Two spheres of radius `a` and `b` respectively are charged and joined by a wire. The ratio of electric field of the spheres is

Options

A. `a/b`
B. `b/a`
C. `(a^(2))/(b^(2))`
D. `(b^(2))/(a^(2))`

Answer

B

Explanation

Let `q_(1)` and `q_(2)` be the charges and `C_(1)` and `C_(2)` be the capacitance of two spheres.
The charge flows from the sphere at higher potential to the other at lower potentials because equal.
After sharing the charge on two sphere would be.
`(q_(1))/(q_(2)) = (C_(1)V)/(C_(2)V)" "......(i)`
Also `(C_(1))/(C_(2)) = (a)/(b) " ".......(ii)`
From (i) `(q_(1))/(q_(2)) = (a)/(b)`
Ratio of surface charge on the two spheres
`(sigma_(1))/(sigma_(2)) = (q_(1))/(4pia^(2)) .(4pib^(2))/(q_(2)) = (q_(1))/(q_(2))(b^(2))/(a^(2)) = (b)/(a)" "("using (ii)")`
`therefore` The ratio of electric fields at the surfaces of tow spheres `(E_(1))/(E_(2)) = (sigma_(1))/(sigma_(2)) = (b)/(a)`