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Q.

A point source of electromagnetic radiation has an average power output of 15 W.  The maximum  value of electric field at a distance of 3m from this source in Vm−1  is

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a

10

b

20

c

15

d

5

answer is A.

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Detailed Solution

I=P4πr2=12∈0E02C ⇒E0=P2πr2∈0c12 ⇒E0=152π×9×3×1084π×9×10912 ⇒E0=10212=10Vm
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