Solution to 1986 Problem 2

We apply Gauss's Law in integral form:
\begin{align*}\int \mathbf{E} \cdot d \mathbf{l} = Q_{enc}/\epsilon_0\end{align*}
We take the closed surface to be a sphere of radius r \leq R concentric with the sphere in the problem statement. Also, let \rho be the charge density, which we know is a constant. Then by Gauss's Law, we have,
\begin{align*}E 4 \pi r^2 = \frac{4}{3} \pi r^3 \rho/ \epsilon	\Rightarrow E = \frac{\rho r}{3 \epsilon_0}\end{align*}
Thus, we see that the relationship between radius and electric field is a linear one, and so curve C must be correct. Hence the answer is (C). For r > R, the electric field falls off like 1/r^2 and is identical to the electric field produced by a point charge with charge \frac{4}{3} \pi R^3 \rho.

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