# Homework Help: Electric Dipoles and potential energy

1. Sep 6, 2008

### StephenDoty

Consider an electric dipole whose dipole moment is oriented at angle theta with respect to the y axis. There is an external electric field of magnitude E pointing in the positive y direction. The positive and negative ends of the dipole have charges + q and - q, respectively, and the two charges are a distance d apart. The dipole has a moment of inertia I about its center of mass. (See Picture)

Using torque = qEdsin(theta), find the potential energy U(theta) associated with the dipole's orientation in the field as a function of the angle theta shown in the figure. Take the zero of the potential to occur when the dipole is at angle pi/2; that is, U(pi/2) = 0.
Express U(theta) in terms of the given quantities.

torque= I * angular acceleration
= I*$$\omega$$ d$$\omega$$/d$$\theta$$
integrate both sides =
w=$$\sqrt{-qEd/I * cos(pi/2- theta 0)}$$

Now what do I do?

Thanks.
Stephen

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2. Sep 6, 2008

### jackiefrost

Imagine drawing a longer line thru the dipole that extends out past the two charges. Draw an E field vector at the center of q+ of some convenient length pointing in the direction of +y. Do the same for -q (only pointing in the -y direction). Project the E vector at +q onto the dipole extension line; i.e. draw E's components in the dir of the dipole line and normal to the line.

Now - looking at the drawing:

1. What does ESin(theta) represent?
2. Then, what is qESin(theta)?
3. Pick a new angle for the dipole, say halfway back to the y axis. What is happening to qESin(theta) as the dipole swings toward the y axis? How is it related to the motion of charge q+?
4. Now, imagine that the dipole is set free to rotate. At regular intervals of arc length along its circular swing a magical instrument of your own design records the torque producing (tangential) force that E is continuously applying to q+. Imagine it supplies 1000 force readings between the starting angle and when it is parallel to the y axis.
5. Since you know that the total arc length was being broken up into 1000 little arc lengths, one for each force reading, and you now possess the associated 1000 tangential force readings, what could you do with that force and distance data to give you an approximation of the original potential energy U?

OK, so now you have an approximation for U(theta). Make it a better approximation.

Suppose you try it again from the same starting angle. This time you set the instrument to take 10,000 readings at 10,000 regular distances of arc travel. Then do it for 100,000. A million! Notice what is happening to the approximated U? How can you mathematically perform the same process but for an infinite number of readings, breaking the arc into an infinite number of differential arcs lengths of travel and thereby obtain something better than an approximation?

[Don't forget - the same thing is happening at q- so your findings need a minor adjustment]

Now, if all has gone well, you should have an expression for U(theta).

jf

3. Sep 6, 2008

### StephenDoty

Esin(theta) becomes smaller as the dipole moves to the positive y axis. qEdsin(theta) is the torque. Would you integrate torque?

4. Sep 6, 2008

### jackiefrost

Sorry - I didn't originally see the d in qEdsin(theta).

The E component of torque producing force for one charge is the qEsin(theta) part of the expression. For each charge that qEsin(theta) force is acting at distance of d/2 from the center of rotation so each torque is qEsin(theta)*d/2. So, for the combined charges its twice that or qEdsin(theta).

So, the torque is a function of angle.

t(theta) = qEdsin(a)

The potential energy is the amount of work that the E field will do in rotating the dipole into parallel alignment with the E field (at angle theta = 0). That work (and therefore the potental energy) is obtained by integrating the product of the torque as a function of angle, t(theta), and the differential element of angular displacement, dtheta. That's equal to the integral of qEdsin(theta)dtheta. Notice that q, E, and d, are constants.

jf

5. Sep 7, 2008

### StephenDoty

the integral of qEdsin(theta) dtheta
=
qEd (-cos(theta))

what would the limits of integration be?

6. Sep 7, 2008

### jackiefrost

As shown in the diagram, theta refers to the angle in relation to the E field direction. So, if we call the initial angle 'a', the limits are from theta=a to theta=0.

Notice something else here. The wording of the problem refers to U(Pi/2)=0 and asks you to find U(theta) which raises an ambiguity. That wording implies that theta is in relation to the X axis, NOT the E field. The angle "theta" we're using is the angle "theta" as shown in the diagram; i.e. relative to the E field. In that case the angle when the dipole is parallel with the E field (and the Y axis) and U(theta)=0 is theta=0, not theta=Pi/2.

jf

7. Sep 7, 2008

### hyperon

Perhaps that's exactly what the question is asking for, it wants you to take U(π/2) = 0 where the dipole is parallel with the x-axis. Then U will be most negative when θ = 0. The limits would then be from wherever U=0 to the current position, from θ = π/2 to θ, if the integral used is U = - int(τ(θ)dθ).

8. Sep 7, 2008

### jackiefrost

It's a bit confusing. The drawing shows theta as the angle the dipole makes with the E field (parallel with the Y axis). If the dipole was lined up parallel to the E field where +q is toward +Y, U should be zero there. According to the diagram, that would be theta = 0. If they are now going to call that angle theta = pi/2 then the angle shown in the diagram should have been (pi/2 - theta).

Or, am I missing something here?

jf