Charge density with proportional radius?

Question

$Q$ is evenly distributed over the volume of a ball of radius $a$ so that the space charge density is proportional to the distance $r$ from the center of the ball. Ie $ρ_v = kr$ where $k$ is a constant. I'm supposed to show that $p_v=kr$

attempt

$$\int E\cdot dA=\int \frac{Q}{4\pi r^2\epsilon_0}\cdot 4\pi r^2=\frac{Q}{\epsilon_0}$$

This is where i'm stuck. Should I use spherical coordinates and just

$$\int \frac{Q}{\epsilon_0}d\tau$$ or $$\iiint \frac{Q}{\epsilon_0}\cdot \rho^2 \sin\phi \, d\rho \, d\theta \, d\phi \, ?$$

Answer 1

Edit: see comments below, there appears to be a misunderstanding of the question set - I shall update this answer to better reflect what OP actually asked when I get the chance later

(I'm not sure what you mean by "space charge density", but I'm assuming you want the $\mathbf{E}$ field as you say in your comment).

You need to use Gauss' law, which is $$\int\int\mathbf{E}\cdot\text{d}\mathbf{S} = \frac{Q_\text{enc}}{\epsilon_0}$$ The easiest way is to exploit the symmetry of the situation and consider a "test sphere" of arbitrary radius $r$, through which $\mathbf{E}$ field lines are passing through perpendicularly. The flux through this test sphere (which is the left hand-side of Gauss' law) is $$\int\int\mathbf{E}\cdot\text{d}\mathbf{S} =|\mathbf{E}|\int\int\text{d}S = A_\text{sphere}|\mathbf{E}|= 4\pi r^2|\mathbf{E}|$$ Where the first step above is true, because the electric field lines are alway parallel to the normal of your test surface sphere.

$Q_\text{enc}$is the charge enclosed in our test sphere, which is $\frac{4}{3}\pi r^3\rho$, where $\rho$ is the constant charge density. From that, you find that the electric field strength is indeed proportional to $r$ (at least inside the sphere).