Potential Due to an Electric Dipole — Question 5

Section 24-7 · Physics pages 3920–3921

Question 5 An equatorial zero voltage retains a nonzero transverse derivative An ideal dipole points along positive z. Its potential is V (x, y, z) = kpz/(x2 + y2 + z2)3/2. Consider P=(R,0,0), R>0. z, p x P Schematic; labels, not drawing lengths, specify the data. (a) Evaluate V at P and calculate Ez = −∂V/∂z there. (b) Find the other field components and compare this result with the axial field magnitude at the same radius.
Show solutionHide solution
Question 5 - Solution An equatorial zero voltage retains a nonzero transverse derivative z, p x P (a) Method and calculation VP = 0, ∂V ∂z = kp ( r−3 −3z2r−5) . Ez(P) = −kp/R3. (b) Interpretation and further calculation Ex = 3kpxz/r5 = 0, Ey = 3kpyz/r5 = 0 at P. Eaxis = 2kp/R3 = 2|EP |. Physical check A function can vanish throughout a plane while varying in the direction normal to that plane.

Question 5

An equatorial zero voltage retains a nonzero transverse derivative

An ideal dipole points along positive z. Its potential is V(x,y,z)=kpz/(x2+y2+z2)3/2V(x,y,z)=kpz/(x^2+y^2+z^2)^{3/2}. Consider P=(R,0,0), R>0.

Diagram shown in the original SVG worksheet.

  1. Evaluate V at P and calculate Ez=−∂V/∂zE_z=-\partial V/\partial z there.

  2. Find the other field components and compare this result with the axial field magnitude at the same radius.

Question 5 - Solution

An equatorial zero voltage retains a nonzero transverse derivative

Diagram shown in the original SVG worksheet.

(a) Method and calculation

VP=0,∂V∂z=kp(r−3−3z2r−5).V_P=0,\quad\frac{\partial V}{\partial z}=kp\left(r^{-3}-3z^2r^{-5}\right).Ez(P)=−kp/R3.E_z(P)=-kp/R^3.

(b) Interpretation and further calculation

Ex=3kpxz/r5=0,Ey=3kpyz/r5=0at P.E_x=3kpxz/r^5=0,\quad E_y=3kpyz/r^5=0\quad\text{at P}.Eaxis=2kp/R3=2|𝑬P|.E_{\mathrm{axis}}=2kp/R^3=2|\mathbf E_P|.

Physical check

A function can vanish throughout a plane while varying in the direction normal to that plane.