Electric Field on the Equatorial Plane of a Dipole Potential: 04-BS-9, December 2016, Q7

Question 7: Electric Field on the Equatorial Plane of a Dipole Potential (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Potential	$V(x,y,z)=kpz/r^3$ — the standard electric-dipole potential, dipole moment $p$ along $z$
Dipole moment $p$	$1.6\times10^{-29}$ C·m
Field point	$(a,0,0)$, $a=10^{-10}$ m — on the equatorial ($z=0$) plane

Find. The components $(E_x,E_y,E_z)$ of $\vec E=-\nabla V$ at $(a,0,0)$.

Field point lies on the equatorial plane ($z=0$) of the dipole, perpendicular to the dipole axis. By symmetry the field here is purely axial, antiparallel to the dipole moment.

Approach. Differentiate $V=kpz\,(x^2+y^2+z^2)^{-3/2}$ directly to get $\vec E=-\nabla V$ as a general function of $(x,y,z)$, then evaluate at $(a,0,0)$.

General gradient components. $$E_x=-\frac{\partial V}{\partial x}=\frac{3kpxz}{r^5},\qquad E_y=-\frac{\partial V}{\partial y}=\frac{3kpyz}{r^5},\qquad E_z=-\frac{\partial V}{\partial z}=-\frac{kp}{r^3}+\frac{3kpz^2}{r^5}$$
Evaluate at $(a,0,0)$, i.e. $x=a,\,y=0,\,z=0,\,r=a$. Every term carrying a factor of $z$ vanishes: $$E_x=\frac{3kp(a)(0)}{a^5}=0,\qquad E_y=0,\qquad E_z=-\frac{kp}{a^3}+0=-\frac{kp}{a^3}$$
Numeric value of $E_z$. $$E_z=-\frac{(8.992\times10^9)(1.6\times10^{-29})}{(10^{-10})^3}$$ $$\boxed{\vec E=(0,\ 0,\ -1.439\times10^{11})\ \text{V/m}}$$

Quantity	Result
$E_x$	$0$
$E_y$	$0$
$E_z$	$-1.439\times10^{11}$ V/m

Concept & Theory

Topic. Electric field of a point dipole, obtained by direct differentiation of the given potential rather than from the memorized closed-form dipole-field formula.

Key relations. Standard dipole potential $V=kp\cos\theta/r^2=kpz/r^3$; equatorial-plane field $E=-kp/r^3$ (antiparallel to $\vec p$), exactly half the magnitude of the on-axis field $E=2kp/r^3$ (parallel to $\vec p$) at the same distance — a useful cross-check on any dipole-field result.

Why this works. On the equatorial plane every point is equidistant from the "+" and "−" halves of the dipole, so the potential is exactly zero there; but the FIELD is not zero because the two halves' field contributions do not cancel — they add along $-\hat z$ (antiparallel to $\vec p$), which is exactly what the direct gradient calculation recovers.

Common pitfall. Assuming the field vanishes on the equatorial plane because the potential does there (potential and field are different objects; a zero of $V$ says nothing about $\nabla V$), or reusing the on-axis field magnitude $2kp/r^3$ at an off-axis point.

Source. Sadiku, Elements of Electromagnetics, Ch. 4 (electric dipole potential and field).

Question 7 of 8: Electric Field on the Equatorial Plane of a Dipole Potential

Question 7: Electric Field on the Equatorial Plane of a Dipole Potential (20 marks)