91Ó°ÊÓ

(a) As given,

Because the electric field inside the sphere is zero and no work is done on a test charge moving from one place inside the sphere to another, the electric potential V at the sphere's surface is

Here and R is the radius of the sphere,

Putting the values of r and R

Hence, the potential at the surface is

(b) As given when two identical raindrops collide, they merge into a single larger raindrop with radius and a new volume.

Before collision the volume of the drop that is equivalent to the volume of sphere is given as

After colliding, the new volume of the raindrop is equal to the sum of the volumes of both spheres; however, because the two drops have the same volume, the new raindrop's volume will be double that of one sphere, therefore

So, the radius of the new drop is

Putting the values;

Radius of the drop

Because the charge is conservative and the volume has been doubled, the charge Q of the new raindrop will be the total of the charges of the two spheres

Putting the values in equation for the electric potential;

Therefore, radius of the dropis and the potential at its surface is.