Torque on Dipole Griffith EM 4.5 p. 165

[stunner5000pt]

Griffith's E&M problem 4.5 page 165

In the figure p1 and p2 are perfect dipoles a disantce r apart. What is the torque on p1 due to p2.? Wjat is the torque on p2 due to p1?
the second part is done in post #4
p1 is located on the right pointing upward
p2 is a distance r from p2 and is oriented poitning right

ok first of all the field at p2 due to p1 is

E = \frac{1}}{4 \pi \epsilon_{0} r^3} (3(\vec{p}\bullet\hat{r})-\vec{p}) = \frac{p_{1}}{4 \pi \epsilon_{0} r^3} (3pr\cos\theta - p)

theta is pi/2 so
E = \frac{-p_{1}}{4 \pi \epsilon_{0} r^3}

then the magnitude of torque on p2 is
N = p_{2} \times \frac{-p_{1}}{4 \pi \epsilon_{0} r^3}

p2 points in the y
p1 in the z
y cross z is positive x
is this correct??

 

[marcusl]

Looks mostly right. Two small comments:
1) you left out an r-hat in the first equation
2) y cross z is plus x as you say but E has a minus sign so torque is along -x

 

[stunner5000pt]

Looks mostly right. Two small comments:
1) you left out an r-hat in the first equation
2) y cross z is plus x as you say but E has a minus sign so torque is along -x

good point i forgot about hte minus sign
it is -x

thanks a lot!

 

[stunner5000pt]

this time r points to the left (correct?)
so the angle between p2 and r is -pi? E =\frac{p_{1}}{4 \pi \epsilon_{0} r^3} (3(\vec{p_{2}}\bullet\hat{-r} \cos\theta - \vec{p}) = \frac{p_{1}}{4 \pi \epsilon_{0} r^3} (-3p_{2}r\cos\theta - \vec{p_{2}}) = \frac{1}{4 \pi \epsilon_{0} r^3} 2p_{2}

is that correct? the electric field points to the right??

now let s see if that makes snese intuitively

suppose you are 'behind' the negative charge of a dipole very far away such that r>> d (d is the separation of the dipole) then the electric field due to the two charges points toward the dipole?? because the negative cahrge 'slightly' dominates over the positive charge?

the torque as a result is

\vec{N} = p_{1} \hat{z} \times \frac{2p_{2}}{4 \pi \epsilon_{0} r^3} \hat{y} = \frac{2p_{1}p_{2}}{4 \pi \epsilon_{0} r^3} (-\hat{x})

the answers are differnet in magnitude but same direction