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#