A Dipole in an Electric Field — Question 7

Section 22-8 · Physics pages 3624–3625

Question 7 An inclined dipole in a saddle field experiences both translation and rotation An ideal permanent point dipole is centered at (x, y) = (0, b) in the static field E = (Gx, −Gy, 0), with G, b > 0. Its instantaneous moment is p = p(ˆx + ˆy)/ √ 2. Use U = −p · E and hold orientation fixed when taking spatial derivatives. x y (0, b) p E Schematic; labels, not drawing lengths, specify the data. (a) Derive the general force for fixed components (px, py) and evaluate it for the stated orientation. (b) Calculate the torque about the center and explain why zero total charge does not imply zero force.
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Question 7 - Solution An inclined dipole in a saddle field experiences both translation and rotation x y (0, b) p E (a) Method and calculation U = −G(pxx −pyy), F = −∇U = (Gpx, −Gpy, 0). F = Gp √ 2(1, −1, 0). (b) Interpretation and further calculation E(0, b) = (0, −Gb, 0), τz = pxEy −pyEx. τ = −pGb √ 2 ˆz. The field differs between the two constituent charge locations, leaving a net force in the dipole limit. The local field also tends to turn the moment clockwise. Physical check The partial derivatives act on the external field, not on a moment allowed to reorient during the derivative.

Question 7

An inclined dipole in a saddle field experiences both translation and rotation

An ideal permanent point dipole is centered at (x,y)=(0,b)(x,y)=(0,b) in the static field 𝑬=(Gx,−Gy,0)\mathbf E=(Gx,-Gy,0), with G,b>0G,b>0. Its instantaneous moment is 𝒑=p(𝒙̂+𝒚̂)/2\mathbf p=p(\hat{\mathbf x}+\hat{\mathbf y})/\sqrt2. Use U=−𝒑⋅𝑬U=-\mathbf p\cdot\mathbf E and hold orientation fixed when taking spatial derivatives.

Diagram shown in the original SVG worksheet.

  1. Derive the general force for fixed components (px,py)(p_x,p_y) and evaluate it for the stated orientation.

  2. Calculate the torque about the center and explain why zero total charge does not imply zero force.

Question 7 - Solution

An inclined dipole in a saddle field experiences both translation and rotation

Diagram shown in the original SVG worksheet.

(a) Method and calculation

U=−G(pxx−pyy),𝑭=−𝜵U=(Gpx,−Gpy,0).U=-G(p_xx-p_yy),\quad\mathbf F=-\boldsymbol\nabla U=(Gp_x,-Gp_y,0).𝑭=Gp2(1,−1,0).\mathbf F=\frac{Gp}{\sqrt2}(1,-1,0).

(b) Interpretation and further calculation

𝑬(0,b)=(0,−Gb,0),τz=pxEy−pyEx.\mathbf E(0,b)=(0,-Gb,0),\quad\tau_z=p_xE_y-p_yE_x.𝝉=−pGb2𝒛̂.\boldsymbol\tau=-\frac{pGb}{\sqrt2}\hat{\mathbf z}.The field differs between the two constituent charge locations, leaving a net force in the dipole limit. The local field also tends to turn the moment clockwise.

Physical check

The partial derivatives act on the external field, not on a moment allowed to reorient during the derivative.