A point charge q is placed at a distance of r from cebtre of an uncharged conducting sphere of rad `R(lt r)`. The potential at any point on the sphere is

Explanation

Remember that the superposition principle is valid even in the presence of conductors.

'q' alone would produce an electric field at all points within the volume of the conductor . The conductor must some how cancel the field due to q at all points interior to it. But there is no charge given to the conductor. The conductor solves this problem in a remarkable way. It pulls out positive and negative charges of equal magnitude from its volume and spreads them on its surface in such a manner that at every point in its interior the field due to q is cancelled by the field due to the surface charges. The potential `phi` within the sphere is uniform, because it is a conducting sphere. If we can find the value of `phi` at any point in the sphere , our problem is solved . Al the points within the sphere, there is one point which is best suited for the calculation of `phi` . This point is the centre C of the sphere . Every element of the surface is at the same distance `'a'` from the centre of the sphere and, therefore, the potential at C due to all the surface charges ( irrespective of their distribution ) is
` oint (sigma.ds)/( 4pi epsilon_(0)a) =(1)/( 4 pi epsilon_(0)a) oint sigma . ds = 0`
Because the total charge on the surface of the sphere is zero in this problem . So only contribution to potential at C comes from q now, and this is equal to `(q)/( 4pi epsilon_(0)R)`
Thus the potential of the sphere is `(q)/( 4pi epsilon_(0)R)`
Note that the above discussion can be extended to the case where the net charge on the spherical conductor has any given value Q. Several charges outside the conductor can also be taken into account.