Question

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

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

Answer 1

Answer: (C)

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