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Homework Help: Electric Potential Energy of More Than Two Charges

  1. Aug 3, 2009 #1
    1. The problem statement, all variables and given/known data


    Questions and solutions are given. However, I have some questions.

    26.47. Can anyone explain to me in more detail this solution? For instance, why do they reverse the order and add them together (equations 2 and 3)? And how does one obtain this obvious Calculus generalization?

    26.48. Can anyone explain this is more detail to me?
  2. jcsd
  3. Aug 3, 2009 #2


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    This is the total electrostatic energy for an assembly of charges. Let's assemble just three at first by bringing them in from infinity one at a time at a point in space where there is nothing to begin with. The total electrostatic energy will be the sum of energies it costs us to bring each charge in. For the first charge q1


    because it costs us nothing since there are no electrical forces, yet. Once the first charge is in place, there is an electrostatic potential V(1) set up by charge 1 and the cost to bring the second charge, q2 is


    Here V2(1) is the electrostatic potential at charge 1 (subscript of V) due to charge 2 (number between parentheses). When we bring charge q3, we have two terms because there are two charges already there so that


    The total electrostatic energy is the sum of all three

    U=0 + q2V2(1) + q3[V3(1)+V3(2)] = q2V2(1)+q3V3(1)+q3V3(2).

    Bringing in the charges in reverse order gives us the same total electrostatic energy, but provides the "missing" subscripts and terms in parentheses. We have

    U3 = 0
    U2 = q2V2(3)

    The total energy is again the sum

    U = q2V2(3) + q1V1(3)+q1V1(2)

    Then when we add,

    2U =q2V2(1)+q3V3(1)+q3V3(2)+q2V2(3) + q1V1(3)+q1V1(2)

    If you divide by two, you get the formula that says that the total energy is the one-half the sum of three terms, each term being a charge times the total electrostatic potential at that charge generated by all the other charges.

    I hope that if you see how this works for three charges you will see how it works for an arbitrary number of them.
  4. Aug 3, 2009 #3


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    26.48 is an extension of the formula to the continuous case. The expression in equation (4)

    [tex]U = \frac{1}{2}\int V \rho dv[/tex]

    tells you that to find U, you need to add a whole of things (integral sign). What kind of things? V is the electrostatic potential at "point charge" ρ dV. So you are adding the charge at a given point multiplied by the potential at that point, just like the previous case in 26.47 where the charges were discrete.
  5. Aug 3, 2009 #4
    Thanks for the response. But I still don't understand this, even with just 3 charges.

    Why do you need to bring them back in reverse order?
  6. Aug 3, 2009 #5


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    Bringing the charges back in reverse order has no deep physical significance. It is just an algebraic trick in the course of the derivation of Equation (4) in 26.47. It is like squaring both sides of an equation to get rid of a radical.

    Any expression for U, no matter in what order you assemble the charges, can be used when you have discrete charges and it will be correct. The problem with this is that, unless you write U in a symmetric form, i.e. a summation of n terms each of which is a charge times the potential at that charge as is done here, the conversion from a discrete summation to an integral will not be obvious. If I plopped Equation (4) in front of you and told you that this is the electrostatic energy of a continuous charge distribution, you might say "How did you get that?" Well, the derivation in 26.47 shows you how I got that.
  7. Aug 3, 2009 #6
    Well, even though that may be how it's done, it doesn't help me at all if it's just a 'trick'.

    I was hoping that the derivation of 26.47 would help me understand 26.48 but it looks like it's just going to be a memorization thing. I understand the equation in 26.48, but it doesn't 'mean' anything to me since I don't understand the derivation of 26.47.
  8. Aug 4, 2009 #7


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    Let's try one more time with a different derivation. For three charges the total electrostatic energy is the sum of three terms, namely the electrostatic energies of all the possible pairs of charges:

    [tex]U = \frac{kq_{1}q_{2}}{r_{12}}+\frac{kq_{2}q_{3}}{r_{23}}+\frac{kq_{3}q_{1}}{r_{31}}[/tex]
    where k = 1/4πε0 and rij is the distance between qi and qj. Clearly, rij=rji.

    Now watch this: I split each term into two halves because 1/2 + 1/2 = 1 no matter what. Why do I do this? Because I have seen this derivation many times in the past. What about the first time? I saw it in my textbook and I have remembered it since then. This is what some people call "learning new things." Anyway, with the split the expression develops into six terms

    [tex]U = \frac{1}{2}[\frac{kq_{1}q_{2}}{r_{12}}+\frac{kq_{2}q_{3}}{r_{23}}+\frac{kq_{3}q_{1}}{r_{31}}+\frac{kq_{2}q_{1}}{r_{21}}+\frac{kq_{3}q_{2}}{r_{32}}+\frac{kq_{1}q_{3}}{r_{13}}][/tex]

    and this can be factored to give

    [tex]U = \frac{1}{2}[q_{1}(\frac{kq_{2}}{r_{12}} +\frac{kq_{3}}{r_{13}})+q_{2}(\frac{kq_{1}}{r_{21}} +\frac{kq_{3}}{r_{23}})+q_{3}(\frac{kq_{1}}{r_{31}} +\frac{kq_{2}}{r_{32}}) ] [/tex]

    Note that the terms between parentheses are the potential at a given charge due to the presence of the other charges. For example the potential at charge 1 is

    [tex]V_{1} = (\frac{kq_{2}}{r_{12}} +\frac{kq_{3}}{r_{13}}) [/tex]

    and likewise for V2 and V3. With these substitutions, you recover

    [tex]U = \frac{1}{2} (q_{1}V_{1} +q_{2}V_{2} +q_{3}V_{3}) [/tex]

    This derivation is more common in E&M textbooks than the one in yours. However, some people may consider the derivation in your textbook as more "elegant" because it is shortened by using the very same algebraic trick that confused you in the first place.
  9. Aug 4, 2009 #8
    U = \frac{1}{2}[\frac{kq_{1}q_{2}}{r_{12}}+\frac{kq_{2}q_{3}}{r_{2 3}}+\frac{kq_{3}q_{1}}{r_{31}}+\frac{kq_{2}q_{1}}{ r_{21}}+\frac{kq_{3}q_{2}}{r_{32}}+\frac{kq_{1}q_{ 3}}{r_{13}}]

    So what's the point in splitting it in half? What makes the above equation any more pretty than if you hadn't split it in two?

    And if you hadn't seen and memorized the splitting in half, would you have ever come up with that idea yourself? That's the problem I'm having. It's not intuitive at all to do that.
  10. Aug 4, 2009 #9


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    It's not the above equation that is more pretty. It is an intermediate step for getting to the final form

    [tex]U = \frac{1}{2}\int V\rho dv [/tex]

    which is a very compact way to say the same thing, not to mention easy to remember and easy to use, therefore "pretty." Furthermore, the expression in this "pretty" form can be used to develop other "pretty" forms. You will see what I mean when you get to Maxwell's Equations.

    To answer your other questions, I saw the "splitting in half" trick in high-school algebra for the first time and then I saw it again in different contexts, including this one. I recognized it as a useful thing to remember and as something to try when doing algebraic manipulations. So I put it in my "bag o' tricks" that contains a whole lot of things of this sort accumulated over the years. If you plan a career in physics (or related science), I suggest that you create such a bag if you have not already done so. Everyone in the professions has one.

    I agree with you, doing all this is not intuitive. Physics is not intuitive in the sense that it asks you to translate the physical reality around you into mathematical form, manipulate the mathematical form algebraically to get a new form, and then translate this new form back into physical reality. This is a counterintuitive (if not unnatural) way to perceive the world for a large segment of the population. I think this is one reason why some students of physics "hate" it. Hate is a very strong emotion if you think about it.
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