Two isolated metallic solid spheres of radii R and 2R are charged such that both of these have same charge density `sigma`. The spheres are located far away from each other and connected by a thin conducting wire. Find the new charge density on the bigger sphere.

Answer

6

Explanation

Let `q_(1) and q_(2)` be the charges on the two spheres before connecting them.
Then, `q_(1)=sigma(4piR^(2)),andq_(2)=sigma(4pi)(2R)^(2)=16sigmapiR^(2)`
Therefore, total charge (q) on both the spheres is
Now, after connecting, the charge is distributed in the ratio of their capacities, which in turn depends on the ratio of their radii `(C=4piepsi_(0)R)`
`therefore" "(q'1)/(q'2)=(R)/(2R)=(1)/(2)" "therefore" "q_(1)=(q)/(3)=(20)/(3)sigmapiR^(2)" "and" "q_(2)=(2q)/(3)=(40)/(3)sigmapiR^(2)`
Therefore, surface charge densities on the spheres are
`sigma_(1)=(q_1)/(4piR^2)=((20//3)sigmapiR^(2))/(16piR^(2))=(5)/(6)sigma" "and" "sigma_(2)=(q_2)/(4pi(2R)^(2))=((40//3)sigmapiR^(2))/(16piR^(2))=(5)/(6)sigma`
Hence, surface charge density on the bigger sphere is `sigma_(2)i.e.(5//6)sigma`.