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Electric Dipole Electric Field.. 
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#1
Sep110, 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 xcomponent, 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
Sep110, 06:13 PM

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P: 40,740

You are looking for the net electric field as a function of the distance out on the xaxis. What direction is the resultant E field pointing? 


#3
Sep110, 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 xcomponent? And then, how do you go about solving for it? 


#4
Sep110, 06:25 PM

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P: 40,740

Electric Dipole Electric Field..
On the figure shown, go to a point out on the xaxis to the right, and draw the two Efield vectors for the two point charges (only at that point). What would you get when you add those two vectors? 


#5
Sep110, 06:41 PM

P: 62

This is what I've come up with.
but now I'm completely lost. 


#6
Sep110, 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 xaxis EYred = red*sin(theta) EXblue = blue*cos(theta) angle between the blue vector and the xaxis EYblue = blue*sin(theta) so you add the X and Y directions, the X cancels out, and youre left with only Y? 


#7
Sep110, 07:32 PM

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