- #1

Pushoam

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- 52

## Homework Statement

## 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 ##

To answer the final part,

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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