Answer:
Explanation:
From the given information.
The charge density of the cylinder can be computed as:
[tex]\rho = \dfrac{Q_t}{V_t}[/tex]
where:
[tex]Q_t[/tex] = total charge on cylinder
[tex]\rho[/tex] = density of the cylinder
[tex]V_t =[/tex] net volume on cylinder
Considering the charge enclosed by the Gaussian surface; we have:
Q = ρV
Now, determining the volume of the cylinder at a distance r from the axis of the cylinder as follows:
[tex]\int Edsn = \dfrac{q}{\varepsilon_o}[/tex]
Here;
[tex]\hat E \ \ and \ \ \hat n[/tex] are in the same direction;
If we replace [tex]\int Edsn[/tex] with [tex]E ( 2 \pi rl)[/tex] and q with [tex]\rho \pi r^2 l[/tex]; Then:
[tex]E ( 2 \pi rl ) = \dfrac{\rho \pi r^2 l}{\varepsilon _o}[/tex]
By rearrangement;
[tex]E = \dfrac{\rho \pi r^2 l}{ ( 2 \pi rl ) \varepsilon _o}[/tex]
[tex]\mathbf{E = \dfrac{\rho r}{ 2 \varepsilon _o}}[/tex]
(b)
Using the same formula:
[tex]\int Edsn = \dfrac{q}{\varepsilon_o}[/tex]
where;
[tex]\hat E \ \ and \ \ \hat n[/tex] are in the same direction;
If we replace [tex]\int Edsn[/tex] with [tex]E ( 2 \pi rl)[/tex] and [tex]Q_t[/tex] with q;
Then:
[tex]E ( 2 \pi rl ) = \dfrac{Q_t}{\varepsilon _o}[/tex]
[tex]E = \dfrac{Q_t}{( 2 \pi R l ) \varepsilon _o}[/tex]
Replacing [tex]\lambda[/tex] for [tex]\dfrac{Q_t}{l}[/tex].
From above [tex]\lambda[/tex] = the charge per unit length
∴
[tex]\mathbf{E = \dfrac{\lambda}{ 2 \pi R \varepsilon _o}}[/tex]
