# Electrodynamics and electrical fields

1. Apr 23, 2008

### Niles

[SOLVED] Electrodynamics and electrical fields

The question is: A long cylinder carries a charge density that is proportional to the distance from the axis ρ =kr, where k is a constant, r is the distance from the axis. Find electric field inside the cylinder.

My attempt: Ok, first of all we know from Gauss' law that $$\oint_S {{\bf{E}} \cdot d{\bf{a}}} = \frac{1}{{\varepsilon _0 }}Q$$.

We also know that $$Q = \int {\rho dV = } \int {k \cdot s \cdot s \cdot dsd\phi dz = \frac{2}{3}\pi kls^3 }$$. Here I have used a Gaussian cylinder of length l and radius s.

Now I must find $$\oint_S {{\bf{E}} \cdot d{\bf{a}}}$$.

First question: The electric field E is given by $${\bf{E}} = \frac{1}{{4\pi \varepsilon _0 }}\frac{q}{{r^2 }}{\bf{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over r} }}$$. Is the unit vector r in this case the spherical radius-vector?

For this cylinder, I write the parametric and differentiate it with respect to phi and z. Then I find the normal-vector as the cross-products of these two.

Question two: When I find the element da as above, what do I do with E, which is in spherical coordinates?

I hope you can help me.

2. Apr 23, 2008

### foxjwill

Ok, you're making this way to hard for yourself. Assume our cylinder has radius a and height h, and choose our Gaussian surface to be a cylinder of radius r co-axial to the original cylinder. The electric field at any point on the Gaussian surface for a radius $$r \leq a$$ has constant magnitude and is directed parallel to the surface normal there, so we can simplify the integral to

$${\int_S E d{A}}$$​

and then to

$$E{\int_S d{A}}.$$​

Now, since $${\int_S d{A}}$$ is just the surface area of the Gaussian surface, which we know to be $$2\pi r h$$, we can substitute that in, and we are left with the very simple equation

$$E 2\pi r h = \frac{q_{\mbox{enc}}{{\varepsilon _0 }}.$$​

I will leave it to you to include the density function in that.

Last edited: Apr 23, 2008
3. Apr 23, 2008

### Niles

Hi foxjwill.

First, thanks for taking the time to reply. I can see that using the definition of the dot-product simplifies things greatly - but still I would like to know how it can be done "my way". I have just studied vector-calculus most of this week, and I am kinda disappointed in myself that I cannot figure it out.

The thing that annoys me is particularly the r unit-vector in E. Is this in Cartesian or Spherical coordinates?

Last edited: Apr 23, 2008
4. Apr 23, 2008

### foxjwill

Whichever is more convenient for the particular problem, as long as the same point is described. I'd suggest actually using cylindrical coordinates.

5. Apr 23, 2008

### Niles

Ok, so $${\bf{E}} = \frac{1}{{4\pi \varepsilon _0 }}\frac{q}{{r^2 }}\rho}$$

I get that the normal-vector has the rho-component 3*cos(theta). I integrate this over theta and z, which are respectively 0..2Pi and 0..l - and this gives 0, which is not correct.

Where is my error?

Last edited: Apr 23, 2008
6. Apr 24, 2008

### Niles

I worked it out. d\vec{A} = \vec{n}*dA, and the vectors go out since they are parallel.

Thanks.