Complete Question:
When the Voyager 2 spacecraft passed Neptune in 1989, it was 4.5×109 km from the earth. Its radio transmitter, with which it sent back data and images, broadcast with a mere 21 W of power. Assuming that the transmitter broadcast equally in all directions,What are the signal intensity recieved at Earth, (I = ?) and What electric field amplitude was detected (E = ? V/m)? The received signal was somewhat stronger than your result because the spacecraft used a directional antenna, but not by much.
Answer:
I= 8.25 * 10-26 W/m2
E= 5.6 * 10-12 V/m
Explanation:
If we assume that the transmitter is broadcasting equally in all directions, which is equivalent to say that the transmitter antenna is omnidirectional, this means that the power spreads out like it the antenna were surrounded by spheres which radius is equal to the distance between the receiver and the source.
So, we can find the intensity, or power density, as the Power irradiated, divided between the area of the sphere, as follows:
a) I = P / (4 *π*r2)
Replacing by the values for P and r, we have:
I = 21 W / 4*π*(4.5*1012 )2 m2 = 8.25* 10-26 W/m2
b) In an electromagnetic wave, the power density and the Electric Field and Magnetic Field, are related for this expression:
I = E * H
Now, when the wave behaves as a plane wave (at a distance larger than two wavelengths from the source, which is called the far-field) there exists a proportion between both fields, in vacuum, as follows:
E/H = 377 Ω, which is called impedance of the free space, and relates also the permittivity of free space (εo) and the permeability of free space (μo) , as follows:
E/H= Zo= √(μo/εo)
Replacing in the expresión for I:
I = E²/ Zo
Solving for E:
E = √(I*Zo) = 5.6 * 10-12 V/m