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Non uniform charge density and electric potential

  1. Jun 12, 2012 #1
    As shown in the figure, a rod of length 9.8 m lies along the x-axis, with its left end at the origin. The rod has a non-uniform linear charge density λ = αx, where α = 0.009 C/m2 and x is the position. Point A lies on the x-axis a distance 3.59 m to the left of the rod, as shown in the figure.
    Q is the total charge on the rod. The Coulomb constant is 8.988 × 109 N m2/C2.

    Find the potentia at A, and answer in units of volts




    the two relevant equations i can find are:
    v=kq/r
    dv=kdq/r and lambda=qx?


    i am completely stuck on this problem and just need to find a way to start it. I though that you might be able to integrate the charge density formula over the total area of the rod and treat that value as Q and then plug that value into v=kq/r where r is the 3.59, but apparently that is worng and i am back to square one with no idea as to what to do
     
  2. jcsd
  3. Jun 12, 2012 #2

    ehild

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    The potential is additive, you need to integrate the contribution dV of a small segment of the rod dx with charge dq=λdx for the whole rod:
    V(A)=∫(kdq/r)=∫(kλdx/r).

    ehild
     

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    Last edited: Jun 12, 2012
  4. Jun 12, 2012 #3
    wait so then what would be the values for the r? cause i have been using just 3.95 since thats the distance of your point away from the rod but it doesn't work
     
  5. Jun 12, 2012 #4
    Maybe this will help: r = 3.59 + x
     
  6. Jun 12, 2012 #5
    ok, so then in the last integral you will have a ln, but would that be the same that you integrate over?
     
  7. Jun 12, 2012 #6
    This equation is wrong. It should be dq = [itex]\lambda[/itex]x dx.

    Integrate this over the total length of the rod to get [itex]\lambda[/itex] expressed in terms of q. Then integrate dv=kdq/r to get total v at A.
     
  8. Jun 12, 2012 #7
    i know that you have to integrate dq over the rod, what I'm saying is what do you integrate dv over?
     
  9. Jun 12, 2012 #8

    ehild

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    dV is the contribution from dq to the potential at A (with respect to infinity). The integral of dV is simply V(A) the potential at A. Without charge, V is zero.

    ehild
     
  10. Jun 13, 2012 #9
    ok, so the integral for dv would be from 0 to 3.59 because you're integrating to the potential at A?
     
  11. Jun 13, 2012 #10

    ehild

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    dV is in volts, not in meters. dV is integrated over the potential belonging to zero charge to the actual potential V(A) of the total charge of the rod. The integral on the right-hand side goes along the length of the rod, from x= to x=9.8 m.
    You need to use the 3.59 m in the term "r", distance from the charge dq to A.

    ehild
     
    Last edited: Jun 13, 2012
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