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Contradiction about Electric Potential / Voltage in my Textbook

  1. Apr 12, 2007 #1
    Of course it's not a contradiction, it's really just my misunderstanding. But here's what I don't get:

    (N.B.: I'll be using the terms "electric potential" and "voltage" interchangeably in this post.)

    Consider a charge moving in the direction of an electric field. If it moves a distance of x meters in the direction of the field, then it goes to a lower electric potential. This makes sense, since it's analogous to gravitational potential energy. What doesn't make sense is this fact paired with the definition of voltage:

    The voltage at any arbitrary point P is the amount of work per unit charge it takes to move a positive test charge from infinity to that point. Consider this diagram.

    ~~~~~~~~~~~~| <--x--> |~~~~~~~~~~~~~~~~
    Infinity ------------ A --------- B ------------ Electric Field

    A charge that goes from A to B goes to a lower potential. The voltage at B is lower than the potential at A. HOWEVER, using the definition of electric potential above, you can say that the electric potential at B is actually greater than the electric potential at A, since it takes more work or equal work (if the electric field is doing the work for you) to move a positive test charge from infinity to B than it takes to move the positive test charge from infinity to A, because B is simply farther away than A is.

    You can argue that A and B are both infinitely away from the point selected at infinity where the electric potential is zero, so thus the fact that B appears farther away from the 0-point doesn't matter. But I still don't get then why it would take more work to move a positive test charge to B than it would to A, since the electric field is only weaker at A than it is at B.

    So can someone tell me why there's a lower potential at B than there is at A using the DEFINITION of voltage (work per unit charge from infinity to P) to explain this?
  2. jcsd
  3. Apr 12, 2007 #2
    Interesting question. I think what we have here is the problem that arises when physicists throw around the word "infinity." Because unlike the mathematicians, we physicists don't really know what we're doing when we play with numbers that don't exist.

    If it helps, think about a more illustrative analogy. Consider the graph of y = 1/x² As it so happens, this represents the electric field function in one dimension, for a point charge. Now consider the points x = 1, and x = 2. From calculus, you know that you can find the area under the graph of y = 1/x² from any point greater than zero all the way to infinity, and you'll get a finite number. If you take the definite integral from 1 to infinity, you get an area of 1. And if you take the definite integral from 2 to infinity, you get 1/2. But seeing as how both of these two regions have infinite "length" along the x-axis, one could qualitatively argue that they should have the same area. Yet when you actually graph the function y = 1/x², it probably makes more sense from the picture.

    The reason I'm using a mathematical analogy is because electric fields and potentials work exactly the same way. In your diagram, both points A and B might be considered the same "distance" from infinity. But then, infinity doesn't exist physically. The fact that the electric field changes more rapidly between A and B than it does between B and infinity is why A and B are not at the same potential.

    Does this make sense? If not, then I think I've got a more physical analogy up my sleeve.
  4. Apr 12, 2007 #3
    Interesting analogy. I still don't get it in the sense that I cannot explain why A has a higher potential than B using the definition of potential (explaining it using the gravitational potential energy analogy makes sense). I always imagined the definition of voltage to be analogous to the work required for a person to push a boulder from some infinity to some point P, where the person must fight friction and therein we see the work done. But why exactly does it take more work to push the boulder to B than it does to A? I'm abandoning the idea that B pushing the boulder to B requires more work because it's farther away -- it's not farther away, both are at the same "distance" as you said -- but the only thing I can think of to explain this is that the electric field is different at B than it is at A. Since B is closer to the charge distribution responsible for the electric field than A is, the electric field is stronger at B. So does it take more work to push the positive test charge to A because you get less help from the electric field?
  5. Apr 12, 2007 #4


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    Staff: Mentor

    You need to be careful about stating who or what does the work.

    If you're talking about the work done by the electric force, then the voltage (electric potential) at P equals the negative of the work done in bringing the charge to P from infinity, divided by the amount of charge.

    If you're talking about the work done by you or some other force acting against the electric force, then the voltage equals the work done, divided by the amount of charge (no extra negative sign). This assumes that the kinetic energy of the charge is the same at the starting point and at the destination, for example if it starts and ends at rest.

    So the electric field is directed to the right.

    The work done by the electric field on the charge is greater when it moves to B than when it moves to A, but the potential depends on the negative of that work, so it's lower (more negative) at B than at A.

    Suppose instead that you carry the charge to A or to B from the left, at constant speed. To keep it moving at constant speed, you have to exert a force to the left, against the electric force, in order to restrain the charge from accelerating under the influence of the electric force. Therefore you have to do negative work. To carry the charge to B, you have to do more of that negative work because the distance is greater. The potential depends directly on the work that you do, so by this analysis also, the potential is lower (more negative) at B than at A.
  6. Apr 13, 2007 #5
    Well this is no THE definition of voltage. This is the definition of voltage is you decide that the voltage at infinity is equal to zero. But this is just an assumption (most usual).

    But there are some problems where this assumption cannot be done.

    Instead of an "aerial" of your potential, draw a "side view":
    ---------------------------------------arbitrary zero

    Now what is the problem?
    If a reasoning does not hold with the drawing, search what is wrong with the reasoning.
    Verify what jtbell wrote about who is doing the work.
    Last edited: Apr 13, 2007
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