# Volumentric or surface charge density

Tags:
1. Jul 31, 2016

### Fernandopozasaura

1. The problem statement, all variables and given/known data
It is known that the potencial is given as V = 80 ρ0.6 volts. Assuming free space conditions, find a) E, b) the volume charge density at ρ=0.5 m and c) the total charge lying withing the closed surface ρ=0.6, 0<z<1

2. Relevant equations
E
=-∇V

3. The attempt at a solution
(this is author's solution)
a) E field: E=-∇V=-∂V/∂ρ aρ = -48ρ-0.4
b)ρv=-28.8 ε0 ρ-1.4 = -673 pC/m3
So, we get an expression for ρv that depends on ρ.
c) the total charge lying within the closed surface ρ = 0.6, 0 < z < 1. The author says "the easiest way to do this calculation is to evaluate Dρ at ρ = .6 (noting that it is constant), and then multiply by the cylinder area."
Using part a, we have Dρ for ρ=0.6
= −480(.6)−.4 = −521 pC/m2. Thus
Q = −2π(.6)(1)521 × 10−12C = −1.96 nC.

And here is where I need some help becouse I understand that the author is saying that the charge is located only in cylinder's surface and so, he uses the *surface* of cylinder to get the total charge.
I worked out the solution thinking that the charge is distributed in the whole volume, not only in the surface. I thought this because the expression that gives us charge density is function or ρ and because it is expressed in C/m3 (charge per unit of volume).
I've done a volume integration from 0 < Φ < 2π, 0 < z < 1 and 0 < ρ < 0.6 meters.
There's nothing in exercise that make you think that the cylinder is conductive in which case the charge would be on surface.
I guess I'm wrong, but why?

Fernando

Author's solution says that

2. Jul 31, 2016

### TSny

Hello and welcome to PF!

Can you relate the author's solution for part (c) to Gauss' law? Can you clarify the notation Dρ? Does this stand for the ρ component of the $\vec{D}$ field?

3. Jul 31, 2016

### Fernandopozasaura

Thank you very much for such a quick answer Tsny,
Well, first of all, let me get familiar with this editor.

It's related through this expression
∫D.dS = Q (this integral is over a closed generic surface containing Q)
On the other hand working out that integral you can get to
Divergence of D = div D = ∇.D = ρv and then, so
Q=∫v∇.D dv
and this is the volume integral I did.

Yes. That it's. The only component of D in this case.

What I don't understand is why author fix ρ to 0.6 (the surface of cylinder) instead of going since the axis of it (ρ=0) to the surface, taking into account the charge (I think) exists inside. Moreover when in part b) the answer for the volume charge density is expressed in C/m3 meaning that there are charge inside the cylinder.

4. Jul 31, 2016

### TSny

Yes, there is a nonuniform volume charge density spread throughout space. Your method should work, but it is easier to use Gauss' law in the form ∫D.dS = Q for part (c).

5. Jul 31, 2016

### Fernandopozasaura

Humm, would you say the exercise is wrongly solved?
I think I would.
To me, most important thing is that there is no weird magic hidden that I didn't notice.
Thanks again Tsny.

6. Jul 31, 2016

### TSny

I think the solution as given is correct. But I did notice a typo:
the 480 should be 48ε0, where the ε0 comes from the relation D = ε0 E.

What part of using Gauss' law do you think is incorrect in the solution?

7. Jul 31, 2016

### Fernandopozasaura

What i think is wrong is, in part c), not making a volume integration from cylinder axis to its surface. Instead supposing all charge is placed on it and then just multiply lateral surface times charge density.
There is nothing wrong with Gauss's las use.
Once again, thanks
Fernando

8. Jul 31, 2016

### TSny

In the solution, the integration over the surface is not an integration of any charge density. It's an integration of the normal component of $\vec{D}$ over the surface. According to Gauss' law, this will give you the total charge Q contained within the volume enclosed by the surface.

9. Aug 1, 2016

### Fernandopozasaura

Good morning Tsny,
I see the way. There are two ways to solve the same problem. One of them gets you to a volumetric integral that is forced in the exercise or experiment you are workin on, and the other makes you to choose a surface, at your own! and that's the good point.
In this case it was possible to solve either way because the volume has cylindrical symmetry, but if not, the second one is better as long as you can choose the surface for your convenience.

Now I'm having a hard time getting to the same numerical result using both ways, but that's my problem.

¡Solved!, Thank you very much.
Fernando