A SPHERICAL CAPACITOR
Two concentric spherical conducting shells are separated by vacuum (Fig. 24.5). The inner shell has total charge +Q and outer radius r_a, and the outer shell has charge -Q and inner radius r_b. Find the capacitance of this spherical capacitor.
IDENTIFY and SET UP:
By definition, the capacitance C is the magnitude Q of the charge on either sphere divided by the potential difference V_{ab} between the spheres. We first find V_{ab}, and then use Eq. (24.1) to find the capacitance C = Q/V_{ab}.
C=\frac{Q}{V_{a b}} (24.1)
EXECUTE:
Using a Gaussian surface such as that shown in Fig. 24.5, we found in Example 22.5 (Section 22.4) that the charge on a conducting sphere produces zero field inside the sphere, so the outer sphere makes no contribution to the field between the spheres.
Therefore the electric field and the electric potential between the shells are the same as those outside a charged conducting sphere with charge +Q. We considered that problem in Example 23.8 (Section 23.3), so the same result applies here: The potential at any point between the spheres is V=Q / 4 \pi \epsilon_{0} r. Hence the potential of the inner (positive) conductor at r = r_a with respect to that of the outer (negative) conductor at r = r_b is
The capacitance is then
C=\frac{Q}{V_{a b}}=4 \pi \epsilon_{0} \frac{r_{a} r_{b}}{r_{b}-r_{a}}As an example, if r_a = 9.5 cm and r_b = 10.5 cm,
\begin{aligned}C &=4 \pi\left(8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)\frac{(0.095 \mathrm{~m})(0.105 \mathrm{~m})}{0.010 \mathrm{~m}} \\&=1.1 \times 10^{-10}\mathrm{~F}=110 \mathrm{pF}\end{aligned}
EVALUATE: We can relate our expression for C to that for a parallel-plate capacitor. The quantity 4 \pi r_{a} r_{b} is intermediate between the areas 4 \pi r_{a}^{2} \text { and } 4 \pi r_{b}^{2} of the two spheres; in fact, it’s the geometric mean of these two areas, which we can denote by A_{gm}.
The distance between spheres is d = r_b – r_a, so we can write C=4 \pi \epsilon_{0} r_{a} r_{b} /\left(r_{b}-r_{a}\right)=\epsilon_{0} A_{\mathrm{gm}} / d This has the same form as for parallel plates: C=\epsilon_{0} A / d. If the distance between spheres is very small in comparison to their radii, their capacitance is the same as that of parallel plates with the same area and spacing.