Question

Calculate the radius of a water drop which would just remains suspended in an electric field of 300V/cm and charge with one electron​

Answer 1

electric field (E) = kq/r^2 = 300 V/cm = 30000 v/m

q = ne

no. of electron = 1

q = 1 x (1.6 x 10^-19)

q = 1.6 x 10^-19

E = (9 x 10^9 x 1.6 x 10^-19)/r^2

r^2 = (9 x 16 x 10^-11)/30000

so r = 2.19 x 10^-7 m

Step-by-step explanation: