The force between the plates of a parallel pl | vedclass

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Question

(A) The total electric field between the plates of a parallel plate capacitor is given by $E = \frac{V}{d}$.The electric field produced by one plate a

Vedclass · Accepted Answer

$\frac{CV^2}{2d}$

Vedclass · Answer

(A) The total electric field between the plates of a parallel plate capacitor is given by $E = \frac{V}{d}$.The electric field produced by one plate alone is half of the total electric field,which is $E_1 = \frac{E}{2} = \frac{V}{2d}$.The charge on the plate is $Q = CV$.The force $F$ exerted by one plate on the other is given by the product of the charge on one plate and the electric field due to the other plate:$F = Q 	imes E_1 = (CV) 	imes \left( \frac{V}{2d} ight) = \frac{CV^2}{2d}$.

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

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