electric dipole in an electric field

jhess12

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.

Dick

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

jhess12

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?

Dick

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 -|p||E(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?

jhess12

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

Dick

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

jhess12

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

Dick

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.

jhess12

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?

Dick

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

jhess12

sorry dont bother

learningphysics

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

jhess12

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.

Dick

Sometimes having somebody else say the same thing makes all the difference. Sorry, I lost patience as well.