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Volts per Meter

  1. Oct 13, 2014 #1
    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 ?
     
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  3. Oct 13, 2014 #2

    jbriggs444

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    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.
     
  4. Oct 13, 2014 #3

    A.T.

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    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
     
  5. Oct 13, 2014 #4

    jbriggs444

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    Thank you for the correction. As you say, it is the change in potential.
     
  6. Oct 14, 2014 #5
    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 ?
     
  7. Oct 15, 2014 #6
    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.
     
  8. Oct 16, 2014 #7
  9. Oct 17, 2014 #8
    electric field 1.JPG 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
  10. Oct 17, 2014 #9
    Ι was wrong. Jbriggs 444 was referred to an uniform electric field. As i told before, "meter" in "volt / meter" do not show the distance between the charge-source and the charge at which the source's force, is applied. If this was the case, then, with increase this "meter", then we would have a dicrease of the source's force apply on the second charge and this means that the work produced from the movement of the charge would be smaller which means a smaller potential (or difference potential). But this "meter" shows the distance that the second charge (the "hypothema") runs, because of the force from source charge, apply on it. And the larger is this distance ( the "meter" in "Volt per meter") the larger the work, and the potential (or potential difference). In a uniform electric field, the Strength is stable. But this refers to potential difference between two point and not in potential in one point (that would be everywhere inside the field, the same). This means that if we imagine, two parallel metal plates, where each of them has equals but oposite charges and the distance between them is λ=1meter, between them we will have a uniform electric field with stable Strength and the potential difference between the two plates will be for example V=20volts. If ,inside this field we have an imaginary object with length 10mm, then at the two ends of it, the potential difference will be 0.2 volts. The electric field is uniform, means that if we move this imaginary object in other positions in the field, always the potential difference at its two ends will be the same (0.2volts). But this do not mean that if we put in the field another imaginary object with different length, would have the same potential difference at the two ends of it, with the first object . It just mean that the potential difference between the two ends of it, whichever is it, will be stable in the field. If for examble put in the field an imaginary object with 5mm length, at it's ends will have a potential difference V=0.1volts. And if we move the object , then the potential difference between the two ends of it, would be the same (0.1volts). If the field were not uniform, then when we move the object from one posiotion to the other, then it's potential difference(between the two ends of it) would change. For somebody to understand what i mean, i add a drawing i made.
    So now i think i can find volts per meter. Jbriggs444 said that 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. But i want to do the oposite. I will measure the total potential difference across a segment, and then i will calculate the Volts per meter, by devide by the distance in meters.
    Then, the question somebody may have is : If you know all these, why you ask ? If all these i told, are truth ( and please if somebody read and find wrong things in my thoughts, tell me, for me to understand),

    this is because jbriggs444 with his answer, made me to search and read again physic, and i want to say thanks for him. What confused me , was that he tried to explain me the electric field with the 1.5 volt battery hooked up to a uniform 100 meter wire . Is there an electric field in a wire at the ends of which is hooked up a battery ? The definition of electric field is that electric field is a space which in every electric charge that is inside this space exert electric forces. I think it is. If we thought the wire as a space, the charges that are inside it, exert electric forces, if a battery hooked up to its ends. But it would be better, in my opinion, if he tried to explain it using an example like the uniform electric field between the plates of a capacitor where the meaning of word "field" is more tangible, than in a wire.
    uniform electric field.JPG
     
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