1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Potential due to a rod

  1. Oct 29, 2005 #1
    A rod of length L lies along the x-axis with the left end at the origin. It has non-uniform charge density given by [tex]\lambda = \alpha x[/tex]. where [tex]\alpha[/tex] is a positive constant. Calculate the potential at a position A, which is a distance d to the left of the origin.

    Here is what i did:

    we have [tex] dV = \frac{k_e dq}{r}[/tex]

    then i made use of the [tex]\lambda = \alpha x[/tex]

    and we have [tex]V=\int\frac{k_e\lambda dx}{x} = \int\frac{k_e\alpha x dx}{x} = k_e\alpha (d+L-d)[/tex] where my limits of integration were L to d+L in the second integral

    however the answer provided says that [tex]V=k_e\alpha (L-d ln(1 + \frac{L}{d})[/tex].

    where did i go wrong?

    thanks
     
    Last edited: Oct 29, 2005
  2. jcsd
  3. Oct 29, 2005 #2

    Astronuc

    User Avatar

    Staff: Mentor

    [tex]V=\int\frac{k_e\lambda dx}{x} = \int\frac{k_e\alpha x dr}{r}[/tex]

    then determine the relationship between r and x, i.e. r = r(x), which will also determine the relationship between dr and dx.

    x is the distance along the rod, r is the separation between the incremental charge dq and the point at which the potential is being measured.
     
  4. Oct 29, 2005 #3
    so i should put r = d + x???
     
  5. Oct 29, 2005 #4

    Astronuc

    User Avatar

    Staff: Mentor

    Given: A rod of length L lies along the x-axis with the left end at the origin.

    Calculate the potential at a position A, which is a distance d to the left of the origin.

    Perhaps one can draw a diagram.

    x [0,L]
    d to left of origin
    r [d,L+d] = d+x

    so what is dr if 'd' is constant?
     
  6. Oct 30, 2005 #5
    if d is constant dr is just d+x, this is correct right? and i have a diagram that came with the question but i still dont see what to put into the integral or how...
     
  7. Oct 30, 2005 #6
    ok i got it it was really simple actually.
     
  8. Oct 30, 2005 #7

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    Incidentally, it's not the "potential due to a rod." It's the electrical potential due to electrical charge that happens to reside on a rod.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?