Why is the charge distribution on the outer surface of a hollow conducting sphere uniform and independent of the charge placed inside it? There is positive charge inside the hollow conducting sphere, closer to the left edge of the sphere. It attracts more electrons on the left, than on the right. 


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*What I can't get is why is the charge distribution on the outer surface uniform, even if negative charge on the inner surface is not uniform due to the off-centred electric charge placed inside the cavity? 

*The field outside of the sphere will be induced by charge $+q$. Will this field be induced by the positive charge on the outer surface, or by the charge inside the sphere? 

 A: When you place a negative charge $-q$ inside a hollow conducting sphere, $+q$ amount of charge is induced on the inner surface of the hollow sphere. This is because of the fact, in electrostatics the net electric field inside a conductor must be zero. If we consider a Gaussian surface as shown in the diagram below:

Image taken from this answer. Original image source: Halliday, Resnick, Walker 10th edition.
As the net electric flux passing through the Gaussian surface is zero, the net electric charge enclosed by the surface must also be zero. This shows why an amount of $+q$ charge is induced on the inner surface when a charge $-q$ is placed anywhere inside the sphere.
If the sphere, is initially uncharged, then in order to maintain electrical neutrality, a charge $-q$ is induced on the outer surface of the sphere. I hope you already knew all this.
Now let's come to your question: Why is the charge distribution on the outer surface uniform when the charge inside the sphere is not placed at it's centre?
When the charge $-q$ is placed at the centre of the sphere, the charge distribution in the inner surface of the sphere is uniform. However, when the negative charge is placed at some other internal point other than the centre of the sphere, the induced charge will be non-uniform. Why? This is because, the electric field inside the conductor must be zero and if the charge distribution on the inner surface must get reorganised in such a way, they effectively act as a positive charge $+q$ placed at the new position of the charge $-q$, thus cancelling it's effect.
It must be noted that the charge on the outer surface of the sphere doesn't give rise to any electric field inside it. Even if you supply or remove charges from the outer surface by some other means, the charge distribution on the inner surface remains the same and is dependant only on the position of the negative charge $-q$ within the sphere.
A: The superposition principle allows for solution construction from different pieces. The first piece of the solution in this problem, it is a system of a $+q$ point charge and $-q$ inner surface charge which induces zero electric fields outside the inner surface. The electric field in the volume of metal is zero. From Newtonian times it is known, that uniform distribution of charges (masses) gives zero electric (gravitational) field inside a sphere. Hence, the second piece of the solution is the uniform $+q$ charge distribution on the outer surface.  And from this solution, it follows that the electric field outside the outer surface is that of the uniformly distributed over the outer sphere $+q$ charge.
A: This is best tackled by first considering potentials.  Because the sphere is a conductor, the outer surface of the sphere is an equipotential, and of course outside the sphere one ought to solve $\nabla^2 V=0$ in spherical coordinates to obtain the potential everywhere, from which we would deduce that the field is everywhere radial.
Having established that $\vec E=k\hat{r}/r^2$ consider now the boundary condition at the interface: the normal part of $\vec D$ suffers a discontinuity proportional to the surface charge density:
$$
\hat r \cdot (\vec D_a-\vec D_c)=\rho_s = \epsilon_0 E_a
$$
where $\vec E_a$ is the field in air and $\vec E_c=0$ is the field in the conductor.  Since $E_a$, the magnitude of the field, is constant on the air side of the conductor-air boundary, it follows that $\rho_s$ must also be constant.
A: The sphere, being a conductor, allows the electrons within it to distribute themselves through its thickness such that by the time the net charge is represented on the exterior of the sphere it is evenly distributed, and there is no net field inside the sphere material at all (i.e. the charges find the configuration with the lowest potential energy, and all their fields cancel out so that any movement of a charge requires an energy input).
Gauss's law is somewhat scale dependent; if you think of a conductor, you usually think of something at least a few microns across; this is still many atoms thick, which allows charge to distribute and neutralize through the thickness via electron dislocations. But, you can imagine a conducting sphere 1 atom thick, and predict that maybe then this would not work.  In fact, you'd be right; it takes a few nanometers of conductor thickness to get to a uniform distribution of charge, which causes problems when you start making nanoscale structures: they no longer behave as Gaussian surfaces.  You can read a bit about it in this research paper: http://web.mit.edu/bazant/www/papers/pdf/Levy_2020_JCIS_electroneutrality_breakdown_nanopores.pdf
But, the short answer is that the equal external distribution happens because the scales across which the charge distribution inside a conductor reach equilibrium are very very tiny, much smaller than the thickness of any conductor Gauss ever had a chance to work with when he was coming up with his rules.
