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Electric Potential at Infinity

  1. May 8, 2004 #1
    Please help me wrap my head around this concept... I don't really need help on a problem in particular.

    If we define electric potential V to be zero at r = infinity for a point charge, how can we determine the potential difference of a test charge that moves from infinity to a specific point near the point charge?

    I guess my real problem is with the concept of infinity. It seems to me that bringing a test charge from infinity to a specified point would have some undetermined change in potential given the nature of infinity.

    Maybe I'm making this harder than it really is...
  2. jcsd
  3. May 8, 2004 #2


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    The problem with your statement is that voltage is an integral based on distance where higher distance is higher voltage. If the voltage is 0 at infinite distance, the voltage is 0 everywhere else.

    Say you have charges, 1 and 2. Say 1 is a referance point. Here is the formula for the force per charge based on that referance point. The symbol is a strange looking E but I can't remember the name so I'll just use E.


    [tex]E = \frac{kq_1}{r^2}[/tex] in N/C

    To get J/C (voltage), we have to integrate with respect to distance. The distance is the radius (r).

    [tex]V = \int \frac{kq_1}{r^2} dr[/tex]

    [tex]V = -\frac{kq_1}{r} |^{\infty}_r[/tex]

    Now you are asking that it moves from so now fill that in

    [tex]V = -\frac{kq_1}{\infty} + \frac{kq_1}{r}[/tex]

    Anything over infinity is 0 (or almost).

    [tex]V = \frac{kq_1}{r}[/tex]

    Right here is the voltage AT infinity. Remember that you said the voltage at infinity was 0. That means this term is equal to 0, which means the voltage at ANY given point r is 0.

    Do you see the problem? No matter where the test charge moves, the change in voltage is ALWAYS 0.
    If you don't have a voltage at infinity distance, you don't have a voltage anywhere.
    Last edited: May 9, 2004
  4. May 8, 2004 #3
    I got into the same problem... What is inf and what is potential...

    say you separate such +/- particle inf distance away. They should have 0 potential. But what if you said that if you let them run they can "gain" an infinte amount of potential by slowly being attracted towards one anther? Well, then I thought... If you put them real close you have to exert a inf amount of energy to pull them apart to go to inf... So its alright, just the concept defined backwards... Anyways... Hope that didnt' confuse you anymore than what you came in here with!
  5. May 8, 2004 #4


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    That's where the problem is. Energy is Fd, right? F was never mentioned in the question so you cannot at any point assume F is infinite. d on the other hand is stated as being infinite distance away. From that, you can conclude that if the distance between the particles is infinite, so is the energy.
  6. May 8, 2004 #5
    I think its a bit subtle to see it, but here is the way I look at it. First of all, it is very important to see that when there is no electric field, the electric potential difference is constant. This means that if they are no electric field lines around, the electric potential at one point is equal to the electric potential at some other point. Now, onto this concept of infinity. It is important to realize that whenever you are dealing with potential, you always need to define a reference point where the potential is defined. For example, when you lift a ball 10 meters above the ground, you say that the ball's potential has increased because it is common practice to define the ground as zero potential. Similarly, when dealign with say a point charge, it is common practice to define it as zero at infinity. Now, if your point charge is negative, and you are coming closer to it, then your potential would decrease, and so you will get a (-) answer. If you charge is positive and you are coming close to it, then your potential will increase.
  7. May 9, 2004 #6


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    You should remember that potential is just a relative scalar.
    It has a relative value and no direction. The value is relative
    because you're free to define the value of the potential
    at some point as you see fit, and once you have that
    reference point you can make all your calculations
    based upon that. You could, for example, define a charge
    as being a zero potential point, if you wish or it helps
    you solve a problem. With point charges, it is easier to
    define the potential at infinity as zero because then it comes
    out naturally from the integral that the potential of a point
    charge is V = k*q/r (like ShawnD said) and you need not
    mess with any additional numbers.

    Live long and prosper.
  8. May 9, 2004 #7


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    I just noticed that my answer is really biased. I assumed that the charges were opposite and moving closer to each other. If the charges are the same and moving away from each other, take away or add a - sign to all of my equations.
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