= Solution
\b[The boundary-value method.] In a charge-free region, the <electric potential> obeys the <Laplace equation>, $\nabla^2\Phi=0$, and $\mathbf E=-\nabla\Phi$. Conductors impose equipotential boundaries, with specified voltages or charges, and a far-field condition fixes the remaining behaviour. The <Uniqueness of the Dirichlet problem> guarantees that a solution meeting the prescribed conductor potentials is the physical one. To see why, the difference $w$ of two candidate solutions is harmonic and vanishes on the boundaries; integration by parts gives $\int|\nabla w|^2\,dV=0$, hence no difference in their fields. An exterior problem also needs the appropriate condition at infinity.
\b[Two axisymmetric harmonic functions.] For $\Phi=f(r)\cos\theta$ in <spherical coordinates>, the <Laplacian> reduces to
$$
\nabla^2\Phi=\frac{\cos\theta}{r^2}
\left[\frac{d}{dr}(r^2f')-2f\right].
$$
For $f=r$, the bracket is $2r-2r=0$. For $f=r^{-2}$, it is $2r^{-2}-2r^{-2}=0$. Therefore
$$
\boxed{\nabla^2(r\cos\theta)=0,\qquad
\nabla^2(\cos\theta/r^2)=0\quad(r>0).}
$$
The second function is singular at the origin, which is allowed here because that point lies inside the excluded conductor. The printed spherical Laplacian has a typo in its azimuthal term: its denominator should be $r^2\sin^2\theta$. Both functions are axisymmetric, so that term vanishes and the typo does not affect these calculations.
\b[Potential and field geometry.] Set the lower conductor to zero potential and the distant upper plate to $V$. The small boss sees approximately uniform background field $-E_0\widehat{\mathbf z}$, where $E_0=V/d$. Combining the two harmonic functions to make $\Phi=0$ on $r=a$ gives the <conducting hemispherical protrusion in a uniform field>,
$$
\boxed{\Phi=E_0\left(r-\frac{a^3}{r^2}\right)\cos\theta
=E_0z\left(1-\frac{a^3}{r^3}\right).}
$$
It also vanishes on the surrounding plane $z=0$. The <electric field> components are
$$
E_r=-E_0\left(1+\frac{2a^3}{r^3}\right)\cos\theta,\qquad
E_\theta=E_0\left(1-\frac{a^3}{r^3}\right)\sin\theta.
$$
Equipotentials crowd near the apex, and field lines meet the conductor normally.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2018/paper-1-b6-hemisphere-field.png]
{title=Original calculated local equipotentials and electric-field lines above the hemispherical boss; this is the small-boss approximation rather than an exact finite-gap solution}
The formula has the correct uniform far field but does not exactly impose a flat upper equipotential at finite $d$. It is the leading approximation justified by $a\ll d$; an exact finite-gap solution would include further corrections.
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