A Dipole in an Electric Field — Question 5

Question 5

A linear field gradient gives an exact force on a finite axial dipole

A rigid dipole lies on the x axis with charge $-q$ at $x-d/2$ and $+q$ at $x+d/2$, where $p=qd>0$. On this axis the external field is $\mathbf E(X)=(E_0+GX)\hat{\mathbf x}$. The orientation is constrained and d remains constant.

1. Sum the two charge forces to obtain the net force, and find the torque about the midpoint.

2. Derive the work when the center moves from $x_1$ to $x_2$ and a corresponding position-dependent potential energy.

Question 5 - Solution

A linear field gradient gives an exact force on a finite axial dipole

(a) Method and calculation

$$F_x=q[E_0+G(x+d/2)]-q[E_0+G(x-d/2)].$$$$F_x=qdG=pG,\quad\boldsymbol\tau=\mathbf0.$$

(b) Interpretation and further calculation

$$W=\int_{x_1}^{x_2}pG\,dx=pG(x_2-x_1).$$$$U(x)=-p(E_0+Gx)+C,\quad-\frac{dU}{dx}=pG.$$For a linear field this result is exact at finite d; a positive gradient pulls a positively oriented dipole toward increasing x.

Physical check

Neutrality cancels the uniform part of the force; it does not cancel the difference in field at the two charges.