# Non-uniform line charge density with r not constant

1. Sep 27, 2007

### bravo340

1. The problem statement, all variables and given/known data

We have a non uniform line charge density $$P_{l}$$ = $$\rho_{l}$$ cos$$\phi$$

It is a spiral line where 0 $$\leq$$ $$\phi$$ $$\leq$$ 4 $$\pi$$

It is on the x-y plane with z=0.

r varies: r ( $$\phi$$ ) = $$\phi$$ * $$r_{0}$$ + a

We need to find the Potential and Electric Field at the origin.

2. Relevant equations

V = (KQ/r)

E = (KQ)/ r$$^{2}$$

E = -$$\nabla$$V

3. The attempt at a solution

The east way would be to find the Potential and then to find the Electric Field by using the relationship between E and the gradient of V.

I think this problem wouldn't be as tough if r was constant.

Last edited: Sep 27, 2007
2. Sep 27, 2007

### Mindscrape

So are you going to use a line integral? I suggest cylindrical coordinates.

Also, I don't know what phi is, the polar angle or some constant? Oh, nevermind I see where r is bounded. You still need to show some work before you get any help.

Last edited: Sep 27, 2007
3. Sep 27, 2007

### JoAuSc

I've tried it, and I have gotten to the point where I have an integral that determines the potential at this point. Here's how I got this far:

For a bunch of separate point charges, we have

$$V = \sum {\frac {-1}{4 \pi \epsilon_{o} } \frac {q_i}{r_i} }$$
where I've replaced $$\rho_l$$ with $$\lambda$$ to make things easier for me. (Sometimes $$\rho$$ denotes the radial distance.)

For a line, we replace $$q$$ with $$d \lambda$$ and integrate. To do this, we need to replace $$d \lambda$$ with something with $$d \phi$$ in it; I'll leave the details up to you. We replace $$r$$ with your formula for $$r$$.

4. Sep 27, 2007

### Mindscrape

Can I fix your formulas too?

$$V = k \int_{\Omega} \frac{\rho(\mathbf{r'})}{||\mathbf{r} - \mathbf{r'}||} d\gamma'$$

Which would specifically be

$$V = k \int_l \frac{\lambda(\mathbf{r'})}{||\mathbf{r} - \mathbf{r'}||} dl'$$

for a line charge.