- #1

moderate

- 21

- 0

This page contains one derivation for the potential energy:

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

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]U

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:

Force

Force

[tex]\Delta[/tex]U

Adding the changes for both charges gives:

[tex]\Delta[/tex]U

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

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

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

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]U

_{1->2}=-W_{1->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:

Force

_{electric}*distance=E*q*[tex]\Delta[/tex]xForce

_{electric}*distance=E*q*(d/2 - d/2*cos([tex]\theta[/tex])[tex]\Delta[/tex]U

_{q+}=-E(d/2)(1-cos[tex]\theta[/tex])Adding the changes for both charges gives:

[tex]\Delta[/tex]U

_{0->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

#### Attachments

Last edited by a moderator: