# Volts per Meter

1. Oct 13, 2014

I measure the Electromagnetic Field Radiation at a point for example x distance from a source, and i am reading for ex. 15 milli volts per meter. Are these 15 millivolts per meter, equals to 15 milli volts ? I mean, if i do the opposite procedure, and receive a reading of 15 milli volts, at a distance of x units, lets say 3 meters, away from an EMF source, are these 15 milli volts, equals to 15 milli volts per meter ? Are these, two units, related each other and what is their relation ?

2. Oct 13, 2014

### jbriggs444

15 millivolts per meter would express the rate of change of the electric field over distance. If you had a 1.5 volt battery hooked up to a uniform 100 meter wire then there would be 15 millivolts of potential difference across each meter of that wire -- 15 millivolts per meter. Given 3 meters of this wire you could multiply 1.5 millivolts per meter by 3 meters and see that the potential difference across this segment would be 4.5 millivolts. Because the wire is uniform, the potential difference per meter is constant throughout.

In the case you describe, we probably do not have a uniform field. With a point source, the potential difference per meter will drop off as you get farther and farther away. The total potential difference across the three meters to the source could be computed by taking the integral of millivolts per meter across the three meters.

3. Oct 13, 2014

### A.T.

No. 15 Millivolts per meter is the electric field, not the change of the electric field over distance. You probably mean change of the potential (voltage) over distance, which equals the electric field strength:

http://en.wikipedia.org/wiki/Electric_field#Electrostatic_fields

4. Oct 13, 2014

### jbriggs444

Thank you for the correction. As you say, it is the change in potential.

5. Oct 14, 2014

you mean 45 millivolts (not 4.5) but if i follow this logic, means that for 4 meters the potential difference across this segment would be 60 millivolts and for 5 meters, 75 millivolts and so on. This means that the potential difference would increase with distance, but this is not logic. Except you mean something different. Can you give me a drawing of this "1.5 volt battery hooked up to a uniform 100 meter wire" to understand what you meant ? But i thing the "meter" in "Volts per meter" unit of electric field strength is not related with the distance between two points, at which measure the potential difference. Maybe my first wording was wrong. What i want to do , is to take a reading on volts, but not between to points. I will use an antenna through which i will take a measurement of volts of the EMF. Then i want to transform this Volts measurement to Volts per meters. But maybe, these Volts are the same with the Volts per meters. Why i am saying this ? Because, the Electric Field strength is the force exerted on a charge located at a point of an electric field, divited by this charge (F/Q=Newton/Coulomb = Volts/meter) It refers to one point. Not between two points. And Potential difference is something else than potential because it refers to the Work from the force (to the charge), when this force makes the charge to move from point A to point B, devided by the charge, and the potential refers to the Work from the force (to the charge), when this force make the charge to move from point A to infinite, devided by the charge. The "meter" on the "Volt per meter" unit, comes from the Joules , because qoulomb= Joule/Volts. Thus, the Newton/Qoulomb turns to N/(Joule/Volts), and because the Joule equals to Nweton * Meter, the N/(Joule/Volts) turns to N/((N*m)/Volts)) wich is equal to Volts per meters. And if this "meters" comes from Joules that is comes from Newton (Joule=meter*Newton), means that do not connect with the distance between to points but has to do with the force, and the work behind the force (applied to the charge), and that has not occurred yet. Because 1 Newton Is the force need to apply to an object of 1 kg to obtain 1m/sec^2 acceleration. I am not sure if all these i am talking aboute are correct, thats why i want your opinion.
As for the integral of millivolts per meter across the three meters, do you mean that i must do this because (Volt / meters)*meters = Volts ? And to do this, i must first find out the function governing my data, and then find the indefinite integral of this function ?

6. Oct 15, 2014

i want to make a small modulation in my previous thoughts. The "meter" in "Volts per meter" for the Strength of an electric field, comes from Joules (..A=Newton/Coulomb=AND/(Joule/Volt)=N/((N*meter)/V)...). 1 Joule is the work from applying 1 Newton for 1 meter. This , 1 meter, is the one that appears on the "Volt per meter" unit of EF strength. The work, i mentioned before, comes from the movement, through a point, of 1 coulomb (6*10^18 electrons) for 1 sec, in a way wich is "produced" 1 volt potential at this point (if we are talking for 1 watt (1 joule per second)). This meter is not a distance between two points, at which we are measure the potential differences, but gives the work we can take from the force as this force, moves the electrons. And the bigger this movement is, the bigger the work. But the function is E=Volt / meter, that means that when distance increase, Volts decrease. And this is truth because with more distance, the force apply to a negative charge from a positive source, in a electric field, is less, wich means that the movement of the negative charge will be less and the work will be also less. Thus , this "meter" declare,in a way ,the movement of the charge, wich will produce work, and not the distance between two points at which there is a potential difference. If we imagine an electrical field, with a positive source and two negative charges in X meters from the source, the force apply at these two charges will be the same, and the strength of the electrical field at this two points will be the same. Lets say that these two charges have a distance between them , but their distance from the source is the same. This can happen, if they are on the same homocentric from the source. Then, the potential difference between the two points (charge's position) will be zero. If now , the two charges start moving to the oposite direction from the source, then the potential in each will be reducing, but it will be the same between them. This means that, always, in this particular movement, the potential different between them, will be zero, but the potential, for each of them, will be redused, until they reach the distance from the source, in which the source cannot affect them, and then, both, their potential and the potential difference (between them) will be zero. If all these are truth, then to transform the "Volts per meter" to "Volts", i must find the increase or decrease of the Volts, inside one meter. I must try to go front and back, to see in wich direction the "Volt" indication, increase or dicrease, and then to measure this change in a range of 1 meter.If for example find a drop of 1 volts, in a distance of 3 meters, means that the strength is 0.33 volts per meter or 333 millivolt per meter. It does not meter the actual volt reading, but the rythm of dicrease or increase. If this is small, means that the EF strength is big, and if it is big means, that the EF strength is small. I attached a drawing i made, for somebody to understand, what i mean.

7. Oct 16, 2014

8. Oct 17, 2014

View attachment 74552 Νew modulation. E=Newton/Coulomb= Newton/(Joule/Volt)= Newton/((Newton*meter)/volt)= Volt/meter. The "meter " in "Volt / meter" comes from Joule, which is the Work produced from f (where f equals to 1 Nweton, or to the force needed for 1 kg mass be accelerated for 1 meter per sec^2) when f moves q for 1 meter. This means that here we have Work produced from charge's movement (moved from f) from, for example, point A to point B, with a distance between A and B , one meter. But this is a potential difference ' that is why i will add a potential difference in my drawing, between point A and A' and between B and B'. Because the field, which I have designed, is, like jbriggs 444 said, not uniform, the drop of potential and potential differences, until they will go to zero, will not be the same. Will not have a drop of, for ex., 1 volt at each of x meters from source, x+1 meter from source, x+2 meters from source etc. As we approach the x+7 meters, where the potential is zero, the potential drop will be less, between the points, at which, there is equal distance between them.

Last edited: Oct 17, 2014
9. Oct 17, 2014