Given this question from the full license exam in the UK:

What is the field strength at a distance of 7m from an antenna with an effective radiated power of 100W?

in which the answer is:

10 V/m

How do you calculate it?

  • $\begingroup$ Related, but requiring slightly different calculations: ham.stackexchange.com/questions/7464/… $\endgroup$
    – Kevin Reid AG6YO
    Commented Mar 18, 2018 at 16:42
  • 3
    $\begingroup$ The information in the question is incomplete in that the frequency was not stated. For some frequencies a path distance of 7 meters from the transmit antenna could still be within its near field, which would require a different analytic method to determine E-field intensity. Also, the propagation environment was undefined, and for some situations that environment could produce a reflection that could affect its field at 7 meters. $\endgroup$ Commented Mar 18, 2018 at 20:00
  • 1
    $\begingroup$ @RichardFry Points well taken! $\endgroup$
    – Glenn W9IQ
    Commented Mar 18, 2018 at 20:06
  • $\begingroup$ @RichardFry I updated my answer to note your points. Let me know if you think this is sufficient. $\endgroup$
    – Glenn W9IQ
    Commented Mar 19, 2018 at 10:37
  • $\begingroup$ I was about to respond, but saw the advice about not answering questions in the Comments area. $\endgroup$ Commented Mar 19, 2018 at 15:58

2 Answers 2


The field strength in volts/meter as a function of effective radiated power and distance is given by:

$$E=\frac{7.01\sqrt{P_{ERP}}}{d} \tag 1$$

where d is the distance from the radiating antenna in meters and PERP is the effective radiated power in watts.

So you can see your sample problem was chosen such that the 7 meter distance cancels out the upper 7.01 term so you are simply left with the square root of 100 which results in 10 volts/m.

Formula 1 is derived from the long form of first calculating the power density on the imaginary sphere that has the transmitting antenna as its center and the point of observation on the surface of the sphere.

The area of the sphere is given as:

$$A=4\pi r^2 \tag 2$$

where r is the radius of the sphere (or in this case the distance from the transmitting antenna).

The power density is then calculated as:

$$S=\frac{P_TG_T}{4\pi r^2} \tag 3$$

where PT is the transmitter power and GT is the linear gain of the transmitting antenna.

Since the problem states "effective radiated power", this means that gain of the transmitting antenna in this case is ~1.64 for a 1/2 wavelength dipole. Thus we find that the power density is ~0.266 watts/m2. Had the problem stated "effective isotropic radiated power", the gain would be 1 to reflect the gain of an isotropic antenna.

Since the power density is assumed to be in free space with an impedance of 377 ohms, we can use a variant of Ohm's law to find the field strength:

$$E=\sqrt{377S} \tag 4$$

This results in the same answer as equation 1 with a field strength of ~10 V/m. Equations 3 and 4 can be combined and reduced to form equation 1. I leave this as an exercise for the reader.

It would also be good to take note of the comments by @RichardFry. For this analysis to be accurate, the observation distance must be in the far field of the antenna and the propagation must satisfy the "free space" condition.


How do you calculate it?

In the general case, you use a numerical EM field solver.

Without a specification of the antenna geometry and exactly which point 7 meters away is under consideration, this is not an answerable question. Even with such specifications the answer is very complex to calculate, and not the sort of thing one would casually do with a pencil for an exam.

Only if 7 meters away is in the far field, meaning several wavelengths away can we use the impedance of free space and power density to calculate the field strength as Glenn W9IQ explains.

If this point 7 meters away is not in the far field, the geometry of the antenna is very relevant. At very close distances, a dipole will have a stronger voltage field than a small loop antenna, for example. The near-field strength has little to do with the radiated power density.

Since nothing in the question establishes that the operating wavelength is significantly less than 7 meters, and there are many amateur bands with a wavelength far in excess of 7 meters, there is insufficient information in the question to make a correct answer.

Apparently, the UK exam has a pattern of asking such grossly oversimplified questions.


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