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Q.

The average power output of a point source of an electromagnetic radiation is 1080 W. The maximum value of the rms value of the electric field at a distance of 3 m from the source is

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a

90 Vm-1

b

20 Vm-1

c

60 Vm-1

d

40 Vm-1

answer is C.

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Detailed Solution

I=P4πR2

I=uavg⋅c=12∈oEo2×c=P4πR2⇒Eo=P2πR2ϵoc⇒Erms=Eo2=P4πR2ϵoc

⇒Erms=10804π×32×8.85×10-12×3×108=60 V/m

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