1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    [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

    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


    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Intergration Dipole Date
Energy dissipated in a resistor Jan 10, 2018
Hard intergration May 6, 2010
Problem with an intergral! Sep 27, 2005
Intergrate e^x^2 Sep 5, 2005