Question #df5d0

1 Answer
Jan 19, 2017

Given that the radius of smaller sphere is R and that of larger sphere is 2R and their common surface charge density is rho

So initial charge on smaller sphere q_1=4piR^2rho

And initial charge on larger sphere Q_1=16piR^2rho

Now we know that the capacitance of a sphere is proportional to its radius. So the ratio of their capacitances (smaller:larger) will be R:2R=1:2

Let the capacitance of smaller one be C_s=C and that of larger one be C_l=2C.

After they are connected by a thin conducting wire remaining at a large distance apart their charge will be redistributed until they gain same potential.

Let this common potential be V.

By principle of conservation of charge we can write

C_sV+C_lV=q_1+Q_1=20piR^2rho

=>CV+2CV=20piR^2rho

=>V=(20piR^2rho)/(3C)

So changed charge on large capacitor =C_lV=2Cxx(20piR^2rho)/(3C)=(40piR^2rho)/3

So the changed surface charge density on larger capacitor
=(40piR^2rho)/(3xx16piR^2)=5/6rho