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Can Electric Field Strength describe the Electric Field?

  1. Oct 24, 2014 #1
    We have two difference electric fields A and B. We are standing at point W1 in field A and the potential at a point M1 which is 0.5 meter left is 10 volts while the potential at point M2, which is 0.5 meter right, is 5volts. The potential difference between M1 and M2 is 5volts and the strength in our point (W1) is 5 volts per meter. Now we are standing at point W2, in field B and the potential at a point M3 which is 0.5 meter left is 500 volts while the potential at point M4,which is 0.5 meter right, is 455volts. The potential difference between M1 and M2 is 5volts and the strength in our point (W) is 5 volts per meter. The strength of both the two fields, are the same, at the point we are standing each time. But is this right, when in field A, next to us there is a potential of 10 volts, and in field B, next to us there is a potential of 500 volts?
    Can, with EF Strength, describe an EF, for example to see if the EF is harmful for our health? I mean that the limits for Electromagnetic Radiation (for them to be harmful) are in Watts per m^2, and there is a relation between Watts per meter ^2 and Volts per meter, which is V/m = square root of ((Watt/meter)^2 * 377).
    thanks EF Strength.JPG EF Strength.JPG
     
  2. jcsd
  3. Oct 29, 2014 #2
    Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Oct 30, 2014 #3
    thanks for reply. First, i want to know if there is something wrong in my example and if my thoughts are right, somebody who knows about electric fields, to give me an answer. I think, my example is very clear. But I really want an answer. It seems , to me , very strange, that nobody has answered, because I think it is a very simple answer, to those who knows physic.
    (Τhe little frog in your picture is "Agalychnis callidryas" or the clown-frog and it is very rare)
     
  5. Oct 30, 2014 #4
    There is nothing wrong. It's just that the value of the potential is completely arbitrary. Only the potential difference is meaningful.
    In both cases you describe the same thing; a potential difference of 5 V. You can make the potentials of the two points 0 and 5 or 1 million and 1 million and 5. it's up to you to pick up the reference for potential.
    Its's similar to falling from a height. It's not more dangerous to fall from 1000 m to 995 m than is to fall from 5 m to the ground (0 m).
     
  6. Oct 31, 2014 #5
    Nasu, your example is quite good, and help me. It seems, that my question is not so simple, but it is complicate. It involves a) unit <watt per meter^2> b) unit <volt per meter> c) potentiale d)potential differences ) Electro Magnetic Field's safe limits for health . May be Greg is right, and I must reword the post. So, let say that I am standing next to an electricity pillar With a gauss meter, I measure the EMF strength, and I receive “A” milli Gauss . If now will move to another place, were next to me, is not the pillar, but an electric device that produce an EMF , but not so strong as the pillar, then I receive from the gauss meter , let’s say “B” milli Gauss. If I am not wrong, it must be “A”>”B”. Now, we have that :

    , “AmGauss = “2678.57 * Amicro Watts per m^2 = , “2678.57 * A *10^-6Watts per m^2 = SQUARE ROOT of [(“2678.57 * A *10^-6”) * 377] Volts per meter and

    BmGauss = “2678.57 * Bmicro Watts per m^2. = , “2678.57 * B *10^-6Watts per m^2 = SQUARE ROOT of [(“2678.57 * B *10^-6”) * 377] Volts per meter .

    If we assume that A = 10* B, then , in the first case were I stand next to the pillar the strength in my point is SQUARE ROOT of [(“2678.57 * A *10^-6”) * 377] Volts per meter. Let’s name it “X”, thus the EMF strength is X volts per meter

    And in the second case were I stand next to the “weak” device, the strength in my point is

    SQUARE ROOT of [(“2678.57 * B *10^-6”) * 377] Volts per meter . or

    SQUARE ROOT of [(“2678.57 * {A/10} *10^-6”) * 377] Volts per meter, and let’s name it “Y”, thus the EMF strength is Y volts per meter

    What we have until now? We have that when I am standing next to a strong source , the potential difference is X volts per meter, which is bigger than Y volts per meter (which is the potential difference, in my point, when I am standing next to a weak source). This mean that in a field that produced by a strong source, the decrease of potential (when the distance from the source, change) is larger than the decrease of the potential in a source that produced from a weak source.

    And my question is: Is this right ? If I am in an EMF, and measure the volts (not volts per meter, but just volts), and find that the drop of volts in distance of 1 meter is Y (from K to K/10 for example), then if I go to another EMF which is stronger, and make the same measurement, will i find that the drop of volts in distance of 1 meter will be X (from K to K/2 for example), were X>Y ? ΕΜF Volt dicrease.jpg
     
  7. Oct 31, 2014 #6
    You make it complicated (if I understand correctly what you say) by assuming that V/m measures a potential difference.
    This is not so. The intensity of the electric field is measured in V/m (or N/C). The unit V/m reflects the fact that the intensity of the electric field is proportional to the gradient of the potential. If the field have the same value, the gradient will be the same.

    So you have two measurements of the field which correspond to two values of the intensity. As the intensity is proportional to the square of the field, the factor of 10 in the field measurements result in a factor of 100 in the intensities.
    Working with numerical values and units instead of quantities is not a good idea.
    The basic formula is quite simple.
    For example,
    [itex]I=\frac{E^2}{2 c \mu_o}[/itex] for electric field.
    Here E is the electric field in V/m and I the average intensity of the electromagnetic wave in W/m^2.
    The numerical coefficients come form the speed of light and the permeability of the vacuum.
     
  8. Oct 31, 2014 #7

    A.T.

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    As nasu hinted, the potential difference is measured in Volts, not in Volts per meter.
     
  9. Nov 2, 2014 #8
    A.T. My question was if the decrease of volts (by distance), in an electric field "A" is bigger than the decrease of volts in an electric field "B" , were "A" is stronger than "B" (at the same distance difference for both fields). If somebody know it, just tell me.

    Who says that the potential difference is not measured in Volts per meters? May be you did not read all my post or you read but you do not understood, and this, may be is my fault. I told <<.... I am standing next to an electricity pillar With a gauss meter, I measure the EMF strength, and I receive “A” milli Gauss>> "milliGauss" can also be written as "Volts per meter", because there is a relationship between "milliGauss" and "μWatt per m^2", and a relationship between "μWatt per m^2" and "Volt per meter". So if the strength of EF described as "X" Volt per meter, it means that there is a potential difference between two points, with 1 meter distance from each other, and is "X" Volts.(if strength is "X" Volts per meter, then between two points in a distance of 0,1 meter, the potential difference will be "X/10" Volts.) If you read my first post again, you will see that it is based in two measurements, both in a distance of one meter. Also you can read the last lines of my second post I said <<…..If I am in an EMF, and measure the volts (not volts per meter, but just volts), and find that the drop of volts in distance of 1 meter is Y (from K to K/10 for example), then……>> When i said <<..the potential difference is X Volts per meter...>> i meant that between two points(with one meter distance between them) the potential difference is X Volts. So if you describe an electric field, with the "Volts per meter" unit, you refer to a potential difference between two points. If you want to refer to potential (when you describe an EF with Volts per meter) then you must say, about the potential of two points, of which, two points, i want to measure the difference between their potentials. The unit "Volt per meter" include the potential difference. We can say that the potential difference is X Volts per peter , if the distance between the points that we are refer is one meter and the potential difference between them is "X" Volts. Or we can say that the potential difference is X volt per 5 meters , if the distance between the points that we are refer is five meters and the potential difference between them is "X" Volts. If you would ask me : <<Tell me the potential difference that you measured please>> And if i said << "X" volts>>, then , could you understand the strength of the field, only with this "X" Volts ? Would not you want me to tell you which is the distance i measured? Because "X" Volts per one meter is quite difference from "X" Volts per five meters. If you see the definitions for potential and potential difference you will see, that the difference is that potential, refers to the Work produced from a force which move a charge from a point A to the infinity (divided by the charge) and potential difference (between point A and B) refers to the Work produced from a force which move a charge from a point A to a point B (divided by the charge). So when we say that the potential difference is "X" Volts, as a matter of fact, we say that the potential difference is "X" volts per "K" meters, were "K" is the distance between the two points that the potential difference is referred to. What you said <<.....the potential difference is measured in Volts, not in Volts per meter>> is a kind of convention. Τhe important is not to understand the conventions but the meanings.

    Nasu did not hinted, that potential difference is measured in Volts and not in Volts per meter. If you read his post you will see that he talks about “assume”, which is quite difference. He told me that the Volt per meter unit reflects the fact that the intensity of the electric field is proportional to the gradient of the potential, and this is truth. He advises me not to think of Volt per meter as potential difference between two points but as a gradient of the potential.
     
  10. Nov 2, 2014 #9
    Nasu, with your answers, you help me to clear up my mind. I, just do not understand two points.

    1)<…As the intensity is proportional to the square of the field…>> Do you mean that the intensity expressed as Watts per m^2 is proportional to the square of the intensity expressed as volts per meter ?

    2) The type you wrote is the same with this : W per m^2 = [(volt per meter) ^2] by 377 were 377 comes from c and vacuum ?

    But generally I think I am starting to understand. The fact that there is a relation between W/m^2 (intensity) and V/m (gradient of the potential), in which relation involves the factor of ten, means that an “X” difference between two fields gradientof potential (V/m) “reflects” an “Y” gradient of intensities (W/m^2) between the two fields, were “Y” = “100*”X” if “X”=A/B=10 (were A and B are gradient of potentials). Thus , there is an exponential increase of the intensity by increase gradientof potential (V/m). Or if I follow the opposite direction, can say that, if I measure the intensity of two fields and I find an “Y” difference between them , then this difference is “created” by a smaller “X” difference of gradientof potentials (V/meters) than the “Y”? And I assume that this is happening because the relation between the, one dimension (per meter) and two dimensions (per meter^2). Ok, until now I think I understood (correct me if I am wrong).

    And now another question that is connected with all these. If we assume that there is a devise which can detect the EM Field, but not in milliGauss, or in Watts per meter, or mTesla or Volts per meters , but as Volts, and let’s say that I am sitting in my desk and as using this device I receive 30 milli Volts. Can I convert this, to Volts per meter, in order to convert this to μWatts per m^2 in order to know what is the intensity of the EMF in my area?
     
  11. Nov 3, 2014 #10

    A.T.

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  12. Nov 3, 2014 #11

    Dale

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    Krakadoros, your units are all messed up.
    AT is correct (as he usually is). Potential difference is measured in V, not V/m.

    No, it cannot. Gauss has SI base units of ##kg/(s^2 A)## and V/m has SI base units of ##kg~m/(s^3 A)##. They are different.

    No, there isn't. W/m^2 has SI base units of ##kg/s^3##. They are different.

    No, they are different.

    Assuming a uniform field and measurements separated in the direction of the field, yes. Neither of these hold in general.

    No, you cannot. Potential difference is always in V, never in V/m. It is like saying that you can measure the area of a puddle in liters because you know the depth. No, area is in m^2 and volume is in L. Even if you have a scenario where knowing the volume allows you to solve for the area they are still different quantities.
     
  13. Nov 3, 2014 #12

    Dale

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  14. Nov 3, 2014 #13
    If the hint was not obvious, then I say it directly: Potential difference is measured in volts and not in V/m.

    Your obstinacy in operating with units rather than quantities is counter-productive. As we can see from your confusions and irrelevant questions.
    I tried to suggest to operate with quantities and the relationships between them.

    I am afraid that I cannot understand all you are asking.
    One thing I can say: if you measure the potential in one point (these mV) you cannot calculate the electric field at that point. Neither the intensity of radiation. You need to know the gradient of the potential at that point so you need the potential distribution around that point.
    And this is true no matter what units you use to measure these quantities.
     
  15. Nov 6, 2014 #14
    you talk about "μWatt per m^2" and "Volt per meter" and that there is not a relationship between them. But there is a relationship between them. On free space conditions (impedance of 377 ohm) it is 1 V/m = 2.653 e^-3 W/m^2 = 2.653e^-4 mW/cm^2 = 2.661 e^-3 A/m = 3.350e^-5 Gauss = 3.350e^-9 Tesla = 70.50 dBpT = 3.350e^+3pT = 68.50 dBuA/m = 120 dBuV/m .
    You are talking about relationship between "milliGauss" and "μWatt per m^2" (i said μW/m^2 not W/m^2. μ stands for "μικρό" Greek word for "small" , "micro") But there is a relationship between them. On free space conditions (impedance of 377 ohm) it is : 1Gauss = 29853.83 V/m = 209.5 dBuV/m = 158 dBua/m = 1.000e^+8 pT = 160 dBpT = 1.000e^-4 Tesla = 7.943e+1 A/m = 2.364e^+5mW/cm^2 = 2.364e^+6 W/m^2, and if there is relationship between Gauss and W/m^2 there is also between milliGauss and W/m^2 and μW/m^2.
    you talk about Gauss and V/m, that milliGauss" cannot also be written as "Volts per meter", like i said. From the above equations you can see that milliGauss can also be written as V/m.(1Gauss = 29853.83 V/m, so 1 milli Gauss = 29.85383 V/m)
    When somebody say that potential difference is for example 5 volts , what the word "difference" in "potential difference" stands for ? Is it 5 volts difference for what ? It means that there are 5 volts difference between two points. If these 2 points have 5 meters distance, from each other, then we have 5 volts per 5 meters (1volt per meter), if these 2 points have 10 meters distance, from each other, we have 5 volts per 10 meters (0.5 volts per meter), if the potential difference is 12 volts and If these 2 points have 8 meters distance, from each other, then the potential difference is 12 volts per 8 meters (1.5 volt per meter), if the potential difference is X Volts and If these 2 points have Y meters distance, from each other, then the potential difference is X volts per Y meters. Why do you find this wrong ? I am talking aboute the meaning, not the convention. I know that if we refer to potential difference we say X volts, but have you understund that this is a convention ? Instead to say " the potential difference between these two points ,which have 10 meters distance, from each other, is 12 Volts (which is the same to say "the potential difference is 12 volts per 10 meters), we say the potential difference is 12 volts. If i would tell you that the potential difference between two points which have 8 meters distance, from each other, is 12 Volts, then will you be satisfied ? Is not the same if i told "12 volts between 8 meters" and "between 8 meters, 12 volts" ? It is the same when we meassure something, let's say the length of a car. What we say ? This car is 4 meter. What is this 4 meters ? It is 4 meters from where ? This "4 meters" is not alone. Like the "12 Volts" is not alone. 4 meters is always accompanied by "from a point at the front edge to a point at the back edge, but these points must belong to a straight line paralel with ground." And 12 volts is always accompanied by "between two points have 10 meters distance each other" (in our example) .No matter that it is not mentioned, It is always there, and if somebody vocalize it , is not wrong. Wrong will be if instead to meassure the distance between the front edge and the back edge, will meassure the distance between the top edge and the bottom, because instead of length would meassure the height, like in the electric field, wrong will be , if instead to meassure the potential difference between two points A and B, meassure the potential difference between poin A and infinity. Or i will use another example comes from your photos. I can see two photos of your face ( i admit very good photos). If there are 2 questions ? 1)"Rated the beauty" and 2)"Rated the cleverness" then if i will say 8, but just 8 without say 8 for the beauty (like 12 volts per meter) or 8 for the cleverness (like 12 volts per meter), then nobody will knows if you are more beautifull or more clever.
     
  16. Nov 6, 2014 #15
    How many are you ?
    give me an example
    What about move back or front ? I mean, if first measure, let's say, 33mV and then move front for 1 meter and have a new measurement that is 36mV (this is because i would approach the source (it will be in my back, let's say)) then is not this a gradient of the potential ? Thus 3mV per meter. Or if i would move front, for 5 meters and have a new measurement 25mV (this is because i would fend off the source) then is not this a gradient of the potential ? Thus 8mV per 5 meters or 1.6 V/m ?
     
  17. Nov 6, 2014 #16
    will advise you not to believe everything wikipedia says. It is better to open a book with bibliography. For example i will say this : Before a month or something i read in wikipedia that , only mammals have hairs. But this is incorrect. Next time a bee or a spider, sit on you, catch it , and then see it closely. You will find that , arthropods also have hairs !!
     
  18. Nov 7, 2014 #17

    A.T.

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    For the basic level of your misconceptions wikipedia will do fine.
    Which book have you opened and what does it say about the units of electric potential?
     
  19. Nov 7, 2014 #18
    Books with bibliographies tend to go out of date very quickly. The 'hair' on a bee is made of chitin, which is also the stuff used by insects for making shells, mammalian hair is made from keratin as with finger nails. As such only mammals do have hair and the 'hair' on a bee is not in fact related to mammalian fur.
     
  20. Nov 7, 2014 #19

    Dale

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    No, your units still don't work out. In base units V/m/ohm -> A/m, which is not the same as W/m^2 -> kg/s^3.

    This also is wrong. In base units gauss/ohm -> s A/m^2, which is not the same as V/m -> kg/(s^2 A)

    The word "difference" stands for "subtraction". So a potential difference is ##\Delta V = V_B-V_A##. Since both ##V_A## and ##V_B## are in volts, then ##\Delta V## must also be in volts. This is just basic and straightforward math of the simplest kind. Your informal attempted justifications are not relevant.
     
  21. Nov 16, 2014 #20

    Doc Al

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    While the difference between chitin hair and keratin hair might prove entertaining to some, it is wildly off topic for this thread. The simple fact is that Krakadoros is wrong about a trivial issue discussed in any elementary physics book. There is nothing controversial or worthy of lengthy discussion here, so this thread is done.
     
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