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Finding electric potential given a cylindrical configuration

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

    You have two concentric cylinders. The inner cylinder has radius a and the external cylinder has radius b. Find the electric potential in the region between the cylinders.

    [Hint: The final equation takes the form V(r) = constant1 - constant2 ln(something) ]


    2. Relevant equations

    I think:
    Gauss's law: [itex]\Phi[/itex]=EA=q/ε
    Coulomb's law: E=q/(4πεr^2)

    3. The attempt at a solution

    I really don't know.
     
  2. jcsd
  3. Feb 5, 2012 #2

    vela

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    Start by finding the electric field E between the two cylinders using Gauss's Law.

    How is E related to the potential?
     
  4. Feb 6, 2012 #3
    Okay

    So

    EA=q/ε

    E=q/(εA)

    A=2πrL

    dA=2πLdr

    dE=q/(ε2πLdr)

    E = ∫(limits a and b) q/(ε2πLdr) = q/(ε2πL) ∫(limits a and b) 1/dr

    Am I on the right track? If so, how to evaluate ∫(1/dr) ?

    Thank you.
     
  5. Feb 6, 2012 #4
    Do you know the answer to this? I am just learning this material also, so I don't know if I am thinking through it correctly. But the outer shell should be irrelevant as the electric field is going to be zero for any point inside it assuming it is infinitely long, correct? For the electric field at r distance from the center of cylinder with radius a, would its magnitude be λ/(2πεr) (derived with Gauss's law)? If so, you would end up with Δv(r) = v(r) (taking v_i = 0) = λln(a)/(2πε) - λln(r)/(2πε) where a <= r <= b, λ = q/l. But I have no idea if that is correct.
     
  6. Feb 6, 2012 #5

    vela

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    No. First off, what's your Gaussian surface and what expression is equal to the amount of charge enclosed in it?
     
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