# Sperical charge dist

1. Mar 7, 2005

### Gale

Ok, so obviously i'm not understand physics so well... so sorry...

i have this problem, says,
For a sperical charge distrubution with radius R the charge density $$\rho(r)=Cr$$ inside the sphere and 0 outside, the point of strongest electric feild is at or closest to...?

I was looking through my book... and i couldn't really find any examples like this. I don't understand how outside the sphere why the electric feild would drop off. Plus i don't get what Cr is supposed to mean...

2. Mar 7, 2005

### Staff: Mentor

Since the charge distribution is symmetric, use Gauss's law to figure out where the field is greatest. (Integrate to find the charge within a sphere of radius = r.)

For the same reason that the field from a point charge drops off the farther you get from it.

It's telling you that the charge density is proportional to the distance from the center.

3. Mar 7, 2005

### Gale

I'm not really understanding how to use gauss's law actually...

ok, right i got that... but it just drops off suddenly outside the sphere... i think i'm missing something...

4. Mar 7, 2005

### Staff: Mentor

Gauss's law says that the flux through a closed surface equals the charge contained within that surface (divided by a constant). So... use an imaginary "Gaussian surface" in the shape of a sphere of radius r. By symmetry, the field is radial and uniform over the sphere. Calculate the flux (field X area) and set it equal to the total charge within the sphere ($Q/\epsilon_0$). To find the charge within the sphere, you'll have to integrate the charge density over the volume. (It's easy.)

Who says it drops off suddenly?

5. Mar 7, 2005

### chroot

Staff Emeritus
Well, you should now understand Gauss' law -- the flux through a surface is proportional to the charge inside it. In the other problem, you knew the field, and had to find the charge. This problem is the reverse: you know the charge, but have to find the field. It's the same law, of course, just Gauss' law.

First, make sure you understand the simplification implied by the symmetry: the field is spherically symmetric, so the field is the same at any point on a surface of constant radius.

Thus, to find the field at a given radius, you just need to find the charge inside that radius. To find the charge inside a given radius, you need to integrate the charge density over the volume inside that radius:

$$Q_\textrm{at radius a} = \int_0^a \rho dV$$

Use spherical shells for your integral -- a stack of spherical shells makes a spherical volume. Each spherical shell has surface area $4 \pi r^2[/tex] and thickness [itex]dr$, giving each spherical shell a volume of $4 \pi r^2 dr$. The integral above becomes

$$Q_\textrm{at radius a} = \int_0^a (\rho) (dV) = \int_0^a (Cr) (4 \pi r^2 dr)$$

Perform that integral, and you'll have an expression for the charge contained inside any arbitrary radius a. From that, it's a simple division to get to the flux, and then another division to get the field.

$$\Phi = EA = \frac{Q_\textrm{enc}}{\epsilon_0}$$

- Warren

6. Mar 7, 2005

### GCT

I'm actually on this chapter myself.

You can further interpret chroot's result further to the final

$$E=k_e r \frac{Q}{a^3}$$

for inside the sphere

based on this it seems that the electric field would at the surface of the sphere. Correct me if I'm wrong. The electric field inside the insulating sphere is defined by the above equation, while outside the sphere we can observe it as a point charge; try creating a gaussian surface which is concentric with the sphere except with a greater radius.

7. Mar 7, 2005

### GCT

"would be strongest at the surface of the sphere"...an error in my post

8. Mar 7, 2005

### Staff: Mentor

You may want to recheck your work. (Express the field in terms of C and a. If you use Q and r, define them.)