Electric Dipole Electric Field

  1. Electric Dipole Electric Field.. URGENT

    1. The problem statement, all variables and given/known data

    For the electric dipole shown in the figure, express the magnitude of the resulting electric field as a function of the perpendicular distance x from the center of the dipole axis in terms of the electric dipole moment, p.

    [​IMG]


    2. Relevant equations

    E = Fq
    P = qd
    E = kq/(r^2)


    3. The attempt at a solution

    R = sqrt[(d/2)^2 + x^2]
    q = P/d

    Thus,

    E = kP/{d*[sqrt(d/2)^2+x^2]^2}

    Since I'm looking for the x-component, multiply E by the cos(theta), which in this case is x/R, or...

    cos(theta) = x/sqrt[(d/2)^2+x^2]

    so the answer i get is....

    E = kP/{d*[sqrt(d/2)^2+x^2]^2} * x/sqrt[(d/2)^2+x^2]

    which comes out to be...

    E = kP/[sqrt(d/2)^2+x^2]^(3/2) * x/d



    The final answer is actually...

    E = kP/[sqrt(d/2)^2+x^2]^(3/2)

    So my question is, what happens to the x/d?
     
  2. jcsd
  3. berkeman

    Staff: Mentor

    Re: Electric Dipole Electric Field.. URGENT

    You are not "looking for the x-component" (Quiz Question -- why not?)

    You are looking for the net electric field as a function of the distance out on the x-axis. What direction is the resultant E field pointing?
     
  4. Re: Electric Dipole Electric Field.. URGENT

    The resultant E field would be a curved path, clockwise from the + to the - charge.

    We're looking for the E field as a function of the perpendicular distance, but what does that mean if it's not the x-component?

    And then, how do you go about solving for it?
     
  5. berkeman

    Staff: Mentor

    Re: Electric Dipole Electric Field.. URGENT

    They ask only for the E-field on the x-axis. That means you will solve for E(x). You do that by going to a point x on the x-axis, and adding up the two E-field vectors from the two point charges. Some of those vector components will cancel when they are added, and others will give a non-zero resultant.

    On the figure shown, go to a point out on the x-axis to the right, and draw the two E-field vectors for the two point charges (only at that point). What would you get when you add those two vectors?
     
  6. Re: Electric Dipole Electric Field.. URGENT

    This is what I've come up with.

    [​IMG]

    but now I'm completely lost.
     
  7. Re: Electric Dipole Electric Field.. URGENT

    This is what else I've come up with...

    EXred = red*cos(theta) --theta being the angle between the red vector and x-axis
    EYred = red*sin(theta)

    EXblue = blue*cos(theta) --angle between the blue vector and the x-axis
    EYblue = blue*sin(theta)

    so you add the X and Y directions, the X cancels out, and youre left with only Y?
     
  8. berkeman

    Staff: Mentor

    Re: Electric Dipole Electric Field.. URGENT

    Very good! Now use the formula for E-field as a function of distance (the distance to the point x for each is the hypoteneus of each triangle, right?), and add up the y-components. You're almost there. Be sure to express the sum as a vector, saying which direction in y the resultant points.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?