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Considering the case of a parallel plate capacitor being charged, show how one is requiredto generalize Ampere's circuital law to include the term due to displacmentcurrent. |
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Answer» Solution :Using GAUSS' law, the electric flux `phi_(epsi)` of a parallel plate capacitor having an area A, and a total charge Q is `phi_(epsi)=EA=(1)/(epsi_(0))(Q)/(A)xxA=(Q)/(epsi_(0))` Where electric field is, `E=(Q)/(Aepsie_(0))`. As the charge Q on the capacitor plates CHANGES with time, so CURRENT is given by `i=dQ//dt` `(phi_(epsi))/(dt)=(d)/dt((Q)/(epsi_(0)))=(1)/(epsi_(0))(dQ)/(dt)` `rArrepsi_(0)(depsi_(0))/(dt)=(dQ)/(dt)=i`This is the missing. Term in Ampere's circuital law. So the total current through the conductor is `i="Conduction current"(ic)+"Displacement current" (id)` `therefore ""i=i_(c)+i_(d)=i_(c)+epsi(dphi_(E))/(dt)` As Ampere's circuital law is given by `therefore phivecB*vecdt=mu_(0)I` After modification we have Ampere's Maxwell law is given as `phiB.dt=mu_(0)i_(c)+mu_(0)epsi_(0)(dphi_(E))/(dt)` The total current passing through any surface, of which the closed loop is the perimeter, is the sum of the conduction and diplacement current. |
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