Two spherical conductors A and B are having charge densities  σA and  σB respectively. If they are joined by a thin wire then what will be the ratio of charge densities finally if the radii are rA and rB. 

Asked by hitanshu04 | 17th May, 2021, 09:09: PM

Expert Answer:

Charged spherical conductors are joined by thin wire , then they are at same potential.
 
K × ( qA / rA ) = K × ( qB / rB ) .........................(1)
 
where K = 1 / ( 4 π εo ) is Coulomb's constant, qA is charge on sphere-A , qB is charge on sphere-B, 
 
rA is radius of sphere-A and rB is radius of sphere-B . 
 
Hence from eqn.(1) , we get ,
 
begin mathsize 14px style q subscript A over q subscript B space equals space r subscript A over r subscript B end style   ...........................(2)
 
if begin mathsize 14px style sigma subscript A apostrophe end style and begin mathsize 14px style sigma subscript B apostrophe end style are charge densities respectively on sphere-A and sphere-B after connetcing by thin wire ,
 
then we have , begin mathsize 14px style q subscript A over q subscript B space equals space fraction numerator 4 pi space r subscript A superscript 2 space sigma subscript A apostrophe over denominator 4 space pi space r subscript B superscript 2 space sigma subscript B apostrophe end fraction space equals space r subscript A over r subscript B end style
Hence we get , begin mathsize 14px style fraction numerator space sigma subscript A apostrophe over denominator space sigma subscript B apostrophe end fraction space equals space r subscript B over r subscript A end style
 

Answered by Thiyagarajan K | 17th May, 2021, 11:13: PM