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Electric Dipole Electric Field..

by pleasehelpme6
Tags: dipole, electric, field, urgent
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pleasehelpme6
#1
Sep1-10, 06:06 PM
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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berkeman
#2
Sep1-10, 06:13 PM
Mentor
berkeman's Avatar
P: 40,932
Quote Quote by pleasehelpme6 View Post
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?
pleasehelpme6
#3
Sep1-10, 06:18 PM
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?

berkeman
#4
Sep1-10, 06:25 PM
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P: 40,932
Electric Dipole Electric Field..

Quote Quote by pleasehelpme6 View Post
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?
pleasehelpme6
#5
Sep1-10, 06:41 PM
P: 62
This is what I've come up with.



but now I'm completely lost.
pleasehelpme6
#6
Sep1-10, 06:44 PM
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?
berkeman
#7
Sep1-10, 07:32 PM
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berkeman's Avatar
P: 40,932
Quote Quote by pleasehelpme6 View Post
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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