Solving Two Point Dipoles - Find Torque of p1 & p2

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Homework Help Overview

The problem involves two point dipoles, p1 and p2, positioned at a distance r apart and oriented perpendicularly. The task is to find the torque on each dipole due to the other, utilizing the relationship between dipole moment and electric field.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the formula for the electric field of a dipole and considers the torque expression. They express confidence in calculating the torque for one dipole but seek clarification on the second case. Other participants suggest that rotating the coordinate system could simplify the calculations, while one participant challenges this approach, indicating a potential misunderstanding.

Discussion Status

The discussion is ongoing, with participants exploring different perspectives on the problem. Some guidance has been offered regarding the relationship between torque and angular momentum, but there is no explicit consensus on the best approach to take.

Contextual Notes

There appears to be some confusion regarding the implications of rotating the coordinate system and how it affects the calculation of torques. The original poster's assumptions about the dipole orientations and their effects on the results are also under scrutiny.

Kolahal Bhattacharya
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Homework Statement


Griffiths offers this problem: two point dipoles p1 and p2 are given.They are r distance away and they are perpendicular.We are asked to find torue of p1 (about p1's centre) due to p2 and vice versa.
Well, the situation is that I knoe it is =p cross E
even I know the formula for a point dipole pointing in the z direction:
E_dip=(1/(4*pi*epsilon))(p/r^3)[2cos(theta) (r^)+sin (theta) (theta^)
where (r^) and (theta^) are the unit vectors in a polar system coincident with the xyz system.
I can surely find the result for 1 part.Assuming p1 points in the z direction,it can be evaluated.What could be done in the second case?


Homework Equations





The Attempt at a Solution

 
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Both torques are equally easy to compute. Just rotate your coordinate system. But you don't even have to. Their sum is the time rate of change of angular momentum, a conserved quantity. Think Newton's third law.
 
Last edited:
You are WRONG.If I rotate the axes,the relative difference of answers will not be visible.
 
Kolahal Bhattacharya said:
You are WRONG.If I rotate the axes,the relative difference of answers will not be visible.

Really? I have no idea what you mean to say, but I guess you know best.
 

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