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This is a question, from a radio amateur, that should have a straightforward answer: but I cannot find one on the Web.

Is there a robust but simple explanation, not needing advanced maths, as to why the field strength in the far field falls off as \$1/r\$ and not \$1/r^2\$ ?

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    \$\begingroup\$ Note that there is an amateur radio StackExchange site in beta now: ham.stackexchange.com \$\endgroup\$ Commented Oct 31, 2013 at 19:51

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Picture a transmitting antenna like a lightbulb. All the power emitted by the bulb can be thought of as hitting a sphere at any particular distance. The surface area of that sphere grows with the square of the distance, to the illumination hitting a piece of paper on that sphere goes down with 1/r2.

However, that was power. Field strength is like voltage in that power is the square of the field strength. If the square of the field strength falls off with 1/r2, then the field strength itself must be falling off with 1/r.

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  • \$\begingroup\$ Check your last sentence Olin \$\endgroup\$
    – Andy aka
    Commented Oct 31, 2013 at 19:31
  • \$\begingroup\$ @Andy: OK, I give up, what's wrong? Power (as in Watts/meter) is inversely proportional to the square of the distance, then field strength (volts/meter) is inversely proportional to distance. What exactly do you see wrong? \$\endgroup\$ Commented Oct 31, 2013 at 19:52
  • \$\begingroup\$ It's me - i read it as "If the field strength falls off" rather than as it is correctly written! \$\endgroup\$
    – Andy aka
    Commented Oct 31, 2013 at 20:28
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If you have an isotropic antenna it pushes power out equally in all directions and this means, at an arbitrary distance of "R" all the power is travelling thru a sphere of radius "R". A receiving antenna is said to have an "effective area" (or aperture) which picks-up a fraction of the transmitted power. At twice "R" the surface area of the sphere is 4x as big hence power received into the same antenna is quartered. Surface area of a sphere is \$4\pi R^2\$

Hence power reduces with distance squared. The two fields associated with EM power are the electric and magnetic fields. These combine to yield "power" according to their mathematical product and, if one halves, then so does the other. There is no compromise on this - free-space has a radiation resistance (just like coax has a characteristic impedance that is resistive) and E and M waves are proportioned to each other with a resistance of 377 ohms (see this wiki article).

Thus, a halving of E field produces a halving of the H (magnetic) field and therefore a quartering of the power. A quartering of power is when distance doubles so therefore when distance doubles the E field (or the H field) halve in amplitude.

All you've got to do is convince yourself that my explanation of power quartering with a doubling of distance is bona fide! Here's a picture of a real antenna radiation pattern that might help: -

enter image description here

The above picture is for the classic dipole antenna - the third diagram (c) shows (like a sort of doughnut with very tiny hole), equal-power levels distributed at a certain distance. As distance increases, the shape remains the same but the power per square metre at the surface reduces as the square of the distance because of the physical area of the doughnut.

For a dish antenna, the same \$\frac{1}{d^2}\$ rule applies with one exception. If the receiving dish is bigger than the transmitting dish it can completely "capture" all the transmitted power if the distance isn't too great - power received will remain theoretically constant up to this point then, as the transmiter's lobe area starts to fall outside the effective aperture of the receiving dish, power will reduce as per distance squared.

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