Question

A 2.50-m -diameter university communications satellite dish receives TV signals that have a maximum electric field strength (for one channel) of $\mathbf{7}\mathbf{.}\mathbf{50}\mathbf{\mu V}\mathbf{/}\mathbf{m}$. (See Figure24.28.) (a) What is the intensity of this wave? (b) What is the power received by the antenna? (c) If the orbiting satellite broadcasts uniformly over an area of $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{13}}{\mathbf{m}}^{\mathbf{2}}$(a large fraction of North America), how much power does it radiate?

Short answer

(a) The intensity is$7.47\times {10}^{-14}\mathrm{W}/{\mathrm{m}}^2$ .

(b) The power is $3.67\times {10}^{-13}\mathrm{W}$.

(c) The power radiated is $1.12\mathrm{W}$.

Step-by-step solution

Define electric field strength

The intensity of an electric field at a certain area is quantified as electric field strength. The volt per meter is the standard unit.

Evaluating the intensity

(a)

The formula for intensity in terms of electric field and light speed is

$$\begin{gathered} {\mathrm{I}}_{\mathrm{avg}}=\frac{{\mathrm{c\epsilon }}_0{\mathrm{E}}_0^2}{2} \\ \mathrm{Where}\mathrm{C}\mathrm{is}\mathrm{the}\mathrm{speed}\mathrm{of}\mathrm{light},{\mathrm{\epsilon }}_0\mathrm{is}\mathrm{the}\mathrm{permittivity}\mathrm{of}\mathrm{free}\mathrm{space},{\mathrm{E}}_0\mathrm{is}\mathrm{maximum} \\ \mathrm{electric}\mathrm{field}\mathrm{intensity},\mathrm{and}\mathrm{I}\mathrm{is}\mathrm{intensity}. \\ \mathrm{Substitute}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation},\mathrm{and}\mathrm{we}\mathrm{get} \\ \mathrm{I}=\frac{\left(3\times {10}^{-8}\mathrm{m}/\mathrm{s}\right)\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}-{\mathrm{m}}^2\right){\left(1.75\times {10}^3\mathrm{V}/\mathrm{m}\right)}^2}{2} \\ =7.47\times {10}^{-14}\mathrm{W}/{\mathrm{m}}^2 \\ \mathrm{Therefore},\mathrm{the}\mathrm{intensity}\mathrm{is}7.47\times {10}^{-14}\mathrm{W}/{\mathrm{m}}^2. \end{gathered}$$

Evaluating the power

(b)

The formula for calculating the power received by an antenna is

$$\begin{gathered} \mathrm{P}=\mathrm{I}\times \mathrm{A} \\ \mathrm{Substitute}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation},\mathrm{and}\mathrm{we}\mathrm{get} \\ \mathrm{P}=\left(7.47\times {10}^{-14}\mathrm{W}/{\mathrm{m}}^2\right)\times \mathrm{\tau \tau }\times {\left(1.25\mathrm{m}\right)}^2 \\ =3.67\times {10}^{-13}\mathrm{W} \\ \mathrm{Therefore},\mathrm{the}\mathrm{power}\mathrm{is}3.67\times {10}^{-13}\mathrm{W}. \end{gathered}$$

Evaluating the power radiated

(c)

The power radiated by satellite is

$$\begin{gathered} \mathrm{P}=\mathrm{I}\times \mathrm{A} \\ \mathrm{Substitute}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation},\mathrm{and}\mathrm{we}\mathrm{get} \\ \mathrm{P}=\left(7.47\times {10}^{-14}\mathrm{W}/{\mathrm{m}}^2\right)\times \mathrm{\tau \tau }\times \left(1.5\times {10}^{13}{\mathrm{m}}^2\right) \\ =1.12\mathrm{W} \\ \mathrm{Therefore},\mathrm{the}\mathrm{power}\mathrm{is}1.12\mathrm{W}. \end{gathered}$$