Spherical Charge Distribution - Electric Field Intensity

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Homework Help Overview

The problem involves a spherical charge distribution defined by a specific charge density function, with the goal of finding the electric field intensity within the distribution. The context is rooted in electrostatics and the application of Gauss's law.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply Gauss's law to find the electric field intensity but encounters a discrepancy between their result and the book's answer. They question whether their method or the book's answer is incorrect.
  • Some participants suggest that the original poster needs to integrate the charge density over the volume rather than using a fixed value at the Gaussian surface to find the total charge.

Discussion Status

The discussion is active, with participants providing guidance on the integration approach needed to resolve the discrepancy. The original poster indicates they have arrived at the correct answer after receiving help, but the conversation continues to explore the reasoning behind the solution.

Contextual Notes

The charge density is noted to vary with radius, which is a critical aspect of the problem that affects the calculation of the total charge within the Gaussian surface.

deedsy
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Homework Statement


A spherical charge distribution is given by [itex]p = p_0 (1- \frac{r^2}{a^2}), r\leq a[/itex] and [itex]p = 0, r \gt a[/itex], 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


[itex]\oint \vec E \cdot d \vec A = \frac{q_{inside}}{\epsilon_0}[/itex]

The Attempt at a Solution


[itex]\oint \vec E \cdot d \vec A = \frac{q_{inside}}{\epsilon_0}[/itex]

[itex]E 4\pi r^2 = \frac{4\pi r^3 p}{3 \epsilon_0}[/itex]

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

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

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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