# Gauss and E-Field

1. Feb 19, 2008

### jesuslovesu

[SOLVED] Gauss and E-Field

1. The problem statement, all variables and given/known data
http://img443.imageshack.us/img443/9452/eletroof9.th.png [Broken]

My main problem with this problem is finding the electric field at Point 2 (P2)

2. Relevant equations

3. The attempt at a solution

I derived the electric field using Gauss's Law for a cylindrical charge distribution.
$$E (2\pi rL ) = \rho(\pi r^2 L) / \varepsilon_0$$
$$E = \frac{\rho * r}{2\varepsilon_0}$$
So I found E at P1 to be 0 V/m

Now onto finding E at P2, according to my equation, it would seem that E is 0, since the radial distance is 0, however that seems strange to me, if the point is at the surface of the charged volume cylinder, wouldn't there be some charge, I know it's not a conductor but still I would think that there would be some charge?

And since I've got it all out here, is the E field at P3 just the integral [3/2L, L/2] of a charged disk?

Last edited by a moderator: May 3, 2017
2. Feb 19, 2008

### michalll

Try working with Ex = k*q*x/r^3.

3. Feb 19, 2008

### jesuslovesu

I can't use Gauss's Law for finding any of the points? I know usually the cylinder is an infinite length if you want to find E, but even for Point 1 I can't use it?

4. Feb 20, 2008

### michalll

Basically not in or outside finite length cylinder the tangential forces will cancel only at the midpoints. Try finding E on axis of a charged disc first.

Last edited: Feb 20, 2008
5. Feb 23, 2008

### jesuslovesu

Ok so I've got the equation for a charged disk
$$2\pi k \rho (1 - \frac{x}{\sqrt{x^2+a^2} ) }$$ how do I integrate that over the length (L) to get the electric field of the cylinder?

Would I say that dE = $$2\pi k \rho (1 - \frac{x}{\sqrt{x^2+a^2} ) }$$ and then integrate dE from 0 to L or do I need to differentiate Edisk before I do that?

Last edited: Feb 23, 2008
6. Feb 25, 2008

### michalll

Just integrate it form 0 to L, since ddQ = 2*pi*rho*a*da*dx and u integrated it along a then your equation should still contain dx.