The flux through a cylinder from a charge at the origin

[hmparticle9]

Homework Statement

Consider a cylinder that has its origin at the centre. It has height 2 and radius 1. Let us place a charge $q$ at the origin. Compute the flux coming from the top and bottom faces. Compute the flux coming from the curved surface. First answer the two questions using Gauss's law then use direct computation.

Relevant Equations

Gauss's law:
$$\int_{S} \mathbf{E} \cdot \text{d}\mathbf{S} = \frac{q}{\epsilon_0}$$

First the things I can figure out.

For the top face:
$$\text{d}\mathbf{S} = r \text{d} \theta \text{d} r \mathbf{e}_z$$
For the curved surface (remember that the radius is 1):
$$\text{d}\mathbf{S} = \text{d} \theta \text{d} z \mathbf{e}_r$$

I am not sure how to apply Gauss's law. Because the top face does not enclose the charge at the origin.

 

[hutchphd]

Think about inscribing the cylinder in a sphere. The rest should follow easilly.

 

[vela]

For the top face:
$$\text{d}\mathbf{S} = r \text{d} \theta \text{d} r$$
For the curved surface (remember that the radius is 1):
$$\text{d}\mathbf{S} = \text{d} \theta \text{d} z$$

$d\mathbf{S}$ is a vector quantity, so you need to include unit vectors to indicate the directions.

I am not sure how to apply Gauss's law. Because the top face does not enclose the charge at the origin.

Gauss's law applies only to the entire cylinder. You have to break the integral into a sum over the three pieces.
$$\oint \vec E \cdot d\vec S = \int_{\rm top face} \vec E \cdot d\vec S + \int_{\rm curved\ surface} \vec E \cdot d\vec S \int_{\rm bottom} \vec E \cdot d\vec S$$ For this problem, however, this isn't really going to help you. You want to think about @hutchphd's hint.

 

[hmparticle9]

Okay. I have figured out that we need a sphere with radius $\sqrt{2}$ for us to fit my cylinder snug.

Gauss's law says that the flux through the sphere is $q/\epsilon_0$.

I think what we are interested in is the flux through the top dome as a ratio. Because the stuff coming through the top dome must have came through the head.

Yea sorry @vela it was a typo. fixed it now

 

[hmparticle9]

So the area of the dome is $2 \pi \sqrt{2} (\sqrt{2} - 1)$, we can divide this by the total area of the sphere and we obtain:
$$\frac{q}{2\epsilon_0}\bigg(1- \frac{1}{\sqrt{2}}\bigg)$$
Obviously if we include the bottom as well we get the correct answer:
$$\frac{q}{\epsilon_0}\bigg(1- \frac{1}{\sqrt{2}}\bigg)$$

For the curved surface, it is clear from the above expression that the flux is
$$\frac{q}{\epsilon_0}\bigg(\frac{1}{\sqrt{2}}\bigg)$$

Now I need to re-derive these expressions without using Gauss's law.

 

[hmparticle9]

From looking at a diagram, the only component we need to worry about is the component of the flux in the $z$-direction. The other directions get cancelled by flux passing through diametrically opposed points. I am going to deal with the top and bottom faces first. So here is what I think the answer should be.

$$F = \frac{q}{4 \pi \epsilon_0} \times \frac{1}{1 + r^2} \times \frac{1}{(1 + r^2)^{\frac{1}{2}}} \times r \text{d}r \text{d}\theta$$

The first element in the above expression is boiler plate. The second term is the one over the squared distance to the patch we are considering. The third term is the fact we want to cosine component of the force. The last term comes from the fact that we are considering a little patch on the top of the cylinder. By integrating over $\theta \in (0,2\pi)$ and $r \in (0,1)$ we obtain the correct result.

For the curved surface:
$$F = \frac{q}{4 \pi \epsilon_0} \times \frac{1}{1 + z^2} \times \frac{1}{(1 + z^2)^{\frac{1}{2}}} \times \text{d}z \text{d}\theta$$

integrating over $z \in (-1,1)$ and $\theta \in (0,2\pi)$ we obtain the result achieved using Gauss's rule above.