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Homework Help: Voltage of cylinder of radius R

  1. Mar 5, 2014 #1
    ε1. The problem statement, all variables and given/known data
    A very long solid cylinder of radius R = 6 cm has:
    -uniform volume charge density ρ = +7 µC/m3.
    -linear charge density λ = 7.9168E-8 C/m

    Outside: What is the electric field at a radial distance of 7 cm from the axis of the cylinder?
    Inside: What is the electric field at a radial distance of 5 cm from the axis of the cylinder?

    Voltage: Setting V=0 at r=3R, find the voltage at the following locations:
    V (r = 0)
    V (r = R/2)
    V (r = R)
    V (r = 2R)
    V (r = 4R)

    2. Relevant equations

    using gauss law to find the electric field inside and outside the cylinder

    inside the cylinder E = ( ρ*r )/(2*εo)

    outside the cylinder E = ( λ )/(2*∏*r*εo)

    Voltage is just integration of E, so V=-∫Edr

    3. The attempt at a solution

    I find the electric field from the axis 0m to 0.07m using E = ( λ )/(2*∏*r*εo), i got 20338.948 N/C

    I find the electric field from the axis 0m to 0.05m using E = ( ρ*r )/(2*εo), i got 19773.9484 N/C

    The problem i have is to find the voltage from the asked distance to distance r=3R
    we set V at 3R = 0.

    i do the integration on E = ( ρ*r )/(2*εo) from r=0 to r=R to find the Voltage inside the cylinder

    then do integration on E = ( λ )/(2*∏*r*εo) from r=R to r=3R to find the voltage outside the cylinder.

    when it ask find the voltage at r=0, it means that i have to find the ΔV, which is from
    V(r=0) to v(r=3R), then i do separate integral on that two different E equation, then add them up. Since V at r=3R is 0, then i pretty much ignored it during the integration. but i got it wrong. how am i supposed to do this. i have try many method and i think this is the correctway to do it. can someone help me out or explaining it to me? Thank you for your time and reply

    Attached Files:

    Last edited: Mar 5, 2014
  2. jcsd
  3. Mar 5, 2014 #2
    For finding the potential at r = 0, you're right in having to add up the two integrals, one from r = 3R to r = R, then another from r = R to r = 0, but in your work there's only one. What happened to the other?
  4. Mar 5, 2014 #3
    yeah, thats what i thought but the webassign said my answer is wrong, so i wonder what is the way to do it then?

    so you say the method im doing it is correct isnt it?

    Or is be because the equation that i use is wrong?
    when r=R, do i use equation of r gauss < R cylinder or r gauss > R cylinder?

    if that is the way to do it, then might be the way i setting up the equation for the integral is wrong.

    what do you thnk?
  5. Mar 5, 2014 #4
    Yes, that method should be correct. Think about it like this:
    $$\Delta V = V(r) - V(3R)$$
    So this means that
    $$V(r) = V(3R) + \Delta V$$
    You know that V(3R) = 0, and also that you start integrating from 3R. So say to find the potential at r = 0:
    $$V(r = 0) = 0 - \int^{R}_{3R}E_{outside}dr - \int^{0}_{R}E_{inside}dr$$
    I think that's what you were doing -- you may have been a little off in the bounds or what field to use perhaps. And when r = R, you don't really pick between the fields -- R is when the field changes, so it's more that you switch the fields between the two integrals.
  6. Mar 5, 2014 #5
    i dont know what to do.
    I just do the easiest thing first, which is find the voltage V at r=R
    based on the integration, i have 2 answers.
    either 711.864 (integrated from r=0 to r=R) or 4005.527 (integrated from r=3R to r=R)
    i put down either answer, and its wrong....

    i find out that its weird, based on the graph of the voltage over radius,
    the voltage from r=0 to r=R should be greater than from r=R to r=3R.
    but based on the integral value, its the opposite.

    Attached Files:

  7. Mar 5, 2014 #6
    Okay, so you evaluated both integrals, correct? But the answer isn't either of those -- look at the expression I posted. Those integrals are added together, right? Also, talking about the voltage at R is misleading -- what that means is the potential with respect to zero (V(3R)), so only the value of V(R) - V(3R) is meaningful. V(0) - V(R) means the potential difference between 0 and R, but says nothing about the potential at R.
  8. Mar 5, 2014 #7
    I finally got it right. Apparently i was stupid to cross out the value during the integration just because they set the V=0 at r=3R. so i ended up get everything wrong.
    thank you for your help :D
    i appreaciate it
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