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Intergration Dipole

  1. Nov 18, 2009 #1
    1. The problem statement, all variables and given/known data

    We know the magnitude of the electric field at a location on the x axis and at a location on the y axis, if we are far from the dipole.

    (a) Find [tex]\Delta[/tex]V= V_p - V_a along a line perpendicular to the axis of a dipole. Do it two ways: from superposition of V due to the two charges and from the integral of the electric field.

    (b) Find [tex]\Delta[/tex]V = V_c - V_d along the axis of the dipole. Include the correct signs. Do it two ways: from the superposition of V due to the two charges and from the integral of the electric field.


    2. Relevant equations

    V_q = (1/4[tex]\pi[/tex][tex]\epsilon[/tex]_0)(+or-q/r)
    k= 1/4[tex]\pi[/tex][tex]\epsilon[/tex]_0

    3. The attempt at a solution

    (a)
    superposition
    [tex]\Delta[/tex]V= (1/4[tex]\pi[/tex][tex]\epsilon[/tex]_0)(+q/(d^2+(s^2)/4))^(1/2))-(1/4[tex]\pi[/tex][tex]\epsilon[/tex]_0)(-q/(d^2+(s^2)/4))^(1/2))

    [tex]\Delta[/tex]V= 0

    BUT
    i dont kno how to do integral
    i think its

    integral from p to a : E times dd

    how do i do it?

    i thinks its....

    int from a to b: kq/r^2

    and i use formula:
    int of 1 / x^2 + a^2 dx = (1/a)tan^-1 (x/a)

    but im not sure how to use it or show the answer


    (b) superposition



    [tex]\Delta[/tex]V = V_c - V_d
    = (E_c)a - (E_d)b = 2kqsa/(a^3) -2kqsb/(b^3)
    = ((sq)/(2[tex]\pi[/tex][tex]\epsilon[/tex]_0))(1/(a^2)-1(b^2))

    but again

    i dont kno how to do integration

    i havent got a clue for part B
     

    Attached Files:

    Last edited: Nov 18, 2009
  2. jcsd
  3. Nov 18, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    At the location on the y-axis
    due to +q charge
    E= k*q/r^2
    E = -dV/dr
    dV = - E*dr . V = - k*q* intg(1/r^2)*dr
    Find the integration. Similarly find V due to - q. Then find net V.
    Follow the same method to find V along x-axis.
     
  4. Nov 18, 2009 #3
    wats the R for V_a?

    wats the R for V_p?

    are they different?

    i think the R for V_a is ((s/2)^2 + d^2 ) )^ (1/2)

    but for V_p....

    is it just s/2?

    if thats true i dont think i will get 0 for [tex]\Delta[/tex]V = V_p - V_A
     
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