Charge Q is distributed uniformly over a non conducting sphere of radius R. Find the electric p-0tential at distance r from the centre of the sphere r `(r lt R)`. The electric field at a distance r from the centre of the sphere is given as `(1)/(4pi epsilon_(0)). (Q)/(R^(3))hatr` . Also find the potential at the centre of the sphere .

Explanation

The electric potential on the surface of such a sphere is
`V(R) = (1)/(4pi epsilon_(0)).(Q)/(R)`
We can use the equation `V(r) : V(R)= : vint_(R)^(r)vecE.dvecr`
`:. V(r): (R) =: int_(R)^(r)(1)/(4piepsilon_(0))(Q)/(R^(3))rdr (hatr.hatr)`
`( because dvecr = dr vecr)`
`=: (Q)/(4pi epsilon_(0)R^(3))int_(R)^(r)rdr`
`=: (Q)/(4piepsilon_(0)R^(2))[(r^(2))/(2)]_(R)^(r)`
`=: (Q)/(4pi epsilon_(0)R^(3))[(r^(2))/(2) -(R^(2))/(2)]`
`:. V(r)=V(R)+(Q)/(4piepsilon_(0)R^(3))[(R^(3))/(2)-(r^(2))/(2)]`
`:. V(r)=(1)/(4piepsilon_(0))(Q)/(R)+(Q)/(8piepsilon_(0)R^(3))(R^(2)-r^(2))`
`:. V(r) = (1)/(4piepsilon_(0))(Q)/(2R) (2+(1)/(R^(2))(R^(2)-r^(2)))`
`=(1)/(4piepsilon_(0))(Q)/(2R) (2+(R^(2))/(R^(2))-(r^(2))/(R^(2)))`
`:. V(r) = (1)/(4piepsilon_(0))(Q)/(2R) (3-(r^(2))/(R^(2)))=r lt R`
`:. V(r)=(kQ)/(2R)[3-(r^(2))/(R^(2))]`
At the centre of sphere r =0
`:. V (0) =(1)/(4pi epsilon_(0)).(3Q)/(2R) = (3kQ)/(2R)`