What Forces Act on a Dipole Near an Infinite Metal Plane?

[cscott]

Homework Statement

Point electric dipole \vec{p}=p_0 \hat{z} is a distance d above an infinite metal plane of surface normal \hat{n}=\hat{z}. What is the force on the dipole. Is the dipole attracted to, or repelled from the surface?

Homework Equations

V(r) = \frac{\hat{n} \cdot \hat{p}}{4 \pi \epsilon_0 r^2}

The Attempt at a Solution

I treated it like an image charge problem but with point dipoles

Potential from image dipole:
V(r) = -\frac{p_0}{4 \pi \epsilon_0 r^2}Force on real dipole:
F = -p_0 \frac{d}{dr}V(r=2d) = -\frac{2p_0^2}{4 \pi \epsilon_0 (2d)^3}

So attracted to the surface.

Is this the right approach? Not sure else how to "deal" with the dipoles.

 

[cscott]

one bump

 

[diazona]

I think the formula you're using for the force on the dipole is a bit off. (The units in the expression you get aren't correct)

There are a couple of ways you could handle this. One way would be to treat each dipole (the real one and the image) as a pair of charges separated by a distance \delta, compute the net force on the top pair, and take the limit as \delta\to 0 and q \to \infty such that the product p = qd (the dipole moment) is constant. This is the "manual labor" way, although in this case there's not all that much tedious math involved.

Another way would be to calculate the force as the gradient of potential energy,
\vec{F} = -\vec{\nabla}U
The potential energy of a dipole in an external electric field is
U = -\vec{p}\cdot\vec{E}
and in this case, the real dipole would be \vec{p} and the external electric field would be the field produced by the image dipole. If you don't already have an expression for the electric field, you can calculate it from the potential V(r).