Question

Maximum value of electric field at a point at a distance of 2 m from the light source of power 60 watt is

• A. $50N/C$
• B. $30N/C$
• C. $2\times {10}^{-3}N/C$
• D. $5\times {10}^{-3}N/C$

Answer 1

The correct option is B $30N/C$

$I=\frac{P}{A}=\frac{P}{4\pi R^2}\&I=\frac{1}{2}{\epsilon }_{0}c{E}_0^2$
$\frac{1}{2}{\epsilon }_{0}c{E}_0^2=\frac{P}{4\pi R^2}\Rightarrow E_0=\sqrt{\frac{2P}{4\pi {\epsilon }_{0}c{R}^2}}$
${\epsilon }_0=\frac{1}{{\mu }_0{c}^2}\Rightarrow E_0=\frac{1}{R}\sqrt{\frac{2P{\mu }_0{c}^2}{4\pi c}}=\frac{1}{R}\sqrt{\frac{2Pc\text{/}4\pi \times {10}^{-7}}{\text{/}4\pi }}$
$E_0=\frac{1}{2}\sqrt{2\times 60\times 3\times {10}^8\times {10}^{-7}}=30N/C$