Question

A parallel-plate capacitor with circular plates of radius 0.19 m is being discharged. A circular loop of radius 0.28 m is concentric with the capacitor and halfway between the plates. The displacement current through the loop is 2.6 A. At what rate is the electric field between the plates changing?

Answer 1

`(dE)/(dt)=2.59* 10^(12)\ V/m.s`

Step-by-step explanation:

Given that

R= 0.19 m

r= 0.28 m

I= 2.6 A

We know that

`I_D=\epsilon _oA(dE)/(dt)`

A= Area of loop

dE/dt= rate of change of electric filed

I=Displacement current

Here r>R

So A=π R²

Now by putting the values

`I=\epsilon _oA(dE)/(dt)`

`2.6=8.85* 10^(-12)* \pi* 0.19^2(dE)/(dt)`

`(dE)/(dt)=2.59* 10^(12)\ V/m.s`