A charge 'Q' is distributed over two concentric hollow spheres of radii 'r' and 'R' (gtr) such that the surface densities are equal. Find the potential at the common centre.

Options

A. `(Q(R^(2)+r^(2)))/(4piepsilon_(0)(R+r))`
B. `(Q)/(R+r)`
C. zero
D. `(Q(R+r))/(4piepsilon_(0)(R^(2)+r^(2)))`

Answer

d

Explanation

Since surface densities are equal
`(q_(1))/(4pir^(2))=(q_(2))/(4piR^(2))`, Where `q_(1)+q_(2)=Q`
`(q_(1))/(r^(2))=(q_(2))/(R^(2))`, therefore `(q_(1)+q_(2))/(r^(2)+R^(2))=(Q)/(r^(2)+R^(2))`
`q_(1)=(Q)/(R^(2)+r^(2))xxr^(2),q_(2)=(Q)/(R^(2)+r^(2))R^(2)`
Potential at the common cnetre , is
`V=(q_(1))/(4piepsilon_(0)r)+(q_(2))/(4piepsilon_(0)R)=(1)/(4piepsilon_(0))((q_(1))/(r)+(q_(2))/(R))`
`=(1)/(4piepsilon_(0))((Q)/(R^(2)+r^(2))xx(r^(2))/(r)+(Q)/(R^(2)+r^(2))(R^(2))/(R))=(Q(R+r))/(4piepsilon_(0)(R^(2)+r^(2)))`