Answer:
![(V_(big))/(V_(small)) = n^(2/3)](https://img.qammunity.org/2020/formulas/physics/high-school/2a9unm7cfndqm8o3elejirr18jianudxl4.png)
Step-by-step explanation:
Let the total charge on the big drop is given as Q
now if the radius of the drop is R then electric potential of the big drop is given as
![V_(big) = (KQ)/(R)](https://img.qammunity.org/2020/formulas/physics/high-school/uykrtns4giy0tkfbv72e1rd603qvezwtbx.png)
Now if it break into n identical drops
then let the charge on each drop is "q" and radius is "r"
by volume conservation
![(4)/(3)\pi R^3 = n((4)/(3)\pi r^3)](https://img.qammunity.org/2020/formulas/physics/high-school/xgwjqtcp1yqv6o0o2de4a5zlhc13dddqpz.png)
![r = (R)/(n^(1/3))](https://img.qammunity.org/2020/formulas/physics/high-school/bo2utej2ejrbbmz7osd38h3xomgo07hj9h.png)
now we have potential of smaller drop given as
![V_(small) = (kq)/(r)](https://img.qammunity.org/2020/formulas/physics/high-school/4xn5kheagz8n85sdf9cowjk9kuk3p3u2t9.png)
![V_(small) = (K(Q/n))/((R)/(n^(1/3)))](https://img.qammunity.org/2020/formulas/physics/high-school/gueqx4ejf0rbt2wcqpohhp7x97jck6jor1.png)
![V_(small) = (1)/(n^(2/3))(KQ)/(R)](https://img.qammunity.org/2020/formulas/physics/high-school/lgu6twxzfsxnnukfbdt4q7n9y3xc7lyj1n.png)
![(V_(big))/(V_(small)) = n^(2/3)](https://img.qammunity.org/2020/formulas/physics/high-school/2a9unm7cfndqm8o3elejirr18jianudxl4.png)