I understand why charge does not distribute itself uniformly on the surface of a conductor of any other shape. I do not understand why charge distributes itself uniformly only on the surface of a spherical conductor.
[Physics] Why does charge distribute itself uniformly only on the surface of spherical conductors
chargeconductorselectrostatics
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You have a constraint system. For every surface-element the tangent component of the electrical field strength must be zero (else the surface charge would start to move).
Consider a perpendicular edge of a cylindrical configuration as shown in the following picture.
Assume that the surface charge distribution would be uniform (constant charge per area).
There we have small surface charge elements $\sigma dA_1$ and $\sigma dA_2$ with $dA_1=dA_2$ generating a force on a small surface charge element $\sigma dA_3$ very close to the edge.
Thereby, $\sigma dA_1$ and $\sigma dA_2$ are just samples of the field generating surface charge elements over which we have to integrate. Nevertheless, we can discuss the main effect exemplary with their help. Both $\sigma dA_1$ and $\sigma dA_2$ have the same distance from $\sigma dA_3$ so that the absolute value of their field strength at the position of $\sigma dA_3$ is the same.
The tangent component $\vec F_{2\parallel}$ of the force $\vec F_2$ caused by $\sigma dA_2$ on $\sigma dA_3$ is small since $\vec F_2$ is directed almost normal to the surface.
On the other side the force $\vec F_1$ caused by $\sigma dA_1$ on $\sigma dA_3$ directly has tangent direction ($\vec F_1 = \vec F_{1\parallel}$).
Therefore, the sum of tangent forces $\vec F_{1\parallel}+\vec F_{2\parallel}$ on $\sigma dA_3$ will point towards the edge. In principle his holds also for the other charge elements contributing to the electrical field at the location of $\sigma dA_3$. Consequently, the surface charge $\sigma dA_3$ will move towards the edge until there is so much charge in the edge that for all surface charge elements the tangent component of the field strength is zero. ("The edge repells further charge.")
Eventually, at edges the surface charge density becomes infinite. Instead of a finite surface density you have a finite line charge density in the edge.
The effect at curved surfaces is similar but not so drastic.
The statement of the book that only spheres admit uniform charge distribution is only right if you restrict your considerations to bounded conductors and fast enough decaying fields at infinity.
If you admit infinite conductors then you also have uniform surface charge on a circular cylinder.
If you admitt outer charge distributions then you can adjust these outer charges such that the charge distribution on a considered smooth surface is uniform. Thereby, no restrictions are made on the shape of the surface.
tl;dr: thats not how conductors behave.
I am guessing that your doubt stems from the following notion: oppositely charged surfaces when connected with a conductor equilibrate.
So the induced charges, being connected by a conductor's bulk, should merge and vanish!
After all, initially, when there wasn't any charge inside the shell, the shell itself was uncharged. So where were these charges then? They must have been in happy pairs.
So how come the presence of a charge inside the shell ripped them apart? It must be that the force experienced by the +ve and -ve charges being opposite, rips them apart, right? The negative ones move to the inner surface, the positive ones to the outer.
But this can't be true as charges which have initially paired would be so strongly held that no external field, let alone of some measely $Q$, would pull them apart.
Also, how many of these pairs of charges are their anyways? Can we support $10Q$ inside? What about $10^{10}Q$? In theory, $\pm10^{10}Q$ would just pop out of the bulk of the conductor, ready for their surface duty, right?
But surely this can't be true, if for example, we took more charge than their was in the conductor to begin with.
You see, we run into all kinds of trouble assuming that a conductor is a free see/source of as many induced* charges as you want with opposite charges also being induced automatically.
So what now?
An ideal argument in electrostatics should be independent of phenomenology but I can't seem to find a simpler way to clarify you query. The crux is that conductors are made of neutral atoms with delocalized electrons. By delocalized we mean free to move within the bulk of the conductor.
So when you put $+Q$ inside, the free $e^-$s just gather as close as they can to it--on the inner surface. Thats the only imperative they have--as opposite charges attract. If you had put $-Q$, they would have moved away, as far as they could, to the outer surface, (or even to $\infty$ if you grounded the outer surface). Note that there isn't any converse movement of positive charges. Since they already were free, there aren't any pairs to rip them off of. So what about the induced positive charges? They are the exact** locations the $e^-$s left--remember the atoms are neutral.
So you see there isn't any reason the electrons should feel the urge to go back to the outer surface as long as they are pinned to the inner one.
So what about the case when two oppositely charged surfaces are connected by a conductor? How is that different? Oh no..there, $e^-$s have been put on the $-ve$ conductor and removed from the +ve one. So upon connecting, them electrons feel attracted to the +ve atoms and go to them.
*Note that induced $\ne$ created $\because$ charge conservation
**upto lattice spacing
Best Answer
The charge distributes uniformly on the surface of a spherical conductor (which is far from any other body) due to the spherical symmetry of the problem. There is no reason why it should accumulate at any location of the surface more than at any other location. Therefore it is distributed uniformly. Also, if it was distributed inhomogeneously, the electric fields would produce currents that redistribute the charge until there is no electric field inside and tangential to the surface.