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Total Electrostatic Energy in a system of charges

  1. Feb 26, 2015 #1
    Suppose there are N charges fixed at arbitrary locations , how do I calculate the total electrostatic energy ?

    I can think of two ways.

    one being the straightforward coulomb's law :

    k=1/(4*pi*ε)

    Q=k/2 * ΣiΣj qiqj / |ri-rj| ; i≠j

    the other method is to integrate energy density with respect to volume , which would most likely depend on the distribution of the charges.

    My questions are, do these methods yield the same result ? If so , how could I show that they are equivalent in both mathematical and physical sense?
     
  2. jcsd
  3. Feb 26, 2015 #2
    Calculate total energy(work), needed to bring them at their locations, one by one, from the point located at infinity.
     
  4. Feb 26, 2015 #3
    Thanks but I wanna focus on comparing the two approaches in the OP. Intuitively they seem very different but I see them as valid methods to calculate energy.
     
  5. Feb 26, 2015 #4
    If you're dealing with problems having some symmetry of charge distribution, nicely described by mathematical means, than other methods would lead more quickly to the result than straightforward application of Coulomb's law.
     
  6. Feb 28, 2015 #5

    Philip Wood

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    Both methods give the same result for a single spherical shell of charge – I've just verified this. [For the Coulomb's law method you consider increments of charge being brought from infinity and deposited on the spherical surface.] If I have time, I'll work on a proof for a collection of charges where each charge is spread over a surface or volume of finite linear dimensions. I can see one difficulty though… The equivalence won't hold for a collection of point charges, because the Coulomb method will give a finite answer (provided the charges have non-zero separations) but the energy density integration will blow up because the energy density approaches infinity as one gets close to any charge.
     
  7. Feb 28, 2015 #6
    I've verified them using a sphere and a circular cylinder of finite length, so I'm convinced it would work for any geometry of continuous charge distribution. The equivalence breaks down because point charges have infinite charge densities, unlike that of continuous charge distributions. and what do you mean by the bold?
     
  8. Feb 28, 2015 #7

    Philip Wood

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    Try evaluating [itex]\int_0^\inf\frac{\epsilon_0}{2}E^2 4\pi r^2 dr [/itex] for [itex]E=\frac{Q}{4 \pi \epsilon_0 r^2}[/itex] and you'll see what I mean by the bold. But you're saying the same thing when you point out that the equivalence breaks down if there are infinite charge densities.
     
    Last edited: Feb 28, 2015
  9. Feb 28, 2015 #8
    I see. it diverges at r=0.
     
  10. Feb 28, 2015 #9

    Philip Wood

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    I've just looked this up in my beloved Abraham and Becker (1944 edition!) A&B expand [itex] E^2[/itex] at a point as a sum of [itex] E^2[/itex] terms due to individual charges, plus a sum of terms of the form [itex] \mathbf E_i . \mathbf E_j[/itex]. The first sum goes to infinity owing to the singularities at the charges themselves. But this sum is also independent of the relative positions of the charges. A&B show that it is the second sum that equals the Coulomb-derived energy formula!
     
  11. Feb 28, 2015 #10
    Care to show me a scan of that page? I want to look at the mathematical expressions in detail
     
  12. Mar 1, 2015 #11

    Philip Wood

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    Your every whim is my command...
     

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  13. Mar 22, 2015 #12

    Philip Wood

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    No doubt the original poster has long since achieved understanding and moved on, but I thought the question so interesting that I amused myself by recasting the A&B argument. Attached.
     

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