Danelle Albright

Answered

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 $\rho \left(r\right)=\alpha (1-\frac{r}{R})$ , where $\alpha$ is a constant with units $\frac{C}{{m}^{3}}$ and r is the perpendicular distance from the center line of the cylinder.

Answer & Explanation

Stuart Rountree

Expert

2022-01-17Added 29 answers

Step 1

Using the Gauss law,

$\varphi \stackrel{\u2015}{E}\cdot d\varphi \stackrel{\u2015}{A}=\frac{{Q}_{enc}}{{\xi}_{0}}$ [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

$\rho =\alpha (1-\frac{r}{R})$ [3]

Step 4

Use equation (3) in (2) to find the electric field as a function of r.

$E\cdot 2\pi rh=\frac{\alpha (1-\frac{r}{R})\pi {r}^{2}h}{{\xi}_{0}}$

$E=\frac{\alpha}{2{\xi}_{0}}(r-\frac{{r}^{2}}{R})$

Using the Gauss law,

Step 2

Use equation (1) to solve for the electric field as a function of r.

Step 3

The volume charge density of the cylinder is

Step 4

Use equation (3) in (2) to find the electric field as a function of r.

Alex Sheppard

Expert

2022-01-18Added 36 answers

Answer:

1.

2.

3.The results from part 1 and 2 agree when

Explanation:

The volume charge density is given as

We will investigate this question in two parts. First

1. Since the cylinder is very long, Gauss’ Law can be applied.

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

where ‘h’ is the length of the imaginary Gaussian surface.

2. For

3. At the boundary where

alenahelenash

Expert

2022-01-23Added 366 answers

Step 1
1. Since the cylinder is very long, Gauss

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