Calculating potential energy of electric dipole in uniform field

In summary, the conversation discusses the derivation of the potential energy for a dipole in an electric field, with one method using the definition of electric potential energy and the other using the work done by the electric field. The resulting relationship is the same, but there is a linear offset due to the different definitions of zero potential.
  • #1
moderate
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This page contains one derivation for the potential energy:

http://www.shef.ac.uk/physics/teaching/phy205/lecture_4.htm"

Result: U=-pEcos[tex]\theta[/tex]

where
E = magnetic field's magnitude
P = dipole moment = d*Q
d= distance between + and - charge
Q= charges' magnitude
[tex]\theta[/tex] = angle between the field and the straight line connecting the + and - charges

But, shouldn't it also be possible to derive the dipole's potential energy in another way?

The definition of electric potential energy is:

[tex]\Delta[/tex]U1->2=-W1->2
where W=work done by electric field.

So, can't one take the 2 charges as separate (connected by a massless, chargeless rod), and then calculate the work done on them by the field while moving from the angled position to equilibrium?

This should then be equal to the initial potential energy (in the angled position).

Attached is an illustration.

Calculating for the positive charge gives:

Forceelectric*distance=E*q*[tex]\Delta[/tex]x
Forceelectric*distance=E*q*(d/2 - d/2*cos([tex]\theta[/tex])
[tex]\Delta[/tex]Uq+=-E(d/2)(1-cos[tex]\theta[/tex])

Adding the changes for both charges gives:

[tex]\Delta[/tex]U0->theta=Ep(1-cos([tex]\theta[/tex])).

This gives me the same relationship, except that (which is frustrating me somewhat) there is a linear offset of (1)Ep.

My question is: the approach is valid, correct? I am not going crazy?

If that is the case, then it is just a matter of adding a linear shift to come to the same zero potential, correct?

-frustrated at physics
 

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  • #2
*bump*

I derived it by integrating the torque between theta=o and theta and got the same result...

Any ideas anyone?
 
  • #3
moderate said:
So, can't one take the 2 charges as separate (connected by a massless, chargeless rod), and then calculate the work done on them by the field while moving from the angled position to equilibrium?

The offset is due to the fact that you are integrating to equilibrium, which is defining U = 0 to be when p is antiparallel with the field. On the other hand, the link defines U = 0 to be where p is parallel to an equipotential and thus perpendicular to the field.
 

1. How do you calculate the potential energy of an electric dipole in a uniform electric field?

The potential energy of an electric dipole in a uniform electric field can be calculated using the formula U = -pE cosθ, where U is the potential energy, p is the magnitude of the dipole moment, E is the magnitude of the electric field, and θ is the angle between the dipole moment and the electric field.

2. What is the unit of measurement for potential energy in this calculation?

The unit of measurement for potential energy in this calculation is joules (J).

3. Can the potential energy of an electric dipole in a uniform electric field be negative?

Yes, the potential energy of an electric dipole in a uniform electric field can be negative if the angle between the dipole moment and the electric field is greater than 90 degrees.

4. How does the distance between the charges of the electric dipole affect the potential energy?

The potential energy of an electric dipole is inversely proportional to the distance between the charges. As the distance increases, the potential energy decreases, and vice versa.

5. What happens to the potential energy of an electric dipole in a uniform electric field if the dipole is rotated?

The potential energy of an electric dipole in a uniform electric field changes as the dipole is rotated. It is at its maximum when the dipole moment is aligned with the electric field and at its minimum when the dipole moment is perpendicular to the electric field.

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