Question

Calculate the peak value of electric field produced by the radiation coming from a 100 watt bulb at a distance of $3m$. Assume that the efficiency of the bulb is $2.5\%$ and it is a point source?

• A. $4.09V/M$
• B. $5V/M$
• C. $4.08V/M$
• D. $6V/M$

Answer 1

Answer: (C)

$I=\frac{\text{ Power }}{\text{ area }}=\frac{100\times (2.5/100)}{4\pi (3{)}^2}=\frac{2.5}{36\pi }W{m}^{-2}$
Half of this intensity (I) belongs to electric field and half of that to magnetic field. Therefore,
$\frac{I}{2}=\frac{1}{4}{\epsilon }_0{E}_0^{2}C$ ,
$I=\frac{{\epsilon }_0{E}_0^{2}C}{2}$
or $E_0=\sqrt{\frac{2I}{{\epsilon }_{0}C}}$
$I=\sqrt{\frac{2\times 2.5/36\pi }{\left(\frac{1}{4\pi \times 9\times 10^9}\right)\times 3\times 10^8}}$
$I=4.08V{m}^{-1}$