Electrostatics

A charged oil drop is suspended in a uniform field of 3×104 v/m so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge = 9.9×10-15 kg and g = 10 m/s2)

  1. 1.6 X 10-18
  2. 3.2 X 10-18
  3. 3.3 X 10-18
  4. 4.8 X 10-18

Answer

Since the ball is hanging, force by gravity is balanced by the electric force.

qE = mg 

q = mg/E

= (9.9 x 10-15 x 10)/(3 x 104)

q = 3.3 x 10-18 C

The correct option is C.