Question

The amplitude of the electric field in a parallel beam of light of intensity 3 W/m^2 is __________

(a) 23.4 N/C

(b) 34.5 N/C

(c) 47.53 N/C

(d) 51.45 N/C

Answer 1

Right answer is (c) 47.53 N/C

To explain I would say: We know, Intensity, I = εE^2c/2

Now, Emax = (sqrt{frac{2I}{ε_0c}})

Here, I = 3 W/m^2, ε = 8.85 X 10^-12 C^2/Nm^2 and c = 3 X 10^8 m/s

We get, Emax = 47.53 N/c.