Gauss' Law: Solid Non-conducting Cylinder


by Bryon
Tags: cylinder, gauss, nonconducting, solid
Bryon
Bryon is offline
#1
Jan31-11, 12:30 PM
P: 99
1. The problem statement, all variables and given/known data

A long, solid, non-conducting cylinder of radius 8 cm has a non-uniform volume density, ρ, that is a function of the radial distance r from the axis of the cylinder. ρ = A*r2 where A is a constant of value 2.9 μC/m5.

What is the magnitude of the electric field 7 cm from the axis of the cylinder?

2. Relevant equations

Volume of a cylinder: pi*L*R^2
Gauss' Law: ∫ EdA = E(pi*L*R^2) = Qinside0

3. The attempt at a solution

ρ = A*r2 = (2.9x10^-6)*0.7^3 = 1.421e-8

ρV = (1.421x10^-8)* pi*0.7^2 = 2.18745955e-10

E = (Qinside*pi*r^2)/ε0 = 5.433109096 N/C

That does not look right at all to me, and I am not sure where my set up went wrong. Its obvious to me that I did not need the length of the cylinder so I ommited L (actually assumed a value of 1). Did I get the volume correct?
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Metaleer
Metaleer is offline
#2
Jan31-11, 01:05 PM
P: 123
Hi, Bryon.

First of all, we have Gauss' law:

[tex]\oint_\mathcal{S} \mathbf{E} \cdot d\mathbf{S} = \frac{Q_\text{int}}{\epsilon_0}[/tex]

The point where they ask you to get the E field is inside the cylinder, so to get this E field, you would need to set up a gaussian cylinder of radius 7 cm (I'd work symbolically until the very end) and use Gauss' law. The evaluation of the LHS is very easy, it is merely the E times the curved outer surface area of the gaussian cylinder (the sides don't add to the flux integral because there, E is perpendicular to the surface). Assuming that the cylinder is of radius [tex]r[/tex] and length [tex]L[/tex], this turns out to be [tex]E 2 \pi r L[/tex].

The RHS is slightly trickier, as this involves the total charge contained in the gaussian cylinder. That is:

[tex]Q_\text{int} = \int_V \rho dv[/tex]

To work out the differential element of volume, we remember that the volume of a cylinder is [tex]v = \pi r^2 L[/tex], and differentiating with respect to [tex]r[/tex], we have [tex]dv = 2 \pi r L dr[/tex]. We also have the charge density [tex]\rho[/tex] as a function of [tex]r[/tex]. Plugging it into the charge equation,

[tex]Q_\text{int} = \int_V \rho dv = \int_0^r A r'^2 2 \pi r' L dr' = \frac{A \pi L r^4}{2}[/tex]

where I added [tex]'[/tex] to the integrand to not confuse it with the upper limit. All you need to do now is set both sides of Gauss' law equal to each other, solve for the E field, and not forgetting to do unit conversion, evaluate it at r = 7 cm.

Notice how when you set them equal, the [tex]L[/tex]'s cancel out, as expected - this is a better method than to assume it's value to be 1. 1 What? cm? m?

Good luck.
Bryon
Bryon is offline
#3
Jan31-11, 01:41 PM
P: 99
Ah I see. I was not sure how to get rid of L. Which I just assumed to be 1m.

Bryon
Bryon is offline
#4
Jan31-11, 06:50 PM
P: 99

Gauss' Law: Solid Non-conducting Cylinder


Ah now that I read your post I makes a lot more sense! Thanks!


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