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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90 Vm^-1

20 Vm^-1

60 Vm^-1

40 Vm^-1

answer is C.

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Detailed Solution

$I = \frac{P}{4 π R^{2}}$

$\begin{array}{l} I = u_{avg} \cdot c = \frac{1}{2} \in_{o} E_{o}^{2} \times c = \frac{P}{4 {πR}^{2}} \\ \Rightarrow E_{o} = \sqrt{\frac{P}{2 π R^{2} ϵ_{o} c}} \\ \Rightarrow E_{r m s} = \frac{E_{o}}{\sqrt{2}} = \sqrt{\frac{P}{4 π R^{2} ϵ_{o} c}} \end{array}$

$\Rightarrow E_{r m s} = \sqrt{\frac{1080}{4 π \times 3^{2} \times 8.85 \times 10^{- 12} \times 3 \times 10^{8}}} = 60 V / m$

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