Question

The radiation energy emitted per second of a point source is $100W$. If the efficiency of the source is $4\%$, then the rms value of the electric field at distance of $2m$ is $[$ Use $\frac{1}{4\pi {ϵ}_0}=9\times 10^9$ in S.I. Unit]

• A. $\sqrt{60}V/m$
• B. $\sqrt{30}V/m$
• C. $\sqrt{50}V/m$
• D. $\sqrt{40}V/m$

Answer 1

Answer: (B)

Average intensity, $I_{\text{avg }}=\frac{P_{\text{avg }}}{A}$
where, $P_{\text{avg }}=$ average power generated by source
$=4\%$ of $100W=\frac{4}{100}\times 100W=4W$
$A=$ area to which energy is transmitted
$=4\pi r^2=4\pi \times (2^2=4\pi \times 4=16\pi m^2$
On putting the values into Eq. (i), we have
$I_{\text{avg }}=\frac{4}{16\pi }$
$\Rightarrow I_{\text{avg }}=\frac{1}{4\pi }$
Also average intensity, $I_{\text{avg }}={\epsilon }_0{E}_{ms}{}^{2}c$
where, ${\epsilon }_0=$ permittivity of free space
and $E_{rms}=$ root mean square (rms) value of electric field.
$c=$ speed of light in vacuum $=3\times 10^{8}m/s$
On putting the values into Eq. (iii), we get
$I_{\text{avg }}={\epsilon }_0\times E_{ms}^2\times 3\times 10^8$
On equating both the Eqs. (ii) and (iv), we get
$\frac{1}{4\pi }={\epsilon }_0\times E_{ms}^2\times 3\times 10^8$ $\Rightarrow \frac{1}{4\pi {\epsilon }_0\times 3\times 10^8}=E_{mss}{}^2$
$E_{rms}{}^2=\frac{1}{4\pi {\epsilon }_0}\times \frac{1}{3\times 10^8}$
$=9\times 10^9\times \frac{1}{3\times 10^8}$
$\left(\because \frac{1}{4\pi {\epsilon }_0}=9\times 10^{9}N{m}^2/C^2\right)$
$=3\times 10=\sqrt{30}V/m$