Question

Question #df5d0

Answer 1

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`