Understanding Torque and Cross Product Calculations for Dipoles

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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
 
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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


[tex]\mathbf{N} = \mathbf{P} \times \mathbf{E}[/tex]

The Attempt at a Solution


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

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

than after the cross product

[tex]\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)[/tex]

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

than the torque on the other particle

[tex]\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)[/tex]

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

why is [tex]\mathbf{p}_2 \cdot \hat{\mathbf{r}} = p[/tex] while [tex]\mathbf{p}_1 \cdot \hat{\mathbf{r}} = 0[/tex]?