Question #df5d0
1 Answer
Given that the radius of smaller sphere is
So initial charge on smaller sphere
And initial charge on larger sphere
Now we know that the capacitance of a sphere is proportional to its radius. So the ratio of their capacitances (smaller:larger) will be
Let the capacitance of smaller one be
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
So changed charge on large capacitor
So the changed surface charge density on larger capacitor