Question

A point source of electromagnetic radiation has an average power output of $1500W$. The maximum value of the electric field at a distance of $3m$ from this source in $V{m}^{-1}$ is

• A. $500$
• B. $100$
• C. $\frac{500}{3}$
• D. $\frac{250}{3}$
• E. $10\sqrt{5}$

Answer 1

The correct option is B

$100$

Step 1: Given data

Power output, $P=1500W$

Speed of light, $c=3\times {10}^{8}m/s$

Step 2: Find the electric field

Average Intensity of an electromagnetic wave is

$I=\frac{P}{A}$

Where, $I=$Average intensity of electromagnetic wave

$P=$Average output power

$A=$Area of the spherical surface around the point source

$\Rightarrow$$\frac{P}{A}=\frac{1}{2}{\epsilon }_0{E}_0^{2}xc$

Where, ${\epsilon }_0=$Permittivity of the free space

$=8\cdot 85\times 10-12C{N}^{-1}{m}^{-2}$

$E_0=$Peak value of the electric field

$\Rightarrow$$E_0^2=\frac{2P}{A{\epsilon }_{0}c}$

$\Rightarrow$$E_0=\sqrt{\frac{2P}{A{\epsilon }_{0}c}}$

$\Rightarrow$$E_0=\sqrt{\frac{2\left(1500\right)}{4(3.14){\left(3\right)}^2(8.85x{10}^{-12})(3x{10}^8)}}$

$\Rightarrow$$E_0=100V{m}^{-1}$

Thus, The maximum value of electric field is ${\mathit{E}}_{\mathbf{0}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{100}\mathbf{}\mathit{V}{\mathit{m}}^{\mathbf{-}\mathbf{1}}$

Hence, option $\left(B\right)$ is correct.