Electric potential of spherical water droplet I'm trying to answer the following: 

1000 spherical water droplets, each of radius $r$ and each carrying a charge $q$, coalesce to form a single bigger spherical drop. If $v$ is the electric potential of each droplet and $V$ that of bigger drop, then find $\frac{V}{v}$.

It is not mentioned here whether $v$ is potential at a point outside or on the surface of sphere, where should I take the potential? 
As water is a good conductor of electricity so water droplet can be regarded as spherical shell.Potential due to spherical shell is * $\frac{kq}{r}$.* where k is constant and q is charge on the shell and r is the distance between the centre of spherical shell and point on which P is  to be calculated and at a point inside or on the surface of shell P is constant and is equal to * $\frac{kq}{r}$.* here r is the radius of shell. So P  depends on the distance between the centre of spherical shell and point on which P is  to be calculated.So in the above problem P on which point should I need to consider.
I am unable to understand what is meant by the potential of water droplet.I know potential at a point due to point charge,system of charge and due to continuous charge distribution.But they are calculated at  point,I mean we calculate potential at point due to different charge distributions.How can we calculate potential of a body!
I want to know what is meant by( " $v$ is the electric potential of each droplet" )this  line of the question.
 A: The potential inside a conductive sphere is the same as the one on it's surface.
All you really need is the self capacitance of a sphere, which is 4*pi*e*r.
In other words twice the radius - twice the capacitance.
So if the volume is 1000 times higher what does that mean for the radius?
And if the capacitance of the drop increases by a factor of x while the charge increases by 1000 then what happens to the potential?
A: For any spherical body whether insulating or conducting,  potential at surface and potential outside body have same formula I.e. 
$$ V = \frac {kQ}{R} $$
Here $ k $ is coulomb's constant, $ Q $ is total charge ib sphere and $ R $ is radius of sphere.
So the question remains whether you should consider potential at center or at surface. 
If you find out the formula for potential at center of an insulated sphere it comes out to be 
$$ V = \frac{3kQ}{2R} $$
All the symbols have same meanings as before. 
Since in both cases $ V \propto \frac {kQ}{R}$, it will not matter which potential you consider.
