EM: Gauss' law for electricity

AI Thread Summary
To find the electric field strength at the surface of a solid sphere with a variable volume charge density, the correct approach involves integrating the charge density over the sphere's volume. The initial attempt mistakenly assumed a uniform charge density, leading to an incorrect calculation of the enclosed charge. The charge density varies with radius, necessitating the use of integration to determine the total charge accurately. After recognizing this mistake, the correct solution was achieved by properly accounting for the varying density. Understanding the implications of Gauss' law in this context is crucial for solving similar problems.
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Homework Statement



A solid sphere of radius R carries a volume charge density \rho = \rho_0e^{r/R}, where \rho_0 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



\int\vec{E}.d\vec{A} = \frac{q}{\epsilon_0}

The Attempt at a Solution


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}

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.
 
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