A spherical capacitor

A spherical conducting shell with radius b is concentric with a conducting ball with radius a, with a<b.

  1. Compute the capacitance C = Q / \Delta \phi when the shell is grounded and the ball has charge Q.
  2. Compute the capacitance when the ball is grounded and the shell has charge Q.
  3. Compute the full matrix of coefficients of capacitance for the two conductors.
  4. Considering these conductors as a capacitor, determine its capacitance. That is, assign equal and opposite charges \pm Q to the shell and the ball, and compute C = Q / \Delta \phi.

Related Problem: Insulating spherical shell with a hole


(a) First, we ground the shell and give the ball a charge Q as shown below.

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This forces the potential at the shell, \phi(b), to be zero. The charge on the ball will create an electric field in the space between the shell and the ball. This creates a potential difference

(1)   \begin{equation*}   \Delta \phi     = \phi(a)-\phi(b)     = -\int {\bf E} \cdot {\rm d} {\bf r}     = -\int_{b}^{a} {\rm d} r \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^{2}}     = \frac{Q}{4 \pi \varepsilon_0} \left(\frac{1}{a}-\frac{1}{b}\right) . \end{equation*}

The capacitance is therefore

(2)   \begin{equation*}   C = \frac{Q}{\Delta \phi} = 4 \pi \varepsilon_0 \frac{ab}{b-a} . \end{equation*}

Check: We can verify the correctness of this expression by making the shell much larger than the ball. That is, we take b \gg a and find that C \approx 4 \pi \varepsilon_0 a which is just the capacitance of an isolated conducting sphere in space, as expected.

(b) If the ball is grounded the potential at the surface of the ball will be zero. Since the ball is a perfect conductor there can be no electric field inside it, which means the potential at the surface will be the potential everywhere inside the ball. In other words \phi(r \leq a) = 0 (this would also be true if the ball were replaced with a conducting shell). However, we are told that there is a charge Q on the outer shell which tries to make the potential everywhere inside it a constant value of Q/b. To compensate the ball must acquire a charge Q_{\rm ball} such that

(3)   \begin{equation*}   \frac{Q_{\rm ball}}{4 \pi \varepsilon_0 a} + \frac{Q}{4 \pi \varepsilon_0 b}=0 \implies Q_{\rm ball}=-\frac{a}{b} Q . \end{equation*}

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The charge Q on the outer shell produces no field inside the shell, as we can see by Gauss’s Law. Therefore, the electric field in the space between the ball and the shell is solely due to Q_{\rm ball},

(4)   \begin{equation*}   {\bf E} = \frac{Q_{\rm ball}}{r} \hat{\bf r}, \quad a<r<b . \end{equation*}

This is the electric field that creates a potential difference between the two conductors. Since Q_{\rm ball} is negative, the field is directed toward the ball, and \phi(b) > \phi(a). Thus,

(5)   \begin{equation*}   \Delta \phi     = \phi(b) - \phi(a)     % = -\int {\bf E} \cdot {\rm d} {\bf r}     = -\int_{a}^{b} {\rm d} r \frac{1}{4 \pi \varepsilon_0} \frac{Q_{\rm ball}}{r^{2}}     = \frac{Q}{4 \pi \varepsilon_0} \frac{a}{b} \left(\frac{1}{a}-\frac{1}{b}\right) . \end{equation*}

The capacitance is then

(6)   \begin{equation*}   C = \frac{Q}{\Delta \phi} = 4 \pi \varepsilon_0 \frac{b^{2}}{b-a} . \end{equation*}

Check: If we shrink the inner sphere to zero radius, we must recover the capacitance due to an isolated sphere of radius b. Indeed, setting a=0 in the above expression gives C = 4 \pi \varepsilon_0 b.

(c) We place a charge Q on the shell and -Q on the ball as stated in the problem. Just like in the previous cases, the field in the region a<r<b is produced by the charge on the ball alone. The ball is at a lower potential, so we find

(7)   \begin{equation*}   \Delta \phi     = \phi(b) - \phi(a)     = -\int_{a}^{b} {\rm d} r \frac{1}{4 \pi \varepsilon_0} \frac{-Q}{r^{2}}     = \frac{Q}{4 \pi \varepsilon_0} \left(\frac{1}{a}-\frac{1}{b}\right) . \end{equation*}

Therefore, the capacitance is

(8)   \begin{equation*}   C = \frac{Q}{\Delta \phi} = 4 \pi \varepsilon_0 \frac{ab}{b-a} , \end{equation*}

which is exactly what we found in part (a). But this shouldn’t be surprising — The ball and the conductor together have a net charge of zero, and by Gauss’s Law there is no electric field in the region r>b. That means the shell is at a potential \phi(b) = 0, which is the scenario we had in (a).

Check: Another interesting limit of (8) is to take a \to b. That is, we make the ball almost as large as the shell and find

(9)   \begin{equation*}   C \approx 4 \pi \varepsilon_0 \frac{a^2}{b-a} = \frac{\varepsilon_0 A}{d} , \end{equation*}

where d \equiv b-a is the (small) spacing between the two conductors and A = 4 \pi a^2 is the surface area of each conductor when a \approx b. Notice that the final expression in (9) is just the capacitance of a parallel plate capacitor. This is because when a approaches b the electric field between the ball and the shell is nearly uniform and the arrangement resembles a parallel plate capacitor with plate area 4 \pi a^2, with equal and opposite charges on each plate.

Bonus problem: A capacitor is made of three conducting concentric thin spherical shells of radii a, b and c, with a<b<c. The inner and outer spheres are connected by a fine insulated wire passing through a tiny hole in the intermediate sphere. Neglecting the effects of the hole,

  1. determine how any net charge Q_{B} placed on the middle shell distributes itself between its inner and outer surfaces, and
  2. considering these conductors as a capacitor, determine its capacitance.

Tip: Take physical limits of your answer as we did above to check for correctness.

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