# Electric field measured in the far field of an antenna

I came across a question:

Electric field measured in the far field of an antenna at a distance r of 50 m = 1 V/m. Find the electric field at a distance of 500 m from the antenna.

The solution starts with - E is proportional to 1/r.

I know, E = V/r. Is it assumed that potential is the same at 500 m, to say that E is proportional to 1/r !? I thought E is proportional to 1/r^2, as per coulombs formula, if the charges are assumed to be constant. Where did I go wrong ?

• Near field = 1/r. Far field = 1/r^2. If E is proportional to 1/r at 500 meter, your antenna is ginormous. Perhaps this? upload.wikimedia.org/wikipedia/commons/thumb/a/ae/… – winny Dec 7 '17 at 18:55
• @winny power is 1/r^2, right? (from conservation of energy considerations) which would make the E field ... – Brian Drummond Dec 7 '17 at 19:11
• Isn't it the other way ? Farfield = 1/r. Does it suggest that potential is equal everywhere in the Farfield? – Mitu Raj Dec 7 '17 at 19:26

Farfield = 1/r. Does it suggest that potential is equal everywhere in the Farfield?

No it does not. The presence of electric field implies that there will be potential difference, in other words, no potential difference or a constant potential would imply no electric field. You are wrong in assuming $E = \frac{V}{r}$ for any r. Instead, $$E = -\frac{dV}{dr},$$ thus $E\ \ \alpha\ \ \frac{1}{r}$ would imply a logarithmic potential field (i.e. logarithmic dependence on r) rather than a constant field as you say.
If you want to know as to why $E\ \ \alpha \frac{1}{r}$, you can imagine a sphere of radius r centered at the source. Then, power leaving the sphere is: $$P_{rad} = (4\pi r^2)P_d,$$ where $P_d$ is the power density of the field. To have finite power radiated away from the source even for large r, $P_d \ \alpha \frac{1}{r^2}$. As, power density is related to squared of the electric field magnitude, thus we get a inverse dependence on distance for electric field. In other words, $$P_d \ \alpha E^2 \ \alpha \frac{1}{r^2} \implies E\ \ \alpha \frac{1}{r}$$

In general, power density will be of the form: $$P_d = \frac{C_1}{r^2} + \frac{C_2}{r^3} + \frac{C_3}{r^4}+...$$ For large r we just get the radiated term (first term) as the others would be negligible, but for small r's we will have small radiating term and we will have field mostly due to rest of the terms and it will constitute near field.

• What about in nearfield ? Why Pd is not proportional to 1/r^2 in near field ? – Mitu Raj Dec 7 '17 at 20:04
• Added in the answer. – sarthak Dec 7 '17 at 20:11
• E = -dV/dr ...that was a good catch :D – Mitu Raj Dec 7 '17 at 20:34

Since you are in far field conditions, the electric field generated by an antenna decreases with 1/r. Do not confuse it with the electric field generated by a point electric charge.

Your problem is due to confusing power and voltage (E field strength).

From basic geometry and conservation of energy, you should be able to see that the power has to go down with the square of the distance from the source. Think of the same power being spread out over the larger surface area of a larger sphere. That surface area goes with the square of the distance (sphere radius).

Everything else follows from there. Since the E field is proportional to the square root of the power of a traveling EM wave, the E field (measured in volts/meter in your example) must go down with the distance, not the distance squared.

Another way to consider this:

The power density existing in the far field of an e-m radiator (a transmit antenna) consists of an electric and a magnetic field, which are always related to each other by the ~ 377Ω impedance of free space.

Both of those fields decay at a 1/r rate, which results in a 1/r² decay rate in radiated power.