Question

An ac voltage source V = V0 sin(wt) is connected across a parallel plate capacitor C. Verify that the displacement current in capacitor is same as conduction current wires ​

Answer 1

Answer:

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Explanation:

Consider a fully charged parallel plate capacitor, no wires connected. Now discharge it by connecting a wire from the center of one plate through its volume to the center of the other. put a resistor in series to slow down this event and make it safer!

Now you can see that the two terms contribute equal and opposite amounts to the B-field just outside the capacitor. Let's see how.

Since Q=CV we have V˙=(1/C)dQ/dt=I/C=V/RC. In turn, the displacement current is the area integral of ϵ0E˙, which is −ϵ0V˙A/d, where A and d are the area and plate separation of the capacitor, as usual, and we used EV/d and the fact that V is decreasing. Using further C=ϵ0A/d, we see that the displacement current is −CV˙=−Q˙=−I

as  we know that if the parallel plate capacitor is charging up, if a different wire that runs through the volume between the plates can carry any current you wish, and then you have both sources for a magnetic field .

Answer 2

The Displacement current(Id) will be equal to the conduction current wires that are "Id = I = ϵoAdE/dt"

• The concept of displacement current was first coined by Maxwell, he postulated that changing electric field in a vacuum or in a dielectric also produces a magnetic field.
• It means that changing the electric field is equivalent to a current which is flowing in the changing electric field. This equivalent current is called Displacement Current.

For a charged capacitor, the electric field between the plates is given by:

E= Q/εo​A

where Q=εoϕE

Now displacement current is given by,

Id=dQ/dt= εodϕE/dt where (ϕE=EA)(Electric Flux)

Hence we get, Id=εoAdE/dt -------------1

Now we find the conduction current,

As we know, E=Q/εo​A differentiate w.r.t. to t we get,

dE/dt= 1/εoA*dQ/dt=I/εoA

Therefore. I = εoAdE/dt ----------2

Hence from eq 1 & 2 we get

Id = I = εoAdE/dt

Hence Proved.

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