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Trouble with self-energy formula

  1. Mar 7, 2014 #1
    Hi everyone,

    I'm having a little difficulty trying to calculate the self-energy of a uniformly charged sphere of radius a. That is, the work done in assembling such a distribution of charge. It is to my understanding that

    0.5*integral(ρ*ø(r)dV)

    should produce the answer, with ρ being the charge density (constant), ø(r) being some potential, and V being the volume of the sphere. However, I can't figure out what this potential is. Should it be the potential of a spherical shell, ie. Q/(4*pi*epsi0*[r^2])? I tried this but can't get the corrent answer out.

    Many thanks in advance,

    Will
     
  2. jcsd
  3. Mar 7, 2014 #2

    vanhees71

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    Science Advisor
    2016 Award

    Of course, you have to solve for the electrostatic potential for the given charge distribution.

    The potential obeys Poisson's equation (in Heaviside-Lorentz units)
    [tex]\Delta \Phi=-\rho.[/tex]
    This we write in spherical coordinates. Due to symmetry we can assume that [itex]\Phi=\Phi(r)[/itex] is a function of [itex]r[/itex] only. That simplifies the equation to
    [tex]\frac{1}{r} \partial_r^2 \left (r \Phi \right)=\rho.[/tex]
    For [itex]r<a[/itex] we have [itex]\rho=\text{const}[/itex]. This gives
    [tex](r \Phi)''=\rho r \; \Rightarrow \; (r \Phi)'=-\frac{\rho}{2} r^2+C_1\; \Rightarrow \; \Phi(r)=-\frac{\rho}{6} r^2 + C_1+\frac{C_2}{r}.[/tex]
    Since the potential is continuous at the origin we must have [itex]C_2=0[/itex].

    For [itex]r>a[/itex] we have [itex]\rho=0[/itex]. There we have
    [tex]\Phi(r)=C_1'+\frac{C_2'}{r}.[/tex]
    Since we want to define the potential to vanish for [itex]r \rightarrow \infty[/itex] we get [itex]C_1'=0[/itex]. Further the total charge must be [itex]Q=4 \pi a^3 \rho/3[/itex] we find
    [tex]C_2'=Q/(4 \pi).[/tex]
    Finally the potential must be continuous at [itex]r=a[/itex] which gives
    [tex]-\frac{\rho}{6} a^2+C_1=\frac{Q}{4 \pi a}=\frac{a^2 \rho}{3} \; \Rightarrow \; C_1=\frac{a^2 \rho}{2}.[/tex]
    Integrating gives
    [tex]E_{\text{em}}=2 \pi \rho \int_0^{a} \mathrm{d} r r^2 \Phi(r)=\frac{4}{15} \pi \rho^2 a^5.[/tex]
    You get the same result using directly the electric field, which is given by
    [itex]\vec{E}_r=-\Phi'(r) \vec{e}_r[/itex]:
    [tex]E_{\text{em}}=\frac{1}{2} 4 \pi \int_0^{\infty} \mathrm{d} r r^2 \vec{E}^2.[/tex]
     
  4. Mar 9, 2014 #3

    Philip Wood

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    Gold Member

    By elementary methods, consider building up the sphere in thin shells. When the sphere's radius is r, the potential at its surface is [itex]\frac{1}{4 \pi \epsilon_0 r} \frac{4}{3}\pi r^3 \rho[/itex]. So the potential energy of the outer shell (all at distance r from centre) is its charge multiplied by the potential we've just written, that is:
    [tex]\frac{1}{4 \pi \epsilon_0 r} \frac{4}{3}\pi r^3 \rho \times 4 \pi r^2 dr\ \rho\ \ =\ \frac{4}{3} \frac{\pi}{\epsilon_0} \rho^2 r^4 dr.[/tex]
    So imagining the sphere to be assembled shell by shell, the total energy needed to do this is
    [tex]\int_{0}^{a}\frac{4}{3} \frac{\pi}{\epsilon_0} \rho^2 r^4 dr \ = \ \frac{4}{15} \frac{\pi}{\epsilon_0} \rho^2 a^5.[/tex]
     
  5. Mar 9, 2014 #4

    jtbell

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    Staff: Mentor

    No, you want the electrostatic potential (as a function of r) produced by a uniformly charged sphere, matching your original problem statement.

    Finding this potential is a standard exercise in intermediate-level undergraduate E&M courses, and you will find a Google search on "potential of a uniformly charged sphere" to be fruitful. I see a whole group of hits on threads from Physics Forums alone! :biggrin:
     
  6. Mar 10, 2014 #5
    Thanks for the responses, these have been more than helpful!
     
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