Inside a ball charged uniformly with volume density `rho` there is a spherical cavity. The centre of the ball by a distance `a`. Find the field strength `E` inside the cavity, assuming the permittively equal to unity.

Explanation

Using Gauss theroem we can easily show that the electric field strength within a uniformly, charged sphere is `vec(E) = ((rho)/(3 epsilon_(0))) vec(r )`
The cavity in out problem, may be considered as the superpositon of two balls, one with the charge density `rho` and the other with `rho`.
Let `P` be a point inside the cavity such that its position vector with respect with respect to the centre of cavity to `vec(r)` adn with respect to the centre of the ball `vec(r)_(+)`. Then from the principle of superpositon, field inside the cavity at an arbitary point `P`,
`vec(E) = vec(E_(+)) + vec(E_(-))`
`= (rho)/(3 epsilon_(0)) (vec(r_(+)) - vec(r_(-))) = (rho)/(3epsilon_(0)) vec(a)`
Note : Obtained expression for `vec(E)` shows that it is valid regarless of the ratio between the radii of the sphere and the distance between their centires.