Question

A very long conducting tube (hollow cylinder) has inner radius a and outer radius b. It carries charge per unit length +α, where a is a positive constant with units of C/m. A line of charge lies along the axis of the tube. The line of charge has charge per unit length +α.
(a) Calculate the electric field in terms of +α and the distance r from the axis of the tube for
(i) r < a; (ii) a < r < b; (iii) r > b. Show your results in a graph of E as a function of r.
(b) What is the charge per unit length on (i) the inner surface of the tube and (ii) the outer surface of the tube?

Answer 1

Answer:

A) i) E =α/ [2πrL(εo)]

ii) E=0

iii) E = α/(πrεo)

The graph between E and r for the 3 cases is attached to this answer ;

B) i) charge on the inner surface per unit length = - α

ii) charge per unit length on the outer surface = 2α

Explanation:

A) i) For r < a, the charge is in the cavity and takes a shape of the cylinder. Thus, applying gauss law;

EA = Q(cavity) / εo

Now, Qcavity = αL

So, E(2πrL) = αL/εo

Making E the subject of the formula, we have;

E =α/ [2πrL(εo)]

ii) For a < r < b; since the distance will be in the bulk of the conductor, therefore, inside the conductor, the electric field will be zero.

So, E=0

iii) For r > b; the total enclosed charge in the system is the difference between the net charge and the charge in the inner surface of the cylinder.

Thus, Qencl = Qnet - Qinner

Qinner will be the negative of Qnet because it should be in the opposite charge of the cavity in order for the electric field to be zero. Thus;

Qencl = αL - (-αL) = 2αL

Thus, applying gauss law;

EA = Qencl / εo

Thus, E = Qencl / Aεo

E = 2αL/Aεo

Since A = 2πrL,

E = 2αL/2πrLεo = α/(πrεo)

B) i) The charge on the cavity wall must be the opposite of the point charge. Therefore, the charge per unit length in the inner surface of the tube will be - α

ii)Net charge per length for tube is +α and there is a charge of - α on the inner surface. Thus charge per unit length on the outer surface will be = +α - (- α) = 2α