# Torque on a dipole near a infinite conducting plane

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1. Aug 5, 2017

### Pushoam

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

I have drawn this diagram using MS paint, could you please tell me some other software in which I can draw and insert greek symbols, too ?

Let me take the origin at O.
$\hat r \left(\alpha\right) ≡ \hat r\text{ at angle }\alpha ~with ~z - axis$
Considering the image problem,
$\vec E \left (O' \right)$ due to the image dipole $= \frac { -kp} { {2 z}^3 } \left( 2 \cos \alpha ~ \hat r \left( \theta = 0 \right) + \sin \alpha ~ \hat \theta \left ( \theta =0 \right) \right )$
$\\ \hat r = \sin \theta ~ \hat s + \cos \theta ~ \hat z \\ \hat \theta = - \sin \theta ~ \hat z +\cos \theta ~ \hat s \\ \text {here } \hat s\text{ is the cylindrical coordinate unit vector, I don't know its name, please tell me this so if it has a name. } \\\text { at } \theta =0, ~ \hat r = \hat z ~ and ~ \hat \theta = \hat s \\ \\ \vec p = p~ \hat r \left ( \alpha \right)$

Now,
torque $\vec N = \vec p \times \vec E$

$= p ~ \hat r$ $\left( \alpha \right ) \times$ $\frac { -kp} { {2z}^3 } \left( 2 \cos \alpha ~ \hat r \left( \theta = 0 \right) + \sin \alpha ~ \hat \theta \left ( \theta =0 \right) \right)$

$\\ = p \hat r\left( \alpha \right ) \times \frac { -kp} {{2z}^3 } \left( 2 \cos \alpha ~\hat z + \sin \alpha ~ \hat s \right)$

$= \frac {-kp^2}{ \left (2z\right) ^3 } \left[ - 2 \cos \alpha \sin \alpha ~\hat \phi + \cos \alpha \sin \alpha ~ \hat \phi \right ]$

$\\ = \frac {kp^2}{16 z ^3 } \sin \alpha ~ \hat \phi$

I guess the orientation of plane is x-y and $\hat \phi$ is also in the x-y orientation. So, the dipole will come to rest in the orientation which is parallel to the plane .
But why will it come to rest? Why won't it oscillate (as there are no other net force acting on the dipole)?

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Last edited: Aug 5, 2017
2. Aug 5, 2017

### TSny

I had a difficult time following all of your unit vector notations. I believe your unit vector $\hat{s}$ could be called the "radial polar unit vector" in cylindrical coordinates.
Are you sure both terms in the parentheses are positive? There should be a set of parentheses in the denominator (which you finally put in near the very end of your calculation.)

Did you mean to type $2 \alpha$ for the argument of the sine function? I don't get the same numerical factor out front due to the possible sign error that I mentioned above.

Yes, it would oscillate. I guess they want you to assume small damping so that it comes to rest at the equilibrium position. I agree that the equilibrium position would be parallel to the conducting plane.

3. Aug 5, 2017

### Pushoam

$\vec E$ due to an ideal dipole \vec p at a point $\hat r \left( \theta \right )$ is given by

Here, r = 2z
$\theta = \alpha$
The electric field at O' due to p $= \frac { kp} { \{2 z\}^3 } \left( 2 \cos \alpha ~ \hat r \left( \theta = 0 \right) + \sin \alpha ~ \hat \theta \left ( \theta =0 \right) \right )$
So, The electric field at O' due to -p $= \frac { -kp} { \{2 z\}^3 } \left( 2 \cos \alpha ~ \hat r \left( \theta = 0 \right) + \sin \alpha ~ \hat \theta \left ( \theta =0 \right) \right )$

Is there anything wrong with it?

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4. Aug 5, 2017

### TSny

Now you're using spherical coordinates.

This is confusing. You say $\theta = \alpha$ but then you say $\theta = 0$ and $\alpha \neq 0$.

There is a nice way to write the electric field of a dipole that is independent of coordinate system. See equation 13 here:
http://phys.columbia.edu/~nicolis/Dipole_electric_field.pdf

You can apply this expression to any coordinate system that you find convenient. Thus, you can apply it to this picture to find E at O' in terms of unit vectors $\hat y$ and $\hat z$.

You can then switch the sign of E to correspond to p pointing in the opposite direction to that shown in this picture.

5. Aug 5, 2017

### Pushoam

$\vec N = \vec p \left ( ~at ~ O' \right ) \times \{$ ${\vec E \left ( ~at ~ O' \right )} = { \frac k { \{ 2 z \}^3 } }$ $\left [ 3 { \left ( - \vec p ⋅ \hat z \right ) \hat z } +\vec p \right ]$ $\}$
$\\ = \frac {- 3 k p^2} { \{ 2 z \}^3 } \cos \alpha \{ \vec p \times \hat z\} \\ = \frac {- 3 k p^2} {16 z^3} \sin 2 \alpha \{- \hat \phi \}$

The book's solution says that the equilibrium position would be perpendicular to the conducting plane or did I iterpret it wrong?

6. Aug 6, 2017

### TSny

Sorry, it was my mistake. I was not orienting the image dipole correctly which led to a sign error on my part. The image dipole is not oriented as shown in your picture (post #5) but as shown in part (a) of the solution of your last post. The expressions for E and N given in the solution look correct. And the direction of the torque is such as to make the equilibrium orientation of the dipole perpendicular to the conducting plane:

equilibrium is $\uparrow$ if initially $-\pi/2 < \theta < \pi/2$
equilibrium is $\downarrow$ if initially $\pi/2 < \theta < 3\pi/2$

where $\theta$ is as shown in the figure for part (a) solution.

Hope I didn't cause you too much of a headache.

Last edited: Aug 6, 2017
7. Aug 6, 2017

### Pushoam

The image dipole given in the book's solution is not -p. I was taking -p as image dipole so that the potential due to p at the plane will get cancelled by potential due to -p and hence we will get the potential on the space containing the plane to be 0.
What is wrong with this argument?

8. Aug 6, 2017

### TSny

In your diagram, the image dipole looks to me to be parallel to the real dipole. But they are not parallel (in general). See the figure in the solution part (a).