Question

A very long, uniformly charged cylinder has radius $\mathrm{R}$and linear charge density$\mathrm{\lambda }$. Find the cylinder's electric field strength (a) outside the cylinder, $\mathrm{r}\ge \mathrm{R}$, and (b) inside the cylinder, $\mathrm{r}\le \mathrm{R}$. (c) Show that your answers to parts a and b match at the boundary, $\mathrm{r}=\mathrm{R}$

Short answer

a.Electric field strength outside the cylinder is $\frac{\mathrm{\lambda }}{2{\mathrm{\pi ϵ}}_{\mathrm{o}}\mathrm{r}}$.

b.Electric field strength inside the cylinder is $\frac{\mathrm{\lambda r}}{2{\mathrm{\pi ϵ}}_{\mathrm{o}}{\mathrm{R}}^2}$.

c.The answers match at the boundary at$\mathrm{r}=\mathrm{R}$.

Step-by-step solution

Calculation for electric field outside the cylinder (part a)

(a).

Electric flux,

${\mathrm{\Phi }}_{\mathrm{e}}=\mathrm{EA}$

For charge,

${\mathrm{\Phi }}_{\mathrm{e}}=\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{\epsilon }}_{\mathrm{o}}}$

The cylinder has linear charge densitylocalid="1648919513082" $\mathrm{\lambda }$and lengthlocalid="1648919518350" $\mathrm{L}$.

the charge surrounded by

${\mathrm{Q}}_{\text{in}}=\mathrm{\pi rL}$

The flux through the top and bottom faces of the cylinder is zero since they are perpendicular to the electric field.

The flux through the cylinder's wall, on the other hand, is the highest.

So,

${\mathrm{\Phi }}_{\mathrm{e}}={\mathrm{\Phi }}_{\text{top}}+{\mathrm{\Phi }}_{\text{bottom}}+{\mathrm{\Phi }}_{\text{wall}}$

$=0+0+\mathrm{E}\mathrm{A}$

$=2\mathrm{\pi rLE}$

${\mathrm{\Phi }}_{\mathrm{e}}=2\mathrm{\pi rLE}=\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{ϵ}}_{\mathrm{o}}}=\frac{\mathrm{\lambda L}}{{\mathrm{ϵ}}_{\mathrm{o}}}$

For localid="1648919554212" $\mathrm{r}\ge \mathrm{R}$,

${\mathrm{E}}_{\mathrm{r}>\mathrm{R}}=\frac{\mathrm{\lambda }}{2{\mathrm{\pi ϵ}}_{\mathrm{o}}\mathrm{r}}$

Calculation of electric field inside the cylinder (part b)

(b).

The cylinder's volume charge density is equal to the cylinder's charge divided by its volume.

${\mathrm{Q}}_{\text{in}}=\mathrm{\rho V}=\mathrm{\rho }\left({\mathrm{\pi r}}^2\mathrm{L}\right)$

The electric field inside the cylinder is formed by

$\mathrm{E}=\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{ϵ}}_{\mathrm{o}}\mathrm{A}}$

$=\frac{\mathrm{\rho }\left({\mathrm{\pi r}}^2\mathrm{L}\right)}{{\mathrm{ϵ}}_{\mathrm{o}}\left(2\mathrm{\pi rL}\right)}$

$=\frac{\mathrm{\rho r}}{2{\mathrm{ϵ}}_{\mathrm{o}}}$

For a point at distance localid="1648919566296" $\mathrm{r}<\mathrm{a}$,

The charge density is ,

$\mathrm{\rho }=\frac{\mathrm{\lambda }}{{\mathrm{\pi R}}^2}$

Wherelocalid="1648919576060" $\mathrm{\lambda }$is the linear charge density.

As a result, given a point inside the cylinder, the electric field is

${\mathrm{E}}_{\mathrm{r}<\mathrm{R}}=\frac{\mathrm{\rho r}}{2{\mathrm{ϵ}}_{\mathrm{o}}}$

$=\left(\frac{\mathrm{\lambda }}{{\mathrm{\pi R}}^2}\right)\left(\frac{\mathrm{r}}{2{\mathrm{ϵ}}_{\mathrm{o}}}\right)$

$=\frac{\mathrm{\lambda r}}{2{\mathrm{\pi ϵ}}_{\mathrm{o}}{\mathrm{R}}^2}$

match at the boundary (part c)

(c).

The radius of the gaussian surface$\mathrm{r}$and the radius of the cylinder$\mathrm{R}$are the same.

In component (b), the electric field will be

${\mathrm{E}}_{\mathrm{r}<\mathrm{R}}=\frac{\mathrm{\lambda r}}{2{\mathrm{\pi ϵ}}_{\mathrm{o}}{\mathrm{R}}^2}$

$=\left(\frac{\mathrm{\lambda }}{{\mathrm{\pi r}}^2}\right)\left(\frac{\mathrm{r}}{2{\mathrm{ϵ}}_{\mathrm{o}}}\right)$

$=\frac{\mathrm{\lambda }}{2{\mathrm{\pi ϵ}}_{\mathrm{o}}\mathrm{r}}$

Because this is the same electric field as in part (a), our responses are identical at thelocalid="1648846691692">r=Rborder.