# 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