A point source heating device of 1000 W with 10% efficiency is 7 m from the observational location. Find the peak value of electric field produced by the radiation at that location.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 11.05\,\mathrm{V/m} \)
Only 10% of the electrical power is radiated. This power spreads uniformly over a sphere, so the intensity at distance r is \( I = P/4\pi r^{2} \), and the intensity of an electromagnetic wave in terms of the peak electric field is \( I = \frac{1}{2}\varepsilon_0cE_0^{2} \).
\( P = 0.10\times1000\,\mathrm{W} = 100\,\mathrm{W} \)
\( \displaystyle I = \frac{100}{4\pi(7)^{2}} = \frac{100}{615.8} \approx 0.162\,\mathrm{W\,m^{-2}} \)
\( \displaystyle E_0 = \sqrt{\frac{2I}{\varepsilon_0c}} \)
With \( \varepsilon_0 = 8.85\times10^{-12}\,\mathrm{F\,m^{-1}} \) and \( c = 3\times10^{8}\,\mathrm{m\,s^{-1}} \), \( \varepsilon_0c = 2.655\times10^{-3} \).
\( \displaystyle E_0 = \sqrt{\frac{2\times0.162}{2.655\times10^{-3}}} = \sqrt{122} \approx 11.05\,\mathrm{V\,m^{-1}} \)