A Uniformly charged solid non-conducting sphere of uniform volume charge density `rho` and radius R is having a concentric spherical cavity of radius r. Find out electric field intensity at following points, as shown in the figure :

(i) Point A (ii) Point B
(iii) Point C (iv) Centre of the sphere

Explanation

C is the centre of cavity . Let us visualise this system as superposition of two spheres. One is a big sphere of radius R and volume charge density `+rho`, without any cavity .The other is equal to the size of cavity with volume charge density `-rho`. The net result will be zero charge density for cavity . Positive and negative charges will neutralise.
Let P be a point inside the cavity . The position of P is shown by position vectors from the centre of sphere and cavity. Then , the net electric field at P :
`vecE=(rhovecr_1)/(3epsilon_0)-(rhovecr_2)/(3epsilon_0)rArr vecE=(rho(vecr_1-vecr_2))/(3epsilon_0)`
Using the triangle law of vectors we may write as follows :
`rArr vecE=(rho vecr)/(3epsilon_0)(vecr=vecr_1-vecr_2)`
`rArr vecE=(rhovecr)/(3epsilon_0)`, parallel to `vecr`
Here one thing should be noted that the field intensity inside this cavity is maintained uniform.