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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?

Respuesta :

Answer:

[tex]\dfrac{dE}{dt}=2.59\times 10^{12}\ V/m.s[/tex]

Explanation:

Given that

R= 0.19 m

r= 0.28 m

I= 2.6 A

We know that

[tex]I_D=\epsilon _oA\dfrac{dE}{dt}[/tex]

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

[tex]I=\epsilon _oA\dfrac{dE}{dt}[/tex]

[tex]2.6=8.85\times 10^{-12}\times \pi\times 0.19^2\dfrac{dE}{dt}[/tex]

[tex]\dfrac{dE}{dt}=2.59\times 10^{12}\ V/m.s[/tex]

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