Question

A oil drop having a mass of 0.002gm and charge equal to 6e- is suspended stationary in a uniform electric field . find the intensity of electric charge.​

Answer 1

Explanation:

F=ma

F=Eq

Eq=ma

E=mg/q

E=0.002×10/6×1.6×10^-19

E=0.2083×10^17 N/C

adjust the value of answer according to requirement.