# The energy of electric interaction between these dipoles? [closed]

I'm a physics tutor. This is not a homework problem. I'm unable to solve this problem.

The energy of electric interaction between these dipoles will be:

I tried taking P1=q1*d1 and P2=q2*d2 then calculated Potential energy between (q1,-q1), (q1,-q2), (q2,-q2), (-q1,q2) and used certain approximations but couldn't get to any of the results shown.

-

## closed as off-topic by David Z♦May 4 '15 at 15:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – David Z
If this question can be reworded to fit the rules in the help center, please edit the question.

The answer is by assuming a short dipole.

There are many ways of going about this (I'm only giving hints to the full method). One is by splitting the lower dipole into parallel and perpendicular components(its a vector, we can do that). Now use the formulae for field from a dipole at axial and equatorial positions (equatorial is $\frac{kp}{r^3}$, axial is double that), and calculate the change in potential while moving two charges $\pm q$ from infinity to a distance $r \pm d$.

Or, you can directly apply the formula $V(r,\theta)=\frac{k\vec{p}\cdot\vec{r}}{r^3}$, on the two charges at $r\pm d$.

-

You may use the electrostatic potential due to electric dipole $P_2$ (see, e.g., http://en.wikipedia.org/wiki/Dipole#Field_from_an_electric_dipole, although the system of units is different there) and find the energy of charges $+q_1$ at $r+d_1$ and $-q_1$ at $r$ under the condition that $d_1\ll r$.

-

## protected by Qmechanic♦May 4 '15 at 13:42

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).