= Solution
Write $R=|\mathbf p|>a$. In SI units, the exterior <electrostatic potential> solves
$$
\Delta\Phi=-\frac q{\varepsilon_0}\delta(\mathbf x-\mathbf p)\quad(|\mathbf x|>a),\qquad
\Phi=0\quad(|\mathbf x|=a),\qquad \Phi\to0\quad(|\mathbf x|\to\infty),
$$
with the point-charge singularity $q/(4\pi\varepsilon_0|\mathbf x-\mathbf p|)$. The <image charge for a grounded conducting sphere> choices
$$
\boxed{\mathbf p'=\frac{a^2}{R^2}\mathbf p,\qquad q'=-\frac aR q}
$$
lie inside the sphere. On $|\mathbf x|=a$, direct squaring gives $|\mathbf x-\mathbf p'|=(a/R)|\mathbf x-\mathbf p|$. Therefore
$$
\Phi(\mathbf x)=\frac1{4\pi\varepsilon_0}\left(\frac q{|\mathbf x-\mathbf p|}+\frac{q'}{|\mathbf x-\mathbf p'|}\right)
$$
satisfies the source equation, grounded boundary value and decay at infinity. The <Uniqueness of the Dirichlet problem>, applied to the harmonic difference with its singularity removed, identifies this with the physical exterior potential. Thus its gradient gives the required exterior <electric field>.
The charge feels the induced field, equivalently the field of the <image charge>, excluding its own singular field. Their separation is $(R^2-a^2)/R$, so <Coulomb's law> gives
$$
\boxed{\mathbf F=-\frac{q^2aR}{4\pi\varepsilon_0(R^2-a^2)^2}\,\frac{\mathbf p}{R}.}
$$
Its magnitude is $q^2aR/[4\pi\varepsilon_0(R^2-a^2)^2]$ and its direction is toward the sphere's center, independent of the sign of $q$. No factor one-half belongs in this force; that factor occurs in induced electrostatic energy instead.
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