Potential Energy of an Electric Dipole in a Uniform Field

[rtareen]

I have a lot of questions about this single concept. You don't have to answer the questions in the order that I ask, if it is convenient to answer them in a different order.

1. When the dipole moment $\vec{p}$ is in the same direction as the electric field (uniform) it has the least potential energy. To me this makes sense because there is no torque acting on the dipole at that moment and it (probably) has the most kinetic energy at that time, like a pendulum. But then when the direction of the moment is exactly opposite that of the field, we define it to have the greatest potential energy. This doesn't make sense because the dipole is in exactly the same situation as when it is in the same direction as the field - there is no torque acting on it? Why when the moment is perpendicular to the field does it not have the greatest potential energy - that is when the biggest torque is applied.

2. When deriving the expression for potential energy we use U = - W. We have that
$-W = -\int_{90}^{\theta}\tau~d\theta = - -\int_{90}^{\theta}(-pEsin\theta)d\theta$

I don't understand why we add the extra minus sign, because in the actual expression for torque there is no minus sign.

3. I have an idea as to why we start with a lower limit of 90 - we want that orientation to have zero potential energy, but what's really going on here? I don't completely understand it.

4. Another thing is that the least potential energy, when we derive it this way is -pE and the max is +pE. This is apparently the most convenient way to define it. But to fully understand something we should do it in multiple ways. What if I wanted to make it go from 0 to 2pE?

Edit: The uniform field is an electric field

 

[kuruman]

1. If the dipole is exactly antiparallel to the E-field, the torque is indeed zero as is the gravitational torque on a small sphere resting at the very top of a larger sphere. If you give a teeny-tiny nudge to the dipole or the small sphere, they will accelerate and gain quit a bit of kinetic energy. Where did that kinetic energy come from? Answer: From the potential energy that is at is highest.

2. You have too many negative signs in front of the last integral. The torque $\tau$ is restoring, i.e. it opposes an increase in angle (and $\sin\theta$). If the angular displacement is positive/negative the torque must be negative/positive. It's just like the spring restoring force $F=-kx$.

3. We start with the lower limit as $\pi/2$ because we choose to define the potential energy as zero when the dipole is perpendicular to the electric field. We could have chosen any angle we please to define the zero of potential, but that would complicate matters. See my answer in 4.

4. You can define the zero of potential as you please. If you define it to be zero at $\theta=0$, then the potential energy will be $U=pE-\vec p\cdot \vec E$. Someone else might decide to add $4.286~\mathrm {J}$ to that and have $U=pE-\vec p\cdot \vec E+4.286~\mathrm {J}.$ My personal preference is the simplest form, but you are free to make your life as complicated as you desire.

 

[rtareen]

1. If the dipole is exactly antiparallel to the E-field, the torque is indeed zero as is the gravitational torque on a small sphere resting at the very top of a larger sphere. If you give a teeny-tiny nudge to the dipole or the small sphere, they will accelerate and gain quit a bit of kinetic energy. Where did that kinetic energy come from? Answer: From the potential energy that is at is highest.

You didn't contrast this case with the opposite case, when the potential energy is at its lowest. I understand that if the uniform field points towards the right if the moment is antiparallel to it and then the positive charge is nudged up a few degrees the field will push both sides of the dipole in the counterclockwise direction. So that's where the potential energy comes from. On the other hand if the moment is parallel to the field and the negative charge is nudged up then the field pushes the positive charge in the counterclockwise direction and pushes the negative charge in the clockwise direction. So they cancel so it essentially a stable equilibrium, and that's why it has the least potential energy. Let me know if I'm right.

[/QUOTE]

2. You have too many negative signs in front of the last integral. The torque $\tau$ is restoring, i.e. it opposes an increase in angle (and $\sin\theta$). If the angular displacement is positive/negative the torque must be negative/positive. It's just like the spring restoring force $F=-kx$.

I understand that the torque goes in the opposite direction as the angular displacement. But that's only true for half of the circle. In the opposite orientation(starting antiparallel and then nudging the negative charge upwards) the torque will be in the same direction as the angular displacement, due to the reasoning above. I don't think I have too many negative signs I just didn't cancel them out to highlight that my book adds a negative sign to the torque expression. I don't understand mathematically what the purpose of this is. Thanks for answering!

 

[kuruman]

You didn't contrast this case with the opposite case, when the potential energy is at its lowest. I understand that if the uniform field points towards the right if the moment is antiparallel to it and then the positive charge is nudged up a few degrees the field will push both sides of the dipole in the counterclockwise direction. So that's where the potential energy comes from. On the other hand if the moment is parallel to the field and the negative charge is nudged up then the field pushes the positive charge in the counterclockwise direction and pushes the negative charge in the clockwise direction. So they cancel so it essentially a stable equilibrium, and that's why it has the least potential energy. Let me know if I'm right.

I understand that the torque goes in the opposite direction as the angular displacement. But that's only true for half of the circle. In the opposite orientation(starting antiparallel and then nudging the negative charge upwards) the torque will be in the same direction as the angular displacement, due to the reasoning above. I don't think I have too many negative signs I just didn't cancel them out to highlight that my book adds a negative sign to the torque expression. I don't understand mathematically what the purpose of this is.
Thanks for answering!

Without a picture I cannot understand what "up" is or how the dipole is oriented relative to the field.

The idea is simple:
Case I. When the dipole moment vector is in the same direction as the electric field it has minimum potential energy. It is in stable equilibrium because if you turn it by a small angle in either direction, the electric torque will bring it back towards equilibrium. A mechanical analogue is a mass at the end of the rod forming a pendulum. The minimum potential energy is when it is hanging straight down. If you displace it by a small angle, it returns to the equilibrium vertical position.

Case II. When the dipole vector is antiparallel to the electric field the dipole is in unstable equilibrium. Nominally the torque is zero but even the smallest displacement will turn on the restoring torque, restoring in the sense that it brings it back to the equilibrium position, not to the angle from which it started. The mechanical analogue it to have the pendulum rod along the vertical but with the mass above the point of support. It won't stay there for long.

The derivation of the potential energy is simple. First you find the work done by the electric torque when when you rotate it from the reference angle, where the electric potential energy is zero, to some arbitrary angle $\theta$, $$W=\int_{\pi/2}^{\theta}\tau~d\theta=\int_{\pi/2}^{\theta}(-pE\sin\theta)~d\theta=pE\cos\theta$$By definition, the potential energy function is the negative of this work. $$U=-W=-pE\cos\theta=-\vec p\cdot\vec E.$$

 

[Steve4Physics]

Summary:: Conceptual questions and questions about the derivation for the potential energy of a dipole in a uniform field.

I have a lot of questions about this single concept. You don't have to answer the questions in the order that I ask, if it is convenient to answer them in a different order.

1. When the dipole moment $\vec{p}$ is in the same direction as the electric field (uniform) it has the least potential energy. To me this makes sense because there is no torque acting on the dipole at that moment and it (probably) has the most kinetic energy at that time, like a pendulum. But then when the direction of the moment is exactly opposite that of the field, we define it to have the greatest potential energy. This doesn't make sense because the dipole is in exactly the same situation as when it is in the same direction as the field - there is no torque acting on it? Why when the moment is perpendicular to the field does it not have the greatest potential energy - that is when the biggest torque is applied.

2. When deriving the expression for potential energy we use U = - W. We have that
$-W = -\int_{90}^{\theta}\tau~d\theta = - -\int_{90}^{\theta}(-pEsin\theta)d\theta$

I don't understand why we add the extra minus sign, because in the actual expression for torque there is no minus sign.

3. I have an idea as to why we start with a lower limit of 90 - we want that orientation to have zero potential energy, but what's really going on here? I don't completely understand it.

4. Another thing is that the least potential energy, when we derive it this way is -pE and the max is +pE. This is apparently the most convenient way to define it. But to fully understand something we should do it in multiple ways. What if I wanted to make it go from 0 to 2pE?

Edit: The uniform field is an electric field

Hi. See if this helps.

The orientations with zero torque do *not* have the same potential energy.

Think of a lead ball attached to one end of a thin rod, standing on a table.
With the rod vertical and the ball at the bottom, you have stable equilibrium (zero torque, min. potential energy).
With the rod vertical and the ball on top, you have unstable equilibrium (zero torque, max. potential energy)

Uniform electric field, E, to the right:
[-q +q] Call this orientation of the dipole ‘S’ (stable).
[+q -q] Call this orientation of the dipole ‘U’ (unstable).

If d is dipole’s length, changing from S to U means moving each charge a distance d, so |work done| on each charge = (qE)d. Total work done = 2qEd. This equals the change in potential energy (=2pE where p = qd).

In your equation W=-U, W is the work done *BY* the dipole. If W’ the work done by an externally applied torque *ON* the dipole then:
W’ = -W = U

The choice for zero potential energy is arbitrary. It is a convention that we take the zero at 90º. That makes the minimum potential energy -pE and the maximum +pE which is nice and symmetrical mathematically.

If you want to take the stable orientation (θ =0) as zero potential, then you simply use zero as your lower limit for integration. Integrating from 0 to 180º will give you 2pE as required.

 

[rtareen]

Now things are a little clearer for me. I see that the difference between parallel and anti-parallel is stable and unstable. And its also a little clearer that the difference between work done by the field and and an external torque is a minus sign. Thanks a lot to both of you!

I love physics forums.

 

[rtareen]

Hey guys I made some diagrams trying to understand why antiparallel has the most potential energy while parallel has the least. From these torque diagrams as far as I understand it should be the opposite. When the moment is parallel to the field (the field points to the right) and then nudged up or down (1) and (3) the torques are in the direction of the angular displacement meaning it will speed up in that direction, meaning it is unstable. And when the moment is antiparaellel and the nudged up or down the torque acts as a restoring force meaning it is stable. So why is it considered the opposite? Can someone explain?

 

[rtareen]

Hey guys I made some diagrams trying to understand why antiparallel has the most potential energy while parallel has the least. From these torque diagrams as far as I understand it should be the opposite. When the moment is parallel to the field (the field points to the right) and then nudged up or down (1) and (3) the torques are in the direction of the angular displacement meaning it will speed up in that direction, meaning it is unstable. And when the moment is antiparaellel and the nudged up or down the torque acts as a restoring force meaning it is stable. So why is it considered the opposite? Can someone explain?

I guess it depends on the kind of field? I assumed a field set up by positive charges. But apparently we are talking about a field set up by a negative charge. Is that the reason I got the opposite result?

 

[rtareen]

Wow nevermind I think I figured it out. The field always assumes a positive test charge and thus points in the direction the postive charge would go. I am so stupid

 

[kuruman]

Wow nevermind I think I figured it out. The field always assumes a positive test charge and thus points in the direction the postive charge would go. I am so stupid

I disagree with the assessment of yourself. You figured it out on your own and that's worth a lot towards building confidence in yourself and what you can do.

 

[rtareen]

I disagree with the assessment of yourself. You figured it out on your own and that's worth a lot towards building confidence in yourself and what you can do.

Thanks a lot for your kind words! and also for your help before. I'm so happy I finally understand this part.

 

[vanhees71]

Maybe it's also worth to look at this from a quite fundamental point of view and build the dipole as the limiting field of two charges $-Q$ and $+Q$ ($Q>0$) that are close to each other (somehow bound together like a hydrogen atom as a bound state of an electron and proton). Let these charges be in an external electrostatic field,
$$\vec{E}(\vec{x})=-\vec{\nabla} \Phi(\vec{x}).$$
The potential energy of the charges in this field (neglecting the interaction energy between the charges) is
$$V=Q [\Phi(\vec{x}_1)-\Phi(\vec{x})_2],$$
where $\vec{x}_1$ and $\vec{x}_2$ are the positions of the charge $+Q$ and $-Q$ respectively. Since now the charges are supposed to be bound together at a pretty small distance we can Taylor expand the first expression around $\vec{x}_2$, i.e., $\vec{x}_1=\vec{x}_2 +(\vec{x}_1-\vec{x}_2)$:
$$\Phi(\vec{x}_1)=\Phi(\vec{x}_2)+(\vec{x}_1-\vec{x}_2) \cdot \vec{\nabla} \Phi(\vec{x}_2).$$
Then we get
$$V=Q(\vec{x}_1-\vec{x}_2) \cdot \vec{\nabla} \Phi(\vec{x}_2).$$
Now the dipole approximation is to make $\vec{x}_1-\vec{x}_2 \rightarrow 0$ but keeping $\vec{P}=Q(\vec{x}_1-\vec{x}_2)=\text{const}$. Setting $\vec{x}_2=\vec{x}$ you get the potential of a dipole
$$V(\vec{x})=\vec{P} \cdot \vec{\nabla} \Phi(\vec{x})=-\vec{P} \cdot \vec{E}.$$
Note that by definition $\vec{P}$, the dipole moment, is pointing from the negative to the positive charge within the dipole.

Now the force on the dipole is given by
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x})=\vec{\nabla} (\vec{P} \cdot \vec{E}).$$
For your case of a homogeneous em. field this vanishes.

To get the intrinsic torque take the position of the $-Q$ as the reference point. Then the torque is simply the torque of the charge $+Q$ in em. field:
$$\vec{\tau}=Q(\vec{x}_1-\vec{x}_2) \times \vec{E}(\vec{x}_1)$$
Now taking the dipole limit as above you simply get
$$\vec{\tau} = \vec{P} \times \vec{E}.$$