Non-uniform line charge density with r not constant

[bravo340]

Homework Statement

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.

Homework Equations

V = (KQ/r)

E = (KQ)/ r^{2}

E = -\nablaV

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.

 

[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.

 

[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.

 

[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.