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

 

[kerimek]

Hello and welcome to the forum!

To find the electric field inside a sphere with charge density proportional to the distance from the origin, we can use Gauss's law. This law states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space.

In this case, we can choose a spherical Gaussian surface with radius r, centered at the origin. The electric field will be constant on this surface, and we can use the formula for electric flux to find the electric field at any point inside the sphere.

Using Gauss's law, we have:

\oint E \cdot da = \frac{Q_{enc}}{\epsilon_0}

Where Q_enc is the enclosed charge within the Gaussian surface. In this case, the charge density is proportional to the distance from the origin, so we can write:

Q_{enc} = \int_{0}^{r}\rho dV = \int_{0}^{r}kr^2dr = \frac{kr^3}{3}

Substituting this into Gauss's law, we have:

\oint E \cdot da = \frac{kr^3}{3\epsilon_0}

Now, the left side of the equation can be simplified to:

\oint E \cdot da = E \cdot 4 \pi r^2

And we can solve for E:

E = \frac{kr}{12\pi\epsilon_0}

Note that this is the electric field at any point inside the sphere, and it is proportional to the distance from the origin, as expected. I hope this helps!