Question 24.12: A SPHERICAL CAPACITOR WITH DIELECTRIC Use Gauss’s law to fin...

IDENTIFY and SET UP:

The spherical symmetry of the problem is not changed by the presence of the dielectric, so as in Example 24.3, we use a concentric spherical Gaussian surface of radius r between the shells. Since a dielectric is present, we use Gauss’s law in the form of Eq. (24.23).

\oint K \overrightarrow{\boldsymbol{E}} \cdot d \overrightarrow{\boldsymbol{A}}=\frac{Q_{\text {encl-free }}}{\boldsymbol{\epsilon}_{0}}                           (24.23)

EXECUTE:

From Eq. (24.23),

where \epsilon=K \epsilon_{0}. Compared to the case in which there is vacuum between the shells, the electric field is reduced by a factor of 1/K.
The potential difference V_{ab} between the shells is reduced by the same factor, and so the capacitance C = Q/V_{ab}  is increased by a factor of K, just as for a parallel-plate capacitor when a dielectric is inserted. Using the result of Example 24.3, we find that the capacitance with the dielectric is

EVALUATE: If the dielectric fills the volume between the two conductors, the capacitance is just K times the value with no dielectric. The result is more complicated if the dielectric only partially fills this volume.