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Electric Potential Energy Between Two Charges

  1. Mar 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Two small metal spheres, each of mass m, are attached to the lower ends of two long insulated strings of negligible mass as shown. Each string is length, l. The spheres are pulled apart until each string makes an angle "theta" with the horizontal. A charge of +q is placed on one sphere and a charge of -q is placed on the other. The spheres are then released from rest, make a perfectly elastic collision, neutralize their charges and rebound (bounce apart). Find the maximum angle each string makes relative to the vertical after the collision.

    The following link is a copy of the problem with solution and diagram provided by my professor.

    2. Relevant equations

    V=kQ/r ........[1]
    U=-kQq/r .....[2]

    3. The attempt at a solution

    I do not believe this problem is solvable without knowing the radius of the hanging masses. However my professor and her colleague disagreed. They claim U=-kQq/r will always give you the correct value for potential energy between two charges regardless where zero potential is defined. It is my understanding that equation [1] is only useful when we have an initial and a final value of r such that neither is zero. It stands to reason that equation [2] must also be used in a similar manner where r-initial and r-final are clearly defined.

    It appears to me that U=kq^2/r will only tell us how much work is done by the charges when we bring one of them from infinity to r distance away from the other. It does not tell us how much work one will do to the other if we are to release them while they are r distance apart. I was told I was wrong in thinking this way. Am I missing something here? Please enlighten me.
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Mar 20, 2010 #2


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    Hello terence108,

    Welcome to Physics Forums!

    After giving it a lot of thought, I think you might be on to something. I think this problem is bogus.

    It's rare that such an seemingly simple electrostatic problem takes up so much of my spare time, but this one caused me to lose sleep. In the end (after contemplating it for awhile, thinking about it some more over a pint of Guinness or two, a little more while eating some tasty snacks, then another pint of Guinness [it was Friday night after all], then over lots of coffee in the morning), I've come to the conclusion that you are right. The answer to this problem is heavily dependent on the spheres' radii, even if one makes appropriate approximations.

    There are a few things that make this problem difficult from the get-go.

    (1) You cannot treat the spheres as point charges when they are close to each other because they are conducting spheres. When they are close to each other, each sphere loses its spherically symmetrical charge distribution. The charges on a given sphere tend to accumulate closer to the other sphere in this case because the sphere's are oppositely charged. The standard equations that you listed above fall apart when the conducting spheres are close to each other. That being said, we can approximate uniform charge distribution if the spheres are small and far apart. We can continue this approximation when they are close together, but that is not a good approximation. I'm hoping you were not originally supposed to take this into account. It would make the problem far more difficult. Later, I'm going to use the point charge approximation anyway, as flawed as it may be.

    (2) The charges are accelerating in both the x and y directions. This means that some energy will be lost to electromagnetic radiation. This is not a simple electrostatics problem, but rather an electrodynamics problem. Granted, not much energy will be lost to electromagnetic radiation, so I suppose we can ignore the energy loss due to this. But the problem statement says nothing about ignoring this effect. So I'm mentioning it because it's just another reason the problem is bogus. (Well, unless you are really supposed to take this into account.)

    (3) Potential and kinetic energy are not conserved in this situation. When the sphere's touch and the charge is neutralized, some energy that was previously stored in the electrical potential will be lost due to heat. Now this particular energy loss does not really present us with a real problem. We can deal with it. All we need to do is calculate the potential energy of the system when the spheres are located at a distance of [tex] 2lsin \theta [/tex] from each other, and then subtract the energy when they are at a distance of [tex] 2R [/tex] from each other. In this region potential and kinetic energies are conserved. But my point of this is that the amount of energy which gets lost to heat is a function of the spheres' radius, R.

    So making the approximations that we can ignore the electromagnetic radiation, and ignore the non-spherical symmetry, let's analyze the problem.

    Because the collision is elastic, the maximum height and angle due only to gravitation alone does not change before and after the collision. We need to look for the extra energy caused by the electrostatic forces, and then add the extra energy to the initial gravitational potential. In other words, let's ignore gravity for now and just find the difference solely due to the electrical potential energy.

    Suppose we have two oppositely charged spheres of radius R, separated by a distance d on the x-axis. The negative charge sits at x = -d/2 and the positive charge sits at x = d/2.

    Start with formulating the electric field,

    [tex] \vec E = kq \left( -\frac{1}{\left|\vec r + \frac{\vec d}{2} \right|^2} \hat {r_-} + \frac{1}{\left| \vec r - \frac{\vec d}{2} \right|^2} \hat {r_+} \right) [/tex]

    Now we formulate the electric potential from one sphere to the other by integrating from -d/2 + R to d/2 -R (note that electric field in a conductor is zero, so we only integrate the electric field between the spheres -- not all the way from -d to d)

    [tex] V = -\int _P \vec E \cdot \vec {dl} [/tex]

    [tex] V = kq \left( \int _{-\frac{d}{2}+R} ^{\frac{d}{2}-R} \frac{1}{\left|\vec r + \frac{\vec d}{2} \right|^2} \hat {r_-} \cdot \vec{dr} - \int _{-\frac{d}{2}+R} ^{\frac{d}{2}-R} \frac{1}{\left| \vec r - \frac{\vec d}{2} \right|^2} \hat {r_+} \cdot \vec {dr} \right) [/tex]

    Before we go any further, lets get rid of those pesky unit vectors. We are integrating along the x-axis, between the negative and positive charges. So here, [tex] \hat {r_-} \cdot \vec {dr} = +1 [/tex] and [tex] \hat {r_+} \cdot \vec {dr} = -1 [/tex]. Also, we don't need to worry about the absolute values of the vector sums in the denominators anymore since we are sticking to the x-axis only.

    [tex] V = kq \left( \int _{-\frac{d}{2}+R} ^{\frac{d}{2}-R} \frac{1}{\left(x + \frac{d}{2} \right)^2} dx + \int _{-\frac{d}{2}+R} ^{\frac{d}{2}-R} \frac{1}{\left(x - \frac{d}{2} \right)^2} dx \right) [/tex]

    You should be evaluate the above integrals on your own. I encourage you to do so.

    Once you find the electric potential between the two sphere's, the next thing you want to do is find the capacitance between the two spheres, by rearranging your potential equation into the form of C = q/V. (Which is simple. Once you have an equation for V, divide both sides by q. Take the inverse of both sides and call that result C).

    Once you have calculated C, the capacitance, and since you already have the potential V, you can calculate the electric potential energy, U by using

    [tex] U = \frac{1}{2}CV^2 [/tex]

    Normally I would never go this far in helping someone on a homework assignment, but I'm only doing so because I think this problem is bogus, and if your instructor is telling you that the answer is independent of the radius, then I have to disagree with the instructor. Check your work against mine (since I just scribbled my calculations out on a napkin this morning), but I came up with

    [tex] U = kq^2 \frac{d}{R(d-R)} [/tex]

    Which makes sense to me anyway. Note that as R --> 0, U goes to infinity. This is because the energy required to "construct" a point charge is infinite. When the charges are held apart and charged up, it takes more energy the smaller the radii of the spheres are. More so, when released, they will come much closer to each other if the radii are smaller. If the spheres were infinitesimally small, not only would it take an incredible amount of energy to charge up the spheres in the first place, but an enormous amount of kinetic energy would be released when the spheres came together. The answer to this problem is heavily dependent on the spheres' radii.

    Anyway, if your instructor insists that the answer to this problem is independent of the radii of the spheres, please post your instructor's solution once you have it. I'd like to take a look at it. :wink:
    Last edited by a moderator: Apr 24, 2017
  4. Mar 20, 2010 #3


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    I just noticed that your attachment is your instructor's solution. Earlier I thought that it was your attempt at a solution. So I guess I should comment on that now.

    I disagree with the instructor's solution. Starting with:

    [tex] (m_1 + m_2) gh_f = (m_1 + m_2)gh_i + \frac{kq_1q_2}{2x} [/tex]

    Here, [tex] kq_1q_2/2x [/tex] is not the potential energy difference between the two spheres.

    If the two sphere's were the *same* polarity, and you were separated at a distance of 2x, and you were to *release* them in space somewhere, you could calculate the final velocity using this relationship. But that's way different from talking about the energy released from a closer distance to a distance 2x. Not the same thing.

    Or, in the case that they are oppositely charge (as they are in this problem), it represents the amount of work the system has done to bring them together from infinity to a distance of 2x. But it does not represent the energy left when going from 2x to some smaller distance!

    So I'm sticking with my original comments. I think this problem is bogus. :smile:
  5. Mar 21, 2010 #4


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    That formula assumes the zero of electric potential is at infinity, so they're wrong if that's indeed what they said. But I'm guessing what they meant is you can use it to calculate differences in potential energy between two points, which is correct; however...
    You're right about this, which is why the problem is horribly flawed. An underlying assumption in the solution is that the electric potential energy simply disappears when the charges cancel each other; consequently, the maximum height the masses achieve should depend solely on their kinetic energy at the time of collision. Ask your professor to calculate the kinetic energy of the masses when they collide. The kinetic energy will depend on the radius of the spheres, and it will diverge if you assume point masses.
  6. Mar 22, 2010 #5

    Could you please explain why you used the energy stored within a capacitor instead of using U=-kq^2(1/a - 1/b) ?
    Last edited: Mar 22, 2010
  7. Mar 22, 2010 #6


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    Treating the system as a capacitor, is an unnecessary step. I did it because since both sphere's are oppositely charged, the thing is working similarly to an open-circuit capacitor (with sphere's instead of plates). And in my mind, I found it easier to conceptualize the thing that way.

    But one can find the same answer by finding the potential differences between the spheres. The tricky part is knowing where to put the test charge when calculating the potentials, and knowing that the potential is constant everywhere within the conducting sphere (since the electric field is zero).

    Treating it as a capacitor means that it is a self-contained system and I can easily find the potential energy of the system simply by using (1/2)CV2. But you get the same end result either way.

    By the way, after double checking coffee-stained napkin scribble, I now find the electric potential energy of the system is (while the spheres are charged up):

    U = kq^2 \left( \frac{1}{R} - \frac{1}{d-R} \right) = kq^2 \frac{d-2R}{R(d-R)}

    But anyway, I didn't put too much effort into figuring out the precise, final solution once I realized the heavy dependence of R. Once I was able to show that, I didn't see much point in continuing further. There was already so many assumptions/approximations we were making from the start, and at least one of them is not a good approximation (namely the spherical charge distribution).
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