A dielectric ball is polarized uniformly and statically. Its polarization equals `P`. Taking into account that a ball ploarized is this way may be represented as a result of a small shift of all positive chagres of the dielectric relative to all negative charges,
(a) find the electric field strength `E` inside the ball,
(b) demonstrate that the field outside the ball is that of a dipole located at the centre of the ball, the potential of that field being equal to `varphi = p_(0) r//4pi epsilon_(0)`, where `p_(0)` is the electric moment of the ball, and `r` is the distance from its centre.

Explanation

In a uniformly chagred sphere,
`E_(r ) = (rho_(0) r)/(3 epsilon_(0))` or, `vec(E) = (rho_(0))/(3 epsilon_(0)) vec( r)`
The total electric field is
`vec(E) = (1)/(3 epsilon_(0)) rho_(0) vec(r ) - (1)/(3 epsilon_(0)) (vec(r ) - sigma vec(r )) rho_(0)`
`= (1)/(3epsilon_(0)) rho_(0) sigma vec( r) = (vec(P))/(3epsilon_(0))`
where `rho sigma vec( r) = -vec(P)` (dipole moment is defined with its direction being fromt the `-ve` charge to `+ve` charge.)
The potentail outside is
`varphi = (1)/(4pi epsilon_(0)) ((Q)/(r ) - (Q)/(|vec(r ) - sigma vec(r )|)), = (rho_(0) vec(r ))/(4pi epsilon_(0) r^(3)), r gt R`
where `vec(P_(0)) = - (4pi)/(3) R^(3) rho_(0) del vec( r)` is the total dipole moment.