6.4 Conductors in electrostatic equilibrium  (Page 4/9)

A conducting sphere

The isolated conducting sphere ( [link] ) has a radius R and an excess charge q . What is the electric field both inside and outside the sphere?

Strategy

The sphere is isolated, so its surface change distribution and the electric field of that distribution are spherically symmetrical. We can therefore represent the field as $\overrightarrow{\text{E}}=E(r)\widehat{\text{r}}$ . To calculate E ( r ), we apply Gauss’s law over a closed spherical surface S of radius r that is concentric with the conducting sphere.

Solution

Since r is constant and $\widehat{\text{n}}=\widehat{\text{r}}$ on the sphere,

For $r<R$ , S is within the conductor, so $q_{\text{enc}}=0,$ and Gauss’s law gives

as expected inside a conductor. If $r>R$ , S encloses the conductor so $q_{\text{enc}}=q.$ From Gauss’s law,

The electric field of the sphere may therefore be written as

Significance

Notice that in the region $r\ge R$ , the electric field due to a charge q placed on an isolated conducting sphere of radius R is identical to the electric field of a point charge q located at the center of the sphere. The difference between the charged metal and a point charge occurs only at the space points inside the conductor. For a point charge placed at the center of the sphere, the electric field is not zero at points of space occupied by the sphere, but a conductor with the same amount of charge has a zero electric field at those points ( [link] ). However, there is no distinction at the outside points in space where $r>R$ , and we can replace the isolated charged spherical conductor by a point charge at its center with impunity.

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