# Homework Help: Spherical Charge Ball, Gauss law

1. Oct 2, 2012

### hansbahia

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

A spherical charged ball of radius a has total charge Q; there is no charge outside the ball and no sheet-charge on its surface. The (radial) field inside the ball has the form
Er(r) = constant x r2 for r between 0 and a.
Use Gauss's Law in integral form to evaluate the constant in this formula in terms of Q and a, and then rewrite the formula in terms of these parameters.

2. Relevant equations

∫E.dA=Q/εo

3. The attempt at a solution

Lets say constant is b and dA=4.pi.r^2

∫Er(r).dA=Q/εo

from 0 to a ∫b.r^2(4.pi.r^2) dr=Q/εo
(4.pi.b) from 0 to a ∫r^4 dr = Q/εo
(4.pi.b)(a^5/5)=Q/εo
b=(5Q)/(4.pi.a^5.εo)

When i replace b (the constant) in Er(r) i get

Er(r) = b x r2
Er(r) = (5Q)/(4.pi.a^5.εo) x r2
Er(r)= (5.Q.r^2)/(4.pi.a^5.εo)

but the answer is Er(r)= (Q.r^2)/(4.pi.a^4.εo)

can someone please tell me where am I miscalculating or i forgot to add something
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 2, 2012

### PhysicsGente

You say that dA = 4.pi.r^2 , but where is the dr coming from in your integral?

There is another way of expressing the differential area element. Look into that ;).

3. Oct 2, 2012

### hansbahia

I used the wrong dA, however even if I use dA=2.pi.r dr I'm still off

from 0 to a ∫b.r^2(2.pi.r) dr=Q/εo
(2.pi.b) from 0 to a ∫r^3 dr = Q/εo
(2.pi.b)(a^4/4)=Q/εo
b=(2Q)/(pi.a^5.εo)

and when i replace b (the constant) in Er(r) i get

Er(r)= (2.Q.r^2)/(pi.a^4.εo)

I'm supposed to get 1/4 and not 2. I'm not familiar with other ways of expressing the differential area element

4. Oct 2, 2012

### PhysicsGente

The problem also says that there is no sheet-charge on its surface. Try using that.

OK. nvm that. There is a simpler way. If you enclose the sphere with a gaussian surface, what can you say about the electric field flux going out.

Last edited: Oct 2, 2012
5. Oct 2, 2012

### hansbahia

But wouldn't that just mean there is a magnetic field all around?

If I enclose the sphere with a gaussian surface the electric flux would just be equal to the electric field times the area of the sphere.
E.(4.pi.r^2)

6. Oct 2, 2012

### PhysicsGente

Exactly. And what does ∫E.dA represent? ;D

7. Oct 2, 2012

### hansbahia

I still can't figure out
I mean..
electric flux= EA=Q/εo
EA=Q/εo
(b.r^2)(4.pi.r^2)=Q/εo
b.4pi.r^4=Q/εo
b=Q/(4.pi.r^4.εo)

8. Oct 2, 2012

### hansbahia

Wait does that mean I can sub "a" for "r"? Therefore I get b=Q/(4.pi.a^4.εo)

Am I right?

9. Oct 2, 2012

### PhysicsGente

Yes. You can make your gaussian surface as big (or small) as you want.
Now, why does letting r = a work? (hint: look at the problem statement).

10. Oct 2, 2012

### hansbahia

Because there is no charge outside the ball and no sheet-charge on its surface

Thankss a lot!

So simple