Potential Energy for dipole in Electric Field

In summary, the derivation of potential energy for a dipole in an electric field involves calculating the work done by an external agent in turning the dipole against the influence of the electric field. This is done by integrating the torque exerted by the field, which is equal to -pEsin(θ), where θ is the angle between the dipole moment and the electric field. The traditional derivation may have a sign error due to conflicting conventions for the direction of torque and angle.
  • #1
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1

Homework Statement



To derive Potential Energy for dipole p in Electric Field E.

2. Homework Equations


Potential Energy is the work done by the external agent in turning the angle of the dipole from the U=0 position to another position against the influence of the electric field applied right?

The Attempt at a Solution



So if Torque exerted by field for a particular θ is given by $$ \tau = pE\sin\theta $$
then when working out potential energy, should we not take the following:
τapp will act in same sense as dθ and opposite sense of τfield right?
So $$ \tau_{app} = - pE\sin\theta $$
And the potential energy is just
$$ U = \int_{\theta_1}^{\theta_2}\tau_{app}\,d\theta $$
$$ U = \int_{\theta_1}^{\theta_2}-pE\sin\theta\, d\theta $$
$$U=-pE(-\cos(\theta_2)+\cos(\theta_1))$$
Now if θ1 =π/2 and θ2
$$U=pE\cos(\theta)$$
$$U=p\cdot E$$
But the traditional derivation outputs $$-p\cdot E$$ and takes τfield and not τapp in the first step. Why is this the case?
 
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  • #2
Curieuse said:

Homework Statement



To derive Potential Energy for dipole p in Electric Field E.

2. Homework Equations


Potential Energy is the work done by the external agent in turning the angle of the dipole from the U=0 position to another position against the influence of the electric field applied right?

The Attempt at a Solution



So if Torque exerted by field for a particular θ is given by $$ \tau = pE\sin\theta $$
then when working out potential energy, should we not take the following:
τapp will act in same sense as dθ and opposite sense of τfield right?
So $$ \tau_{app} = - pE\sin\theta $$
And the potential energy is just
$$ U = \int_{\theta_1}^{\theta_2}\tau_{app}\,d\theta $$
$$ U = \int_{\theta_1}^{\theta_2}-pE\sin\theta\, d\theta $$
$$U=-pE(-\cos(\theta_2)+\cos(\theta_1))$$
Now if θ1 =π/2 and θ2
$$U=pE\cos(\theta)$$
$$U=p\cdot E$$
But the traditional derivation outputs $$-p\cdot E$$ and takes τfield and not τapp in the first step. Why is this the case?
The sign error is caused by the controversial signs of the angle and torque.
upload_2016-5-29_8-44-54.png


You see that the torque of the electric field causes decreasing angle, clockwise (negative) direction of rotation of the dipole. So τ(field) = -p x E.
 
  • #3
ehild said:
The sign error is caused by the controversial signs of the angle and torque.
View attachment 101364

You see that the torque of the electric field causes decreasing angle, clockwise (negative) direction of rotation of the dipole. So τ(field) = -p x E.

But isn't it only in the work integral that it becomes a - wrt +dθ? because Wfield=∫τfielddθcos π?
 
  • #4
Curieuse said:
But isn't it only in the work integral that it becomes a - wrt +dθ? because Wfield=∫τfielddθcos π?
What do you mean?
The electric field tends to align the dipole, decreasing the angle θ. The "natural" processes tend to make the potential energy minimum. If we set the zero of the potential energy at angle θ=pi/2, the potential energy is negative at θ=0.
 
  • #5
ehild said:
What do you mean?
The electric field tends to align the dipole, decreasing the angle θ. The "natural" processes tend to make the potential energy minimum. If we set the zero of the potential energy at angle θ=pi/2, the potential energy is negative at θ=0.
0Hq2tdM.png

τapp is -p x E will mean that it points out of the plane in this figure right? But how can that be? How do I derive potential energy in this scenario? the violet arrow is the dipole moment vector p. :cry: I know I am missing something big but i am not able to figure out why..
 
  • #6
Curieuse said:
0Hq2tdM.png

τapp is -p x E will mean that it points out of the plane in this figure right? But how can that be? How do I derive potential energy in this scenario? the violet arrow is the dipole moment vector p. :cry: I know I am missing something big but i am not able to figure out why..

The electric field turns the dipole in the opposite direction as in your figure, clockwise. So the torque points into the plane. Calculate the potential energy as you did in the first post, only the sign of torque was wrong.
upload_2016-5-29_16-16-54.png
 
  • #7
ehild said:
The electric field turns the dipole in the opposite direction as in your figure, clockwise. So the torque points into the plane. Calculate the potential energy as you did in the first post, only the sign of torque was wrong.
View attachment 101371

Then i'd be getting the work done by the field right? I am trying to get work done by external agent in turning the dipole as in the figure in my previous post.. The turning i indicated was that produced by external torque..this work is stored in the system right?
 
  • #8
Curieuse said:
Then i'd be getting the work done by the field right? I am trying to get work done by external agent in turning the dipole as in the figure in my previous post.. The turning i indicated was that produced by external torque..this work is stored in the system right?
You need a positive external torque to turn the dipole by an angle dθ.
 
  • #9
ehild said:
You need a positive external torque to turn the dipole by an angle dθ.
since τfield= p x E is intrinsically negative(?) i put the negative sign in front of it to make it positive in τapp..
 
  • #10
Curieuse said:
since τfield= p x E is intrinsically negative(?) i put the negative sign in front of it to make it positive in τapp..
upload_2016-5-30_7-15-59.png

The electric field tends to rotate the dipole in the direction of the green arrow, clockwise, to make it aligned with the field. To move the dipole by the red angle dθ, the external agent has to exert a torque, which would rotate the dipole in anti-clockwise direction, according to the thin blue arrows. The initial angle of the dipole with respect to the electric field is shown by the blue angle.
The elementary work of the external agent is dW=PEsin(θ)dθ, positive, as the displacement and force have the same sign. ( Both the angular displacement and the torque are positive). The work between θ1 and θ2 is equal to the potential difference U(θ2)-U(θ1) = -PE(cos(θ2)-cos(θ1)). If θ1=0 and θ2=pi/2, U(π/2)-U(0) = -PE(cos(π/2)-cos(0)). What is U(0) if U(π/2)=0?
 
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  • #11
ehild said:
View attachment 101413
The electric field tends to rotate the dipole in the direction of the green arrow, clockwise, to make it aligned with the field. To move the dipole by the red angle dθ, the external agent has to exert a torque, which would rotate the dipole in anti-clockwise direction, according to the thin blue arrows. The initial angle of the dipole with respect to the electric field is shown by the blue angle.
The elementary work of the external agent is dW=PEsin(θ)dθ, positive, as the displacement and force have the same sign. ( Both the angular displacement and the torque are positive). The work between θ1 and θ2 is equal to the potential difference U(θ2)-U(θ1) = -PE(cos(θ2)-cos(θ1)). If θ1=0 and θ2=pi/2, U(π/2)-U(0) = -PE(cos(π/2)-cos(0)). What is U(0) if U(π/2)=0?

Ah so I have been confusing angles and magnitudes.. τapp.dθ has to be positive yes! :O
U(θ2)-U(θ1)=-pE(cos θ2-cosθ1)
so if we take U(π/2)=0, then U(0)=-pE.
Is this correct?
 
  • #12
Curieuse said:
Ah so I have been confusing angles and magnitudes.. τapp.dθ has to be positive yes! :O
U(θ2)-U(θ1)=-pE(cos θ2-cosθ1)
so if we take U(π/2)=0, then U(0)=-pE.
Is this correct?
yes.
 
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  • #13
ehild said:
yes.
Thanks very much sir! :D
 
  • #14
You are welcome.:smile:
 
  • #15
Curieuse said:
So $$ \tau_{app} = - pE\sin\theta $$
And the potential energy is just
$$ U = \int_{\theta_1}^{\theta_2}\tau_{app}\,d\theta $$
$$ U = \int_{\theta_1}^{\theta_2}-pE\sin\theta\, d\theta $$

This is where you are getting it wrong .

The confusion you are having is that even though $$ \vec{\tau}_{app} = - \vec{p} \times \vec{E} $$ , $$W_{app} = U = \int (pEsin\theta) d\theta $$ .
 
  • #16
ehild said:
So τ(field) = -p x E.

There should be no minus sign . Except this , I agree with everything you have written :smile: .
 
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  • #17
conscience said:
There should be no minus sign . Except this , I agree with everything you have written :smile: .
Yes, τ(field) = p x E, but with the angle shown, it is -pEsin(θ).
The vector product uses the sin of the angle the second vector encloses with the first, anti-clockwise. In this problem, we used the angle the first vector (p) made with second (E). In this way, the field produced negative torque.
 
  • #18
ehild said:
Yes, τ(field) = p x E, but with the angle shown, it is -pEsin(θ).

If by putting a minus sign you are representing a clockwise torque ,then it is alright . But if the - sign has something to do with the way angle is measured , then I do not understand .

ehild said:
The vector product uses the sin of the angle the second vector encloses with the first, anti-clockwise.

o_O Why complicate things ?

The cross product uses the smaller of the two angles between the vectors when joined from tail to tail .
 
  • #19
conscience said:
o_O Why complicate things ?

The cross product uses the smaller of the two angles between the vectors when joined from tail to tail .
Yes, calculating the magnitude of the torque. But from τ = pxE, it does not follow that τ=pEsinθ, where 0 ≤ θ ≤ π. The torque has sign, in addition that its direction is perpendicular to both p and E, which has to be taken into account when calculating work.
 
Last edited:
  • #20
ehild said:
The torque has sign, in addition that its direction is perpendicular to both p and E, which has to be taken into account when calculating work.

I believe that the sign of torque is not required for calculating work because work ( dot product ) requires magnitude of torque . All that matters is that whether the torque tries to increase or decrease the angle . When calculating work , the minus sign is placed when torque tries to decrease the angle between the vectors .
 
  • #21
conscience said:
I believe that the sign of torque is not required for calculating work because work ( dot product ) requires magnitude of torque . All that matters is that whether the torque tries to increase or decrease the angle . When calculating work , the minus sign is placed when torque tries to decrease the angle between the vectors .
Trying to increase or decrease angle is connected to the direction of torque.
 
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1. What is potential energy in the context of a dipole in an electric field?

Potential energy is the amount of energy that a dipole possesses due to its position in an electric field. It is the energy that is required to move the dipole from its current position to a reference point, usually at infinity.

2. How is the potential energy of a dipole in an electric field calculated?

The potential energy of a dipole in an electric field is calculated using the equation U = -pEcosθ, where U is the potential energy, p is the magnitude of the dipole moment, E is the strength of the electric field, and θ is the angle between the dipole moment and the electric field.

3. How does the potential energy of a dipole change in an electric field?

The potential energy of a dipole in an electric field changes depending on the orientation of the dipole with respect to the electric field. When the dipole is aligned with the electric field, the potential energy is at a minimum. When the dipole is perpendicular to the electric field, the potential energy is at a maximum.

4. Can potential energy be negative for a dipole in an electric field?

Yes, potential energy can be negative for a dipole in an electric field. This occurs when the dipole is aligned opposite to the direction of the electric field. In this case, the potential energy is at its minimum value, which is negative.

5. How does the potential energy of a dipole in an electric field relate to its stability?

The potential energy of a dipole in an electric field is directly related to its stability. A dipole with lower potential energy is more stable in the electric field, while a dipole with higher potential energy is less stable. This means that a dipole aligned with the electric field is the most stable, while a dipole perpendicular to the electric field is the least stable.

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