Potential Due to an Electric Dipole — Question 6

Question 6

Rotating a dipole in an external voltage gradient changes its interaction energy

A small rigid dipole has charges $\pm q$ separated by vector d from minus to plus. An external uniform field is $E\hat{\mathbf z}$ with potential $V_{\mathrm{ext}}=-Ez+C$. Its moment magnitude is p=qd.

1. Sum the two source-potential energies and obtain the orientation-dependent interaction energy.

2. Find the quasistatic external work rotating it from $30^\circ$ to $120^\circ$ relative to the field.

Question 6 - Solution

Rotating a dipole in an external voltage gradient changes its interaction energy

(a) Method and calculation

$$U=qV_{\mathrm{ext}}(z_+)-qV_{\mathrm{ext}}(z_-)=-qE(z_+-z_-).$$$$U=-pE\cos\theta;\quad qC-qC=0.$$

(b) Interpretation and further calculation

$$W_{\mathrm{ext}}=U_f-U_i=pE(\cos30^\circ-\cos120^\circ).$$$$W_{\mathrm{ext}}=\frac{\sqrt3+1}{2}pE\approx1.36603pE.$$

Physical check

The internal energy of the rigid dipole is unchanged; only its interaction with the external field contributes to the rotation work.