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Question

Figure represents a capacitor made of two circular plates each of radius r=12 cm and separated by d=5.0 mm. The capacitor is being charged by an external source. The charging current is constant current, I=0.15 A.

Calculate the rate of change of electric field between the plates.

A
3.14×1011Vm1s1
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B
2.68×1011Vm1s1
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C
1.87×1011Vm1s1
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D
5.12×1011Vm1s1
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Solution

The correct option is A 3.14×1011Vm1s1
We know that
Q=CV
dQdt=CdVdt
As we know that
V=E.d
where d is the distance between the plates
Putting this in above equation
dQdt=CddEdtI=CdEdtddEdt=ICddEdt=Iε0AdddEdt=Iε0AdEdt=0.158.85×1012×π(12×102)2dEdt=3.14×1011 Vm1s1


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