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An isotropic point source emits electromagnetic waves with an average power of 1500 W. The peak electric field intensity at a point 3 m away from the source is V m⁻¹.

1. 500
2. 100
3. 500/3
4. 250/3

Correct answer: 100

Solution

I = P/(4*pi*r^2) = 1500/(4*pi*9) = 13.3 W/m^2. With I = (1/2)*epsilon0*c*E0^2, E0 = sqrt(2I/(epsilon0*c)) = sqrt(2*13.3/(8.85e-12*3e8)) ~ 100 V/m.

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