Suposse the surface charge density over a sphere of radius `R` dendends on a polar angle `theta` as `sigma = sigma_(0) cos theta`, where `sigma_(0)` is a positive constant. Show that such a charge distribution can be represented as a result of a small relative shift of two uniformly charge balls of radius `R` whose charges are equal in magnitude and opposite in sign. Restoring to this representation, find teh electric field strength vector inside the given spehre.

Explanation

We start from two charged spherical balls each of radius `R` with equal and opposite charge densities `+ rho` and `-rho`. The centre of the balls are at `+ (vec(a))/(2)` and `- (vec(a))/(2)` respectively so the equaction of their surfaces are `|vec(r ) - (vec(a))/(2)| = R` or `r - (a)/(2) cos theta = R` and `r + (a)/(2) cos theta = R`, considering `a` to be small. The distance between the two surface in the radial direction at angle `theta` is `|acos theta|` and does not depend on teh azimuhal angle. It is seen from the diagram that the surface of the sphere has in effect a surface density `sigma = sigma_(0) cos theta` when
`sigma_(0) = rho a`.
Inside any uniformly charged spherical ball, the field is radial and has the magnitude given by Gaussain's theorem
`4pi r^(2) E = (4pi)/(3) r^(3) rho//epsilon_(0)`
or `E = (rho r)/(3 epsilon_(0)`
In vector notation, using the fact the `V` myst be measured from the centre of the ball, we get, for the present case
`vec(E) = (rho)/(3 epsilon_(0)) (vec(r ) - (a)/(2)) - (rho)/(3 epsilon_(0)) (vec(r ) + (vec(a))/(2))`
`= -rho vec(a)//3 epsilon_(0) = (sigma_(0))/(3 epsilon_(0)) vec(k)`
When `vec(k)` is teh unit vector along the polar axis from which `theta` is measured.