Gauss' Law: Solid Non-conducting Cylinder

[Bryon]

Homework Statement

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?

Homework Equations

Volume of a cylinder: pi*L*R^2
Gauss' Law: ∫ E·dA = E(pi*L*R^2) = Q_inside*ε₀

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 = (Q_inside*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?

 

[Metaleer]

Hi, Bryon.

First of all, we have Gauss' law:

$$\oint_\mathcal{S} \mathbf{E} \cdot d\mathbf{S} = \frac{Q_\text{int}}{\epsilon_0}$$​

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 $$r$$ and length $$L$$, this turns out to be $$E 2 \pi r L$$.

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

$$Q_\text{int} = \int_V \rho dv$$​

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

$$Q_\text{int} = \int_V \rho dv = \int_0^r A r'^2 2 \pi r' L dr' = \frac{A \pi L r^4}{2}$$​

where I added $$'$$ 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 $$L$$'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]

Ah I see. I was not sure how to get rid of L. Which I just assumed to be 1m.

 

[Bryon]

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

 

[rwisz]

Your attempt at a solution is on the right track, but there are a few errors in your calculations. Here is a corrected solution:

First, you are correct that the length of the cylinder is not needed for this problem, so you can omit it from your calculations.

Next, to find the volume of the cylinder, you need to use the correct formula, which is V = πr^2h. Since the cylinder is solid, h = 2r (the height of the cylinder is equal to twice the radius). So the volume should be V = πr^2(2r) = 2πr^3.

Now, to find the charge inside the cylinder, you need to multiply the volume by the charge density: Qinside = ρV = (2.9x10^-6)(2πr^3) = 5.8x10^-6πr^3.

Finally, you can use Gauss' Law to find the electric field at a distance r from the axis of the cylinder: E = Qinside/(ε0πr^2) = (5.8x10^-6πr^3)/(ε0πr^2) = (5.8x10^-6)/(ε0r).

Plugging in r = 7 cm and the value of ε0 (8.85x10^-12), we get E = 5.24 N/C, which is the correct magnitude of the electric field at a distance of 7 cm from the axis of the cylinder.