# Force on a charge from a cylinder of charge

1. Feb 4, 2010

### kpou

1. The problem statement, all variables and given/known data
Find the force on a charge q a distance r > a away from a cylinder of charge with radius a. The cylinder has a charge density per volume $$\rho$$

2. Relevant equations
E(r)=$$\frac{1}{4\pi\epsilon_{0}}$$$$\int_{V}$$ $$\frac{\rho(r')}{\eta^2}$$$$\eta[hat]$$d$$\tau'$$

3. The attempt at a solution
I am not sure how on earth to start the problem without thinking about the cylinder as infinite length or having q at a midpoint in the cylinder. I don't even know if I am supposed to know how to do that yet :/ What I have done is

$$\eta$$=$$\sqrt{x^2+y^2+z^2}$$

Does this make
E(r)=$$\frac{1}{4\pi\epsilon_{0}}$$ $$\int_V$$ $$\frac{\rho}{\sqrt{x^2+y^2+z^2}}$$ $$\frac{xdx+ydy+zdz}{\sqrt{x^2+y^2+z^2}}$$

Any nudge in the right direction would be great. Thanks :D

Note: formatting gets very easy after you mess up a whole bunch :)

Edit: This problem is near the Dirac delta function section. Perhaps this may help? Unfortunately I don't see how it could fit just for the fact that the dimensions of the cylinder will affect the point charge.

Last edited: Feb 4, 2010
2. Feb 4, 2010

### jdwood983

Is the charge density of the cylinder uniform or is there a function given for it?

3. Feb 4, 2010

### kpou

It is uniform I assume.

4. Feb 4, 2010

### jdwood983

And the problem does not indicate that it is an infinite cylinder?

Also, this problem will likely be loads easier if you work in cylindrical coordinates, rather than Cartesian.

5. Feb 4, 2010

### UgOOgU

This equation is not OK. The $$\eta$$ in the upper part of the fraction do not exists.

6. Feb 4, 2010

### kpou

I'm sorry it is supposed to have a hat over it.

7. Feb 4, 2010

### UgOOgU

No, the equation is really without it.
$$E(r)=\frac{1}{4\pi\epsilon_{0}}\int_{V}\frac{\rho(r')}{\eta^2} d\tau'$$

8. Feb 4, 2010

### kpou

And the problem does not state it is infinite, I actually copied it word for word in case I missed something.

9. Feb 4, 2010

### jdwood983

Unless someone suggests otherwise, I'd assume that it is infinite. In this case, you can use Gauss's Law to find the electric field:

$$\oint \mathbf{E}\,d\mathbf{a}=\frac{Q_{enc}}{\varepsilon_0}\rightarrow\mathbf{E}=\frac{Q_{enc}}{2\pi\varepsilon_0r}\hat{\mathbf{r}}$$

10. Feb 4, 2010

### kpou

I am just going by what my book has written for the E equation. It does include $$\widehat{\eta}$$.

11. Feb 4, 2010

### kpou

bah, I just noticed the next question specifies an infinitely long solenoid. Maybe it wasn't just forgotten like I was hoping.

Last edited: Feb 4, 2010
12. Feb 4, 2010

### UgOOgU

Ok, now I understand, $$\widehat{\eta}$$ is the versor of the vector $$\vec{\eta}$$.

Sorry, I think that it was wrong because of your substitution in this equation
$$\vec{E}(r)=\frac{1}{4\pi\epsilon_{0}}\int_V\frac{\rho}{x^2+y^2+z^2}dx dy dz$$.