Spherical Charge Distribution - Electric Field Intensity

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SUMMARY

The discussion focuses on calculating the electric field intensity within a spherical charge distribution defined by the charge density function \( p = p_0 (1- \frac{r^2}{a^2}) \) for \( r \leq a \) and \( p = 0 \) for \( r > a \). The correct formula for the electric field intensity derived from the integration of the variable charge density is \( E = \frac{p_0 r}{\epsilon_0}(\frac{1}{3} - \frac{r^2}{5a^2}) \). The initial miscalculation stemmed from not integrating the charge density over the volume within the Gaussian surface. The book's answer is confirmed to be accurate.

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  • Understanding of Gauss's Law in electrostatics
  • Familiarity with spherical coordinate systems
  • Knowledge of charge density functions and their implications
  • Proficiency in performing volume integrals in physics
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  • Learn about spherical coordinate integration techniques
  • Explore the derivation of electric fields from varying charge densities
  • Review the properties and applications of electric field intensity in electrostatics
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deedsy
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Homework Statement


A spherical charge distribution is given by p = p_0 (1- \frac{r^2}{a^2}), r\leq a and p = 0, r \gt a, where a is the radius of the sphere.

Find the electric field intensity inside the charge distribution.

Well I thought I found the answer until I looked at the back of the book...

Homework Equations


\oint \vec E \cdot d \vec A = \frac{q_{inside}}{\epsilon_0}

The Attempt at a Solution


\oint \vec E \cdot d \vec A = \frac{q_{inside}}{\epsilon_0}

E 4\pi r^2 = \frac{4\pi r^3 p}{3 \epsilon_0}

E = \frac{p_0 r}{3\epsilon_0}(1 - \frac{r^2}{a^2})

but my book has as the answer:
E = \frac{p_0 r}{\epsilon_0}(\frac{1}{3} - \frac{r^2}{5a^2})

I have no idea where the extra factor came from... Did I do something wrong or is it a mis-print?
 
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The book's answer is correct.

Since the charge density is changing with radius you'll have to do the integration over the interior volume (within the Gaussian surface) to find the total charge. You can't just multiply the volume by the fixed value of charge density at the Gaussian surface.
 
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gneill said:
The book's answer is correct.

Since the charge density is changing with radius you'll have to do the integration over the interior volume (within the Gaussian surface) to find the total charge. You can't just multiply the volume by the fixed value of charge density at the Gaussian surface.

got the right answer now - thanks for the help
 
Just as an aside, if you happen to be looking at this thread again, to put a "rho" (##\rho##) in your LaTeX equations, for charge density, type "\rho". (And similarly for other Greek letters. For capital Greek letters, capitalize the word, e.g. "\Sigma".)
 

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