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Question

An observer is at2mfrom an isotropic point source of light emitting40w power. The RMS value of electric due to the source at the position of the observer is_________


A

5.77x10-8Vm-1

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B

17.3Vm-1

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C

57.7x10-8Vm-1

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D

1.73Vm-1

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Solution

The correct option is B

17.3Vm-1


Step 1. Given data :

An observer is at adistance=2m

Power=40w

Step 2. Find the intensity I :

Intensity=power/4πr2

=40/(4x3.14x22)

=0.7961w/m2

Step 3.Find the value of Emax

Emax=2Iεc

Emax=2×0.79618.85×10-12×3×108

=598.87

=24.47

Step 4. Find the RMS value of electric :

Erms=E/2

=24.47/2

=24.47/2

=17.3V/m

Hence, (B) is the correct option.


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