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These notes are meant to finish class on 28 January 2010. We were talking about the force on an electric dipole in a non-uniform electric field E. The dipole is made from charges ±q separated by a distance b. The vector b runs along the line from −q to +q. If x is the position of the center of the diple, the x-component of the force is given by Fx = qEx (x + b/2) − qEx (x − b/2) = qb · ∇Ex (x) = p · ∇Ex (x) for p ≡ qb. See page 65 of your textbook. The key step here is a Taylor expansion of the electric field Ex about the point x. In three dimensions, you need to include the derivatives with respect to x, y, and z, so you end up with the gradient. Combining all three components of the electric field, we can then write F(x) = (p · ∇)E(x) where the parentheses only emphasize the fact that p · ∇ = px ∂ ∂ ∂ + py + pz ∂x ∂y ∂z acts individually on each component of the electric field E. We could do exactly the same thing with the potential energy of the dipole. That is U = qΦ(x + b/2) − qΦ(x − b/2) = qb · ∇Φ(x) = p · ∇Φ(x) = −p · E(x) for a dipole located at the position x. (The last step just makes use of the definition of the electric field in terms of the gradient of the electric potential.) This is what I was trying to explain on the board when class was ending. Interestingly the force on the dipole can be written in terms of the gradient of the potential energy, that is F(x) = −∇U (x) = ∇ [p · E(x)] which doesn’t quite look like the equation above. However, the two expressions are exactly equivalent. See Exercise 7 in Chapter 2 of your textbook for an explanation. You can carry out the same steps, essentially, to show that the torque τ on an electric dipole in an electric field E(x) is given by τ =p×E Note that even for a uniform electric field, there is a torque. (The result here does not involve the gradient of the electric field.) This expression is necessary for your homework assignment, due Monday.