# Gauss Law? Average charge density

1. Apr 6, 2013

### Roodles01

1. The problem statement, all variables and given/known data
Hi. A cylinder of radius r & length L whose charge density distribution is given by
ρ = C/2 * r3
where r = radial distance in cylindrical coordinates
C = constant
show that the average charge density ρbar = a3 C / 5

2. Relevant equations
Gauss differential law div E = ρ / ε0

div E = 1/r * ∂/∂r * (rEr) + 1/r ∂E∅/∂∅ + ∂Ez/∂z

3. The attempt at a solution

Hmmm! Not really sure where to begin with this.
Not sure if it's to do with stuff as "difficult" as equations above.

Last edited: Apr 6, 2013
2. Apr 6, 2013

### Andrew Mason

What is a?

Also it is not clear whether: ρ = C/(2*r3) or ρ = (C/2)*r3

What does average charge density mean? Can you determine the total charge in the cylinder?

AM

3. Apr 6, 2013

### Roodles01

Ah. Detail.
Get it right, me.

r is radial distance in cylindrical coordinates
first equation is
ρ = (C/2)*r3

As far as "average" goes I'm assuming to be the mean.

Hope this clears things up.

4. Apr 6, 2013

### Andrew Mason

Ok. But give us an expression for it.

AM

5. Apr 6, 2013

### Andrew Mason

Hint: average charge density is: total __________/ total _________

AM

6. Apr 7, 2013

### Roodles01

I'll try total charge / total area.

7. Apr 7, 2013

### Andrew Mason

Why area? This is a solid cylinder. Hint: what are the units of mass density? By analogy, what are the units of charge density?

AM

Last edited: Apr 7, 2013
8. Apr 8, 2013

### Roodles01

Aha! It's volume.
Density has units of mass per unit volume such as g/ml
so charge density must be units of charge per unit volume such as coulombs/ m3 for volume such as the Gaussian surface (cylinder)
Thanks I shall post again soon.

9. Apr 8, 2013

### Andrew Mason

I am not sure why you are referring to Gauss' law here. You are not calculating the electric field so there is no need to think of the cylinder surface as a Gaussian surface for this problem. You are just trying to find the total charge and total volume.

AM

10. Apr 8, 2013

### Roodles01

Ah! It all clears.
Clouds have gone. It's not just about algebra after all.
ρ = charge density distribution which is distribution at a single point, so have double integral

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11. Apr 8, 2013

### Andrew Mason

You don't really need the double integral. You know that the cylindrical surface with radius r and dr thick has volume 2πrLdr. So you just have to integrate from r=0 to r=a to find the charge.

$q = \int_0^a ρdV =\int_0^a ρ2\pi rLdr$

Then it is just a matter of dividing by the total volume.

AM