The electric field between the plates of a parallel plate capacitor is 'E'. If the charge on the plates is Q then the force on each plate is

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The electric field between the plates of a parallel plate capacitor is the superposition of the electric fields produced by each plate.

The total electric field is $E = \frac{Q}{A ϵ _{0}}$ .

The electric field produced by a single plate is $E^{'} = \frac{Q}{2 A ϵ _{0}} = \frac{E}{2}$ .

The force experienced by one plate is due to the electric field of the other plate.

Force $F = Q \times E^{'} = Q \times \frac{E}{2} = \frac{QE}{2}$ .

Answer: $\frac{QE}{2}$

Physics - Capacitor Question with Solution | TestHub

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