Nonuniform line charge density with r not constantby bravo340 Tags: charge, constant, density, line, nonuniform 

#1
Sep2707, 09:26 PM

P: 1

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 xy 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. 



#2
Sep2707, 10:05 PM

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. 



#3
Sep2707, 10:11 PM

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 [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]. 



#4
Sep2707, 10:14 PM

P: 1,877

Nonuniform line charge density with r not constant
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. 


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