Answer :
Answer:
[tex]V = 689.4 Volts[/tex]
Explanation:
charge on the drop is given as
[tex]Q = 0.0000000018 C[/tex]
V = 434.21 Volts
now we know that
[tex]V = \frac{kQ}{r}[/tex]
so we have
[tex]434.21 = \frac{(9 \times 10^9)(1.8 \times 10^{-9})}{r}[/tex]
[tex]r = 3.73 cm[/tex]
now when two such drops is merged and forms a single drop then in that case
[tex]V = V_1 + V_2[/tex]
[tex]\frac{4}{3}\pi R^3 = 2(\frac{4}{3}\pi r^3)[/tex]
so we have
[tex]R = 2^{1/3} r[/tex]
[tex]R = 4.7 cm[/tex]
now after merging two drops we have total charge on both drops is
[tex]Q = 2q = 1.8 nC + 1.8 nC = 3.6 nC[/tex]
now potential of two drops is given as
[tex]V = \frac{KQ}{R}[/tex]
[tex]V = \frac{(9 \times 10^9)(3.6 \times 10^{-9})}{0.047}[/tex]
[tex]V = 689.4 Volts[/tex]
