Connected metal equalizes potential, not charge

Connect two isolated conductors with a thin conducting wire and charge moves until it stops. It stops when no free charge feels a force, which is the same statement as before: the field inside the conducting path is zero, so there is no potential difference along it. The equilibrium condition is

$$V_1 = V_2.$$

Not $Q_1 = Q_2$. Not $\sigma_1 = \sigma_2$. Potential is what a charge responds to when deciding whether to move down the wire, so potential is what equalizes.

Two bookkeeping rules go with it:

1. Charge is conserved. $Q_1 + Q_2 = Q_{\text{total}}$, using signed values, both before and after.
2. The wire is assumed thin and long. That is what lets you keep using the isolated-sphere formula $V = kQ/R$ for each conductor: each is far enough from the other that neither distorts the other's distribution, and the wire itself holds negligible charge.

Two conducting spheres of radii $R_1$ and $R_2$, far apart, joined by a wire, carrying total charge $Q$. Setting the potentials equal:

$$\frac{kQ_1}{R_1} = \frac{kQ_2}{R_2} \quad\Longrightarrow\quad \frac{Q_1}{Q_2} = \frac{R_1}{R_2}.$$

With $Q_1 + Q_2 = Q$ that gives

$$Q_1 = Q\,\frac{R_1}{R_1 + R_2}, \qquad Q_2 = Q\,\frac{R_2}{R_1 + R_2}.$$

Charge splits in proportion to radius. A 1 cm sphere joined to a 3 cm sphere ends with one quarter of the charge, not one half. The even split is not a general rule; it is the special case $R_1 = R_2$, which is why identical spheres brought into contact really do end up with $(Q_1 + Q_2)/2$ each.

The common potential is worth computing once, because it is a compact check:

$$V = \frac{kQ_1}{R_1} = \frac{kQ}{R_1 + R_2},$$

the potential of a single sphere of radius $R_1 + R_2$ carrying $Q$. If a candidate split does not give the same $V$ from both spheres, it is wrong, and this is the fastest way to catch an even split: $Q/2$ on a 1 cm sphere and $Q/2$ on a 3 cm sphere give potentials differing by a factor of three, so the system is not in equilibrium.

The other tempting rule, equal surface charge density, gives $Q \propto R^2$ and is also wrong. Only equal potential is the equilibrium condition.

After connection, the field just outside each sphere is

$$E_i = \frac{kQ_i}{R_i^2} = \frac{V}{R_i},$$

using $V = kQ_i/R_i$. Same $V$, so

$$\frac{E_1}{E_2} = \frac{R_2}{R_1}, \qquad \frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}.$$

The sphere with less charge has the stronger field at its surface. That is not a paradox; it is the curvature result of Topic 10.1 stated with a wire in it, and it is the mechanism behind the lightning rod and behind point discharge from any sharp electrode.

The habit worth building: when a question asks "which one is more likely to spark", compute $E$, not $Q$. When it asks "which one holds more charge", compute $Q$, not $E$. The two answers point at opposite ends of the system.

Grounding connects a conductor to a reservoir so large that adding or removing charge does not change its potential, and by convention that potential is zero. The consequence is

$$V_{\text{conductor}} = 0,$$

and nothing else. It is not a statement that the conductor ends up neutral. Whether charge flows, how much, and in which direction, is set by asking: given everything else nearby, what charge on this conductor makes its potential zero?

1. Isolated charged sphere, nothing else around. $V = kQ/R = 0$ requires $Q = 0$. It does drain completely, which is where the false general rule comes from.
2. Neutral sphere near a fixed $+Q$. The point charge alone raises the sphere's potential above zero, so electrons must flow IN from ground to pull it back down. Cut the wire and the sphere is left negatively charged. This is charging by induction, and grounding is the step that makes it work.
3. Shell with $+q$ in its cavity. Grounding leaves the outer surface with no net charge, so the external field vanishes. The cavity charge and the $-q$ on the cavity wall are untouched.

One more standard result belongs here. Touch a small charged sphere to the inner surface of a hollow conductor and all of its charge transfers, every time, however much the shell already carries. The reason is the cavity argument: the inner wall must end with charge that makes the enclosed total zero, and once the small sphere is part of the wall it can only satisfy that by giving everything up. Repeating the operation is how a Van de Graaff generator piles charge onto a dome long after the dome is strongly charged.