Charge on hollow spherical surface

Question

Let a point charge $+Q$ is placed in center of hollow spherical conductor of inner radius $a$ and outer surface $b$. Then the charge on the inner surface of radius $a$ is $-Q$ and outermost surface has charge $+Q$ (using $E=0$ in conductor and Gauss' theorem).

My question is why doesn't the charge $-Q$ of inner surface travel all the way to outermost surface as a conductor is between them so as to be equipotential? I know it would violate the law of conservation of charge but what is preventing the charge from moving to the outer surface?

Answer 1

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.

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*Note that induced $\ne$ created $\because$ charge conservation
**upto lattice spacing

Answer 2

The positive charge at the center would like its outgoing field to exist in the conductor, but any field in the conductor will cause charges to move until there is no field in the conductor. The charge distribution on the surfaces of the conductor is what is required to neutralize the field from the inner charge.

Answer 3

First A), Suppose you have a spherical shell and you add identical particles of total charge +Q randomly within the shell. Do you understand why this will redistribute itself uniformly on the outer surface of the shell?

Now, if we have a charge in the center of the cavity, we still require a zero electric field within the shell. We know by Gauss's Law the discontinuity in the electric flux through a boundary is proportional to the surface charge density on the boundary surface. The electric flux is zero just within the conductor. The central charge gives rise to a spherically symmetric electric field throughout the cavity, including just within the cavity. Since the flux inside the shell is 0 and the flux just within the cavity is positive and we have this symmetry in the field, we have constant negative surface charge density on the interior surface. By Gauss' Law, this must in total be equal and opposite to the internal charge, hence we have charge -Q evenly distributed on the interior surface.

Now, there is no net charge on the shell. So requiring total zero charge, we have charge of +Q distributed somewhere within the shell. A Gaussian spherical surface centered at the center of the cavity with radius radius between and b can have no net flux passing through it, so the surface can't have any net charge. So we have no charge within the volume of the shell.

Alternatively we also know the fields of the +Q charge in the center and the -Q on the interior surface produce cancelling Electric fields. The particles giving the -Q charge had to come from somewhere within the shell leaving total charge of +Q located arbitrarily within the conducting shell. Because of the other charges cancel each other, these positive charges only feel the field of each other and which results in the case addressed by A).

Answer 4

My question is why doesn't the charge $−Q$ of inner surface travel all the way to outermost surface as a conductor is between them so as to be equipotential?

If the negative charge on the inner surface were to go to the outer surface holding the positive charge, there would be no net charge on the metallic sphere. So the field from the $+Q$ charge would penetrate the bulk of the metal but that's not possible because we are talking about metals which always try to have $0$ electric flux passing through them.

So basically due to the electric field of the positive charge at the center, an equal amount of charge $-Q$ will appear on the inner surface of the hollow sphere and that will leave behind a charge equal to $+Q$ on the outer surface.

Now those induced positive charges will try move as far as possible from each other and hence move to the outer surface. These charges don't feel the central charge because the charges on the outer surface are effectively shielded from that by the negative charges on the inner surface.

The induced positive charges on the outer surface will behave as if nothing were there inside because the negative charges have shielded them from the central charge.

I know it would violate the law of conservation of charge but what is preventing the charge from transferring?

It wouldn't violate the law of conservation of charges because we are neither creating nor destroying any charges.

The central positive charge is stopping the induced negative charge from transferring to the surface because it's attracting those negative charges.

Now you might be thinking why don't the induced positive charges attract the induced negative charges. It's so because there is no electric field inside the sphere due to the the positive charges on the outer surface.