A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b and outer radius c. The outer cylinder is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length λ
Calculate the electric field
(a) At any point between the cylinders a distance r from the axis and
(b) At any point outside the outer cylinder.
(c) Graph the magnitude of the electric field as a function of the distance r from the axis of the cable, from r = 0 to r = 2c.
(d) Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder.

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Explanation:

Part a)

By Guass law we know that

[tex]\int E. dA = \frac{q}{\epsilon_0}[/tex]

[tex]E. 2\pi rL = \frac{\lambda L}{\epsilon_0}[/tex]

[tex]E = \frac{\lambda}{2\pi \epsilon_0 r}[/tex]

Part b)

Outside the outer cylinder we will again use Guass law

[tex]\int E. dA = \frac{q}{\epsilon_0}[/tex]

[tex]E. 2\pi rL = \frac{\lambda L}{\epsilon_0}[/tex]

[tex]E = \frac{\lambda}{2\pi \epsilon_0 r}[/tex]

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.