# Capacitance of two conducting spheres

1. Sep 19, 2009

### AxiomOfChoice

1. The problem statement, all variables and given/known data
We're supposed to find the capacitance of a system of two conducting spheres, one of radius $r_1$ and charge $Q$, the other of radius $r_2$ and charge $-Q$, separated by a distance $L$ (this is the distance between their centers) that's very large compared to either $r_1$ or $r_2$.

2. Relevant equations
We define the capacitance by $C = Q/V$, where $V$ is the potential difference between the spheres.

3. The attempt at a solution
Really, my only question, as of right now, is what approximations or assumptions we can make based on the $L >> r_1,r_2$ assumption. Is it just that the charge distribution on eiter sphere is unaffected by the presence of the other sphere? Such that we can assume the potential is just the superposition
$$\dfrac{Q}{4 \pi \epsilon_0 R_1} - \dfrac{Q}{4 \pi \epsilon_0 R_2},$$
where $R_1$ is the distance from the center of the first sphere and $R_2$ is the distance from the center of the other sphere?

Last edited: Sep 19, 2009
2. Sep 19, 2009

### AxiomOfChoice

Here's a new attempt at a solution:

(1) The potential due to a sphere is $kQ/r$, where $r$ is the distance from the center of the sphere.

(2) If we consider the two spheres mentioned above, set the origin at the center of the left-most, and let the x-axis join the centers of the spheres, then the potential at any point along this axis is given by the linear superposition of the potentials of sphere 1 and sphere 2:
$$V = V_1 + V_2 = \frac{kQ}{x} - \frac{kQ}{L-x}.$$

(3) This means the potential difference between the spheres is given by $\Delta V = V(r_1) - V(r_2)$, which is just
$$\frac{kQ}{r_1} - \frac{kQ}{L-r_1} + \frac{kQ}{r_2} - \frac{kQ}{L-r_2}.$$

I guess we can reduce this, and put $Q/\Delta V$ to get the capacitance.

3. Sep 20, 2009

### AxiomOfChoice

...anyone? Does it look like I'm heading in the right direction here?