Prove that displacement current is equal to conduction current
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Assuming that we have a uniform electric field between the plates of a capacitor, we can find its intensity to be (using Gauss' Law) E(t)=σ(t)ϵ0=q(t)ϵ0AE(t)=σ(t)ϵ0=q(t)ϵ0A, where q(t)q(t) is the absolute value of the charge on each plate, and which is not constant in time since the capacitor is being charged. Now, considering a circular surface between the plates, with same radius R and the same axis as the plates, the electric flux will be
ΦE(t)=E(t)S=q(t)ϵ0AπR2=q(t)ϵ0.ΦE(t)=E(t)S=q(t)ϵ0AπR2=q(t)ϵ0.
So, the displacement current is
ID=ϵ0∂ΦE∂t=dq(t)dt=I(t),ID=ϵ0∂ΦE∂t=dq(t)dt=I(t),
hence, even if the current varies in time, the displacement current will be equal to it.
I HOPE IT WILL HELP YOU AND MARK ME AS A BRAIN LIST ANSWER PLZ.
Assuming that we have a uniform electric field between the plates of a capacitor, we can find its intensity to be (using Gauss' Law) E(t)=σ(t)ϵ0=q(t)ϵ0AE(t)=σ(t)ϵ0=q(t)ϵ0A, where q(t)q(t) is the absolute value of the charge on each plate, and which is not constant in time since the capacitor is being charged. Now, considering a circular surface between the plates, with same radius R and the same axis as the plates, the electric flux will be
ΦE(t)=E(t)S=q(t)ϵ0AπR2=q(t)ϵ0.ΦE(t)=E(t)S=q(t)ϵ0AπR2=q(t)ϵ0.
So, the displacement current is
ID=ϵ0∂ΦE∂t=dq(t)dt=I(t),ID=ϵ0∂ΦE∂t=dq(t)dt=I(t),
hence, even if the current varies in time, the displacement current will be equal to it.
I HOPE IT WILL HELP YOU AND MARK ME AS A BRAIN LIST ANSWER PLZ.
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