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Show that the force on each plate of a parallel plate capacitor has a magnitude equal to 1/2 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 1/2. |
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Answer» Solution :Consider a parallel plate capacitor with one plate having + Q charge and the other having - Q charge. Let F be the force between them. If the plates be displaced by a SMALL distance dl, the work done dW = Fdl. This work results in the change in ELECTROSTATIC POTENTIAL energy of the capacitor. `:.` Energy density of capacitor `U = 1/2 epsi_0 E^2` and charge in volume= A.dl `:.` Change in potential energy `du = 1/2 epsi_0 E^2. Adl` Thus, we have `Fdl = 1/2 epsi_0 E^2 Adl` `rArrF = 1/2 epsi_0 E^2 A = 1/2 epsi_0 E.A.E = 1/2 (epsi_0 (sigma)/(epsi_0)A) E = 1/2(sigmaA)E = 1/2 QE` The physical origin of the factor `1/2 ` in the above formula for force lies in the FACT that just between the plates of capacitor electric field is E but just outside E = 0. Hence, average value `E/2` contributes to the force. 
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