Equipotential cylinder

  • Thread starter neworder1
  • Start date
65
0

Main Question or Discussion Point

An infinite string with linear charge density [tex]\lambda[/tex] is put parallel to the axis of an infinite conducting grounded cylinder (V=0) of radius [tex]R[/tex], the distance between the string and the center of the cylinder is [tex]l[/tex]. Find the potential outside the cylinder.

Is it possible to solve this problem using the method of images, i.e. placing an image string with some charge density [tex]\lambda_{1}[/tex] inside the cylinder so that the cylinder will be an equipotential surface? If V weren't 0, I could place a string of equal charge (i.e. [tex]-\lambda[/tex]) somwehere inside and it should work, but for V=0, the charges can't be equal.
 

Answers and Replies

clem
Science Advisor
1,308
15
Yes, it is the 2D example of a charge outside a sphere.
 
65
0
Ok, but how do I calculate this? If I place an image string with charge [tex]\lambda_{1}[/tex] inside the cylinder, at distance [tex]a[/tex] from the center,I get the following equation for equipotential surfaces:
[tex]\lambda ln(r)-\lambda_{1} ln(r_{1})=C[/tex], where [tex]r, r_{1}[/tex] are distances form resp. 1st and 2nd string. Now if I put C=0 (since on the surface of the cylinder V=0, and I want the cylinder to be an equipotential surface), I get:
[tex]\lambda ln(r)=\lambda_{1} ln(r_{1})[/tex]. Now I need to calculate [tex]\lambda_{1}[/tex] and [tex]a[/tex] from this. Any help?
 
pam
455
1
It's done like the 3D problem.
Write [tex](R\hat{r}-{\vec L})e^\lambda=(R\hat{r}-{\vec L}')e^{-\lambda'}[/tex].
Then let L'=R^2/L.
 
Last edited:

Related Threads for: Equipotential cylinder

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
2
Views
1K
  • Last Post
2
Replies
26
Views
13K
  • Last Post
Replies
4
Views
1K
Replies
1
Views
2K
Replies
1
Views
8K
Replies
1
Views
2K
Top