The amplitude of the electric field is 0.9 v/m 42 km from a radio transmitter. what is the total power emitted by the transmitter, if one assumes that the radiation is emitted isotropically (uniformly in all directions)? the impedence of free space is 377 ω . answer in units of w.

The power is found to be P = 14250600 W

Explanation:

First we find the voltage or potential difference at the position:

Voltage = Amplitude of Field x Distance

Voltage = V = (0.9 V/m)(42000 m)

V = 37800 volts

Now, we know the formula of electric power is:

Power = Voltage x Current

P = V I

but, from Ohm's Law:

V = I R

I = V/R

Therefore,

P = V²/R

Resistance = real part of impedence = 377 Ω

Therefore,

P = (37800 V)(377 Ω)

P = 14250600 W

divide the right side by the left side

0.024÷1.2=0.02 for example

the entire statement is true and correct.  

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