Argument for symmetry of potential Consider the following electrostatic charge configuration of a spherically symmetric, perfect conductor with total charge $Q = 2q$, where $q > 0$. A point charge $q$ is placed at the position shown.

We are asked to calculate the electric potential $V(r)$ a distance $r$ from the centre for $r > c$ and $b < r < c$ with the convention $V(\infty) = 0$. This question seems to presuppose that the potential $V$ is even symmetric. It is clear that this is the case $r = c$, since a conductor is always an equipotential surface. However, I don't see why there could't be asymmetries for $r > c$, considering the fact that the charge distribution is not symmetric. So what is the argument for this symmetry?
If anyone answers using Maxwell's equations, could you use the integral versions?
 A: Suppose the potential on the outside metal surface is $V_s$. Then the potential outside the sphere is given by the solution of Laplace's equation on that domain. However, it's easy to show that the potential 
$$V(\mathbf{r})=V_s\frac{c}{|\mathbf{r}|}$$ satisfies Laplace's equation on the domain $|\mathbf{r}|\geq c$.
Since solutions to Laplace's equation are unique, it follows that the potential is spherically symmetric outside the sphere. The only remaining task is to determine what $V_s$, which I'll leave for you (I actually don't remember how to compute it).
A: 
I don't see why there could't be asymmetries for r>c, considering the fact that the charge distribution is not symmetric. So what is the argument for this symmetry?

Look at the figure below,

In the initial stage, there is an electric field within the conductor( electric field created by the the charge $q$ ) which lasts for an $infinitesimal$ $period$ $of$ $time$. That is because, the free electrons within the conductor move under the influence of the electric field. Under electrostatic conditions, the electric field inside the conductor becomes zero due to the redistribution of these free electrons. The charge distribution is shown in the diagram above( look at $Final$ ). You clearly see that $-q$ appears on the inner surface of the conductor as a result of the redistribution( non-uniformly distributed ). Since charge is conserved, a charge of $3q$ appears on the outer surface of the conductor. The charge $3q$ distributes uniformly on the outer surface of the conductor.This result is consistent with Gauss's law. 

Consider the $Gaussian Surface$ as shown above. The net charge enclosed by this surface is $zero$ and the $flux$ associated with this surface is also $zero$ which tells us that the $electric field$ inside the conductor is $zero$.
Mathematically,
$$\iint \vec{E} \cdot \vec{\mathrm{d}A} = \frac{q}{\epsilon_0}$$
Since $q$ = 0, $\vec{E} = 0$.
Now, if I move the charge $q$ anywhere inside the cavity, the charge distribution on the outer surface will not be affected. Although the charge distribution on the inner portion of the conductor will change as result of the change in position of $q$( remember that there is an electric field in the cavity ), the charge distribution on the outer surface will remain uniform. Why? It's because there is no electric field within the conductor and we know that static charges respond to electric fields. For the charges present on the inner walls of the conductor to interact with the charges present on the outer surface, an electric has to be there which in this case is $zero$. So, charge $3q$ remains uniformly distributed on the outer surface of the conductor. This phenomenon is called $electrostatic$ $shielding$.
For $r > c$, $\mathbf{V} = \frac{3q}{4\Pi \epsilon_0(r)}$
For $r = c$, $\mathbf{V} = \frac{3q}{4\Pi \epsilon_0(c)}$
For $b < r < c$, $\mathbf{V} = \frac{3q}{4\Pi \epsilon_0(c)}$ (Why?)
That is because, there is no electric field inside the conductor which implies $\Delta\boldsymbol{V} = 0$. So, under electrostatic conditions, the potential at the surface of the conductor is equal to the potential anywhere inside the conductor. Also, the surface of the conductor is an $equipotential$ $surface$.
