Question

An observer is at 2m from an isotropic point source of light emitting 40w power. The RMS value of the electric field due to the source at the position of the observer is_________.
A. \[5.77 \times {10^{ - 8}}{\text{ V/m}}\]
B. \[17.3{\text{ V/m}}\]
C. \[57.7 \times {10^{ - 8}}{\text{ V/m}}\]
D. \[1.73{\text{ V/m}}\]

Answer 1

Hint:In this question, we need to determine the RMS value of electricity due to the source at the position of the observer. For this, we need to use the following formulae to get the desired result.

Formula used:
The formula for the intensity is given by
\[I = \dfrac{P}{{4\pi {r^2}}}\]
Where, \[I\] is the intensity, \[P\] is the power and \[r\] is the distance from the source.
Also, electric field that is \[{E_{\max }} = \sqrt {\dfrac{{2I}}{{{\varepsilon _0}c}}} \]
Where, \[I\] is the intensity and \[c\] is the speed of light.

Complete step by step solution:
The intensity of light is given by
\[I = \dfrac{P}{{4\pi {r^2}}}\]
But we know that\[P = 40,r = 2\]
\[I = \dfrac{{40}}{{4\pi {{\left( 2 \right)}^2}}}\]
By simplifying, we get
\[I = \dfrac{5}{{\pi \times 2}}\]
\[\Rightarrow I = 0.795{\text{ w/}}{{\text{m}}^2}\]
Now, we will find the value of the electric field \[{E_{\max }}\].
\[{E_{\max }} = \sqrt {\dfrac{{2I}}{{{\varepsilon _0}c}}} \]
Here, \[{\varepsilon _0} = 8.85 \times {10^{ - 12}}\] and \[c = 3 \times {10^8}\]
Thus, we get
\[{E_{\max }} = \sqrt {\dfrac{{2 \times 0.795}}{{8.85 \times {{10}^{ - 12}} \times 3 \times {{10}^8}}}} \]
By simplifying, we get
\[{E_{\max }} = \sqrt {598.87} \]
\[\Rightarrow {E_{\max }} = 24.47{\text{ V/m}}\]
Here, we can say that \[{E_{\max }} = {E_0}\]
\[{E_0} = 24.47{\text{ V/m}}\]
Now, we will find the value of \[{E_{RMS}}\] that is the root mean square value of an electric field.
\[{E_{RMS}} = \dfrac{{{E_0}}}{{\sqrt 2 }}\]
Put \[{E_0} = 24.47{\text{ V/m}}\] in the above equation.
Thus, we get
\[{E_{RMS}} = \dfrac{{24.47}}{{\sqrt 2 }}\]
By simplifying, we get
\[{E_{RMS}} = 17.3{\text{ V/m}}\]
Hence, the RMS value of the electric field due to the source at the position of the observer is 17.3 V/m.

Therefore, the correct option is (B).

Note: Many students make mistakes in writing the formula for the intensity of light and the electric field. The students may get confused in finding the value of the RMS means the root mean square value of an electric field. It is essential to find these parameters to get the desired result.