The amplitude of the electric field in a parallel beam of light of intensity 2.0 Wm−2 is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 38.8 NC−1
The intensity of an electromagnetic wave is the average energy density times the speed of light. The energy is shared equally between the electric and magnetic fields, so in terms of the electric field amplitude \( E_0 \):
\( \displaystyle I = \frac{1}{2}\,c\,\varepsilon_0 E_0^{2} \)
\( \displaystyle E_0 = \sqrt{\frac{2I}{c\,\varepsilon_0}} \)
\( \displaystyle E_0 = \sqrt{\frac{2 \times 2.0\,\mathrm{W\,m^{-2}}}{3\times10^{8}\,\mathrm{m\,s^{-1}} \times 8.85\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}}}} \)
\( \displaystyle E_0 = \sqrt{\frac{4}{2.655\times10^{-3}}}\,\mathrm{N\,C^{-1}} = \sqrt{1506.6}\,\mathrm{N\,C^{-1}} \)
\( \displaystyle E_0 \approx 38.8\,\mathrm{N\,C^{-1}} \)