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 electric field between the plates. Explain the origin of the factor ½.

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Answer Text

Solution :
Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of `Deltax`. Hence , work done by the force to do so =`FDeltaX` ltBrgt As a result, the potential energy of the capacitros increases by an amount given as uA`Deltax`. <br> Where, u=energy density <br> A=area of each plate <br> d=distance between the plates <br> V=potential difference across the plates <br> the work done will be equal to the increases in the potential energy i.e,. <br> `F Detlax=uA Deltax `<br> `F=uA=(1/2 in_(0)E^(2))A` <br> Electric intensity is given by <br> `E=V/d` <br> `:. F=1/2 in_(0)(V/d)EA=1/2 (in_(0)AV/d)E` <br> however, capacitance, `C=(in_(0)A)/d` <br> `:. F=1/2 (CV)E` <br> charge on the capacitor is given by <br> `Q=CV` <br> `:. F=1/2 QE` <br> The physical origin of the factors , `1/2` , in the force formula lies in the fact that just outside <br> the conductor, field is E and inside it is zero. it is the avearge value, `F/2`, of the field that contributes to the force.

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