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Example 4.1: Electric field of a uniformly charged sphere

Question: An insulating sphere of radius $a$ carries a total charge $Q$ which is uniformly distributed over the volume of the sphere. Use Gauss' law to find the electric field distribution both inside and outside the sphere.
 
Solution: By symmetry, we expect the electric field generated by a spherically symmetric charge distribution to point radially towards, or away from, the center of the distribution, and to depend only on the radial distance $r$ from this point. Consider a gaussian surface which is a sphere of radius $r$, centred on the centre of the charge distribution. Gauss' law gives

\begin{displaymath}
A(r)\,E_r(r) = \frac{q(r)}{\epsilon_0},
\end{displaymath}

where $A(r) = 4\pi\, r^2$ is the area of the surface, $E_r(r)$ the radial electric field-strength at radius $r$, and $q(r)$ the total charge enclosed by the surface. It is easily seen that

\begin{displaymath}
q(r) =
\left\{\begin{array}{lcl}
Q&\mbox{\hspace{1cm}}&r\geq a\\ [0.5ex]
Q\,(r/a)^3&&r<a
\end{array}\right..
\end{displaymath}

Thus,

\begin{displaymath}
E_r(r) =
\left\{\begin{array}{lcl}
\frac{Q}{4\pi\epsilon_0\,...
...5ex]
\frac{Q\,r}{4\pi\epsilon_0\,a^3}&&r<a
\end{array}\right..
\end{displaymath}

Clearly, the electric field-strength is proportional to $r$ inside the sphere, but falls off like $1/r^2$ outside the sphere.


next up previous
Next: Electric Potential Up: Gauss' Law Previous: Worked Examples
Richard Fitzpatrick 2007-07-14