Question

In reading the instruction manual that came with your garagedoor opener, you see that the transmitter unit in your car produces a$250\mathrm{mW},350\mathrm{MHz}$signal and that the receiver unit is supposed to respond to a radio wave at this frequency if the electric field amplitude exceeds$0.10\mathrm{V}/\mathrm{m}$. You wonder if this is really true. To find out, you put fresh batteries in the transmitter and start walking away from your garage while opening and closing the door. Your garage door finally fails to respond when you're $42\mathrm{m}$away. What is the electric field amplitude at the receiver when it first fails$?$$?$

Short answer

At receiver the first fails of electric field amplitude is$0.092\frac{\mathrm{V}}{\mathrm{m}}$.$0.092\frac{\mathrm{V}}{\mathrm{m}}$

Step-by-step solution

Calculation of intensity

Given info:

Power is$250\mathrm{mW}\to 250\times {10}^{-3}\mathrm{W}$.

Distance is$42\mathrm{m}$.

Intensity is,

$\mathrm{I}=\frac{\mathrm{P}}{\mathrm{A}}$

$\mathrm{I}=\frac{\mathrm{P}}{4{\mathrm{\pi r}}^2}$

$\mathrm{I}=\frac{250\times {10}^{-3}\mathrm{W}}{4\mathrm{\pi }(42\mathrm{m}{)}^2}$

$\mathrm{I}=1.13\times {10}^{-5}\mathrm{W}/{\mathrm{m}}^2$

Calculation for electric field amplitude

Intensity,

$\mathrm{I}=\frac{{\mathrm{c\epsilon }}_{\mathrm{o}}}{2}{\mathrm{E}}_{\mathrm{o}}^2$

$2\mathrm{I}={\mathrm{c\epsilon }}_{\mathrm{o}}{\mathrm{E}}_{\mathrm{o}}^2$

${\mathrm{E}}_{\mathrm{o}}^2=\frac{2\mathrm{I}}{{\mathrm{c\epsilon }}_0}$

${\mathrm{E}}_{0=}\sqrt{\frac{2\mathrm{I}}{{\mathrm{c\epsilon }}_0}}$

role="math" localid="1649962031410" ${\mathrm{E}}_{\mathrm{o}}=\sqrt{\frac{2\left(1.13\times {10}^{-5}\mathrm{W}/{\mathrm{m}}^2\right)}{\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)\left(8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2\right)}}$

${\mathrm{E}}_{\mathrm{o}}=0.092\frac{\mathrm{V}}{\mathrm{m}}$