Question

The peak electric field produced by the radiation coming from the $8W$ bulb at a distance of $10m$ is

Answer 1

Step 1: Given data

The power of the bulb$\left(P\right)=8W$

Distance$\left(r\right)=10m$

Step 2: Formula used

The intensity of electromagnetic radiation is computed using the following formula:

$I=\frac{1}{2}c{\epsilon }_0{E}^2$, where $I=$intensity, ${\epsilon }_0=$permittivity, $E=$electric field and $c=$speed of light.

We know that $Intensity\left(I\right)=\frac{Power\left(P\right)}{Area\left(A\right)}$. So,

$$\begin{gathered} \frac{P}{A}=\frac{1}{2}c{\epsilon }_{\circ }{E}^2 \\ E=\sqrt{\frac{2P}{Ac{\epsilon }_{\circ }}}·····\left(1\right) \end{gathered}$$

Area $A=4\pi r^2$

Step 3: Compute the peak electric field

Substitute $P=8W,A=4{\mathrm{\pi r}}^2=\left(4\times 3.14\times {10}^2\right)m^2,\mathrm{c}=3.0\times {10}^{8}m/sec$ and ${\epsilon }_{\circ }=8.85\times {10}^{-12}{C}^2{N}^{-1}{m}^{-2}$ in equation $\left(1\right)$.

$$\begin{gathered} E=\sqrt{\frac{2\times 8}{\left(4\times 3.14\times {10}^2\right)\times \left(3.0\times {10}^8\right)\times \left(8.85\times {10}^{-12}\right)}} \\ E=2V/m \end{gathered}$$

Hence, the peak electric field produced by the radiation is $2V/m$.