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volt per meter (V/m)

The standard unit of electric field (E-field) strength is the volt per meter (V/m). An E field of 1 V/m is represented by a potential difference of 1 V existing between two points that are 1 m apart. Reduced to base SI units, 1 V/m is the equivalent of one meter kilogram per second cubed per ampere (m ? kg ? s -3 ? A -1 ).

The volt per meter, or some fractional unit based on it, is used as a means of specifying the intensity of the electromagnetic field (EM field) produced by a radio transmitter. Although an EM field contains a magnetic (M) component as well as an electric (E) component, the relative field strength of radio signals is easier to measure in free space by sampling only the E component. The magnitude of the E component from a distant radio transmitter is often much less than 1 V/m, and in such cases, fractional units are preferred. One millivolt per meter (mV/m) is equal to 10 -3 V/m; one microvolt per meter (? V/m) is equal to 10 -6 V/m; one nanovolt per meter (nV/m) is equal to 10 -9 V/m; one picovolt per meter (pV/m) is equal to 10 -12 V/m.

The magnitude of the E component of a radio wave varies inversely with the distance from the transmitter in a free-space, line-of-sight link. If the distance is doubled, the E-field intensity is cut in half; if the distance increases by a factor of 10, the E-field intensity becomes 1/10 (0.1 times) as great. The E component of an EM field is measured in a single dimension, so the intensity-versus-distance relation is a straight inverse rule, not the inverse-square law.

When expressing the intensity of EM fields at infrared ( IR ), visible, ultraviolet (UV), X-ray, and gamma-ray wavelength s, the watt per meter squared, or one of the fractional units based on it, is more commonly used.

Also see electromagnetic field .

This was last updated in March 2011

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I hate to say this but i am old school. I remember when things were simpler. When any RF transmitter that emitted RF it was rated in watts. If you had an amateur radio station you knew how many watts you were transmitting.

So if you had a portable RF watt meter with an antenna then the closer you got to the transmitter the more watts the meter would read. And vice versa.

So if a cell tower was transmitting 40 watts of RF and you were at a distance of about 100 meters it would read a very low wattage. Simple not complex math.
You could still get a ratio of the RF output to the RF watts as you went further away. And we know that Rf at high frequency should be at a safe level of about 5 milliwats (that is what it used to be). Simple you get an RF watt meter and test the equivalent watts at a distance and determine what is a safe distance. All this other science and math is great for the actual engineers who have to build this stuff.
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Can someone perhaps ‘speculate’ but ideally provide facts as to the possible effect a handheld GPS receiver will have on a magnetic compass?
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increasing the distance 10 times from the transmitter would not be 0.1 of the output power of the transmitter. Electric field decreases according the inverse square law. that is the power d meters away from the source is P=1/(d^2)
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