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The charge on a parallel plate capacitor varies as q=q_(0)cos 2pi v t. The plates are very large and close together (area = A, separation = d). Neglecting the edge effects, find the displacement current through th capacitor. |
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Answer» SOLUTION :Displacement current in capacitor, `I_(d)=I_(c )=(dq)/(dt) ""`….(1) Here, `Q=q_(0)cos 2PI vt` Substituting in EQUATION (1), `I_(d)=I_(c )=(d)/(dt)(q_(0)cos 2pi vt)` `therefore I_(d)=I_(c )=-q_(0)sin 2pi vt xx 2pi V` `therefore I_(d)=I_(c )=-2pi vq_(0)sin 2pi vt` |
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