Find the energy of an ideal dipole in an electric field

[zezima1]

Homework Statement

Show that the energy of an ideal dipole in an electric field E is given by:
U = -p \bullet E

Homework Equations

The energy required to bring to charges together, their electrostatic energy, is:

W = ½ Ʃq_iV(r_i)

The Attempt at a Solution

Well I want to know what the energy cost is of taking in the dipole. Because if you take in one charge first then the energy cost will be negative (as they attract each other). But if you see the dipole as already bounded before taking it in then the cost will be nothing. Hence: Is the above formula even correct to apply?

 

[gabbagabbahey]

Homework Statement

Show that the energy of an ideal dipole in an electric field E is given by:
U = -p \bullet E

Homework Equations

The energy required to bring to charges together, their electrostatic energy, is:

W = ½ Ʃq_iV(r_i)

The Attempt at a Solution

Well I want to know what the energy cost is of taking in the dipole. Because if you take in one charge first then the energy cost will be negative (as they attract each other). But if you see the dipole as already bounded before taking it in then the cost will be nothing. Hence: Is the above formula even correct to apply?

Keep in mind, that there is an electric field present when you bring the dipole in from infinity. You don't know the potential that gives rise to that field (except for the general integral form), so your equation for W probably won't help you much. Also, an ideal dipole is not the same as a physical dipole (two opposite charges sperated by some finite distance), but rather it's a very special limiting condition of such a dipole (the limit as the distance between the charges goes to zero, but it's dipole moment remains fixed), so even if you were given the potential, you would have to carefully take that limit after calculating the energy of the physical dipole.

Instead, you likely have seen derived an equation for the force on an ideal dipole in a electric field. If so, why not plug that into the definition of work and integrate?