What Is the Electrostatic Force Between an Ideal Dipole and a Conducting Plate?

[millifarads]

Homework Statement

Find the electrostatic attraction force F between a point dipole and an infinite metallic conducting plate. The dipole moment p is perpendicular to the plate and the distance from the dipole to the plate is h. Find the work that has to be done in order to move dipole from position h1 to h2.

Homework Equations

$$\vec{F}^{\,} = \vec{p}^{\,} \frac{\partial E}{\partial x}$$
$$\vec{E}^{\,} = - \frac {\vec{p}^{\,}}{4\pi\epsilon_0r^3}$$
$$W = qV$$
$$V = \frac {\vec{p}^{\,} \cdot \hat{r} } {4\pi\epsilon_0r^2}$$
$$r = 2h$$

The Attempt at a Solution

Calculating the force seems to be straightforward. I've set the problem up using images, with a dipole of the opposite dipole moment a distance -h away from the plate, directly under it. ie. r = 2h. The dipole moment of the real dipole should point away from the plate, and same with the image dipole moment. This seems arbitrary, though, as I think the dipole could be pointing either way. Then:

$$\vec{E}^{\,} = - \frac {\vec{p}^{\,}}{4\pi\epsilon_0(2h)^3} = - \frac {\vec{p}^{\,}}{32\pi\epsilon_0h^3}$$

So, $$\vec{F}^{\,} = \vec{p}^{\,} \frac{\partial E}{\partial h} = \vec{p}^{\,} \frac{3\vec{p}^{\,}}{32\pi\epsilon_0h^4} = \frac{3 p^2}{32\pi\epsilon_0h^4}$$

Now at this point, assuming I've done everything correctly and the formulas I've listed are right, I'm not sure how to evaluate the potential. The r hat unit vector should point the same way as the dipole, right? So in the case I've set up, away from the plate and perpendicular to it. Is the potential then just:

$$V = \frac {p} {4\pi\epsilon_0r^2} = \frac {p} {16\pi\epsilon_0h^2}$$

Then, evaluate this at the distances of h1 and h2 and take the difference?

 

[millifarads]

Anyone?