= Solution
\b[Point-charge field and superposition]. <Coulomb's law> gives the <Coulomb force> on a test <electric charge> $q_t$ a displacement $\mathbf R$ from a source <point charge> $Q$:
$$
\mathbf F=\frac{Qq_t}{4\pi\epsilon_0}\frac{\mathbf R}{R^3}.
$$
The <electric field> is force per infinitesimal test charge, so
$$
\boxed{\mathbf E=\frac{\mathbf F}{q_t}
=\frac{Q}{4\pi\epsilon_0R^2}\hat{\mathbf R}.}
$$
It points outward for positive $Q$ and inward for negative $Q$. For several <point charges>, the <superposition principle> adds their fields as <vectors>:
$$
\mathbf E(\mathbf r)=\frac1{4\pi\epsilon_0}\sum_i
q_i\frac{\mathbf r-\mathbf r_i}{|\mathbf r-\mathbf r_i|^3}.
$$
This formula excludes the source points themselves, where the ideal point-charge field is singular.
\b[Exact field components of the pair]. Put
$$
R_+=\sqrt{(x-a)^2+y^2},\qquad
R_-=\sqrt{(x+a)^2+y^2}.
$$
The displacement vectors from the positive and negative charges to the observation point are $(x-a)\hat{\mathbf i}+y\hat{\mathbf j}$ and $(x+a)\hat{\mathbf i}+y\hat{\mathbf j}$. Including the signs of the charges gives
$$
\mathbf E(x,y)=\frac q{4\pi\epsilon_0}
\left[\frac{(x-a)\hat{\mathbf i}+y\hat{\mathbf j}}{R_+^3}
-\frac{(x+a)\hat{\mathbf i}+y\hat{\mathbf j}}{R_-^3}\right].
$$
Reading off both components,
$$
\boxed{E_x=\frac q{4\pi\epsilon_0}
\left[\frac{x-a}{((x-a)^2+y^2)^{3/2}}
-\frac{x+a}{((x+a)^2+y^2)^{3/2}}\right],}
$$
$$
\boxed{E_y=\frac{qy}{4\pi\epsilon_0}
\left[\frac1{((x-a)^2+y^2)^{3/2}}
-\frac1{((x+a)^2+y^2)^{3/2}}\right].}
$$
For example, on the perpendicular bisector $x=0$, the two $y$ components cancel and the $x$ components add toward the negative charge, checking both relative signs.
\b[Far-field polar components]. The pair's <electric dipole moment> points from negative to positive charge:
$$
\mathbf p=2qa\hat{\mathbf i},\qquad p=2qa.
$$
Use <polar coordinates> $x=r\cos\theta$, $y=r\sin\theta$, with $\theta$ measured from that positive dipole axis. The supplied leading <electric dipole field> becomes
$$
E_x=\frac{p}{4\pi\epsilon_0r^3}(3\cos^2\theta-1),\qquad
E_y=\frac{p}{4\pi\epsilon_0r^3}(3\cos\theta\sin\theta).
$$
Taking the <dot products> with the radial and angular unit vectors gives
$$
\begin{aligned}
E_r&=E_x\cos\theta+E_y\sin\theta
=\frac p{4\pi\epsilon_0r^3}\left[3\cos\theta(\cos^2\theta+\sin^2\theta)-\cos\theta\right],\\
E_\theta&=-E_x\sin\theta+E_y\cos\theta
=\frac p{4\pi\epsilon_0r^3}\sin\theta.
\end{aligned}
$$
Therefore
$$
\boxed{E_r=\frac{2p\cos\theta}{4\pi\epsilon_0r^3},\qquad
E_\theta=\frac{p\sin\theta}{4\pi\epsilon_0r^3}.}
$$
The unit vectors for <polar coordinates> are orthonormal, so the magnitude is
$$
\boxed{|\mathbf E(r,\theta)|=\sqrt{E_r^2+E_\theta^2}
=\frac{p}{4\pi\epsilon_0r^3}\sqrt{1+3\cos^2\theta},\qquad r\gg a.}
$$
The units are $\mathrm{N\,C^{-1}}$, since the <electric dipole moment> has units $\mathrm{C\,m}$.
\b[Angular sketch and comparison]. At fixed large $r$, the magnitude has period $\pi$. It is largest on the dipole axis, $\theta=0,\pi$, with value $2p/(4\pi\epsilon_0r^3)$, and smallest on the perpendicular directions, $\theta=\pi/2,3\pi/2$, with value $p/(4\pi\epsilon_0r^3)$. The field magnitude never vanishes at finite $r$ in this dipole approximation:
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019-ia-physics-q10-dipole-angular-field.png]
{title=The angular dependence of the far-field electric dipole magnitude at a fixed radius}
The neutral charge pair has no net <electric charge>, so its leading far field falls as $r^{-3}$ and depends on direction. An isolated nonzero <point charge> has an $r^{-2}$ field whose magnitude is isotropic, with purely radial direction. The dipole also generally has a nonzero $E_\theta$ component.
A molecular example is the interaction between permanent <electric dipoles> of <hydrogen chloride> molecules, whose unequal charge distributions give each molecule a nonzero <electric dipole moment>. The <electric dipole-dipole interaction> depends on their relative orientations:
$$
U=\frac{\mathbf p_1\cdot\mathbf p_2
-3(\mathbf p_1\cdot\hat{\mathbf r})(\mathbf p_2\cdot\hat{\mathbf r})}
{4\pi\epsilon_0r^3}.
$$
For two parallel dipoles pointing along their line of separation, $U=-2p_1p_2/(4\pi\epsilon_0r^3)<0$, illustrating an attractive molecular configuration.
Back to article page
