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A punctual dipole $\overrightarrow{p}$ is located a distance $d$ from a metalic grounded plate.

What is the work required to turn de dipole from a perpendicular orientation (pointing towards the plate) to a parallel one?

So the fact that is a punctual dipole bothers me a little because the torque equation deductions that I've seen so far were deduced with a physical dipole, anyway I get that it is a vector and you can rotate it.

So the torque equation for the dipole is:

\begin{equation} \overrightarrow{N}=\overrightarrow{p}\times\overrightarrow{E} \end{equation}

So:

$$ W=\int _0 ^\theta N d\theta=\int _0 ^\theta \left( pE\sin\theta \right) d\theta=pE(1-\cos\theta)$$

The real problem for me is $E$. $E$ due to the dipole, the grounded plate or both? And... the plate is grounded, so isn't the charge density $\sigma$ zero?

Anything you can comment on this would be appreciated.

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  • $\begingroup$ Please don't abuse MathJax to emphasize your posts. Use *italic* for italic and **bold** for bold, and ***both*** for both. $\endgroup$ Commented Aug 27, 2017 at 18:12
  • $\begingroup$ I think you're in the right track, and I also wonder about "punctual dipole". The $E$ in your expression is the external field in which the dipole is plunged. $\endgroup$ Commented Aug 27, 2017 at 18:12
  • $\begingroup$ -1. Not clear. What is a "punctual dipole"? $\endgroup$ Commented Aug 27, 2017 at 19:42
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    $\begingroup$ @sammygerbil, punctual: 3. pertaining to or of the nature of a point $\endgroup$ Commented Aug 27, 2017 at 21:04

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In this case the electric field you worry about is the electric field produced by the image dipole on the other side of the grounded plate. Because of that, $\vec{E}$ will depend on the angle the dipole makes, $\theta$, so you need to figure out what that is in order to do the integral. I would suggest you also draw a picture to keep the angles straight (angle of rotation vs. angle between dipole and electric field).

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