitor A.
∴ `C_"net" = C_A + C_B`
= C + 2C
= 3C
∴ Total electrostatic energy in A and B is given by
`U_"net" = 1/2 Q^2/C_"net"`
= `1/2(Q^2/(3C))` ....(since total change Q remains constant)
= `1/3(1/2 Q^2/C)`
= `1/3U_A` .....[By equation (i)]
`U_"net"/U_A = 1/3`
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