Two spherical conductors A and B of radii 1mm and 2mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is

Options

A. `4:1`
B. `1:2`
C. `2:1`
D. `1:4`

Answer

C

Explanation

When A and B are joined by a conducting wire, they acquire a common potential V and the charges are redistributed as `Q_(A) and Q_(B)`. Then the intensity of the electric field at the surface of A
`(E_(A))=(1)/(4piepsi_(0))(Q_(A))/(R_(A)^(2))`
but `Q_(A)=C_(1)V=(4piepsi_(0)R_(A))V`
the charges remain on the outer surface. hence the electric field at the surface is given by
`therefore E_(A)=(1)/(4piepsi_(0))xx(4piepsi_(0)R_(A)V)/(R_(A)^2)=(V)/(R_(A))`
Similarly, `E_(B)=(V)/(R_(B))`
`therefore (E_(A))/(E_(B))=(R_(B))/(R_(A))=(2)/(1)`