= Solution
The <method of images> replaces the earthed plane by an image charge $-q$ at $(-a,0,0)$. The two Coulomb potentials cancel at $x=0$, so uniqueness for the <Dirichlet problem> makes the resulting field the physical field in $x>0$. With $\mathbf r=(x,y,z)$ and $\mathbf e_x=(1,0,0)$,
$$
\boxed{\Phi(\mathbf r)=\frac{q}{4\pi\epsilon_0}
\left(\frac1{|\mathbf r-a\mathbf e_x|}
-\frac1{|\mathbf r+a\mathbf e_x|}\right)}
$$
and
$$
\boxed{\mathbf E(\mathbf r)=\frac{q}{4\pi\epsilon_0}
\left(
\frac{\mathbf r-a\mathbf e_x}{|\mathbf r-a\mathbf e_x|^3}
-\frac{\mathbf r+a\mathbf e_x}{|\mathbf r+a\mathbf e_x|^3}
\right)}.
$$
The <electric multipole expansion> from part (a) gives, for $r\gg a$,
$$
\Phi(\mathbf r)
=\frac{q}{4\pi\epsilon_0}\frac{2ax}{r^3}+O(r^{-4})
=\frac{\mathbf p\mathbin{\cdot}\mathbf r}{4\pi\epsilon_0r^3}+O(r^{-4}),
\qquad
\boxed{\mathbf p=2qa\,\mathbf e_x}.
$$
Thus the leading field is that of an <electric dipole>.
On the plane,
$$
E_x(0,y,z)
=-\frac{2qa}{4\pi\epsilon_0(a^2+y^2+z^2)^{3/2}}.
$$
Taking the plane normal to be $+\mathbf e_x$ and using <polar coordinates>,
$$
\int_{\mathbb R^2}E_x\,dy\,dz
=-\frac{qa}{\epsilon_0}\int_0^\infty
\frac{\rho\,d\rho}{(a^2+\rho^2)^{3/2}}
=\boxed{-\frac q{\epsilon_0}}.
$$
With the outward normal of the region $x>0$, the sign is reversed. This is consistent with <Gauss's law>: all electric flux from the real charge terminates on the grounded conductor.
The <electrostatic boundary condition> gives the induced <surface charge density>
$$
\boxed{\sigma(y,z)=\epsilon_0E_x(0,y,z)
=-\frac{qa}{2\pi(a^2+y^2+z^2)^{3/2}}},
$$
and its integral is
$$
\boxed{Q_{\rm induced}=-q}.
$$
In the plane $z=0$, the field lines leave the positive charge, meet the conductor normally, and are the right-half-plane portions of the field lines joining the real charge to its negative image.
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