Physics - Modern Physics Question with Solution | TestHub

Total power emitted $=100\times \frac{5}{100}=5 \mathrm{W}$

Now intensity due to electric field will be half of the total intensity. Therefore,

$$\begin{gathered} I_E=\frac{1}{2}\times \frac{\text{power}}{\text{area}} \\ =\frac{1}{2}\times \frac{5}{4\mathrm{\pi }\times 5^2} \end{gathered}$$

$=\frac{1}{40\mathrm{\pi }}\frac{\mathrm{W}}{{\mathrm{m}}^2}$