Finding E field inside a sphere with charge proportional to radius

  • Thread starter Vapor88
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  • #1
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First time poster here! EDIT: SOLVED!!

Thanks, I figured out from the related links at the bottom of the page. >_>b

Homework Statement


Find the electric field inside a sphere which carries a charge density proportional
to the distance from the origin, [tex]\rho[/tex] = kr, for some constant k.


Homework Equations


[tex]\oint E.da[/tex]
[tex]a = 4 \pi r^2/3[/tex]
[tex]da = 4 \pi r^2[/tex]
[tex]\rho = kr[/tex]
[tex]E = q/(r^2 4 \pi \epsilon _0)[/tex]
Where q = charge inside


The Attempt at a Solution


[tex]\oint E \bullet da = \int q/(r^2 4 \pi \epsilon _0) \bullet 4 \pi r^2[/tex]

The 4 pi r^2 terms cancel, leaving on the right

[tex]q/ \epsilon_0[/tex]

Substitute rho into the eqn. as to integrate all dimensions of the sphere

[tex] \int \rho d \tau / \epsilon_0[/tex]

Here's where I get stuck, I know that

[tex] \rho = kr [/tex]

What do I do with [tex]d \tau[/tex]? I'd imagine that it'd be easiest to do in spherical coordinates, so do I just add dr, dtheta, drho?

Also... How do I put a dot into this LaTex thing?

Thank you!
 
Last edited:

Answers and Replies

  • #2
ehild
Homework Helper
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1,912
You do not know the electric field, but you can calculate it by applying Gauss' law.

Because of symmetry, E depends only on r and is the same along a sphere of radius R. The surface integral for a sphere of radius R is equal to Q/epszilon, where Q is the charge confined inside the sphere

[tex]4 \pi R^2 E(R) = \frac{4 \pi}{\epsilon}\int_0^R{kr*r^2dr=\frac{4\pi}{\epsilon}k R^4/4 \rightarrow E(R)=\frac{kR^2}{4\epsilon}[/tex]

ehild
 

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