Torque on a dipole near a infinite conducting plane

• Pushoam
In summary: 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
Pushoam

Homework Equations

3. The Attempt at a Solution [/B]

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 { 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)?

Attachments

1.6 KB · Views: 524
1.4 KB · Views: 490
Last edited by a moderator:
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.
Pushoam said:
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 )##
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.)

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

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)?
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.

TSny said:
Are you sure both terms in the parentheses are positive?
##\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?

Attachments

1.8 KB · Views: 505
Pushoam said:
##\vec E ## due to an ideal dipole \vec p at a point ## \hat r \left( \theta \right ) ## is given by

View attachment 208392
Now you're using spherical coordinates.

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 ) ##
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.

##
\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 \} ##

TSny said:
I agree that the equilibrium position would be parallel to the conducting plane.
The book's solution says that the equilibrium position would be perpendicular to the conducting plane or did I iterpret it wrong?

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:
TSny said:
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 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 canceled 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?

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).

1) What is torque on a dipole near an infinite conducting plane?

The torque on a dipole near an infinite conducting plane is the rotational force experienced by the dipole due to the interaction with the electric field of the plane. This torque is perpendicular to both the dipole moment and the electric field, and it tends to align the dipole with the field.

2) How is torque on a dipole near an infinite conducting plane calculated?

The torque on a dipole near an infinite conducting plane can be calculated using the formula:
τ = p x E
Where τ is the torque, p is the dipole moment, and E is the electric field.

3) What factors affect the magnitude of torque on a dipole near an infinite conducting plane?

The magnitude of torque on a dipole near an infinite conducting plane is affected by the strength of the electric field, the distance between the dipole and the plane, and the orientation of the dipole with respect to the plane's surface. The torque is maximum when the dipole is perpendicular to the plane and decreases as the dipole becomes parallel to the plane.

4) How does the torque on a dipole near an infinite conducting plane affect the dipole's motion?

The torque on a dipole near an infinite conducting plane causes the dipole to rotate and align itself with the electric field of the plane. This rotation can result in the dipole moving towards or away from the plane, depending on the orientation of the dipole.

5) Can the torque on a dipole near an infinite conducting plane be zero?

Yes, the torque on a dipole near an infinite conducting plane can be zero if the dipole is parallel to the plane's surface. In this case, the dipole experiences an equal and opposite torque in opposite directions, resulting in no net torque.

Replies
12
Views
595
Replies
1
Views
906
Replies
7
Views
1K
Replies
24
Views
1K
Replies
3
Views
1K
Replies
19
Views
1K
Replies
25
Views
605
Replies
2
Views
503