# Capacitance of a system of spheres

Consider a system of two conducting spheres, as follows

The outer sphere contains a charge $$+Q$$ and the inner surface is neutral.

What would be the capacitance of the system?
We know $$C=\frac{q}{V}$$. But here the potential difference between the spheres is 0, so will it be correct to say, that the capacitance of the system is infinity. but if I consider this whole system as a single conductor and see its potential difference with respect to infinity, I get $$C=4\pi \epsilon _0(3R)$$. Now suppose I earth the inner sphere. then to make the potential of inner sphere=$$0$$, $$\frac{Q}{3}$$ charge will flow into the earth. So now a potential difference will occur between the spheres. Then the capacitance of the system will not be $$\infty$$. and there will be a potential difference with respect to infinity. so how exactly can we calculate the capacitance? Please help me solve this doubt. I am really confused here. I think I have not understood what the term "capacitance" means exactly.

• You will have two capacitors in parallel. Commented Aug 20, 2021 at 18:55
• If the inner sphere is uncharged, there is no potential difference between them. Commented Aug 20, 2021 at 21:27

In general, for concentric spheres $$(a,

$$\begin{pmatrix} V_a \\ V_b \end{pmatrix}= \frac{1}{4\pi \epsilon_0} \begin{pmatrix} \frac{1}{a} & \frac{1}{b} \\ \frac{1}{b} & \frac{1}{b} \end{pmatrix} \begin{pmatrix} Q_a \\ Q_b \end{pmatrix}$$

• The matrix is known as elastance matrix which is the inverse of the capacitance matrix.

The mutual capacitance is given by

$$C=\frac{Q_a}{V_a-V_b}=\frac{4\pi \epsilon_0ab}{b-a}$$

• Note that there's no dependence on charges on the outer sphere due to Gauss' Law.

• In usual practice, the outer sphere is earthed so as $$V_b=0$$.