Question

The amplitude of electric field is $0.5\text{V/m}$, $5\text{km}$ from a radio transmitter. What is the total power emitted by the transmitter?

• A. 104 KWatt
• B. 216 KWatt
• C. 318 KWatt
• D. 84 KWatt

Answer 1

The correct option is A 104 KWatt

$I={ϵ}_0{E}^{2}C$
$={ϵ}_0{\left(\frac{0.5}{\sqrt{2}}\right)}^2\times 3\times {10}^8$
$P=I.\left(4\pi r^2\right)=4\pi r^2{ϵ}_0{\left(\frac{0.5}{\sqrt{2}}\right)}^2\times 3\times {10}^8$
$=4\pi {ϵ}_0\times 25\times {10}^6\times \frac{1}{8}\times 3\times {10}^8$
$=\frac{1}{9\times {10}^9}\times 25\times {10}^6\times \frac{1}{8}\times 3\times {10}^8=\frac{75}{72}\times {10}^5=104\text{kW}$