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Potential difference of a conducting pipe

  1. Mar 4, 2012 #1
    1. The problem statement, all variables and given/known data

    A conducting cylindrical pipe has an inner radius Ra and an outer radius Rb. a charge q is released inside it. Assume the length of a cylinder L>>Rb. Find the potential difference between the axis of the cylinder and outer surface.

    2. Relevant equations

    Vb-Va=-∫E*dl (vector multiplication and the integral is bounded between a and b)

    3. The attempt at a solution
    I found the electric field in the outer surface to be E=q/ε2πRL
    Vb-Va= q/ε2πL ∫1/rdr= q/ε2πL (lnRb-lnRa) (note: the integral is definite between Ra and Rb)
    Since the question asks for the potential difference between the axis and the outer surface I need to run R between 0 and R, so i get a ln(0) which does not make sense...

    Pleaseeeeeeeeee helppppp
     
  2. jcsd
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