= Solution
The original PDF places this complete question before B6 without a visible question number. It is included here under a descriptive heading.
\b[Why images work.] In the <method of images>, place fictitious <image charges> outside the region where the physical <electric field> is required. Choose them so that the combined <electric potential> satisfies the conductor <boundary conditions>. The <Uniqueness of the Dirichlet problem> then proves that this construction gives the physical exterior field. The images do not describe the actual surface-charge locations.
Two examples are a charge $q$ above a grounded conducting plane, represented by $-q$ at the mirror point, and the <image charge for a grounded conducting sphere> used below. In each case the image lies inside the excluded conductor and makes its surface an equipotential of the required value.
\b[Potential of the two charges.] By the <law of cosines>, the distances from the observation point to charges at $x_j$ are $s_j=(r^2+x_j^2-2rx_j\cos\theta)^{1/2}$. The <superposition principle> and <Coulomb's law> therefore give
$$
\boxed{\phi(r,\theta)=\frac{q}{4\pi\varepsilon_0}
\left[\frac1{\sqrt{r^2+x_2^2-2rx_2\cos\theta}}
-\frac{\alpha}{\sqrt{r^2+x_1^2-2rx_1\cos\theta}}\right].}
$$
\b[The grounded sphere.] Set $r=a$ and require $s_1=\alpha s_2$ at every polar angle. Matching the coefficients of $\cos\theta$ and the constant terms gives
$$
x_1=\alpha^2x_2,\qquad a^2+x_1^2=\alpha^2(a^2+x_2^2).
$$
The useful noncoincident solution is
$$
\boxed{x_1=\frac{a^2}{x_2},\qquad \alpha=\frac a{x_2},\qquad
q'=-\frac{aq}{x_2}.}
$$
Indeed $s_1=(a/x_2)s_2$ on the sphere, so the two potential terms cancel there. A <Gaussian surface> enclosing the sphere but not the external charge measures the total image charge in the exterior representation. Thus the physical induced net charge is
$$
\boxed{Q_{\rm sphere}=-\frac{aq}{x_2}.}
$$
\b[An isolated neutral sphere.] The grounded image would give nonzero net charge. Add a central image $q_0=+aq/x_2$ to cancel it. A central image changes the surface potential by a constant, preserving the equipotential condition, whereas an arbitrary off-centre image would generally spoil it. The <image charges for a neutral conducting sphere> give
$$
\boxed{q_0=\frac{aq}{x_2},\qquad
\phi_{\rm sphere}=\frac{q_0}{4\pi\varepsilon_0a}
=\frac{q}{4\pi\varepsilon_0x_2}.}
$$
The potential is measured relative to zero at infinity for each finite $x_2$.
\b[Fixed applied field and dipole moment.] For positive $q$ on the positive $x$ axis, the applied central field is $\mathbf E_0=-q\widehat{\mathbf x}/(4\pi\varepsilon_0x_2^2)$. The central image contributes no <electric dipole moment>, so
$$
\mathbf p=q'x_1\widehat{\mathbf x}
=-\frac{qa^3}{x_2^2}\widehat{\mathbf x}
=\boxed{4\pi\varepsilon_0a^3\mathbf E_0.}
$$
Its units are coulomb metres. Keeping $E_0$ fixed makes $q/x_2^2$ constant, so this induced moment is independent of the distance.
\b[The distant-charge limit.] As $x_2\to\infty$ with $q/x_2^2$ fixed, the applied field becomes uniform over any fixed neighbourhood of the sphere. The neutral pair of opposite images approaches a point dipole of the finite moment above. The external potential contains a growing spatial constant $q/(4\pi\varepsilon_0x_2)$, so the infinity-referenced conductor potential does not have a finite limit. Subtracting this irrelevant constant gives the usual <induced charge on a conducting sphere in a uniform electric field>; the exterior potential becomes
$$
\boxed{\phi=-\mathbf E_0\cdot\mathbf r
+\frac{a^3\,\mathbf E_0\cdot\mathbf r}{r^3},\qquad r\ge a.}
$$
This has zero potential on the sphere, but a uniform applied field cannot simultaneously have its potential vanish at spatial infinity.
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