A large charged spherical liquid drop has an electric potential of...

Let the radius of the large drop be $R$ and its charge be $Q$.

The electric potential at the surface of the large drop is $V = \dfrac{kQ}{R} = 320\text{ V}$.

Let the radius of each small droplet be $r$ and its charge be $q$.

By conservation of charge, $Q = Nq \Rightarrow q = \dfrac{Q}{N}$.

By conservation of volume, $\dfrac{4}{3}\pi R^3 = N \left(\dfrac{4}{3}\pi r^3\right) \Rightarrow r = \dfrac{R}{N^{1/3}}$.

The electric potential at the surface of each small droplet is $v = \dfrac{kq}{r}$.

Substituting the values of $q$ and $r$, we get:

$v = \dfrac{k(Q/N)}{R/N^{1/3}} = \dfrac{kQ}{R} \times \dfrac{1}{N^{2/3}} = \dfrac{V}{N^{2/3}}$

Given $v = 20\text{ V}$ and $V = 320\text{ V}$, we have:

$20 = \dfrac{320}{N^{2/3}}$

$N^{2/3} = \dfrac{320}{20} = 16$

$N = (16)^{3/2} = (4^2)^{3/2} = 4^3 = 64$

Answer: $64$