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Electric potential of a rod

  1. Aug 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Here's the problem:

    pmyig.png


    2. Relevant equations

    V = ke ∫ dq/r
    V is the electric potential, ke Coulomb's constant, q the charge and d the distance.

    λ = q / L , where λ is the charge density, q the charge and L the length of the rod.

    3. The attempt at a solution

    I have one solution for the problem. What I want to know is why is my answer incorrect.
    Since electric potential is a scalar and not a potential what I calculated was the electric potential produced by the left part of the rod, from 0 to L/2, and multiplied the resulted for two due to the symmetry of the problem (I thought that the electric potential produced by the rod from L/2 to L was the same as from 0 to L/2)

    q = λ*L ⇔ dq = λ*dx ⇔ dq = α*x*dx

    V = 2*ke*α ∫0L/2 x/sqrt(x²+b²) dx

    I solved the integral and got:

    V = 2*α*ke*[ sqrt( (L/2)² + b² ) - b ]

    It is incorrect tho.
    Any help will be appreciated!
    Thanks in advance.
     
  2. jcsd
  3. Aug 5, 2012 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Read the problem text: λ=αx, the left end of the rod is at x=0. There is no symmetry.

    ehild
     
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