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Considering the Case of a Parallel Plate Capacitor Being Charged, Show How One is Required to Generalize Ampere'S Circuital Law to Include the Term Due to Displacement Current. - CBSE (Science) Class 12 - Physics

Question

Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere's circuital law to include the term due to displacement current.

Solution

Using Gauss’ law, the electric flux ΦE of a parallel plate capacitor having an area A, and a total charge Q is

phi_E=EA=1/in_0Q/AxxA

=Q/in_0

Where electric field is

E=Q/(Ain_0)

As the charge Q on the capacitor plates changes with time, so current is given by

i = dQ/dt

:.(dphi_E)/dt=d/dt(Q/in_0)=1/in_0(dQ)/dt

=>in_0(dphi_E)/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)

:.i=i_c+i_d=i_c+in_0(dphi_E)/dt

As Ampere’s circuital law is given by

:.ointvecB.vec(dl)=mu_0I

After modification we have Ampere−Maxwell law is given as

ointB.dl=mu_0i_c+mu_0in_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 displacement current.

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Solution Considering the Case of a Parallel Plate Capacitor Being Charged, Show How One is Required to Generalize Ampere'S Circuital Law to Include the Term Due to Displacement Current. Concept: The Parallel Plate Capacitor.
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