A charge `Q=(5)/(100)nC` is distributed over two concentric hollow spheres of radii `r=3` cm and `R=6cm` such that the densities are equal .Find the potential in volt at the common centre.

Answer

9

Explanation

Let `q_(1)` and `q_(2)` be the respective charges distributed over two concentric spehres of radii `r` and `R` such that
`q_(1)+q_(2)=Q`
As surface densities are given to be equal therefore
`sigma_(1)=sigma_(2)rArr (q_(1))/(4pir^(2))=(q_(2))/(4piR^(2))rArr (q_(1))/(q_(2))=(r^(2))/(R^(2))`
`rArr (q_(1))/(q_(2))+1=(r^2)/(R^(2))+1 rArr (q_(1)+q_(2))/(q_(2))=(r^(2)+R^(2))/(R^(2))`
using (1) &(2) we get,
`(Q)/(q_(2))=(r^(2)+R^(2))/(R^(2))` this given `q_(2)=[(R^(2))/(r^(2)+R^(2))]+Q`
Therfore `q_(1)=Q-q_(2)`
` rArr q_(1)=Q-((R^(2))/(r^(2)+R^(2)))Q=((r^(2))/(r^(2)+R^(2)))`
` rArr q_(1)=Q-((R^(2))/(r^(2)+R^(2)))Q=((r^(2))/(r^(2)+R^(2)))Q`
the potential `V_(1)` at common centre due to charge `q_(1)` is given by
`V_(1)=(1)/(4piepsilon_(0))(q_(1))/(r) rArr V_(1)=(1)/(4piepsilon_(0))((r^(2))/(r^(2)+R^(2)))(Q)/(r)`
`rArr V_(1)=(1)/(4piepsilon_(0))(Qr)/(r^(2)+R^(2))`
the potential `V_(2)` at common centre due to
charge `q_(2)` is `V_(2)=(1)/(4piepsilon_(0))(QR)/(r^(2)+R^(2))`
`:.` Net potential at common centre `V=V_(1)+V_(2)`
`rArr V=(1)/(4piepsilon_(0))(QR)/((r^(2)+R^(2))[r+R]`
`rArr V=(1)/(4piepsilon_(0))(Q(R+r))/(R^(2)+r^(2))`
`rArr V=9xx10^(9)xx(5)/(100)xx10^(-9)xx(9)/(45)xx100`
`rArr V=9V`