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Prove that the force acting on one plate due to the other in a parallel plate capacitor isF =(1)/(2) (CV^(2))/(d) . |
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Answer» SOLUTION :The electric field due to ONE PLATE is `E_(1) = (sigma)/(2 epsilon_(0))` A second plate having CHARGE AA is present in the above electric field. `:.` The force acting on second plate `F= F_(1) (sigmaA)` Substituting the value of `E_(1)` from (1) we get `F = (sigma^(2)A)/(2 epsilon_(0))` but `sigma= (Q)/(A)` `:.F =((Q^(2))/(A^(2)).A)/(2epsilon_(0))=(Q^(2))/(2epsilon_(0)A)=((Q^(2))/(d))/(2epsilon_(0)(A)/(d))=(Q^(2))/(2dC)` `[because (epsilon_(0)A)/(d) =C]` `:. F =(1)/(2) (CV^(2))/(d ) "" ( because Q = CV)` |
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