(i) Consider two conducting spheres A and B of radii `r_(1)` and `r_(2)` respectively connected to each other by a thin conducting wire as shown in the Figure. The distance between the spheres is much greater than the radii of either spheres.
(ii) If a charge Q is introduced into any one of the spheres, this charge Q is redistributed into both the spheres such that the electrostatic potential is same is both the spheres. They are now uniformly charged and attain electrostatic equilibrium. Let `q_(1)` be the charge residing on the surface of sphere A and `q_(2)` is the charge residing on the surface of sphere B such that `Q=q_(1)+q_(2)`. The charges are distributed only on the surface and there is no net charge inside the conductor.
The electrostatic potential at the surface of the sphere A is given by
`V_(A)=(1)/(4piepsi_(0))(q_(1))/(q_(2))" "......(1)`
(iii) The electrostatic potential at the surface of the sphere B is given by
`V_(B)=(1)/(4piepsi_(0))(q_(2))/(r_(2))" "......(2)`
(iv) The surface of the conductor is an equipotential. Since the spheres are connected by the conducting wire, the surfaces of both the spheres together form an equipotential surface. This implies that
`V_(A)=V_(B)`
or `(q_(1))/(r_(1))=(q_(2))/(r_(2))" ".......(3)`
(v) Let us take the charge density on the surface of sphere A is `sigma_(2)`. This implies that `q_(1)=4pir_(1)^(2)sigma_(1)andq_(2)=4pir_(2)^(2)sigma_(2)` Substituting these values into equation (3), we get
`sigma_(1)r_(1)=sigma_(2)r_(2)" "......(4)`
from which we conclude that
`sigmar` =constant " "........(5)
Thus the surface charge density `sigma` is inversely proportional to the radius of the sphere. For a smaller radius, the charge density will be larger and vice versa.
