The peak electric field produced by the radiation coming from the 80W bulb at a distance of 10m is x10μ0CπVm. the efficiency of the bulb is 10% and it is a point source. The value of x is
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answer is 2.
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Detailed Solution
Intensity=I=P4πr2Also, Intensity=energy density×speed of light⇒uc=P4πr2⇒12ϵoEo2c=P4πr2Given : Output power=10% of input power=8 W⇒Eo=2P4πr2·ϵoc =2×84π×100×ϵoc=16μoC4π×100=210μoCπv/m