Question

A long, thin straight wire with linear charge density ? runs down the center of a thin, hollow metal cylinder of radius R. The cylinder has a net linear charge density
2?.

Assume ? is positive. Find expressions for the electric field strength. In what direction does the electric field point in each of the cases? (Use any variable or symbol stated above along with the following as necessary: r and ?0.)

(a) inside the cylinder, r < R

Magnitude E =

Direction=

(b)outside the cylinder, r > R

Magnitude E=

Direction=

Answer 1

Answer:

a. E = $\frac{\lambda }{(2\epsilon * r) }$

b. $E = \frac{3(lambda)}{2\pi*\epsilon(outside) }$

Explanation:

The important thing to remember is to use Gauss Law. This is a relation that describes the distribution of electric charge to the resultant electric field.

Linear charge density means charge per unit length of material.

Data:

The metal cylinder is hollow.

The unit length is L.

a.The expression will be as follows:

for charge inside the cylinder, where r < R, the expression is:

E = $\frac{\lambda }{(2\epsilon * r) }$

b. Let's assume that the cylinder is a coaxial cylinder with a radius r > R, then the electrical field strength is given as:

$E = \frac{Q ( enclosed)}{A*\epsilon }$

E = $\frac{\lambda*L+2(\lambda)*L }{(A)*\epsilon(outside) }$

E  = $\frac{3(\lambda)L }{(2\pi*R*L*\epsilon(outside) }$

This gives:

$E = \frac{3(lambda)}{2\pi*\epsilon(outside) }$

The solution informs us that there is a surface change taking place on the cylinder. Therefore, there will not be a magnetic field across it.