Sphere with non-uniform charge density

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SUMMARY

The discussion focuses on calculating the electric field inside a sphere with a non-uniform charge density described by ρ(r) = k/r. The user successfully derived the expression for k using the integral of charge density over volume, leading to the equation 2kπR² = Q. However, they encountered difficulties applying Gauss's Law to find the electric field E for r < R, mistakenly canceling r in their calculations. Ultimately, they acknowledged their oversight in not determining the value of k, marking the problem as solved.

PREREQUISITES
  • Understanding of Gauss's Law in both differential and integral forms
  • Familiarity with charge density functions and their implications
  • Knowledge of vector calculus, particularly divergence
  • Basic principles of electrostatics and electric fields
NEXT STEPS
  • Explore advanced applications of Gauss's Law in non-uniform charge distributions
  • Study the derivation of electric fields from varying charge densities
  • Learn about the implications of charge density on electric potential
  • Investigate the relationship between charge density and electric field strength in spherical coordinates
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This discussion is beneficial for physics students, electrical engineering majors, and anyone studying electrostatics, particularly those dealing with non-uniform charge distributions and their effects on electric fields.

Bonewheel
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Homework Statement


A sphere of radius R carries charge Q. The distribution of the charge inside the sphere, however, is not homogeneous, but decreasing with the distance r from the center, so that ρ(r) = k/r.

1. Find k for given R and Q.

2. Using Gauss’s Law (differential or integral form), find the electric field E inside the sphere, i.e., for r < R.

Homework Equations


\int_V \rho \, dV = Q
\oint \vec E \cdot d \vec A = \frac {Q_{enclosed}} {\epsilon_0}
\vec {\nabla} \cdot \vec E = \frac {\rho} {{\epsilon}_0}

The Attempt at a Solution


1. \int_V \rho \, dV = \int_{0}^{2 \pi} \int_{0}^{\pi} \int_{0}^R \frac k r r^2 \sin \theta \, dr \, d \theta \, d \phi = 2 k \pi R^2 = Q
The units check out here.

2.
Here's where I ran into a problem. I tried using both the differential and integral forms of Gauss's Law, and in both cases the r canceled out, leaving me with an expression for the electric field I know is wrong. Oddly, the units work out here as well.

\oint \vec E \cdot d \vec A = 4 \pi r^2 E = \frac {Q_{enclosed}} {\epsilon_0} = \frac {2 k \pi r^2} {{\epsilon}_0}
\vec {\nabla} \cdot \vec E = \frac 1 {r^2} \frac {\partial} {\partial r}(r^2 E_r)= \frac {\rho} {{\epsilon}_0} = \frac k {r {{\epsilon}_0}}
E = \frac k {2{{\epsilon}_0}}

Thank you so much for any help! Please let me know if you need any further information or edits for clarification.
 
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I just realized that I made a very stupid mistake and never found k... Marking as solved.
 

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