P20B2014 – Solid Insulating Cylinder

Given the symmetry of the charge distribution and its infinite length, we use Gauss’s Law to determine the electric field that it generates. As our Gaussian surface, we choose a cylinder of radius $r$ and height $L$ centered on the axis of revolution of the insulating cylinder. We then derive the electric field inside the insulating cylinder $\left(rR\right)$.

Electric field inside the insulating cylinder $\left(r < R\right)$:

We choose a Gaussian cylinder of radius $r<R$ and height $L$, centered on the axis of revolution of the insulating cylinder and with an outward normal $d\overrightarrow{A}$.

The electric field is constant in magnitude and direction on each surface of the Gaussian cylinder and, since it is orthogonal to the normal $d\overrightarrow{A}$ of the top and bottom (end caps), we conclude that the electric flux through the Gaussian cylinder is equal to

\begin{aligned}
\oiint_A{\overrightarrow{E}\cdot d\overrightarrow{A}}&=\iint_{A_1}{\overrightarrow{E}\cdot d\overrightarrow{A}}+\iint_{A_2}{\overrightarrow{E}\cdot d\overrightarrow{A}}+\iint_{A_3}{\overrightarrow{E}\cdot d\overrightarrow{A}} \\
&=0+\iint_{A_3}{Ecos\left(0\right)dA}+0 \\
&=E2\pi rL
\end{aligned}

The cylinder carries a non-uniform charge density and we therefore must use integration to derive the charge enclosed by the Gaussian cylinder. Recalling that the infinitesimal volume for cylindrical coordinates is given by $dV=rdrd\theta dz$, we compute the infinitesimal charge $dQ_{enc}=\rho \left(r\right)dV$ enclosed in $dV$ and then integrate over the enclosed volume of charge.

Thus, the charge enclosed by the Gaussian surface is derived by integration as follows

\begin{aligned}
Q_{enc}&=\iiint_V{\rho \left(r\right)dV} \\
\\
&=\iiint_V{\frac{\alpha }{r}e^{-\ r/a}rdrd\theta dz} \\
\\
&=\int^L_0{\int^{2\pi }_0{\int^r_0{\alpha e^{-\ r/a}drd\theta dz}}} \\
\\
&=2\pi L\alpha \left(-a\right){\left[e^{-\ r/a}\right]}^r_0 \\
\\
&=2\pi L\alpha a\left(1-e^{-\ r/a}\right)
\end{aligned}

Finally, by Gauss’s Law we derive the electric field inside the insulating cylinder

E2\pi rL=\frac{2\pi L\alpha a}{{\varepsilon }_0}\left(1-e^{-\ r/a}\right)\ \ \ \ \Rightarrow \ \ \ \ \ \boxed{{\overrightarrow{E}}_1=\frac{a\alpha }{{\varepsilon }_0r}\left(1-e^{-\ r/a}\right)\ \hat{r}}

Electric field outside the insulating cylinder $\left(r>R\right)$:

We choose a Gaussian cylinder of radius $r>R$ and height $L$, centered on the axis of revolution of the insulating cylinder and with an outward normal $d\overrightarrow{A}$.

The electric field is constant in magnitude and direction on each surface of the Gaussian cylinder and, since it is orthogonal to the normal $d\overrightarrow{A}$ of the top and bottom (end caps), we conclude that the electric flux through the Gaussian cylinder is equal to

\begin{aligned}
\oiint_A{\overrightarrow{E}\cdot d\overrightarrow{A}}&=\iint_{A_1}{\overrightarrow{E}\cdot d\overrightarrow{A}}+\iint_{A_2}{\overrightarrow{E}\cdot d\overrightarrow{A}}+\iint_{A_3}{\overrightarrow{E}\cdot d\overrightarrow{A}} \\
&=0+\iint_{A_3}{Ecos\left(0\right)dA}+0 \\
&=E2\pi rL
\end{aligned}

With a similar process, we compute the charge enclosed by the Gaussian surface by integrating $dQ_{enc}=\rho \left(r\right)dV$ over the solid cylinder (since only the solid cylinder encloses charge).

The charge enclosed by the Gaussian surface is therefore derived as follows

\begin{aligned}
Q_{enc}&=\iiint_V{\rho \left(r\right)dV} \\
\\
&=\iiint_V{\frac{\alpha }{r}e^{-\ r/a}rdrd\theta dz} \\
\\
&=\int^L_0{\int^{2\pi }_0{\int^R_0{\alpha e^{-\ r/a}drd\theta dz}}} \\
\\
&=2\pi L\alpha \left(-a\right){\left[e^{-\ r/a}\right]}^R_0 \\
\\
&=2\pi L\alpha a\left(1-e^{-\ R/a}\right)

\end{aligned}

Thus, by Gauss’s Law, the electric field inside the insulating cylinder is given by

E2\pi rL=\frac{2\pi L\alpha a}{{\varepsilon }_0}\left(1-e^{-\ R/a}\right)\ \ \ \ \Rightarrow \ \ \ \ \ \boxed{\overrightarrow{E_2}=\frac{a\alpha }{{\varepsilon }_0r}\left(1-e^{-\ R/a}\right)\ \hat{r}}