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What is the force of attraction between the two plates of a parallel plate capacitor? Assume that, area of each plate of the capacitor is A and one plate is charged with +Q and the other with -Q. |
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Answer» Solution :Intensity of electric field at a point near a CHARGED plate having surface DENSITY of charge `sigma` is, `E = (sigma)/(2in_0)` Now, `sigma = (Q)/(A)` `:. E = (Q)/(2A in_0)` MAGNITUDE of the charge on the other plate of the CAPACITOR = Q. So, the force experienced by the plate is, `F = QE = Q*(Q)/(2A in_0)= (Q^2)/(aA in_0)`. |
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