Question

An observer is at$2m$from an isotropic point source of light emitting$40w$ power. The RMS value of electric due to the source at the position of the observer is_________

• A. $5.77x{10}^{-8}V{m}^{-1}$
• B. $17.3V{m}^{-1}$
• C. $57.7x{10}^{-8}V{m}^{-1}$
• D. $1.73V{m}^{-1}$

Answer 1

The correct option is B

$17.3V{m}^{-1}$

Step 1. Given data :

An observer is at a$dis\tan ce=$$2m$

$Power=$$40w$

Step 2. Find the intensity $I$ :

$Intensity=power/4\pi r^2$

$=40/(4x3.14x{2}^2)$

$=0.7961w/m^2$

Step 3.Find the value of $E_{max}$

$E_{max}=\sqrt{\frac{2I}{{\epsilon }_{\circ }c}}$

$E_{max}=\sqrt{\frac{2\times 0.7961}{8.85\times {10}^{-12}\times 3\times {10}^8}}$

$=\surd 598.87$

$=24.47$

Step 4. Find the RMS value of electric :

$E_{rms}=E_{\circ }/\surd 2$

$=24.47/\surd 2$

$=24.47/\surd 2$

$=17.3V/m$

Hence, (B) is the correct option.