# Capacitance of a Single Conductor

## The potential of a charged sphere

Suppose I place some positive charge Q on a metal sphere with radius R. Positive source charges act as "peaks", so the sphere is at a higher potential than its surroundings. But what is its potential? I know the charge will distribute itself on the surface of the sphere, and symmetry suggests that it will do so evenly. Thus we have a spherical shell of charge, and the electric field outside the metal sphere is the same as the electric field of a point charge: $$E=k{Q\over r^2}$$ pointing outward (where r is the distance from the center of the sphere).

Given the field, we can find the potential difference between the surface of the sphere $$(V_s)$$, and infinity $$(V_\infty)$$: $$\begin{eqnarray} \Delta V=V_s-V_\infty&=&-\int_\infty^R E_r\,dr\cr &=&-\int_\infty^R k{Q\over r^2}\,dr\cr &=&-kQ\left[-\frac1{r}\right]_\infty^R\cr V_s-V_\infty&=&k{Q\over R}\cr \end{eqnarray}$$ Because the potential is the same everywhere on a conductor in equilibrium, $$V_s$$ isn't just the potential at the surface of the sphere, but the potential V of the entire sphere.

## Definition of Capacitance

The potential is easy to calculate for a sphere: it wouldn't be so easy for a metal cube, or a cone, or a blob. However, all of these shapes have one thing in common: just like the sphere, the potential difference between each object and infinity $$V-V_\infty$$ is proportional to the charge $$Q$$ placed on it. Thus we can write $$Q=C(V-V_\infty)$$ where $$C$$ is called the capacitance of the metal object. Capacitance is only defined for conductors: only conductors are guaranteed to have a single potential. For the metal sphere above, we saw that $$Q={R\over k}(V-V_\infty)$$ and so $$C=R/k\hbox{ for a sphere}$$
Consider two spheres: sphere A has radius 0.5 m, and sphere B has radius 1 m. Both spheres are 9 volts above the potential at infinity.
Find the capacitance of both spheres.
For sphere A, $$C_A={R\over k}={0.5\u{m}\over 9\ten9\u{Nm^2/C^2}}=5.6\ten{-11}\u{C^2/Nm}$$ Sphere B has twice the radius, and so has twice the capacitance: $$C_B=1.1\ten{-10}\u{C^2/Nm}$$.
How much charge does each sphere contain?
Because $$V=V_\infty+9\u{V}$$ for both spheres, sphere A contains $$Q_A=C_A(V-V_\infty)=(5.6\ten{-11}\u{C^2/Nm})(9\u{V})=0.5\u{nC}$$ while sphere B contains $$Q_B=C_B(V-V_\infty)=(1.1\ten{-10}\u{C^2/Nm})(9\u{V})=1.0\u{nC}$$ Note that the sphere with twice the capacitance contains twice as much charge, when both are at the same potential.
Now suppose both spheres contain $$3\u{\mu C}$$ of charge, and the potential at infinity is $$V_\infty=0$$. What is the potential of each sphere?
We reverse the formula to find that $$V-V_\infty={Q\over C}$$. For the smaller sphere, $$V_A={3\ten{-6}\u{C}\over 5.6\ten{-11}\u{C^2/Nm}}=54\u{kV}$$ and for the larger sphere, $$V_B={3\ten{-6}\u{C}\over 1.1\ten{-10}\u{C^2/Nm}}=27\u{kV}$$ When both contain the same charge, the larger sphere has a lower potential.

The capacitance of arbitrary conductors may be difficult to calculate, but it will generally grow with the size of the object, and will depend only on the size, shape, and material of the conductor. Often it's easier to measure the capacitance of a particular conductor in the lab, than to calculate it from first principles.

Suppose the $$0.5\u{m}$$ sphere from before is connected to the $$1.0\u{m}$$ sphere by a long thin wire. $$6\u{\mu C}$$ of charge is placed on the small sphere. Some of that charge will flow through the wire into the larger sphere.
When the charges stop flowing, how much charge is on each sphere?

When you connect multiple metal pieces together, they act as a single conductor, and the charges in a conductor stop moving when it is in equilibrium. And the conductor is in equilibrium when the electric potential is the same everywhere. In this case, both spheres will end up at the same potential.

Now the potential on a sphere is $$V-V_\infty={Q\over C}$$ and because the potentials are the same for both, we must have $${Q_A\over C_A}={Q_B\over C_B} \implies {Q_A\over Q_B}={C_A\over C_B}$$ The charge distributes itself across the spheres according to their capacitances: because sphere B has twice the capacitance, it ends up with twice the charge: $$Q_B=4\u{\mu C} \quad\mathrm{and}\quad Q_A=2\u{\mu C}$$