Question

An infinitely long insulating cylinder of radius R has a volume charge density that varies with the radius as $\rho = \rho_0(a-\frac{r}{b})$ where $\rho_0$, a and b are positive constants and r is the distance from the axis of the cylinder. Use Gauss's Law to determine the magnitude of the electric field at radial distances (a) $r < R$ (b) $r>R$

Answer 1

For $r < R$: $E = \frac{\rho_0 r}{6\epsilon_0 b} (3ab - 2r)$. For $r > R$: $E = \frac{\rho_0 R^2}{6\epsilon_0 b r} (3ab - 2R)$.

Answer 2

Due to cylindrical symmetry, we choose a coaxial Gaussian cylinder of radius $r$ and length $L$. Gauss's Law is $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$. The electric flux is $\Phi = E \cdot (2\pi r L)$. The volume charge density is $\rho(r') = \rho_0(a - r'/b)$.

For $r < R$ (inside): The enclosed charge is $Q_{enc} = \int_0^r \rho(r') (2\pi r' L) dr' = 2\pi \rho_0 L \int_0^r (ar' - \frac{r'^2}{b}) dr' = 2\pi \rho_0 L (\frac{ar^2}{2} - \frac{r^3}{3b})$. Applying Gauss's Law: $E \cdot (2\pi r L) = \frac{2\pi \rho_0 L}{\epsilon_0} (\frac{ar^2}{2} - \frac{r^3}{3b})$. Solving for $E$: $E = \frac{\rho_0}{\epsilon_0} (\frac{ar}{2} - \frac{r^2}{3b}) = \frac{\rho_0 r}{6\epsilon_0 b} (3ab - 2r)$.

For $r > R$ (outside): The enclosed charge is the total charge within the cylinder of radius $R$: $Q_{enc} = \int_0^R \rho(r') (2\pi r' L) dr' = 2\pi \rho_0 L \int_0^R (ar' - \frac{r'^2}{b}) dr' = 2\pi \rho_0 L (\frac{aR^2}{2} - \frac{R^3}{3b})$. Applying Gauss's Law: $E \cdot (2\pi r L) = \frac{2\pi \rho_0 L}{\epsilon_0} (\frac{aR^2}{2} - \frac{R^3}{3b})$. Solving for $E$: $E = \frac{\rho_0}{\epsilon_0 r} (\frac{aR^2}{2} - \frac{R^3}{3b}) = \frac{\rho_0 R^2}{6\epsilon_0 b r} (3ab - 2R)$.