antennaelectromagnetic
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 ?
An antenna produces fields that fall into three categories: -
And, the generally accepted formulas for these are: -
So, if the largest dimension were 0.75 metres and the frequency was 100 MHz (i.e. a quarter wave monopole), the reactive near field is done at 24 cm and the far field begins at about 38 cm.
However, for a dish with diameter 7.5 metres (at 100 MHz), the reactive near field extends to about 7.4 metres and the far field begins at about 37.5 metres.
This site has a calculator and is where I took the formulas from. The formulas are generally accepted as being meaningful but, there are no hard and fast rules or limits.
This site also uses the same formulas and has a nice picture that shows how the fields gradually converge in amplitude to produce a proper EM wave at an impedance of 377 ohms (free space impedance). This is for an electrically "short" antenna: -
Wiki(Near and far field) also uses the same system for calculating the near and far fields: -
Once a proper EM wave is formed by an antenna, the power density (watts per square metre) reduces with distance squared. Hopefully this can be seen but if not, think about the conceptual isotropic antenna - it takes in a signal power from the feed wires and emits that power equally in all directions. This is just like a light bulb emitting approximately equally in all directions and, if you measured the light power passing through a fixed area at some distance you would find the light power measured falls with distance squared.
Given that the power density is simply the electric field (volts/m) multiplied by the magnetic field (amps per metre) either of these terms reduces linearly with distance so, it's not surprising to see this formula.
The "A" bit refers to the effective aperture of any antenna. If an antenna is to liberate "watts" from "watts per sq metre" then it must have an effective area so that it can "net" the power. For any antenna this is: -
Where lambda is wavelength and G is antenna gain. An isotropic antenna has G set to 1. Also note that an isotropic antenna is just a mathematical concept for comparing performances of different antennas.
Best Answer
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.