UY1: Energy Stored In 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 electric potential energy stored in the capacitor.

There are two ways to solve the problem – by using the capacitance, by integrating the electric field density.

Using the capacitance, (The capacitance of a spherical capacitor is derived in Capacitance Of Spherical Capacitor.)

$$C = 4 \pi \epsilon_{0} \frac{r_{a}r_{b}}{r_{b}-r_{a}}$$

$$\begin{aligned} U &= \frac{Q^{2}}{2C} \\ &= \frac{Q^{2}}{8 \pi \epsilon_{0}} \frac{r_{b}-r_{a}}{r_{a}r_{b}} \\ &= \frac{Q^{2}}{8 \pi \epsilon_{0}} \left( \frac{1}{r_{a}} – \frac{1}{r_{b}} \right) \end{aligned} $$

We’re done.

Now, we try using the integration of electric field energy density method.

Electric field in the volume between the two conducting spheres:

$$E = \frac{Q}{4 \pi \epsilon_{0} r^{2}}$$

Energy density:

$$\begin{aligned} u &= \frac{1}{2} \epsilon_{0} E^{2} \\ &= \frac{Q^{2}}{32 \pi^{2} \epsilon_{0} r^{4}} \end{aligned}$$

$$\begin{aligned} U &= \int u \, dV \\ &= \int\limits_{r_{a}}^{r_{b}} u 4 \pi r^{2} \, dr \\ &= \frac{Q^{2}}{8 \pi \epsilon_{0}} \int\limits_{r_{a}}^{r_{b}} \frac{1}{r^{2}} \, dr \\ &= \frac{Q^{2}}{8 \pi \epsilon_{0}} \left( \frac{1}{r_{a}} – \frac{1}{r_{b}} \right) \end{aligned}$$


Next: Dielectrics In Capacitors

Previous: Transferring Charge And Energy Between Capacitors

Back To Electromagnetism (UY1)

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