InterviewSolution
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InterviewSolution
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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» 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`. 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 increases in the potential energy i.e,. `F Detlax=uA Deltax ` `F=uA=(1/2 in_(0)E^(2))A` Electric intensity is given by `E=V/d` `:. F=1/2 in_(0)(V/d)EA=1/2 (in_(0)AV/d)E` however, capacitance, `C=(in_(0)A)/d` `:. F=1/2 (CV)E` charge on the capacitor is given by `Q=CV` `:. F=1/2 QE` The physical origin of the factors , `1/2` , in the force formula lies in the fact that just outside 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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