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Say we have an uncharged, ungrounded conductor in a vacuum. We now turn on a uniform electric field that will rearrange the charges on the surface of the conductor, so that the potential inside (and on the surface) remains constant.

Now, since the total charge inside the conductor is still zero, would it be correct to say that the self-interaction energy of these charges is zero? If so, would it be correct to say that the total potential energy of the system will be due to the uniform field interacting with the surface charges?

I am trying to solve a different problem, finding the shape of a droplet of mercury in a uniform field with surface tension involved, and I'm trying to write down the Lagrangian for the problem. From what I've gathered, the self-interaction energy of a conductor is proportional to its charge.

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  • $\begingroup$ Even an uncharged conductor contains charged elements, whose field energy is non zero. Think about a positive and negative charge close together, there is still some potential energy between them, and depending on how you model it, (ie as ball of uniform charge density) there is field energy associated with the construction of the charge itself) where the balls have a radius of the classical electron radius $\endgroup$ Commented Jun 18, 2022 at 16:45

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You are correct.

One way to show this is to write the total electrostatic energy in a large spherical volume of radius $R$ that contains the conductor as the integral over the total electric field squared, replace for statics $|\vec E|^2= -\vec E \cdot \vec \nabla \Phi$, and use the divergence theorem you will get \begin{equation} W = -\frac{\epsilon_0}{2} \int_S d\vec S\cdot \vec E \Phi +\frac{1}{2}\int_V d^3r \rho \Phi \end{equation} Since the conductor is neutral and $\Phi$ is constant over it, the second term is zero as you said.

If you now write the field/potential as sums of the external terms and conductor terms, the external-external part gives the uninteresting constant external field energy. The conductor-conductor parts go to zero (even for a charged conductor the potential drops off like $1/R$, the field like $1/R^2$ and the surface area goes like $R^2$ and the result goes to zero like $1/R$.) So only the external-conductor parts contribute. You can expand the potential of the conductor as $\sum_{\ell>0,m} \frac{A_{\ell m}}{r^{\ell+1}} Y_{\ell m}(\theta,\phi)$ Since the external potential has $r Y_{1m}$ dependence only, only the $Y_{1m}$ terms will give a nonzero surface integral. The surface integral will then be $-\vec E_0 \cdot \vec d$, where $\vec d$ is the dipole moment of the charge distribution. That is of course the same as integrating the external potential $\Phi_0 = -zE_0$ with $\rho(\vec r)$ which is the result you wanted.

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  • $\begingroup$ I am not that well-versed with the mathematics involved in your proof, would it be possible to dumb down the argument? Oh an also, are you saying that even if the conductor was charged, the self-interaction energy of the charges would be zero? $\endgroup$
    – datoad
    Commented Jul 11, 2022 at 23:28

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