Electric Field of a Point-Charge Dipole Off-Axis: 04-BS-9, May 2016, Q1

Question 1: Electric Field of a Point-Charge Dipole Off-Axis (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 $+e$ location	$x=-d/2$, with $d/2=10^{-10}$ m
Charge $-e$ location	$x=+d/2$
Field point $A$	$(0,\,d/2,\,0)$
Field point $B$	$(0,\,0,\,d/2)$

Find. The magnitude and direction of $\vec E$ at points $A$ and $B$.

Dipole on the x-axis; both field points sit at perpendicular distance $d/2$ from the midpoint $O$ — the same distance as the half-separation of the charges.

Approach. Superpose the Coulomb fields of the two point charges by vector components at each field point; because $A$ and $B$ are symmetric images of each other (swap $y\leftrightarrow z$) at equal distance from both charges, evaluate one in full and infer the other by symmetry, then confirm numerically.

Distance from each charge to point $A$. $A=(0,d/2,0)$ is displaced $(d/2,d/2,0)$ from $+e$ (at $-d/2,0,0$) and $(-d/2,d/2,0)$ from $-e$ (at $d/2,0,0$); both have the same magnitude: $$r=\sqrt{(d/2)^2+(d/2)^2}=\frac{d}{2}\sqrt2=(10^{-10})\sqrt2=1.4142\times10^{-10}\ \text{m}$$
Superpose the two Coulomb fields at $A$ by components. Each charge contributes magnitude $ke/r^2$ along its own displacement direction (away from $+e$, toward $-e$): $$E_1=\frac{ke}{r^2}=\frac{(8.9918\times10^9)(1.6\times10^{-19})}{(1.4142\times10^{-10})^2}=7.193\times10^{10}\ \text{V/m per charge}$$ The field of $+e$ points away from $+e$, along $(+\hat x+\hat y)/\sqrt2$; the field of $-e$ points toward $-e$, along $(+\hat x-\hat y)/\sqrt2$. Adding component-by-component, the $y$-components ($\pm E_1/\sqrt2$) cancel exactly, while the $x$-components (each $E_1/\sqrt2=5.086\times10^{10}$ V/m) add: $$E_A=2\left(\frac{1}{\sqrt2}\right)(7.193\times10^{10})=\boxed{1.017\times10^{11}\ \text{V/m, along }+x}$$
Point $B$ by symmetry. $B=(0,0,d/2)$ is related to $A$ by simply swapping the roles of $y$ and $z$ — the geometry relative to the two charges (both on the $x$-axis) is identical, so the same cancellation/addition pattern applies with $z$ replacing $y$: $$E_B=\boxed{1.017\times10^{11}\ \text{V/m, along }+x\ \text{(identical to }E_A\text{)}}$$
Cross-check via the closed-form perpendicular-bisector formula. For a point at perpendicular distance $s$ from the midpoint of a $\pm e$ pair of separation $d$, $E=2\cdot\dfrac{ke}{s^2+(d/2)^2}\cdot\dfrac{d/2}{\sqrt{s^2+(d/2)^2}}=\dfrac{ke\,d}{(s^2+(d/2)^2)^{3/2}}$; both $A$ and $B$ have $s=d/2$, giving $E=\dfrac{ke}{\sqrt2\,(d/2)^2}=1.017\times10^{11}$ V/m — matching Step 2/3 exactly.

Quantity	Result
$\vec E$ at $A=(0,d/2,0)$	$1.017\times10^{11}$ V/m, along $+x$
$\vec E$ at $B=(0,0,d/2)$	$1.017\times10^{11}$ V/m, along $+x$ (identical)

Concept & Theory

Topic. Coulomb superposition for a two-charge (dipole) system, evaluated off the perpendicular-bisector far zone.

Key relations. Point-charge field $E=kq/r^2$; equatorial-plane dipole field $E=\dfrac{ke\,d}{(s^2+(d/2)^2)^{3/2}}$, directed antiparallel to the dipole moment $\vec p$ (which points from $-e$ to $+e$, i.e. along $-x$, so $\vec E$ here points along $+x$).

Why this works. Both field points are equidistant from both charges (an isosceles configuration), so the components perpendicular to the dipole axis cancel by symmetry while the components along the axis reinforce — and because $A$ and $B$ occupy geometrically identical positions relative to the $x$-axis (only the labelling of the transverse coordinate differs), the two answers must be identical, which is a useful self-check.

Common pitfall. Using the far-field dipole approximation $E\approx p/(4\pi\varepsilon_0 s^3)$, which requires $s\gg d$; here $s=d/2$ is comparable to the charge separation, so only the exact superposition (or the exact perpendicular-bisector closed form used above, not the small-angle far-field limit) is valid.

Source. Sadiku, Elements of Electromagnetics, Ch. 4 (Coulomb's law, superposition, dipole fields).

Question 1 of 8: Electric Field of a Point-Charge Dipole Off-Axis

Question 1: Electric Field of a Point-Charge Dipole Off-Axis (20 marks)