= Solution
For this <space-charge-filled parallel-plate capacitor>, place the first plate at $x=0$ and the second at $x=d=x_o$, ignoring <fringing field of a capacitor>. With <vacuum permittivity> between the plates and fixed <charge density> $\rho$, <Poisson's equation> and the prescribed boundary <electric potentials> give
$$
\frac{d^2V}{dx^2}=-\frac{\rho}{\epsilon_0},\qquad V(0)=V_1,\quad V(d)=V_2.
$$
Integrating twice and imposing both values yields
$$
\boxed{V(x)=V_1+\frac{V_2-V_1}{d}x+\frac{\rho}{2\epsilon_0}x(d-x).}
$$
The <electric field> is minus the <gradient> of the <electric potential>, hence
$$
\boxed{E_x(x)=\frac{V_1-V_2}{d}+\frac{\rho}{\epsilon_0}\left(x-\frac d2\right),\qquad
|\mathbf E(x)|=\left|\frac{V_1-V_2}{d}+\frac{\rho}{\epsilon_0}\left(x-\frac d2\right)\right|.}
$$
Thus the <electric potential> is a parabola, concave downward for $\rho>0$ and upward for $\rho<0$. The signed <electric field> is a straight line. Its magnitude has a V-shaped corner if its zero
$$
x_* =\frac d2-\frac{\epsilon_0(V_1-V_2)}{\rho d}
$$
lies inside the gap; otherwise its magnitude is a single straight segment. An interior zero requires $|V_1-V_2|<|\rho|d^2/(2\epsilon_0)$. At that point the <electric potential> has its interior extremum. For $\rho=0$, these reduce to a linear <electric potential> and a uniform <electric field>.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2019-resit-space-charge-capacitor.png]
{title=Potential and electric-field magnitude in a parallel-plate capacitor with fixed space charge}
{description=Illustrative dimensionless curves for positive <charge density>, using $u=x/d$, $W=\epsilon_0(V-V_2)/(\rho d^2)$ and $\delta=\epsilon_0(V_1-V_2)/(\rho d^2)$. The two choices show an interior field reversal and a field with one sign throughout. They do not assume numerical source data.}
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