Torque that a grounded infinite sheet exerts on a dipole

[PhDeezNutz]

Homework Statement

Find the torque that an infinite grounded sheet exerts on a dipole $\vec{p}_1$ at a height $z$ above it where the dipole makes an angle $\theta$ with the vertical to the right. Additionally in what direction will the dipole come to rest?

Relevant Equations

Generally these are the equations.

$\vec{N} = \vec{p} \times \vec{E}$
$\vec{E}_{dip} = \frac{1}{4 \pi \epsilon_0 r^3} \left( 2 \cos \theta \hat{r} + \sin \theta \hat{\theta} \right)$

The formula for $\vec{E}_{dip}$ is in spherical coordinates and assumes the dipole is oriented in the positive $z$ direction.

The situation is shown directly below

(Figure 1)

The correct image configuration that mimics (Figure 1) is shown below.....that is to say it maintains the boundary conditions. I'd imagine the torque that the grounded sheet exerts on the real dipole is the same as the torque the image dipole exerts on the real dipole. I want to calculate everything in the frame where the image dipole $\vec{p}_2$ is aligned along the $z-\text{axis}$. Let's call it $z_2$ so as now to confuse it with the distance from the plate to the real dipole. The frame is included in the figure below.

Where of course $\vec{p}_1$ and $\vec{p}_2$ have the same magnitude. Might as well call it $p$.

(Figure 2)

I'd like to compute all of it in the local spherical vector basis of $\vec{p}_2$ at the location of $\vec{p}_1$.


I contend that according to the picture below that $\vec{p}_1 = p \cos \theta \hat{r}_2 + p \sin \theta \hat{\theta}_2$

(Figure 3)

and $\vec{E}_2$ at the location of $\vec{p}_1$ is of course

$\vec{E}_{2} = \frac{1}{4 \pi \epsilon_0 (2z)^3} \left( 2 \cos \theta \hat{r}_2 + \sin \theta \hat{\theta}_2 \right)$

So

$\vec{E}_{2} = \frac{1}{32 \pi \epsilon_0 z^3} \left( 2 \cos \theta \hat{r}_2 + \sin \theta \hat{\theta}_2 \right)$

Take the cross product $\vec{N}_1 = \vec{p}_1 \times \vec{E}_2$

$\vec{N}_1 = \frac{p^2}{32 \pi \epsilon_0 z^3} \left( \cos \theta \hat{r}_2 + \sin \theta \hat{\theta}_2 \right) \times \left( 2 \cos \theta \hat{r}_2 + \sin \theta \hat{\theta}_2 \right)= -\frac{p^2}{ 32 \pi \epsilon_0 z^3} \left( \sin \theta \cos \theta \right) \hat{\phi}_2$

Using double angle identity for $\sin 2 \theta = 2 \sin \theta \cos \theta$ we get

So $\vec{N}_1 = -\frac{p^2}{ 64 \pi \epsilon_0 z^3} \left( \sin 2 \theta \right) \hat{\phi}_2$ which is into the page as expected using the right hand rule.

I think this answer is right.

My guess for finding equilibrium orientations would be to set the torque equal to 0 which is weird because we'd get that $\theta = 0$ and $\theta = \frac{\pi}{2}$ and I don't think that either aligns or anti-aligns with the field from the pictures drawn above.

I guess I could plot it and see.

 

[kuruman]

Why would the dipole come to rest? Are we assuming that the dipole undergoes underdamped harmonic motion because of some dissipative force? If so, and if the point dipole is constrained to a fixed height above the plane, then its equilibrium orientation should be at $\theta =0.$ You can see that because the low energy configuration of the dipole and its image is "ferromagnetic."

 

[haruspex]

Your figures for $\vec p_2$ look wrong. It should be a true reflection of $\vec p_1$, with the arrow pointing down and right, not up and left.

Edit: Forget that, I'm wrong.

How do you get that expression for $\vec E_2$? Shouldn’t it be zero when $\theta=0$?

 

[PhDeezNutz]

@kuruman

Honestly it’s been a few years since I’ve touched the material so I don’t remember what diamagnetic and ferromagnetic are. And I don’t think it’s covered until way later in Griffiths. This question is in Griffiths.

It is not given that the system undergoes undamped harmonic motion.

@haruspex interesting. I know Wikipedia is not a valid source around these parts but

https://en.m.wikipedia.org/wiki/Method_of_image_charges

It seems to corroborate in the section “Reflecting in a conducting plane”

I’ll look through more legitimate sources if I can find them.

Also the expression for E2 is given in Griffiths.

Edit: I hadn’t read the entire wiki page before but it actually has the result for torque and it corroborates mine. Previously I had just googled “method of images for a dipole” and just looked at the google image results without clicking on the page.

 

[kuruman]

Honestly it’s been a few years since I’ve touched the material so I don’t remember what diamagnetic and ferromagnetic are. And I don’t think it’s covered until way later in Griffiths.

Sorry for being unintentionally mysterious. All I meant to convey is that the low potential energy orientation for a dipole is when its moment is parallel to the local field. That's because the potential energy of a dipole in an electric field is $U=-\mathbf p \cdot \mathbf E$ and has a minimum value $U_{\text{min}}=-pE$ when the vectors are parallel. In that orientation the torque exerted by the image charge on the "real" charge is zero. Here, this means that $\theta =0.$

 

[kuruman]

Your figures for $\vec p_2$ look wrong. It should be a true reflection of $\vec p_1$, with the arrow pointing down and right, not up and left.

The arrowheads are hard to read but I think the picture is correct. If you consider the dipole as two point charges separated by some distance, the image charges will appear in the reflected position with a change of sign (see figure on the right.) The dipole moment points along the separation distance from negative to positive charge.

 

[PhDeezNutz]

Sorry for being unintentionally mysterious. All I meant to convey is that the low potential energy orientation for a dipole is when its moment is parallel to the local field. That's because the potential energy of a dipole in an electric field is $U=-\mathbf p \cdot \mathbf E$ and has a minimum value $U_{\text{min}}=-pE$ when the vectors are parallel. In that orientation the torque exerted by the image charge on the "real" charge is zero. Here, this means that $\theta =0.$

Honestly I am pretty stupid for not immediately remembering that.

I guess the question is what is the direction of the field ( I guess relative to the vertical because I haven't really plotted the field lines) at the location of $\vec{p}_1$ and then there is our answer.

Let me work on it. Thank you for your help. @haruspex I'm really tempted to dive in and "prove" the method of images configuration for the dipole instead of just taking it on faith. I'll let you know if I come up with anything and thank you for your help as well.

 

[kuruman]

I did this slightly differently to verify your result.

I wrote the dipolar field of the image dipole $\mathbf p_1$ in the form $$\mathbf E_1=\frac{1}{4\pi\epsilon_0}\left[\frac{3(\mathbf p_1\cdot \mathbf r)\mathbf r}{r^5}-\frac{\mathbf p_1}{r^3}\right].$$ Then the torque on the real dipole is $$\mathbf N=\mathbf p_2\times \mathbf E_1=\frac{1}{4\pi\epsilon_0}\left[\frac{3(\mathbf p_1\cdot \mathbf r)(\mathbf p_2\times \mathbf r)}{r^5}-\frac{\mathbf p_2\times \mathbf p_1}{r^3}\right]. $$Now $\mathbf r=2z~\mathbf {\hat z}$ which gives the direction of both cross products out of the screen. Expanding the magnitudes of the cross products and the dot product we have $$ |\mathbf p_2\times \mathbf r|=p~r~\sin\!\theta~;~~|\mathbf p_2\times \mathbf p_1|=p^2\sin(2\theta)~;~~\mathbf p_1\cdot \mathbf r=p~r~\cos\!\theta.$$ Then the magnitude of the torque is $$\begin{align} N= & \frac{1}{4\pi\epsilon_0}\left[\frac{3(p~r\cos\!\theta)(p~r\sin\!\theta)}{r^5}-\frac{p^2\sin(2\theta)}{r^3}\right] \nonumber \\
& = \frac{p^2~}{4\pi\epsilon_0r^3}\left[3\sin\!\theta \cos\!\theta-2\sin\!\theta\cos\!\theta)\right]
= \frac{p^2~\sin(2\theta)}{8\pi\epsilon_0r^3}. \nonumber
\end{align}$$ With $r=2z$, the expression agrees with yours. The direction of the torque on the "real" dipole is out of the screen, i.e. it is a restoring torque given Figure 1 in post #1.

Note: My notation is 1 for the image and 2 for the "real" dipole which is the reverse of yours. I noticed the discrepancy after I wrote all the LaTeX and I don't think it is a good idea to change it now.

 

[PhDeezNutz]

I really like how you started with the coordinate free form of the field..….it saves a lot of busy work.

 

[kuruman]

My guess for finding equilibrium orientations would be to set the torque equal to 0 which is weird because we'd get that $\theta = 0$ and $\theta = \frac{\pi}{2}$ and I don't think that either aligns or anti-aligns with the field from the pictures drawn above.

I guess I could plot it and see.

When a point dipole is at the origin of coordinates and is oriented along the z-axis,
(a) at point P on the z-axis the electric field points in the +z direction, i.e. parallel to the dipole.
(b) at point Q in the xy-plane the electric field is in the -z direction, i.e. antiparallel to the dipole.

When you place a second dipole at point P, that corresponds to $\theta =0$
When you place a second dipole at point Q, that corresponds to $\theta =\frac{\pi}{2}.$

Do you see why it is so?

 

[PhDeezNutz]

When a point dipole is at the origin of coordinates and is oriented along the z-axis,
(a) at point P on the z-axis the electric field points in the +z direction, i.e. parallel to the dipole.
(b) at point Q in the xy-plane the electric field is in the -z direction, i.e. antiparallel to the dipole.

When you place a second dipole at point P, that corresponds to $\theta =0$
When you place a second dipole at point Q, that corresponds to $\theta =\frac{\pi}{2}.$

Do you see why it is so?

I think so. Just had to turn my head sideways.

 

[kuruman]

I think so. Just had to turn my head sideways.

Exactly.