Question

Consider a parallel plate capacitorwhose plates are closely spaced. Let Rbethe radius of the plates and the currentin the wire connected to the plates is5 A, calculate the displacement currentthrough the surface passing betweenthe plates by directly calculating therate of change of flux of electric fieldthrough the surface.​

Answer 1

Answer:

Rate of change in electric flux is given as

\frac{d\phi}{dt} = 5.65 \times 10^{11} Vm/sdtdϕ=5.65×1011Vm/s

Explanation:

Let the charge at any moment of time on the plate is given as q

so the electric field between the plates is given as

E = \frac{q}{\pi R^2 \epsilon_0}E=πR2ϵ0q

now we have

q = \pi R^2 \epsilon_0 Eq=πR2ϵ0E

now we have

i = \frac{dq}{dt}i=dtdq

so we have

i = \epsilon_0 \frac{d\phi}{dt}i=ϵ0dtdϕ

Now we have

i = 5Ai=5A

so we have

5 = (8.85 \times 10^{-12})\frac{d\phi}{dt}5=(8.85×10−12)dtdϕ

so we have

\frac{d\phi}{dt} = \frac{5}{8.85 \times 10^{-12}}dtdϕ=8.85×10−125

\frac{d\phi}{dt} = 5.65 \times 10^{11} Vm/sdtdϕ=5.65×1011Vm/s