Danelle Albright

2022-01-16

A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribution has cylindrical symmetry but varies with perpendicular distance from the axis of the cylinder. The volume charge density is $\rho \left(r\right)=\alpha \left(1-\frac{r}{R}\right)$, where $\alpha$ is a constant with units $\frac{C}{{m}^{3}}$ and r is the perpendicular distance from the center line of the cylinder.

Stuart Rountree

Expert

Step 1
Using the Gauss law,
$\varphi \stackrel{―}{E}\cdot d\varphi \stackrel{―}{A}=\frac{{Q}_{enc}}{{\xi }_{0}}$ [1]
Step 2
Use equation (1) to solve for the electric field as a function of r.
$E\cdot 2\pi rh=\frac{{Q}_{enc}}{{\xi }_{0}}$
$=\frac{\rho V}{{\xi }_{0}}$ [2]
Step 3
The volume charge density of the cylinder is
$\rho =\alpha \left(1-\frac{r}{R}\right)$ [3]
Step 4
Use equation (3) in (2) to find the electric field as a function of r.
$E\cdot 2\pi rh=\frac{\alpha \left(1-\frac{r}{R}\right)\pi {r}^{2}h}{{\xi }_{0}}$
$E=\frac{\alpha }{2{\xi }_{0}}\left(r-\frac{{r}^{2}}{R}\right)$

Alex Sheppard

Expert

1. $E\left(r\right)=\frac{\alpha }{4\pi {\xi }_{0}}\left(2-\frac{r}{R}\right)$
2. $E\left(r\right)=\frac{1}{4\pi {\xi }_{0}}\frac{\alpha R}{r}$
3.The results from part 1 and 2 agree when $r=R$.
Explanation:
The volume charge density is given as
$\rho \left(r\right)=\alpha \left(1-\frac{r}{R}\right)$
We will investigate this question in two parts. First . We will show that at $r=R$, the solutions to both parts are equal to each other.
1. Since the cylinder is very long, Gauss’ Law can be applied.
$\int \stackrel{\to }{E}d\stackrel{\to }{a}=\frac{{Q}_{enc}}{{\xi }_{0}}$
The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is
$\int da=2\pi rh$
where ‘h’ is the length of the imaginary Gaussian surface.

$E2\pi rh=\alpha h\frac{2Rr-{r}^{2}}{2R{\xi }_{0}}$
$E\left(r\right)=\alpha \frac{2R-r}{4\pi {\xi }_{0}R}$
$E\left(r\right)=\frac{\alpha }{4\pi {\xi }_{0}}\left(2-\frac{r}{R}\right)$
2. For $r>R$, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

$E\left(r\right)=\frac{1}{4\pi {\xi }_{0}}\frac{\alpha R}{r}$
3. At the boundary where $r=R$:
$E\left(r=R\right)=\frac{\alpha }{4\pi {\xi }_{0}}\left(2-\frac{r}{R}\right)=\frac{\alpha }{4\pi {\xi }_{0}}$

alenahelenash

Expert