EM: Gauss' law for electricity

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Homework Statement



A solid sphere of radius R carries a volume charge density [tex]\rho = \rho_0e^{r/R}[/tex], where [tex]\rho_0[/tex] is a constant and r is the distance from the center.

Find an expression for the electric field strength at the sphere's surface.

Homework Equations



[tex]\int\vec{E}.d\vec{A} = \frac{q}{\epsilon_0}[/tex]

The Attempt at a Solution


[tex]E * 4 \pi R^2= \frac{\rho \frac{4}{3}\pi R^3}{\epsilon_0} = \frac{\rho_0e \frac{4}{3}\pi R^3}{\epsilon_0} = \frac{\rho_0 e R}{3*\epsilon_0}[/tex]

This is not correct. Why not? With the gaussian surface right at the sphere's surface, the enclosed charge is the volume charge density times the volume, no? What am I missing?
 
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The charge density is not uniform; it varies with radius r within the sphere. You must integrate the density over the sphere's volume to calculate the total charge.
 
Can't believe I missed that. Thanks, got the correct solution now.