Two large, parallel plates are located in vacuum. One of them serves as a cathode, a source of electrons whose initial velocity is negligible. An electron flow directed towards the opposite plate produces a space charge causing the potential in the gap between the plates to vary as $V\mathit{=}a{x}^{\frac{\mathit{4}}{\mathit{3}}}$ , where $a$ is a positive constant, and $x$ is the distance from the cathode.
The volume density of space charge as a function of $x$ is $\frac{\mathit{-}\mathit{k}}{\mathit{9}}a{\epsilon }_{\mathit{0}}{x}^{\frac{\mathit{-}\mathit{2}}{\mathit{3}}}$  where ‘$k$’ is

Consider a volume $dv=l\left(dx\right)b$ at a distance $x$   from left plate. $l,b$ = mutually perpendicular dimensions of the element normally into the paper. Electric flux entering into this volume is due to the electric field $E+dE$ and that leaving is due to the electric field  $E$. Net electric flux through that closed surface,
$d\varphi =-\left(dE\right)A$

$V=a{x}^{\frac{4}{3}}\Rightarrow E=\frac{-dV}{dx}=-\frac{4a}{3}{x}^{\frac{1}{3}}$

$\frac{dE}{dx}=\frac{4a}{9}\frac{1}{x^{\frac{2}{3}}}$

Magnitude of electric field is $\frac{4a}{3}{x}^{\frac{1}{3}}$ which increases with $x$ . Hence electric flux entering the differential volume from right is more than that leaving the volume from left. So, charge inside the volume is negative.
$d\varphi =-\left(dE\right)A$$=-\frac{4a}{9}\frac{dx}{x^{\frac{2}{3}}}lb$  …(1)
From Gauss law, $d\varphi =\frac{dq}{{\epsilon }_0}$  …(2)
(1) = (2) $\Rightarrow \frac{dq}{{\epsilon }_0}=\frac{-4a}{9}\frac{dx}{x^{\frac{2}{3}}}lb=-\frac{4a}{9}\frac{dv}{x^{\frac{2}{3}}}$
Volume density of charge, $\frac{dq}{dv}=-\frac{4}{9}a{\epsilon }_0{x}^{\frac{-2}{3}}$