Electric Field of a Point-Charge Pair (Dipole): 04-BS-9, May 2013, Q6

Question 6: Electric Field of a Point-Charge Pair (Dipole) (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
Charge magnitude	$1.6\times10^{-19}$ C (elementary charge)
Negative charge position	$(-0.5,\,0,\,0)\times10^{-10}$ m
Positive charge position	$(+0.5,\,0,\,0)\times10^{-10}$ m
Field point $P$	$(0,\,0.707,\,0.707)\times10^{-10}$ m

Find. $\vec E$ (magnitude and direction) at $P$.

Both charges lie on the x-axis, so the system is axially symmetric about it; $P$ can be drawn in the plane containing the x-axis and $P$, at radial distance $\rho=\sqrt{y^2+z^2}$.

Approach. Because both charges sit on the $x$-axis, the configuration is symmetric about that axis, so $P$'s field depends only on its axial coordinate ($x=0$) and its radial distance from the axis, $\rho=\sqrt{0.707^2+0.707^2}\times10^{-10}=1.0\times10^{-10}$ m. Superpose the Coulomb fields of the two point charges as vectors.

Distance from each charge to $P$. By symmetry both distances are equal: $$r_+=r_-=\sqrt{(0.5)^2+(0.707)^2+(0.707)^2}\times10^{-10}=\sqrt{1.25}\times10^{-10}=1.118\times10^{-10}\ \text{m}.$$
Vector superposition. Writing $\vec r_+=P-\vec r_{q+}=(-0.5,0.707,0.707)\times10^{-10}$ and $\vec r_-=P-\vec r_{q-}=(+0.5,0.707,0.707)\times10^{-10}$, Coulomb's law gives $$\vec E=\frac{q}{4\pi\varepsilon_0 r_+^3}\vec r_+ + \frac{(-q)}{4\pi\varepsilon_0 r_-^3}\vec r_-.$$ The $y$ and $z$ components of the two terms are equal in sign and magnitude (same $r_+=r_-$, same $y,z$), so they add for a field of a purely +/- pair evaluated in the perpendicular ($x=0$) plane... in fact substituting numbers shows the $y,z$ terms exactly cancel, leaving only an $x$-component (see Step 3) — the classic result that a dipole's field in its equatorial plane is purely axial, antiparallel to the dipole moment.
Numerical evaluation. With $k=1/4\pi\varepsilon_0=8.99\times10^9$ N·m²/C²: $$\vec E \approx (-1.030\times10^{11},\ 0,\ 0)\ \text{V/m}.$$
Magnitude and direction. $$E=\boxed{1.03\times10^{11}\ \text{V/m}},\quad\text{direction: along }-\hat x\text{ (the equatorial dipole field points from + toward − charge).}$$

Quantity	Result
$r_+=r_-$	$1.118\times10^{-10}$ m
$\vec E$	$(-1.03\times10^{11},\,0,\,0)$ V/m
$\|\vec E\|$	$1.03\times10^{11}$ V/m
Direction	$-\hat x$ (parallel to axis, from + toward − charge)

Concept & Theory

Topic. Superposition of Coulomb fields — the electric dipole's equatorial-plane field.

Key relations. Coulomb's law $\vec E=\dfrac{q}{4\pi\varepsilon_0 r^2}\hat r$; superposition; dipole equatorial field $\vec E=-\dfrac{k\vec p}{r^3}$ for $r\gg$ separation (here used exactly, without the far-field approximation).

Why this works. Any point in the plane perpendicular to the dipole axis through its midpoint is equidistant from both charges, so the axial ($y,z$) components of the two Coulomb fields are identical while the axial-direction ($x$) components add — the mirror symmetry of the charge pair about that plane forces the net field to lie exactly along the axis.

Common pitfall. Assuming the far-field dipole approximation $E\approx kp/r^3$ (valid only for $r\gg$ separation) applies here — with $\rho=1\times10^{-10}$ m and a $1\times10^{-10}$ m charge separation, the field point is comparable in distance to the separation, so the exact vector superposition (not the small-dipole-moment shortcut) is required.

Source. Sadiku, Elements of Electromagnetics, Ch. 4 (Electrostatic Fields — Coulomb's Law and Field Intensity).

Question 6 of 8: Electric Field of a Point-Charge Pair (Dipole)

Question 6: Electric Field of a Point-Charge Pair (Dipole) (20 marks)