# Non-uniform line charge density with r not constant

by bravo340
Tags: charge, constant, density, line, nonuniform
 P: 1 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.
 P: 1,877 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.
 P: 200 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 $$r$$.
P: 1,877

## Non-uniform line charge density with r not constant

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.

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