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+### Statement
+
+$6.6.14.$ [Insert the problem statement]
+
+### Solution
+
+The electric field between plates is:
+
+
+\begin{equation}
+E=\frac{q}{\varepsilon\varepsilon_0S}
+\end{equation}
+
+
+Plugging in the $\varepsilon$ from the problem gives us:
+
+\begin{equation}
+E=\frac{q}{\varepsilon_0^2S}(1+\frac{x}{d})
+\end{equation}
+
+
+
+To find $\rho$ we use Gauss's law in the differential form:
+
+\begin{equation}
+\nabla\cdot\vec E=\frac{\rho}{\varepsilon_0}
+\end{equation}
+
+
+where $\nabla\cdot\vec E$ is:
+
+\begin{equation}
+\nabla\cdot\vec E=\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}
+\end{equation}
+
+
+But we know that $E_y=0$,$E_z=0$ and $E_x=-E$,so:
+
+
+\begin{equation}
+\frac{dE_x}{dx}=\frac{\rho}{\varepsilon_0}
+\end{equation}
+
+
+Differentiating eq(2) with respect to $x$ and substituting it into eq(5), we get:
+
+\begin{equation}
+\rho=-\frac{q}{\varepsilon_0Sd}
+\end{equation}
+
+
+#### Answer
+
+[Insert a concise answer or boxed result]