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Energy of homogenously charged sphere

  1. Jan 15, 2012 #1
    hi

    i know that the energy of a conducting charged sphere is E=Q²/(8epsilon_0*pi*r) and i know how to calculate this.

    now i found that a homogenously charged sphere has E=3/5*Q^2/(epsilon_0*pi*r).

    does anybody have an idea how to calculate this? i don't know where this 3/5 come from?
     
  2. jcsd
  3. Jan 15, 2012 #2

    vanhees71

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    What you have there is the energy in the electrostatic field of the sphere. This one you need first. You get it from the corresponding Maxwell equations for static fields,

    [tex]\vec{\nabla} \cdot \vec{E}=\frac{\rho}{\epsilon_0}, \quad \vec{\nabla} \times \vec{E}=0.[/tex]

    [The above is corrected for typos according to #3; thanks for pointing them out!]

    From the second equation you see that the electrostatic field is a potential field, i.e.,

    [tex]\vec{E}=-\vec \nabla \Phi.[/tex]

    Substituting this into the 2nd equation yields\

    [tex]\Delta \Phi=-\frac{\rho}{\epsilon_0}.[/tex]

    Now you have

    [tex]\rho(\vec{x})=\begin{cases} \rho_0=\frac{3Q}{4 \pi R^3} & \text{for} \quad r=|\vec{x}<R|\\ 0 & \text{for} \quad r \geq R. \end{cases}.[/tex]

    Due to symmetry, [itex]\Phi[/itex] depends only on [itex]r[/itex] and thus writing the Laplacian in spherical coordinates for [itex]r<R[/itex] leads to

    [tex]
    \frac{1}{r} \frac{\mathrm{d}}{\mathrm{d} r^2} [r \Phi(r)]=-\rho_0.
    [/tex]

    From this you get

    [tex]\epsilon_0 [r \Phi(r)]''=-\rho_0 r \; \Rightarrow \; \epsilon_0 [r \Phi(r)]'=-\frac{\rho_0}{2} r^2+C_1 \; \Rightarrow \; \Phi(r)=-\frac{\rho_0}{6 \epsilon_0} r^2+C_1+\frac{C_2}{r}.[/tex]

    Since for [itex]r=0[/itex] the potential should be regular, we must have [itex]C_2=0[/itex].

    For [itex]r>R[/itex] the same calculation leads to

    [tex]\Phi(r)=\frac{C_3}{r},[/tex]

    where we have determined one constant such that [itex]\Phi(r) \rightarrow 0[/itex] for [itex]r \rightarrow \infty[/itex]. The constant, [itex]C_3[/itex], is determined by Gauss's Law to be [itex]C_3=Q/(4 \pi \epsilon_0)[/itex].

    Continuity at [itex]r=R[/itex] leads to

    [tex]-\frac{Q}{8 \pi \epsilon_0 R}+C_2=\frac{Q}{4 \pi \epsilon_0 R}[/tex]

    i.e.

    [tex]C_2=\frac{3 Q}{8 \pi \epsilon_0 R}.[/tex]

    For the total energy of the electric field you find

    [tex]\mathcal{E}=\frac{\epsilon_0}{2} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \vec{E}^2(\vec{x})=\frac{1}{2} \int_{\mathbb{R^3}} \mathrm{d}^3 \vec{x} \rho(\vec{x}) \Phi(\vec{x}).[/tex]

    This gives from our above result

    [tex]\mathcal{E}=2 \pi \rho_0 \int_0^R \mathrm{d} r \Phi(r) \; r^2 =\frac{3Q^2}{20 \pi \epsilon_0 r}.[/tex]

    In your solution, obviously a factor of [itex]4 \pi[/itex] is missing.
     
    Last edited: Jan 15, 2012
  4. Jan 15, 2012 #3

    Simon Bridge

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    Um ... I always thought:
    [tex]\vec{\nabla}\cdot\vec{E} = \frac{\rho}{\epsilon_0}[/tex][tex]\nabla^2 \varphi = -\frac{\rho}{\epsilon_0}[/tex]... for the potential formulation.

    The approach is basically correct though.
    Compare:
    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html
     
  5. Jan 15, 2012 #4

    vanhees71

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    Sure, that were typos in the 1st and 3rd equation (I hate the SI because of this damn [itex]\epsilon_0[/itex]...). The rest is fine. I'll correct it also in the original posting.
     
  6. Jan 15, 2012 #5

    Simon Bridge

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    No worries - there is a thread in PF someplace about the why's and why-nots of that [itex]\epsilon_0[/itex] but if you trained in SI the pain is the other way around.

    At first I thought maybe the typo would affect the calculation but I was the epsilon vanished at the spherical coordinates but was recovered in the correct place on the next line. I was concerned this may puzzle someone.

    I think we need to hear from Gavroy about the source of his equation.
     
  7. Jan 15, 2012 #6
    you are totally correct, i accidentally lost 4 in the denominator. thank you for your help!
     
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