Finding E field inside a sphere with charge proportional to radius

[Vapor88]

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, $$\rho$$ = kr, for some constant k.

Homework Equations

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

The Attempt at a Solution

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

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

$$q/ \epsilon_0$$

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

$$\int \rho d \tau / \epsilon_0$$

Here's where I get stuck, I know that

$$\rho = kr$$

What do I do with $$d \tau$$? 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!

 

[ehild]

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

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

ehild