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Show that the force on each plate of a parallel platecapacitor has a magnitudeequalto (1)/(2) QE, where Q is the chargeon the capacitorand E si the magnitudeof electricfield between the plates. Explain the orginof factor (1)/(2). |
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Answer» Solution :If F is the force on each plate of a parallel plate capacitor, then WORK donein INCREASINGTHE separation betweenthe plates by `Delta x = F. Delta x`. This must be the increase in potential energyof the capacitor. Now, increase in volume of capacitor `=A. Delta x` If u = energy density = energy stored/ volume, then increase in pot, energy`= u.A Delta x :. F Delta x = u. A Delta x` `F = u.A = ((1)/(2) in_(0) E^(2)) A = (1)/(2) (in_(0) AE)E , F = (1)/(2) (in_(0) A(V)/(d))E (( :. C = (in_(0) A)/(d))` `F = (1)/(2) (CV) E = (1)/(2) QE` This origin of factor `1//2` in force can be explained by the FACT that inside the conductor, field is zero and outside the conductor, the field is E. Thereforethe averagevalue of the field `(i.e. E//2)` constribuites to the force. |
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