electric dipole in E field

Does an electric dipole in a uniform E field precess? PhL 3.24.17 Basically no web hits on this subject, lots on mag dipole in B field. 1. What do we know about a dipole in an E field? Start in kindergarten with this picture Upper right is 1, lower left is 2. Then F1 = qE = qE r1 = rsinθ + rcosθ F2 = (-q)E = -qE = - F1 r2 = - r1 N1 = r1x F1 = [rsinθ + rcosθ ] x [ qE] = rsinθqE N2 = r2x F2 = (-r1)x (-F1) = N1 = rsinθqE We have a massless stick connecting the charges. The total torque on this dumbbell is (about origin) N = N1+ N2 = 2N1 = 2rsinθqE Now define the dipole moment p as p ≡ 2rq p ≡ 2rq Then N = 2rsinθqE = pEsinθ Now consider p x E = [ 2rq] x [E] = [p] x [E] = pE x = pE sinθ Therefore we have shown that N = p x E where torque is about the origin. You can imagine increasing q and decreasing r in a way that the product is constant, then you could have p very close to the origin. So kindergarten class, we know that there is a torque on a dipole in an electric field given by N = p x E . This will be true in general, not just for the above picture. 2. Now suppose the two charges in the above picture are in motion at t = 0 where q is going into paper at v, and -q is coming out of paper at v. Then v1 = -v v2 = v Since we are introducing "motion" into the problem, assume the two changes have same mass m. Then L1 = r1 x (mv1) = r x (-mv) = -rmv x = -rmv [sinθ + cosθ ] x = -rmv [ -sinθ + cosθ ] = rmv [ sinθ - cosθ ] L2 = r2 x (mv2) = (-r1) x (-mv1) = L1 So just for interest, the dumbbell has total angular momentum L = L1 + L2 = = 2L1 = 2rmv [ sinθ - cosθ ] We do not have torque free motion of the dumbbell because E makes a torque on it, N = p x E . The equation of motion is then N = Now write L = 2 r1 x (mv1) = 2m r1 x 1 We then have equation of motion 2m r1 x 1 = p x E