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A parallel-plate capacitor has plates of area `A` and separation `d` and is charged to a potentialdifference `V`. The charging battery is then disconnected, and the plates are pulled apart until their separation is `2d`. Derive expression in terms of `A, d` and `V` for (a) the new potential difference (b) the initial and final stored energies, `U_i` and `U_f` and (c) the work required to increase the separation of plates from d to `2d`. |
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Answer» Correct Answer - A::B::D a. `q=C_iV=((epsilon_0A)/d)V` `V_f=q/C_f=((epsilon_0AV//d))/((epsilon_0A//2d))=2V` b. `U_i=1/2C_iV^2=1/2((epsilon_0A)/d)V^2` `U_f=1/2C_fV_f^2=1/2((epsilon_0A)/(2d))(2V)^2` `=((epsilon_0A)/d)V^2` c. `W=U_f-U-i=1/2((epsilon_A)/d)V^2` |
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