Question

The radiation energy emitted per second of a point source is $100\, W$. If the efficiency of the source is $4 \%$, then the rms value of the electric field at distance of $2\, m$ is $\left[\right.$ Use $\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9}$ 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 $100 W =\frac{4}{100} \times 100\, W =4\, W$
$A=$ area to which energy is transmitted
$=4 \pi r^{2}=4 \pi \times\left(2^{2}=4 \pi \times 4=16 \pi m^{2}\right.$
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, $I_{\text {avg }}=\varepsilon_{0} E_{m s}{ }^{2} c$
where, $\varepsilon_{0}=$ permittivity of free space
and $E_{r m s}=$ 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
$I_{\text {avg }}=\varepsilon_{0} \times E_{m s}^{2} \times 3 \times 10^{8}$
On equating both the Eqs. (ii) and (iv), we get
$\frac{1}{4 \pi}=\varepsilon_{0} \times E_{m s}^{2} \times 3 \times 10^{8}$ $\Rightarrow \frac{1}{4 \pi \varepsilon_{0} \times 3 \times 10^{8}}=E_{m s s}{ }^{2}$
$E_{r m s}{ }^{2}=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{1}{3 \times 10^{8}}$
$=9 \times 10^{9} \times \frac{1}{3 \times 10^{8}}$
$\left(\because \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} / C ^{2}\right)$
$=3 \times 10=\sqrt{30} V / m$