UY1: Capacitance Of Spherical Capacitor

Two concentric spherical conducting shells are separated by vacuum. The inner shell has total charge +Q and outer radius $r_{a}$, and outer shell has charge -Q and inner radius $r_{b}$. Find the capacitance of the spherical capacitor.

Consider a sphere with radius r between the two spheres and concentric with them as Gaussian surface. From Gauss’s Law,

$$\begin{aligned} EA &= \frac{q}{\epsilon_{0}} \\ E \times 4 \pi r^{2} &= \frac{Q}{\epsilon_{0}} \\ E &= \frac{Q}{4 \pi \epsilon_{0} r^{2}} \end{aligned}$$

To find V, we use integration on E:

$$\begin{aligned} E &= \frac{Q}{4 \pi \epsilon_{0} r^{2}} \\ – \frac{dV}{dr} &= \frac{Q}{4 \pi \epsilon_{0} r^{2}} \\ – \int\limits_{0}^{V} dV’ &= \int\limits_{0}^{r} \frac{Q}{4 \pi \epsilon_{0} r^{2}} \, dr’ \\ V &= \frac{Q}{4 \pi \epsilon_{0} r} \end{aligned}$$


$$\begin{aligned} V_{ab} &= V_{a} – V_{b} \\ &= \frac{Q}{4 \pi \epsilon_{0} r_{a}} – \frac{Q}{4 \pi \epsilon_{0} r_{b}} \\ &= \frac{Q}{4 \pi \epsilon_{0}} \left( \frac{r_{b} – r_{a}}{r_{a}r_{b}} \right) \end{aligned}$$

The capacitance is then:

$$\begin{aligned} C &= \frac{Q}{V_{ab}} \\ &= 4 \pi \epsilon_{0} \frac{r_{a} r_{b}}{r_{b} – r_{a}} \end{aligned}$$


Next: Capacitance Of A Cylindrical Capacitor

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2 thoughts on “UY1: Capacitance Of Spherical Capacitor”

  1. It is of when inner sphere is charged….What happen when we charged outer sphere and earth the inner sphere…

  2. When you subtract the voltages, how do you account for one of the spheres being negatively charged so that the equation doesn’t become kr_a + kr_b instead?


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