The radiation energy emitted per second by a point source is 100 ~W . If the efficiency of the source is 4 % , then the rms value of the electric field at distance of 2 ~m is [use 1 4 ₀ =9 10^9 in SI unit]

Step-by-step solution

where, P_ avg = average power generated by source =4 % of 100 ~W = 4 100 100 ~W =4 ~W A= area to which energy is transmitted =4 r^2=4 (2)^2=4 4=16 m ^2 On putting the values into Eq. (i), we have where, ₀= permittivity of free space and E_ rms = root mean square ( rms ) value of electric field. c= speed of light in vacuum =3 10^8 ~m / s On putting the values into Eq. (iii), we get I_ avg = ₀ E_ rms ^2 3 10^8 On equating both the Eqs. (ii) and (iv), we get gathered 1 4 = ₀ E_ rms ^2 3 10^8 1 4 ₀ 3 10^8 =E_ rms ^2 E_