Question

what is the value of electric field to balance a drop of oil of mass 0.002 mg having 6 electrons cha​

Answer 1

Answer:

$qe = mg \\ so \: puting \: the \: values \\ m = 2 \times 10 ^{ - 6} \\ q = ne = 6 \times 1.6 \times 10 ^{ - 19} \\ e = mg \div q = 1.25 \times 10 ^{13}$

e=electric field =1. 25 ×20^13N/C