Potential of a dipole in E field

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raggle
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Homework Statement



I'm going over some EM notes and I found a derivation for the potential energy of a dipole in an E field which first found the torque on the dipole and then integrated.
I had a go at a derivation that goes the other way, starting from the potential of the 2 charges. I've ran into a bit of a problem midway through and I can't see how to get past it.

Homework Equations



Want to show:
U = -p.E

The Attempt at a Solution


U = q(VB-VA)

now I don't know how to get from here to what I want to show. Both of the potentials look like [itex]\frac{q}{4πε<sub>0</sub>r}[/itex] , where the r is different in either potential.
The only thing I can think of is to use an approximation to r, but that doesn't give an acosθ term in the numerator.
 
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raggle said:

Homework Statement



I'm going over some EM notes and I found a derivation for the potential energy of a dipole in an E field which first found the torque on the dipole and then integrated.
I had a go at a derivation that goes the other way, starting from the potential of the 2 charges. I've ran into a bit of a problem midway through and I can't see how to get past it.

Homework Equations



Want to show:
U = -p.E

The Attempt at a Solution


U = q(VB-VA)

now I don't know how to get from here to what I want to show. Both of the potentials look like [itex]\frac{q}{4πε_0 r}[/itex] , where the r is different in either potential.
The only thing I can think of is to use an approximation to r, but that doesn't give an acosθ term in the numerator.

The charges of the dipole are in an external electric field. By changing the angle, that potential energy gained from the field changes, the potential energy of their interaction does not. You ignored the contribution of the external field to the potential energy.

As for the potential energy of the interaction between two point charges, it is [itex]\frac{q_1 q_2}{4 \pi \epsilon_0 r}[/itex] where r is the distance between the charges.

ehild