• #1
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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  • #2
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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  • #3
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
 
  • #4
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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