• Support PF! Buy your school textbooks, materials and every day products Here!

Electric potential on an infinite line of charge

  • Thread starter Bryon
  • Start date
  • #1
99
0

Homework Statement


An infinite, uniform line charge with linear charge density λ = +5 µC/m is placed along the symmetry axis (z-axis) of an infinite, thick conducting cylindrical shell of inner radius a = 3 cm and outer radius b = 4 cm. The cylindrical shell has zero net charge.

The electrical potential is chosen to be zero at the outer surface of the cylindrical shell, V(b) = 0. (In this problem, it is not possible to chose the potential to be zero at infinity because the charge distribution extends to infinity.)

(c) Calculate the potential at a radial distance r = a/2.

???


Homework Equations



Vp-Vref = -∫E•dl = 2kλ*ln(Rref/R)

The Attempt at a Solution



Since the point b = 0.4 is the reference point I substituted that into the equation. Here is what I did:

R = 0.03/2 = 0.015

V = 2*(8.99x10^9)*(5x10^-6)*ln(0.04/0.015) = 88176.54985V

What did I miss?
 

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,303
998
For a < r < b, E(r) = 0 .

∴ [tex]\int_{r=b}^{r}\vec{E}\cdot \vec{d\ell}=\int_{r=b}^{r=a}\vec{0}\cdot\vec{d\ell}+\int_{r=a}^{r}\vec{E}(r)\cdot\vec{d\ell}[/tex]

The non-zero part of the integral begins at r=a.
 
  • #3
99
0
I understand where the potential is zero but I cannot figure out the potential at r=a/2. Did I calculate it wrong?
 
  • #4
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,303
998
I understand where the potential is zero but I cannot figure out the potential at r=a/2. Did I calculate it wrong?
Yes, it's wrong.

I showed where your "formula": Vp-Vref = -∫E•dl = 2kλ*ln(Rref/R) comes from.

So, it's as if Rref = a, not b. Technically it's because V(b) = V(a).
 

Related Threads on Electric potential on an infinite line of charge

  • Last Post
Replies
4
Views
15K
Replies
3
Views
718
Replies
2
Views
9K
  • Last Post
Replies
2
Views
567
Replies
1
Views
1K
Replies
1
Views
21K
  • Last Post
Replies
6
Views
2K
Replies
6
Views
2K
  • Last Post
Replies
2
Views
347
Top