I solved the first question and got the right answer, but I just can't seem to get the correct answer for the second part.
(a) An AM radio transmitter broadcasts 52.0 kW of power uniformly in all directions. Assuming all of the radio waves that strike the ground are completely absorbed, and that there is no absorption by the atmosphere or other objects, what is the intensity 31.0 km away? (Hint: Half the power will be spread over the area of a hemisphere.)
Answer: 4.31×10-6 W/m^2
(b) What is the maximum electric field strength at this distance? (in mV/m)
I solved and got an equation for Emax, but I still don't get the right answer when I plug in my intensity above.
Emax=sqrt(2*I/(3e8*8.85e-12)
Any help would be appreciated!!
To find the maximum electric field strength (Emax) at a certain distance (d), you can use the equation:
Emax = sqrt(2 * I / (c * ε))
Where:
- I is the intensity of the radio waves at that distance (found in part (a)),
- c is the speed of light (approximately 3 × 10^8 m/s),
- ε is the permittivity of free space (approximately 8.85 × 10^-12 F/m).
Since you have the intensity from part (a), you can substitute the values into the equation to find Emax.
Let's plug in the given values:
I = 4.31 × 10^-6 W/m²
c = 3 × 10^8 m/s
ε = 8.85 × 10^-12 F/m
Emax = sqrt(2 * (4.31 × 10^-6) / (3 × 10^8 * 8.85 × 10^-12))
Now, let's simplify the equation:
Emax = sqrt(8.62 × 10^-6 / (2.655 × 10^-3))
Emax = sqrt(3.24 × 10^-3)
Taking the square root:
Emax ≈ 0.057 V/m
Thus, the maximum electric field strength at a distance of 31.0 km is approximately 0.057 V/m.