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A parellel-plate capacity whose electrodes are shaped as round disc is changed slowly. Demonstrate that the flux of the Poynting vector across the capacitor's lateral surface is equal to the increment of the capcitor's enegry per unit time. The dissipation of field at theedge is to be neglected in calculations. |
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Answer» Solution :If the charge on the capacitor is `Q`, the rate of increase of the capcitor's energy `=(d)/(dt) ((1)/(2)(Q^(2))/(c ))= (Q Q)/(C ) = (d)/(epsilon_(0)PIR^(2))Q Q` Now electric field betweeb the pates (INSIDE it) is, `E = (Q)/(piR^(2)epsilon_(0))`. So discplacement CURRENT `= (delD)/(delt) = (Q)/(piR^(2))` This will lead to a magnetic field, (circuital) insidethe plates . At a radial distance `r` `2pirH_(theta)(r ) =pir^(2)(Q)/(piR^(2))` or `H_(theta) = (Qr)/(2piR^(2))` Hence `H_(theta)(R) = (Q)/(2piR)` at the EDGE. Thus inward polynting vector `=S = (Q)/(2piR) xx (Q)/(piR^(2)epsilon_(0))` Total flow `= 2piR d xx S = (Q Qd)/(piR^(2)epsilon_(0))` Proved |
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