# Electric Dipole Electric Field..

Tags: dipole, electric, field, urgent
 P: 62 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. 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?
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P: 41,127
 Quote by pleasehelpme6 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. 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?
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?
 P: 62 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?
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P: 41,127
Electric Dipole Electric Field..

 Quote by pleasehelpme6 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?
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?
 P: 62 This is what I've come up with. but now I'm completely lost.
 P: 62 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?
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P: 41,127
 Quote by pleasehelpme6 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?
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.

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