Question 24.11: ENERGY STORAGE WITH AND WITHOUT A DIELECTRIC Find the energy...

IDENTIFY and SET UP:

We consider the ideas of energy stored in a capacitor and of electric-field energy density. We use Eq. (24.9) to find the stored energy and Eq. (24.20) to find the energy density.

U=\frac{Q^{2}}{2 C}=\frac{1}{2} C V^{2}=\frac{1}{2} Q V                           (24.9)

u=\frac{1}{2} K \epsilon_{0} E^{2}=\frac{1}{2} \epsilon E^{2}                           (24.20)

EXECUTE:

From Eq. (24.9), the stored energies U_0 and U without and with the dielectric in place are

The final energy is one-third of the original energy.
Equation (24.20) gives the energy densities without and with the dielectric:

The energy density with the dielectric is one-third of the original energy density.

EVALUATE: We can check our answer for u0 by noting that the volume between the plates is V=\left(0.200 \mathrm{~m}^{2}\right)(0.0100 \mathrm{~m})= 0.00200 \mathrm{~m}^{3}. Since the electric field between the plates is uniform, u_{0} is uniform as well and the energy density is just the stored energy divided by the volume:

This agrees with our earlier answer. You can use the same approach to check our result for u.

In general, when a dielectric is inserted into a capacitor while the charge on each plate remains the same, the permittivity ε increases by a factor of K (the dielectric constant), and the electric field E and the energy density u=\frac{1}{2} \epsilon E^{2} decrease by a factor of 1/K. Where does the energy go? The answer lies in the fringing field at the edges of a real parallel-plate capacitor. As Fig. 24.16 shows, that field tends to pull the dielectric into the space between the plates, doing work on it as it does so. We could attach a spring to the left end of the dielectric in Fig. 24.16 and use this force to stretch the spring. Because work is done by the field, the field energy density decreases.