InterviewSolution
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InterviewSolution
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A parallel plate capacitor is charged by an exteranl ac source straight the displacement current inside the capacitor is the same as the current charging the capacitor. |
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Answer» Solution :Electric FIELD between the capacitor PLATES `E = (SIGMA)/(epsilon_(0)) = (q)/(epsilon_(0)A)` Where q is the charge accumulated on the positive plate. The electric flux through this plate `phi_(E ) = EA = (q)/(epsilon_(A))cdot A = (q)/(epsilon_(0))` `therefore` Dispacement current `I_(d) = epsilon_(0)cdot (d PHI )/( dt ) = epsilon_(0) (d)/(dt) [(q)/(epsilon_(0))] = (dq)/(dt)` `(dq)/(dt)` is the rate at which charge flows to positive plate through the conducting wire. `I_(d) = I_(C )` i.e., displacement current between the capacitor plates = conduction current through wire. |
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