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gnegnegne

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## Homework Statement

A dipole ##\textbf{p}=p\mathbf{\hat{z}}## (oriented along the z axis for example) is parallel to a conducting grounded plane (xz plane) and is placed at a distance ##d## from it. I have to find the interaction energy and the force between the dipole and the plane.

## Homework Equations

Method of image charges,

Electric field of a dipole: [itex]\textbf{E}=\frac{1}{4\pi \epsilon_0}\Big[\frac{3(\textbf{(pr)r}}{r^5} - \frac{\textbf{p}}{r^3} \Big][/itex]

Potential energy of a dipole in an electric field: [itex]U=-\textbf{pE}[/itex]

Force on a dipole in an electric field: [itex]\textbf{F}=\nabla\textbf{(pE)}[/itex]

## The Attempt at a Solution

The image charges is a dipole antiparallel to the original one and on the same axis. The electric field along the y-axis generated by the image dipole is [itex]\mathbf{E}=\frac{p}{4\pi\epsilon_0(2y)^3}\mathbf{\hat{z}}=\frac{p}{32\pi\epsilon_0y^3}\mathbf{\hat{z}}[/itex]. The force acting on the original dipole is therefore [itex]\mathbf{F}=-\frac{3p^2}{32\pi\epsilon_0y^3}\mathbf{\hat{y}}[/itex] evaluated at ##y=d##. To calculate the potential energy I can use the other equation [itex]U=-\textbf{pE}[/itex] or I could also calculate the work done by the force from ##y=\infty## to ##y=d##. The result is the same, which is [itex]U=-\frac{p^2}{32\pi\epsilon_0y^3}[/itex]. But is this the correct potential energy? In the example of the method of image charges with a single charge and a conducting plane it is often shown that if you calculate the potential energy [itex]U=-\frac{q^2}{4\pi\epsilon_02y}[/itex] you get twice the correct energy because you are including also the energy of the image charge. Should I consider half the potential energy I calculated? How are the two examples different?