Radiation from a source is striking a surface at a rate of 50 W/m². The peak value of the electric field is A) 110 V/m. B) 150 V/m. C) 190 V/m. D) 250 V/m.

Question

Quizplus · Accepted Answer

The intensity (I) of the radiation is given as 50 W/m^2. The peak value of the electric field (E_0) can be found using the relation between the intensity and the peak electric field in electromagnetic waves, given by $I = \frac{1}{2} \epsilon_0 c E_0^2$, where $\epsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} F/m$) and $c$ is the speed of light ($3 \times 10^8 m/s$). Solving for $E_0$ gives $E_0 = \sqrt{\frac{2I}{\epsilon_0 c}}$. Substituting the given values, $E_0 = \sqrt{\frac{2 \times 50}{8.85 \times 10^{-12} \times 3 \times 10^8}}$ yields a peak electric field of approximately 190 V/m.

Radiation from a Source Is Striking a Surface at a Rate