Inside a indinitely long circular cylinder cavity. The distance between the axes of the cylinder and the cavity is equal to `a`. Find the electric field strength `E` inside the cavity. The permittivity is assumed to eb equal to unity.

Explanation

Let us consider a cylindrical Gaussain surface of radius `r` and height `h` inside an infinitely long charged cylinder with charge density `rho`. Now from Gauss theroem :
`E_(r) 2pi r h = (q_("inclosed"))/(epsilon_(0))`
(where `E_(r)` is the field inside the cylinder at a distance `r` from its axis.)
or, `E_(r) = 2pi rh = (rho pi r^(2) h)/(epsilon_(0))` or `E_(r) = (rho r)/(2 epsilon_(0))`
Now, using the method of `3.28` field at point `P`, inside the cavity, is
`vec (E) = vec(E_(+)) + vec(E_(-)) = (rho)/(2epsilon_(0)) (vec(r_(+)) - vec(r_(-))) = (rho)/(2epsilon_(0)) vec(a)`