A conducting sphere `S_(1)` of radius `r_(1)` is connected by a conducting wire to another conducting sphere `S_(2)` of radius `r_(2)`, where `r_(1) = 3 cm and r_(2) = 2` cm. Before they are connected, `S_(1)` carries charge of 10 units. The electric potential at the point which is at a distance 4 cm from the centre of `S_(1)` and a distance 3 cm from the centre of `S_(2)` is

Options

A. `(1)/(4pi epsi_(0)) (17)/(6)`
B. `(1)/(4pi epsi_(0)) (3)/(2)`
C. `(1)/(4pi epsi_(0)) (1)/(6)`
D. `(1)/(4pi epsi_(0)) (17)/(12)`

Answer

A

Explanation

When sphere are connected, charges are redistributed such that potential on their surfaces are same.

So, we have
`q_(1) + q_(2) = 10`
and `(kq_(1))/(r_(1)) = (kq_(2))/(r_(2)) rArr (q_(1))/(3) = (q_(2))/(2)`
Hence, `q_(1) =6` units and `q_(2)` = 4 units.
Now, potential at some point P distance 4 cm from `s_(1) and 3` cm from `s_(2)` is
`V = (kq_(1))/(r_(1)) + (kq_(2))/(r_(2)) = k (6)/(4) + k(4)/(3)`
`= k ((3)/(2) + (4)/(3)) = k (17)/(6)V = (1)/(4pi epsi_(0)) (17)/(6) V`