Electrostatics Part 1

Given, number of mercury drops, n=512
Voltage of each drop, V=2V Let r,R be the radius of drop small and combined spherical drop, respectively.
Now, when all drops are joined into single drop, volume remains constant,
i.e. ‌‌512×‌

4

3

πr3=‌

4

3

πR3
∵‌‌V=‌

kq

r

=2. . . (i)
∴‌‌V‌net ‌=‌

kQ

R

where, Q be the charge of bigger sphere.
⇒‌‌V‌net ‌=‌

k×512q

8r

. . . (ii)
On dividing Eq. (ii) by Eq. (i), we get
‌

V‌net ‌

V

=‌

k×512q×r

8r×kq

=‌

512

8

‌V‌net ‌=2×‌

512

8

=128V