Question

A very long, solid cylinder with radius R has positive charge uniformly distributed throughout it, with charge per unit volume r. (a) Derive the expression for the electric field inside the volume at a distance r from the axis of the cylinder in terms of the charge density r. (b) What is the electric field at a point outside the volume in terms of the charge per unit length l in the cylinder

Answer 1

To derive the expression for the electric field inside the volume of the cylindrical shell at a distance r from the axis, we can consider an infinitesimally thin cylindrical shell of radius r and height dx. The electric field inside the cylindrical shell can be expressed as E=(1/(4*pi*epsilon_0))*(dq/r^2). For a point outside the cylindrical shell, the electric field can be found using the formula E=Q/(4*pi*epsilon_0*r^2).

Step-by-step explanation:

To derive the expression for the electric field inside the volume of the cylindrical shell at a distance r from the axis, we can consider an infinitesimally thin cylindrical shell of radius r and height dx. The charge enclosed within this cylindrical shell is given by dq = r*dx. The electric field produced by this cylindrical shell at a point inside the shell can be found using Coulomb's law, which states that the electric field at a point due to a charged particle is directly proportional to the charge and inversely proportional to the square of the distance. Thus, at a point inside the cylindrical shell, the electric field can be expressed as:

E = (1/(4*pi*epsilon_0))*(dq/r^2)

where epsilon_0 is the permittivity of free space (a constant).

For a point outside the cylindrical shell, the electric field can be found by considering the entire charge distribution of the cylinder. Since the charge is uniformly distributed, we can calculate the total charge Q enclosed within a cylindrical shell of radius r and height h. The electric field outside the cylinder can then be calculated using the formula:

E = Q/(4*pi*epsilon_0*r^2)

Answer 2

Step-by-step explanation:

From the given information.

The charge density of the cylinder can be computed as:

`\rho = (Q_t)/(V_t)`

where:

`Q_t` = total charge on cylinder

`\rho` = density of the cylinder

`V_t =` 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:

`\int Edsn = (q)/(\varepsilon_o)`

Here;

`\hat E \ \ and \ \ \hat n` are in the same direction;

If we replace
`\int Edsn` with
`E ( 2 \pi rl)` and q with
`\rho \pi r^2 l`; Then:

`E ( 2 \pi rl ) = (\rho \pi r^2 l)/(\varepsilon _o)`

By rearrangement;

`E = (\rho \pi r^2 l)/( ( 2 \pi rl ) \varepsilon _o)`

`\mathbf{E = (\rho r)/( 2 \varepsilon _o)}`

(b)

Using the same formula:

`\int Edsn = (q)/(\varepsilon_o)`

where;

`\hat E \ \ and \ \ \hat n` are in the same direction;

If we replace
`\int Edsn` with
`E ( 2 \pi rl)` and
`Q_t` with q;

Then:

`E ( 2 \pi rl ) = (Q_t)/(\varepsilon _o)`

`E = (Q_t)/(( 2 \pi R l ) \varepsilon _o)`

Replacing
`\lambda` for
`(Q_t)/(l)`.

From above
`\lambda` = the charge per unit length

∴

`\mathbf{E = (\lambda)/( 2 \pi R \varepsilon _o)}`