- #1

stunner5000pt

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Griffith's E&M problem 4.5 page 165

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

[tex] 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)[/tex]

theta is pi/2 so

[tex]E = \frac{-p_{1}}{4 \pi \epsilon_{0} r^3} [/tex]

then the magnitude of torque on p2 is

[tex] N = p_{2} \times \frac{-p_{1}}{4 \pi \epsilon_{0} r^3} [/tex]

p2 points in the y

p1 in teh z

y cross z is positive x

is this correct??

**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

[tex] 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)[/tex]

theta is pi/2 so

[tex]E = \frac{-p_{1}}{4 \pi \epsilon_{0} r^3} [/tex]

then the magnitude of torque on p2 is

[tex] N = p_{2} \times \frac{-p_{1}}{4 \pi \epsilon_{0} r^3} [/tex]

p2 points in the y

p1 in teh z

y cross z is positive x

is this correct??

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