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This question already has an answer here:

While calculating electric potential due to a spherical shell at a point outside the shell, why do we say that the entire charge is present at the centre of shell? Is this true even if the charge on the shell is distributed non-uniformly?

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marked as duplicate by ACuriousMind, Danu, Brandon Enright, John Rennie, BMS Oct 12 '14 at 4:44

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We may find the potential by integrating $E\cdot dr$ . When we calculate $E$ due to a conducting sphere using Gauss's law, we get $E A = \frac{q}{\epsilon_0}$. Here $A$ = the area of the sphere, so $A = 4 \pi r^2$. We get $E = \frac{q}{4 \pi \epsilon_0 r^s}$, which is same as if all charge is placed at the center. Hence for finding $V$ we may as well take all $q$ to be at the center.

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    $\begingroup$ I suggest your merge your two answers into a single one. Furthermore, it might be appropriate to comment on what happens if one wants to consider a charge distribution that is no spherically symmetric, i.e. $\rho=\rho(r,\theta)$. $\endgroup$ – Danu Jun 1 '14 at 13:11

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