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Electric dipole in an electric field 
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#1
Oct507, 01:53 PM

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1. The problem statement, all variables and given/known data
A small object with electrc dipole moment [tex]\overrightharpoonup{p}[/tex] is placed in a nonuniform electric field [tex]\overrightarrow{E}[/tex] =E(x)[tex]\hat{i}[/tex]. That is, the field is in the x direction and its magnitude depends on the coordinate x. Let [tex]\theta[/tex] represent the angle between the dipole moment and the x direction. (a) Prove that the net force on the dipole is F=p([tex]\frac{dE}{dx}[/tex])cos[tex]\theta[/tex] acting in the direction of the increasing field. 2. Relevant equations U=pEcos[tex]\theta[/tex] p[tex]\equiv[/tex]2aq 3. The attempt at a solution im not asking for anyone to do the problem, but I dont even know where to start. if someone could just please maybe help me understand the problem better and help me get started i would much appreciate it. 


#2
Oct507, 02:15 PM

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You have the potential U. The force is the negative of the gradient of the potential. Is that a good starting point?



#3
Oct507, 03:15 PM

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so F=U, so F=pE[tex]cos\theta[/tex], and (dE/dx) is the direction of the electric field as it changes with the coordinate x because the electric field is nonuniform right?



#4
Oct507, 03:23 PM

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Electric dipole in an electric field
The potential is a scalar function. It's p.E ('.'=dot product). In your case since the directions of the vectors are fixed, you can as you have, write this as pE(x)cos(theta). So U is a function of x. F is not equal to U. It's equal to minus the GRADIENT of U. How do you compute a gradient?



#5
Oct507, 03:38 PM

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i just dont know. i think that it might be E=[tex]\delta[/tex]V/[tex]\delta[/tex]x



#6
Oct507, 03:41 PM

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The gradient of a function U is (dU/dx,dU/dy,dU/dz) (where the derivatives are partial derivatives).



#7
Oct507, 03:57 PM

P: 11

im sorry, i just dont understand, i see all these equation for W(work)=U and i can see that there should be some way for me to solve this problem because as you have explained it to me, it actually seems very simple, except that i dont know how to put the gradient into the equation. i know that W=[tex]\int[/tex]F.ds=[tex]\int[/tex]qE.ds where F and E and ds are vectors but...im sorry



#8
Oct507, 04:30 PM

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Ok, so if W=integral(F*ds) then dW/ds=F. Or F=dU/ds. Apply that to this problem. In three dimensions you want to think of a gradient rather than a simple derivative, but if that is driving you crazy, forget about it for now. Treat it as a one dimensional problem, but afterwards think about why dU/dy=0 and dU/dz=0 mean F_y=0 and F_z=0.



#9
Oct607, 12:18 AM

P: 11

are du/dy=0 and du/dx=0 because they are perpendicular to the electric field? but i think i understand some. and since i am trying to find F_x right? then i would evaluate W=[tex]\int[/tex]pEsin[tex]\theta[/tex]find the derivative and i would have my answer right? i worked it out except for the p part of the equation. i know that p is a constant so to find the derivative?



#10
Oct607, 12:56 AM

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I give up. Can somebody else take this post please? I really give up.



#11
Oct607, 01:33 AM

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sorry dont bother



#12
Oct607, 02:11 AM

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Given the potential U... the force is:
[tex]\vec{F} = \bigtriangledown{U}[/tex] In other words: Fx = dU/dx. Fy = dU/dy. Fz = dU/dz. Once you know U, you can immediately get the force... What is U? As Dick mentions: [tex]U = \vec{p}\cdot\vec{E}[/tex] so [tex]U = (pcos(\theta), psin(\theta),0)\cdot (E(x),0,0)[/tex] so that gives [tex]U = E(x) pcos(\theta)[/tex] now you can directly get the components of the force using Fx = dU/dx. Fy = dU/dy. Fz = dU/dz. the important thing here is that E(x) is only a function of x... it is independent of y and z. So you should be able to get your result directly using Fx = dU/dx 


#13
Oct607, 02:30 AM

P: 11

thank you so much, im sorry i didnt get it before, now that you explain it, it seems so simple like i thought, i just dont know why i didnt get it before. sorry dick.



#14
Oct607, 12:07 PM

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