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Derive an expression for the potential

  1. Apr 23, 2015 #1
    1. The problem statement, all variables and given/known data
    A positive charge +q is located at the point x= 0, y = -a and a negative charge -q is located at the point x=0, y=a (a) Derive an expression for the potential at points on the y-axis as a function of the coordinate y. Take V to be zero at an infinite distance from the charges.

    2. Relevant equations
    ΔV = -∫E⋅dl
    E= kq/r2
    V = kq/r
    3. The attempt at a solution
    I don't understand how my book solved this problem. Can someone explain this solution to me?:
    http://imgur.com/vIEK7ql [Broken]

    I understand that the potential is calculated at points between, below, and above the two charges, but how was the final equation produced from the first 3? Some have a's in the numerator while other had y's and they randomly disappeared in the answer.

    Additionally, is it possible to solve this by finding the electric field of an arbitrary point between the charges then integrating it from that point, y, to infinity with respect to y.

    I've been given multiple methods of calculating electric potential in my book and I can't seem to figure out which equation is the appropriate one to use. This one seemed like it could be solved using ΔV = -∫E⋅dy, but when I attempted it, it became very complicated. The solution my book provided seem simple, but I'm just not sure what method they are using to obtain the solution.
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Apr 23, 2015 #2


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    Show what you have so far, and maybe we can help point you in the right direction where you are stuck.

    If, for a single point charge, you define V = 0 at r = ∞, then [itex] V = k \frac{q}{|r|} [/itex], assuming that the point charge is at the origin. But what if the point charge is not at the origin? What if it's at some displacement [itex] \vec a [/itex]? In that case, the potential is inversely proportional to the distance away from point [itex] \vec a [/itex]. In other words, it becomes [itex] V = k \frac{q}{| \vec r - \vec a|} [/itex].

    It's not really random. Notice the variables in each denominator are arranged such that the given denominator is always positive. That you you can get rid of the absolute values. The rest is a matter of which terms do not cancel in the numerator after finding a common denominator.

    Yes, it is possible! But it can be a huge headache, (particularly if you stray off the y-axis for a problem like this). Why the headache, you may ask: because of those pesky unit vectors. When working with electric fields, you can add the individual electric fields from each point charge, at any given point in space -- but you must make sure to add them as vectors, meaning you must first break everything up into the vector components before adding them.

    On the other hand, potentials are scalar fields. You don't have to worry about those pesky unit vectors.

    Do you recall way back when, when you first learned about how to solve problems with conservation of energy? Where if you wanted to find the speed of a block sliding down a frictionless slope, you knew you could calculate the acceleration of the block, solve the the time it takes to reach the bottom, then use of your kinematics equations to solve the the velocity. Then you realized that with conservation of energy you could get there in one easy step?

    Well, working with potential's isn't quite like that, but it's a full leap in that direction.

    With potential fields, you can simply add individual potential fields from different charges together (you can add the fields of each point charge together) as scalars. You don't need to worry about breaking them up into their individual vector components; in potential fields there are no vector components.

    In problems where there are a lot of charges around, working with potentials can make the problems much easier.

    I think now you are beginning to see what I mean. It's possible to do it that way, but hairy.

    A simpler approach is to find the potential contribution of one charge. Then find the potential contribution of the other charge. Then add the two potential contributions together, and there ya go.*

    *[Edit: of course both charges must be represented by the same coordinate system, and both potential fields (each corresponding to one the two charges) must share the same definition of what location has 0 potential. In other words you need to be consistent, is all I'm saying.]
    Last edited by a moderator: May 7, 2017
  4. Apr 23, 2015 #3
    Thank you that was extremely useful, but I'm still confused about the general solution for the electric potential that my book provided. You said that the denominator is always positive, but for |y|<a, the denominator is negative. I just don't see how the general expression can be applied to all points along the y-axis, from observing the three expressions above it. I understand how the numerator and the constant k would apply to all situations on the y-axis, but I don't know where the values for the denominators came from.

    I don't get the purpose of having y2-a2, as the denominator for each of the three equations. Even though the denominators are the same, the numerator is either multiplied by an additional y,-a, or a, which means all three of them would yield different results. So how is it that the general expressions can be derived from three different equations that provide different results.
  5. Apr 23, 2015 #4


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    y2 - a2 comes from the fact that it's the common denominator which allows combining the two rational expressions into one . It's simply some basic algebra.

    Using a range of specific values for y for some specific value of a may help you see what's going on.
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