Dipole in an Uniform Electric Field concept?

fallen186

1. The problem statement, all variables and given/known data
Dipole in an Uniform Electric Field:
torque is calculated about the position of either charge has the magnitude FLsin(x) = qELsin(x) = pEsin(x). The direction of the torque vector is into the paper such that it tends to rotate the dipole moment vector p so it aligns with the direction of E. The torque can be expressed most concisely as the cross product: T = p x E

I don't know why the magnitude is F * L sin(x) or why qE turns into pE. And I don't understand what the concept iit s trying to tell me. It would also be helpful if someone could explain the hand thing for torque. I understand most of the stuff about torque but not the hand thing.
2. Relevant equations
x = theta
L = distance between charges in dipole
p = vector of the dipole movement that points from negtive charge to positive.
p =q*L

3. The attempt at a solution

buffordboy23

This is all given in a standard textbook. Did you overlook it? Perhaps, the magnitude of the torque may be clearer this way: T = F*L*sin(x) = 2*[F*(L/2)*sin(x)] The L/2 is the distance from the center of the dipole to one of its ends, and since there are two ends that experience a torque in the same direction we multiply by 2. Does this help at all?

fallen186

Thanks, that does clear things up

buffordboy23

After I logged off, I realized that I could have been clearer. My description in the previous post is only valid if the center of the mass is equidistant from both of the ends of the dipole, which may not always be the case. Therefore, it's derivation is a "special" case of the more "general" case. There is a better way to describe the magnitude of the torque.

Let "L" be the length of the dipole from end to end. Now suppose that the center of the mass is not equidistant from both ends of the dipole. Then one end lies "y" units away from the center of mass and the other end must lie "L-y" units away. The net torque on the dipole is the sum of the torques: T = F*y*sin(x) + F*(L-y)*sin(x) = F*y*sin(x) + F*L*sin(x) - F*y*sin(x) = F*L*sin(x). Notice that this result is of the same magnitude as before but we got here by a different route.