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:

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

Hence,

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

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…

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  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?

    Reply

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