Answer:
0.0006091222 m
Explanation:
q = Charge = 42 pC
V = Voltage = 620 V
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
Electric potential is given by (at r = R)
[tex]V=\dfrac{q}{4\pi\epsilon R}\\\Rightarrow R=\dfrac{q}{4\pi\epsilon V}\\\Rightarrow R=\dfrac{42\times 10^{-12}}{4\pi\times 8.85\times 10^{-12}\times 620}\\\Rightarrow R=0.0006091222\ m[/tex]
The radius of the drop is 0.0006091222 m
