Question

A point source of electromagnetic radiation has an average power output of $1500{\text{W}}$ . Find the maximum value of the electric field at a distance of $3{\text{m}}$ from this source (in ${\text{V}}{{\text{m}}^{ - 1}}$ ).
A) $500{\text{V}}{{\text{m}}^{ - 1}}$
B) ${\text{73V}}{{\text{m}}^{ - 1}}$
C) $\dfrac{{500}}{3}{\text{V}}{{\text{m}}^{ - 1}}$
D) $\dfrac{{250}}{3}{\text{V}}{{\text{m}}^{ - 1}}$
E) $100{\text{V}}{{\text{m}}^{ - 1}}$

Answer 1

The electromagnetic wave from the point source at the given distance can be viewed to create a spherical surface. The average intensity of an electromagnetic wave is defined as the average power of the wave over the spherical area described by the EM radiation from the point source. The average intensity is also proportional to the square of the peak value of the electric field.

Formula used:
-The average intensity of an electromagnetic wave is given by, ${I_{avg}} = \dfrac{P}{A}$ or ${I_{avg}} = \dfrac{1}{2}{\varepsilon _0}{E_0}^2c$ where $P$ is the average output power, $A$ is the area of the spherical surface around the point source, ${\varepsilon _0}$ is the permittivity of free space, ${E_0}$ is the peak value of the electric field and $c$ is the velocity of the EM wave in vacuum.

Complete step by step answer.
Step 1: List the parameters known from the question.
The average output power of the EM wave is given to be $P = 1500{\text{W}}$ .
The peak electric field ${E_0}$ at a distance of $3{\text{m}}$ is to be determined.

Hence the radius of the spherical surface generated by the EM wave from the point source can be taken as $r = 3{\text{m}}$ .
Then the area of the spherical surface will be $A = 4\pi {r^2} = 4\pi \times {3^2} = 36\pi {{\text{m}}^2}$ .
Step 2: Express the relations for the average intensity of the EM wave.
The average intensity of the electromagnetic wave from the point source is given by,
${I_{avg}} = \dfrac{P}{A}$ -------- (1)
The average intensity of the EM wave can also be expressed as ${I_{avg}} = \dfrac{1}{2}{\varepsilon _0}{E_0}^2c$ ----------(2)
where ${\varepsilon _0} = 8 \cdot 85 \times {10^{ - 12}}{\text{C}}{{\text{N}}^{ - 1}}{{\text{m}}^{ - 2}}$ is the permittivity of free space and $c = 3 \times {10^8}{\text{m}}{{\text{s}}^{ - 1}}$ is the velocity of the EM wave in vacuum.
Equating equations (1) and (2) we get, $\dfrac{P}{A} = \dfrac{1}{2}{\varepsilon _0}{E_0}^2c$
$\Rightarrow {E_0}^2 = \dfrac{{2P}}{{A{\varepsilon _0}c}}$
$\Rightarrow {E_0} = \sqrt {\dfrac{{2P}}{{A{\varepsilon _0}c}}}$--------- (3)
Substituting for $P = 1500{\text{W}}$ , $A = 113 \cdot 04{{\text{m}}^2}$ , ${\varepsilon _0} = 8 \cdot 85 \times {10^{ - 12}}{\text{C}}{{\text{N}}^{ - 1}}{{\text{m}}^{ - 2}}$ and $c = 3 \times {10^8}{\text{m}}{{\text{s}}^{ - 1}}$ in equation (3) we get, ${E_0} = \sqrt {\dfrac{{\left( {2 \times 1500} \right)}}{{\left( {36\pi \times 8 \cdot 85 \times {{10}^{ - 12}} \times 3 \times {{10}^8}} \right)}}} \cong 100{\text{V}}{{\text{m}}^{ - 1}}$
Thus the peak value of the electric field at the given distance is ${E_0} \cong 100{\text{V}}{{\text{m}}^{ - 1}}$ .

So the correct option is E.

Note: The electromagnetic wave has an electric field component and a magnetic field component which are perpendicular to each other. So half of the intensity of the EM wave is contributed by the electric field and the other half is by the magnetic field. Hence we can express the average intensity as ${I_{avg}} = \dfrac{1}{2}{\varepsilon _0}{E_0}^2c$ or ${I_{avg}} = \dfrac{1}{2}{\varepsilon _0}{B_0}^2c$ where ${B_0}$ is the peak value of the magnetic field.