The radiation energy emitted per second by a point source is 100 ~W...

where, $P_{\text {avg }}=$ average power generated by source $=4 \% \text { of } 100 \mathrm{~W}=\frac{4}{100} \times 100 \mathrm{~W}=4 \mathrm{~W}A=$ area to which energy is transmitted $=4 \pi r^2=4 \pi \times(2)^2=4 \pi \times 4=16 \pi \mathrm{m}^2$ On putting the values into Eq. (i), we have
where, $\varepsilon_0=$ permittivity of free space and $E_{\mathrm{rms}}=$ root mean square $(\mathrm{rms})$ value of electric field. $c=$ speed of light in vacuum $=3 \times 10^8 \mathrm{~m} / \mathrm{s}$ On putting the values into Eq. (iii), we get $I_{\mathrm{avg}}=\varepsilon_0 \times E_{\mathrm{rms}}^2 \times 3 \times 10^8$ On equating both the Eqs. (ii) and (iv), we get $\begin{gathered} \frac{1}{4 \pi}=\varepsilon_0 \times E_{\mathrm{rms}}^2 \times 3 \times 10^8 \Rightarrow \frac{1}{4 \pi \varepsilon_0 \times 3 \times 10^8}=E_{\mathrm{rms}}^2 \\ E_{\mathrm{rms}}^2=\frac{1}{4 \pi \varepsilon_0} \times \frac{1}{3 \times 10^8}=9 \times 10^9 \times \frac{1}{3 \times 10^8} \\ \quad\left(\because \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2\right) \\ =3 \times 10=\sqrt{30} \mathrm{~V} / \mathrm{m} \end{gathered}$