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Griffiths example 3.8 says

An uncharged metal sphere of radius $R$ is placed in an otherwise uniform electric field $\mathbf{E}=E_{0} \hat{\mathbf{z}} .$ The field will push positive charge to the "northern" surface of the sphere, and-symmetrically -negative charge to the "southern" surface ...

What's the proof that we have a symmetrical distribution of charge?

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The only difference between positive and negative charges is the sign of the force that they experience in an external field. The external field is along $\mathbf{\hat{z}}$, and so the positive charges will be pushed "up", and exactly the same thing will happen to the negative charges, except in the opposite direction.

Forget the sphere for a moment, and consider a simple dipole and convince yourself that the positive and negative charges will behave "symmetrically" in a constant external field. Now imagine your "uncharged" sphere to be composed entirely of such "dipoles".

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  • $\begingroup$ Thank you, I would like an explanation that assumes that only the negative charges move while the positive remain stationary. $\endgroup$ – Yasir Sadiq Sep 30 '20 at 8:27
  • $\begingroup$ What a strange thing to want! But ok, I'll bite. :) Assume that you have two identical spheres, one with only positive charges and one with only negative charges. If you superpose them, then you get a single uncharged sphere. Now, imagine what happens to each of these spheres independently. Clearly, what happens to the first is exactly the inverse of what happens to the second. (Positive charges move up, negative charges move down). Now superpose the resulting charge densities. Can you see why the result is "symmetric"? $\endgroup$ – Philip Sep 30 '20 at 8:36
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Imagine that, before considering the charge that's on the sphere, you flip the universe in the z-direction. This leaves the sphere the same but flips the electric field because it depends on $ \mathbf{\hat z}$. Now flip all the postive charges with all the negative charges (this is called charge conjugation). This again flips the electric field. We are now exactly where we started. The transformation flip + charge conjugation gives you the same initial conditions so it should give you the same final charge density. This means that whatever charge density you get should remain the same after you perform the flip + charge conjugation which is the same as saying the charge distribution is symmetric.

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    $\begingroup$ Thank you this made me think of yet another method $\endgroup$ – Yasir Sadiq Oct 1 '20 at 5:12
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To begin with, the electric field is defined as the negative gradient of the potential, the electric field at any point $(x, y, z)$ is $$ \begin{array}{c} E_{1}=E_{0} \hat{\mathbf{x}}+E_{0} \frac{\sigma_{1}-\sigma_{0}}{\sigma_{1}+2 \sigma_{0}} \frac{R^{3}}{r^{5}}\left[\left(2 x^{2}-y^{2}-z^{2}\right) \hat{\mathbf{x}}+(3 x y) \hat{\mathbf{y}}+(3 x z) \hat{\mathbf{z}}\right](r>R) \\ E_{2}=E_{0} \frac{3 \sigma_{0}}{\sigma_{1}+2 \sigma_{0}} \hat{\mathbf{x}}(r<R) \end{array} $$

According to Gauss's Law for Electric Fields, the electric charge accumulated on the surface of the sphere can be quantified by $$ \begin{array}{c} \int_{V} \boldsymbol{\nabla} \cdot \mathbf{e} \mathrm{d} V=\int_{V} \frac{\rho}{\varepsilon_{0}} \mathrm{d} V=Q \end{array} $$ and \begin{equation} \mathbf{e}=-\nabla V \end{equation} Based on Gauss's theorem, surface charge density at the interface is given by

$$ \mathbf{e}_{1} \cdot \mathbf{n}-\mathbf{e}_{2} \cdot \mathbf{n}=\frac{\rho_{s}}{\varepsilon_{0}} $$ Then, the charge quantities accumulated at the surface is $$ \oint_{S} \rho_{s} \mathrm{d} a=\varepsilon_{0} \oint_{S}\left(\mathbf{e}_{1 n}-\mathbf{e}_{2 n}\right)=\varepsilon_{0} \oint_{S} 3 \mathbf{E}_{0} R^{2} \frac{\sigma_{1}-\sigma_{0}}{\sigma_{1}+2 \sigma_{0}} \cos \theta \sin \theta \mathrm{d} \phi \mathrm{d} \theta $$

After a painfull of calculation, you get a symmetrical distribution.

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