# 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. - Physics

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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.

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#### 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.

Concept: The Parallel Plate Capacitor
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2013-2014 (March) All India Set 2
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