A charged drop of oil of radius  2.76 micro C and density 920 kg / m³ is held stationary in a vertically downward electric field of 1.65 * 10⁶ NC^-1 .

(1) Find the magnitude and sign of charge on the drop.

(2)  If the drop captures two electrons in the same electric field , what would be  its acceleration ?

Ignore viscous drag.

Hello,

m = mass of the drop

r = radius of the drop

 = ?

F_e = Force on the drop to be in equilibrium due to electric field

E = 1.65 × 10⁶ N/C

ρ = density of drop = 920kg/m³

So mass of the drop = m = ρ V

V = Volume of drop =

Find m = ?

(1) All the equilibrium 

F_e = mg

⇒ 9E = mg

Please find it, because you did not mention correctly the unit of radius of the drop.

Charge on drop would be negative because electric force on it is opposite to the electric field to balance the weight of the drop.

(2) If drop capture two more electrons. Then net charge on it would be

Q = (q + 2e)  = ?

Where e = charge on electron

So, electric force on drop increase and accelerate in upward direction.

Apply Newton's 2nd law of motion.

F'_e – mg = ma  (a = acceleration of drop)

⇒ QE – mg = ma

Here find the value of a, by plug–in the values of variables Q, E, m and g.