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Elan Coil [88]
3 years ago
10

A parallel-plate capacitor with circular plates of radius 0.19 m is being discharged. A circular loop of radius 0.28 m is concen

tric 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?
Physics
1 answer:
aliina [53]3 years ago
7 0

Answer:

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

Explanation:

Given that

R= 0.19 m

r= 0.28 m

I= 2.6 A

We know that

I_D=\epsilon _oA\dfrac{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\dfrac{dE}{dt}

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

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

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