TWO WAYS TO CALCULATE ENERGY STORED IN A CAPACITOR
The spherical capacitor described in Example 24.3 (Section 24.1) has charges +Q and -Q on its inner and outer conductors. Find the electric potential energy stored in the capacitor (a) by using the capacitance C found in Example 24.3 and (b) by integrating the electric-field energy density u.
IDENTIFY and SET UP:
We can determine the energy U stored in a capacitor in two ways: in terms of the work done to put the charges on the two conductors, and in terms of the energy in the electric field between the conductors. The descriptions are equivalent, so they must give us the same result. In Example 24.3 we found the capacitance C and the field magnitude E in the space between the conductors. (The electric field is zero inside the inner sphere and is also zero outside the inner surface of the outer sphere, because a Gaussian surface with radius r < r_a or r > r_b encloses zero net charge. Hence the energy density is nonzero only in the space between the spheres, r_a < r < r_b.) In part (a) we use Eq. (24.9) to find U. In part (b) we use Eq. (24.11) to find u, which we integrate over the volume between the spheres to find U.
U=\frac{Q^{2}}{2 C}=\frac{1}{2} C V^{2}=\frac{1}{2} Q V (24.9)
u=\frac{1}{2} \epsilon_{0} E^{2} (24.11)
EXECUTE:
(a) From Example 24.3, the spherical capacitor has capacitance
C=4 \pi \epsilon_{0} \frac{r_{a} r_{b}}{r_{b}-r_{a}}where r_a and r_b are the radii of the inner and outer conducting spheres, respectively. From Eq. (24.9) the energy stored in this capacitor is
U=\frac{Q^{2}}{2 C}=\frac{Q^{2}}{8 \pi \epsilon_{0}} \frac{r_{b}-r_{a}}{r_{a} r_{b}}(b) The electric field in the region r_a < r < r_b between the two conducting spheres has magnitude E=Q / 4 \pi \epsilon_{0} r^{2}. The energy density in this region is
u=\frac{1}{2} \epsilon_{0} E^{2}=\frac{1}{2} \epsilon_{0}\left(\frac{Q}{4 \pi \epsilon_{0} r^{2}}\right)^{2}=\frac{Q^{2}}{32 \pi^{2} \epsilon_{0} r^{4}}The energy density is not uniform; it decreases rapidly with increasing distance from the center of the capacitor. To find the total electric-field energy, we integrate u (the energy per unit volume) over the region r_a < r < r_b. We divide this region into spherical shells of radius r, surface area 4 \pi r^{2}, thickness dr, and volume d V=4 \pi r^{2} d r. Then
\begin{aligned}U &=\int u d V=\int_{r_{a}}^{r_{b}}\left(\frac{Q^{2}}{32 \pi^{2}\epsilon_{0} r^{4}}\right) 4 \pi r^{2} d r \\&=\frac{Q^{2}}{8 \pi \epsilon_{0}}\int_{r_{a}}^{r_{b}} \frac{d r}{r^{2}}=\frac{Q^{2}}{8 \pi \epsilon_{0}}\left(-\frac{1}{r_{b}}+\frac{1}{r_{a}}\right) \\&=\frac{Q^{2}}{8 \pi \epsilon_{0}} \frac{r_{b}-r_{a}}{r_{a} r_{b}}\end{aligned}EVALUATE: Electric potential energy can be associated with either the charges, as in part (a), or the field, as in part (b); the calculated amount of stored energy is the same in either case.