Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) QE, where Q is the charge on the capacitor, and E is the magnitude of the electric field between the plates. Explain the origin of the factor ½
Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of △x. Hence, work done by the force to do so =F△x As a result, the potential energy of the capacitor increases by an amount given as uA△x Where, u= Energy density A= Area of each plate d= Distance between the plates V= Potential difference across the plates The work done will be equal to the increase in the potential energy i.e., F△x=uA△x F=uA=(21∈0E2)A Electric intensity is given by, E=dV ∴F=21∈0(dV)EA=21(∈0AdV)E However, capacitance, C=d∈0A ∴F=21(CV)E Charge on the capacitor is given by, Q=CV ∴F=21QE The physical origin of the factor, 21, in the force formula lies in the fact that just outside the conductor, field is E and inside it is zero. Hence, it is the average value, 2E, of the field that contributes to the force.
