Two conducting spheres of radius `r_(1)` =8 cm and `r_(2)` =2 cm are separated by a distance much larger than 8 cm and are connected by a thin conducting wire as shown in the figure . A total charge of Q=+100 nC is placed on one of the spheres . After a fraction of a second the charge Q is redistributed and both the spheres attain electrostatic equilibrium .

(a) Calculate the charge and surface charge density on each sphere.
(b) Calculate the potential at the surface of each sphere .

Explanation

( a) The electrostatic potential on the surface of the sphere A is `V_(A)=(1)/(4piepsillon_(0))(q_(1))/(r_(1))`
The electrostatic potential on the surface of the sphere A is `V_(B) = (1)/(4piepsilon_(0))(q_(2))/(r_(2))`
Since `V_(A)=V_(B)`. We have `(q_(1))/(r_(1))=(q_(2))/(r_(2))implies q_(1)=((r_(1))/(r_(2)))q_(2)`
But from the conservation of total charge Q`=q_(1)+q_(2)` we get `q_(1)=Q-q_(2).` By substituting this in the above equation ,
`Q-q_(2)=((r_(1))/(r_(2)))q_(2)" " "so that" " "q_(2)=Q((r_(2))/(r_(1)+r_(2)))`
Therefore `q_(2)=100xx10^(-9)xx((2)/(10))=20nC` and `q_(1)=Q-q_(2)=80nC`
The electric charge density for sphere A is `sigma_(1)=(q_(1))/(4pir_(1)^(2))`
The electric charge density for sphere B is `sigma_(2)=(q_(2))/(4pir_(2)^(2))`
Therefore `sigma_(1)=(80xx10^(-9))/(4xx64xx10^(-4))=0.99xx10^(-6)cm^(-2)`
and `sigma_(2)=(20xx10^(-9))/(4pixx4xx10^(-4))=3.9xx10^(-6)cm^(-2)`
Note that the surface charge density is greater on the smaller sphere compared to the larger sphere `(sigma_(2)~~4sigma_(1)) ` which confirms the result `(sigma_(1))/(sigma_(2))=(r_(2))/(r_(1))` .
The potential on both sphere is the same . So we can calculate the potential on any on of the spheres .
`V_(A)=(1)/(4piepsilon_(0))(q_(1))/(r_(1))=(9xx10^(9)xx80xx10^(-9))/(8xx10^(-2))=9kV`