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

Answer 2

(a) Electric field at the given distances is:

    (i) E = α/ [2πε₀r]

    (ii) E = 0

    (iii) E = α/(πε₀r)

(b) charge per unit length on

     (i) the inner surface = - α

    (ii) outer surface = 2α

Calculating charge and electric field:

(a)(i) For r < a, from Gauss law;

EA = Q(enclosed) / ε₀

where, Q(enclosed) = Q(cavity) = αL

So, E(2πrL) = αL/ε₀

E =α/ [2πε₀r]

(ii) For a < r < b;

Inside a conductor, the electric field will is zero.

So, E=0

(iii) For r > b;

Q(enclosed) = Q(net) - Q(inner)

Now, Q(inner) = -Q(net), since an opposite charge will be induced inside.

Q(enclosed) = αL - (-αL) = 2αL

EA = Q(enclosed) / ε₀

E(2πrL) = 2αL/ε₀

E = α/ [πε₀r]

(b) (i) The charge per unit length in the inner surface of the tube will be -α, since an opposite charge will be induced in the cavity walls of the conductor due to the point charge.

(ii) Net charge per length 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

Learn more about Gauss Law:

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