# Basic Gauss's Law problem

1. Jan 29, 2009

### Seraph404

1. The problem statement, all variables and given/known data

We have solid ball of radius rb and charge density $$\rho$$.

I need to find the magnitude of electric field E(r) at a distance r < rb from the center of the ball and express my answer in terms of $$\rho$$, r, rb, and $$\epsilon$$ 0.

2. Relevant equations

Gauss's Law
$$\int$$ EA = Q/$$\epsilon$$0

3. The attempt at a solution

E*(4*$$\pi$$*rb2) = 4*$$\pi$$*r3*$$\rho$$/(3*$$\epsilon$$0)

E(r)= (r3*$$\rho$$)/(3*rb2*$$\epsilon$$0)

But the website I check my answers with tells me it's wrong. Can someone help me to understand, please? The hint it gave me was that my answer should not depend on rb. Why wouldn't it?

2. Jan 29, 2009

### Mr.Amin

This looks right on first glance.
The integral on the left should yield

E*(area enclosed by a sphere) = E*(4*pi*r_b^2) .

The right hand side is simply, the charge density times the volume of the arbitrary sphere whose r < r_b.
Q = (4/3 pi r^3) * rho.

Therefore the resulting ratio should go as

E(r < r_b) = r^3 * rho /(3 * r_b^2 *epsilon).

Check with the instructor. The computer's solution may have a typo.

3. Jan 29, 2009

### tiny-tim

Hi Seraph404!

(have a pi: π and a rho: ρ )

Hint: E is the same for all points on the sphere of radius r < rb

so what is the flux through the surface of that sphere?

4. Jan 29, 2009

### Seraph404

Ah.. ><
I'm almost 99% sure that the problem is me. But I'll ask anyway if nobody else on here can spot the mistake.

5. Jan 29, 2009

### Seraph404

Um. I guess radially outward? But how does that remove my need for rb in the expression?

I should also have mentioned that I'm struggling with the material. My professor is absolutely no help. I've started skipping lecture just to hang out in the Physics helpdesk and work problems, since that's the only way I'm going to learn any of this.

6. Jan 29, 2009

### Mr.Amin

Ok, I did the math on paper instead of in my head.

Here is the answer. Got a pen and paper? Nope that pen is dry. Try another one.
Ahh one that works.

Left hand side = E * (area of a sphere ENCLOSING THE CHARGE)
= E * (4*pi*r^2)

Right hand side = Q/epsilon
= (volume of ENCLOSED CHARGE) * (Charge density)
= (4/3) * pi * r^3 * rho / epsilon

Result => E = (r * rho) / (3 * epsilon)

You can check this at the boundary where r = r_b. Let me show you.

7. Jan 29, 2009

### tiny-tim

Hi Mr.Amin! Welcome to PF!

Yes, but how much is it? E(r) times what?

And then what should that flux be equal to?

(what is flux through a surface always equal to? )

8. Jan 29, 2009

### Mr.Amin

The continues from my previous post, I can't type as fast as you!!!!!!!!!

r > r_b

Left hand side = E * (4 * pi * r^2)

Right hand side = (4/3) * pi * r^2 * rho / epsilon

Solve for E

E = (r^3 * rho) / (3 * epsilon * r^2)
= (r * rho) / (3 * epsilon).

You could have gotten here if you used E = 1/(4*pi*epsilon) * Q/r^2 .

Now check E(r = r_b) for both equations.

for r < r_b
E(r = r_b) = (r_b * rho) / (3 * epsilon)

for r > r_b
E(r = r_b) = (r_b * rho) / (3 * epsilon).

The reason why r_b does not come into your previous question is because a uniform distribution of charge creates essentially no net electric field. If you pretend your sensor has penetrated into the charged sphere, everything at the radius above it will add to zero field. ONLY the charge in the sphere below the probed radius will generate the non-zero electric field.

This works for gravity too since Gauss's law works for g-fields as well.

9. Jan 29, 2009

### Seraph404

Oh.. I guess I didn't realize that the area in Gauss's law was for enclosed charge only. Anyway,Thanks guys! I will definately refer back to this post when I work future problems! Hopefully, the more of these I work, the better my understanding of this subject will be.

My main problem is that our prof began by throwing equations at us (not even bothering to derive them) and scribbling down example problems here and there that hardly any of us could make out. So conceptually, I have little to no idea what is going on in this subject. The book is a big help, but sometimes I need things drawn out for me before I'll really understand. Anyway, thanks again!