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A uniform electric field $E_{0} \hat{z}$ is produced by fixed charges located at a large distance from a particular spherical region in the field. Find a distribution of charge density and dipole layer density on the surface of the spherical region that will make the field zero in the interior of this region without changing it in the exterior.

The potential due to the charge distributions of the sphere should be $E_{0}z$ inside the sphere (to cancel out the potential $-E_0 z$ created by the uniform field) and a constant outside the sphere (so as to not change the field). Laplace's equation gives the volume charge density as zero (second derivative of $z$ is zero) so we only have surface charge. Using $\partial \phi/\partial r = \sigma/\epsilon_0$ at the boundary with $\phi = E_0 r \cos \theta$ gives $\sigma = -\epsilon_0 E_0 \cos \theta$; this is the surface charge density required to have a uniform field inside the sphere. The problem is, I'm not sure how the dipole layer enters into this. It must have something to do with the second part, which is to make sure that the field outside the sphere is unchanged, which doesn't seem to be true.

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To cancel out the field inside the sphere and leave the field outside the same, the charge distribution of the sphere should create electric field $-E_0 \hat{z}$ inside and none outside. However, I argue that such a field is not conservative, so no charge distribution can create such an electric field configuration.

Here is an example (see the figure below): $V_{A\rightarrow C \rightarrow B} = E_0 r$ and $V_{A\rightarrow D\rightarrow B} = 0$. Since $V_{A\rightarrow C \rightarrow B} \neq V_{A\rightarrow D \rightarrow B}$, the potential is not path independent, so it is not conservative. We know that electric field is conservative, this field is invalid.

Uniform Electric Field in the Spherical Region

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  • $\begingroup$ Agreed, and very nice, the field is invalid $\endgroup$ – Bob Bee Oct 6 '16 at 0:48
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If i understand your question correctly, the dipole layer simply refers to the variatiom of positive and negative charge over the surface of the sphere. One can create exactly such a distribution, by superposing 2 spheres having uniform volume charge densities, but opposite sign, their centres being separated by a very small distance d(just like a dipole)

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  • $\begingroup$ Thanks, that's sort of what I was thinking, but if you just make a dipole, wouldn't you just have a dipole field outside the sphere? The exterior field is supposed to be unchanged $\endgroup$ – Mr. G Sep 26 '16 at 14:44
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The Poisson equation is $$\nabla ^{2}\phi = -\frac{\rho}{\epsilon_{o}}$$ For a spherical coordinate it assumes $$\nabla^2 \phi = \frac{\partial^2 \phi}{\partial r^2} + \frac{1}{r^2}\frac{\partial^2 \phi}{\partial \theta^2} + \frac{1}{r^2 sin^{2}\theta}\frac{\partial^2 \phi}{\partial \psi^2} + \frac{2}{r}\frac{\partial \phi}{\partial r} + \frac{cot\theta}{r^2}\frac{\partial \phi}{\partial \theta} =-\frac{\rho}{\epsilon_{o}} $$

Outside the sphere, the electric potential remains to be $E_{o}r$.

Inside the sphere, the potential looks something like $$\phi = ar^2 + b$$ Substituting into the Poisson above gives $a=-\frac{\rho}{6\epsilon_{o}}$.

At the surface, R=r, $\phi=E_{0}R$, give, $b= \frac{\rho R^2}{6\epsilon_o} + E_{o}R $ Inside the sphere, then $$\phi = \frac{\rho (R^2 -r^2)}{6\epsilon_o} + E_{o}R $$ For much further inside the sphere, $r<<R$, the above solution becomes, $$\phi = \frac{\rho R^2 }{6\epsilon_o} + E_{o}R =0$$, which gives a solution of the charge density, $$\rho =-\epsilon_{o}E_{o}/R $$. This is the charge density at the surface that would make zero field inside the sphere. This is negative charge distribution at the surface of the sphere and originally a positive charge distribution, which have produced $E_{o}$ inside the sphere make a dipole. That's what I am thinking of but I am not dead sure though.

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