Non-uniform line charge density with r not constant

bravo340

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

We have a non uniform line charge density [tex]P_{l}[/tex] = [tex]\rho_{l}[/tex] cos[tex]\phi[/tex]

It is a spiral line where 0 [tex]\leq[/tex] [tex]\phi[/tex] [tex]\leq[/tex] 4 [tex]\pi[/tex]

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

r varies: r ( [tex]\phi[/tex] ) = [tex]\phi[/tex] * [tex]r_{0}[/tex] + a

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

2. Relevant equations

V = (KQ/r)

E = (KQ)/ r[tex]^{2}[/tex]

E = -[tex]\nabla[/tex]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.

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

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

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

Mindscrape

Can I fix your formulas too?

[tex] V = k \int_{\Omega} \frac{\rho(\mathbf{r'})}{||\mathbf{r} - \mathbf{r'}||} d\gamma'[/tex]

Which would specifically be

[tex] V = k \int_l \frac{\lambda(\mathbf{r'})}{||\mathbf{r} - \mathbf{r'}||} dl'[/tex]

for a line charge.