Sol. Let λ be the charge per unit length on the cylinder with radius r 1 . From Gauss's law we have

dA = ….. (1)

where the surface S is defined to be a cylinder of radius r and of unit length. Since E r is a function of r only, (1) immediately leads to

E r = . ….. (2)

The potential difference between the two cylinders is

V = = ….. (3)

from which we find the capacitance C

C = = …. (4)

From (3), we can solve for the charge density

λ = . ….. (5)

From Gauss's law we get

E = for r 1 < r < r 3

= for r 3 < r < r 2 ….. (6)

from which we obtain the displacement in the dielectric medium

When the potential difference between the two cylinders is kept constant, the system can no longer be isolated. There has to be some source of charge (battery) to supply energy. Let C' be the capacitance of the system without the dielectric material. From (4) we find

C' = . ….. (7)

The work needed is

work needed = C'V 2 – CV 2 – (Q' – Q)V

where Q' is the final total charge on one cylinder. Using Q' = C'V, we get

Work needed = (C –C')

= 2π ε 0 L

where we have used (4) and (7).