Electrostatics > The electric field in the presence of a dielectric > Problem 6.6.14

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]