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The force between the plates of a parallel plate capacitor of capacitance `C` and distance of separation of the plates `d` with a potential difference `V` between the plates, is.A. `(CV^(2))/(2d)`B. `(C^(2)V^(2))/(2d^(2))`C. `(C^(2)V^(2))/(d^(2))`D. `(V^(2)d)/(C )` |
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Answer» Correct Answer - A `F=(QE)/(2)=(CV)/(2)[(V)/(d)]`. |
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