Understanding Torque and Cross Product Calculations for Dipoles

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SUMMARY

This discussion focuses on calculating the torque on dipoles using the equation N = P x E. The participants analyze the torque generated by two dipoles, p1 and p2, positioned in the same plane with p1 pointing upwards and p2 pointing to the right. Key equations include E = 1/(4πε0 r3)[3(p1·r^)r^ - p1] and the resulting torque equations for both dipoles. The discussion clarifies why p1·r^ = 0 and p2·r^ = p, emphasizing the geometric relationships in the torque calculations.

PREREQUISITES
  • Understanding of vector calculus, specifically cross products
  • Familiarity with dipole moments in physics
  • Knowledge of electric field equations, particularly E = 1/(4πε0 r3)
  • Basic grasp of torque and its mathematical representation
NEXT STEPS
  • Study the mathematical derivation of torque for dipoles in different orientations
  • Learn about the implications of dipole interactions in electric fields
  • Explore the use of TeX for formatting physics equations
  • Investigate the role of ε0 in electrostatics and its impact on dipole calculations
USEFUL FOR

Physics students, educators, and researchers interested in electromagnetism and dipole interactions will benefit from this discussion. It provides insights into torque calculations and the underlying principles of electric fields.

leonne
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Homework Statement


This is a physics problem. need to find torque on dipoles
the
the 2 dipoles are on the same plane with distance r p1 is pointing up while p2 is pointing right


Homework Equations


N=PxE


The Attempt at a Solution


I know how to do a cross product, you make the matrix with xyz, but not sure about this

E=1/(4pieEo r3)[3(p1*r^)r^-p1)

than after the cross product
n=1/(4pieEo r3)[3(p1*r^)p2x r^-p2xp1)
Why is this? Than they go about saying that p1*r^=0 than (p2 x p1^)= p1p2

than the torque on the other particle
n=1/(4pieEo r3)[3(p2*r^)p1x r^-p1xp2)
but this time they saild that (p2*r^)=p

why is p2*r^=p while p1*r^=0?


also this other problem, where it has a point p on z axis and same p2 point on -z axis and ask for torque on p the dipole is pointing perpendicular to the plane in a angle@

n=1/(4pieEo r3)[3(p2*r^)px r^-pxp2)
than somehow they get
n=1/(4pieEo r3)[3(pcos@)(psin@)-p2sin2)

Thanks
 
Last edited:
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This is very difficult to read. I'm going to attempt to fill in what the equations should be. Please correct them if they're wrong, and then someone may try to answer. Take a look at the TeX source and you can learn how to type this kind of thing.

leonne said:

Homework Statement


This is a physics problem. need to find torque on dipoles
the
the 2 dipoles are on the same plane with distance r p1 is pointing up while p2 is pointing right

Homework Equations


\mathbf{N} = \mathbf{P} \times \mathbf{E}

The Attempt at a Solution


I know how to do a cross product, you make the matrix with xyz, but not sure about this

\mathbf{E} = \frac1{4\pi\epsilon_0 r^3}(3(\mathbf{p}_1 \cdot \hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{p}_1)

than after the cross product

\mathbf{n} = \frac1{4\pi\epsilon_0 r^3}(3(\mathbf{p}_1 \cdot \hat{\mathbf{r}})\mathbf{p}_2 \times \hat{\mathbf{r}} - \mathbf{p}_2 \times \mathbf{p}_1)

Why is this? Than they go about saying that \mathbf{p}_1 \cdot \hat{\mathbf{r}} = 0 than (p2 x p1^)= p1p2 [I can't tell what this one is supposed to be]

than the torque on the other particle

\mathbf{n} = \frac1{4\pi\epsilon_0 r^3}(3(\mathbf{p}_2 \cdot \hat{\mathbf{r}})\mathbf{p}_1 \times \hat{\mathbf{r}} - \mathbf{p}_1 \times \mathbf{p}_2)

but this time they saild that \mathbf{p}_2 \cdot \hat{\mathbf{r}} = p

why is \mathbf{p}_2 \cdot \hat{\mathbf{r}} = p while \mathbf{p}_1 \cdot \hat{\mathbf{r}} = 0?
 
lol thanks
 

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