Question

09. A negatively charged oil drop is prevented from failing under gravity by
applying a vertical electric field 100 V m-1. If the mass of the drop is 1.6 x 10
g, the number of electrons carried by the drop is (g = 10 ms)
a) 10^18
b) 10^6
c) 10^9
d) 10^12​

Answer 1

Answer:

10^(19) electrons (if mass of drop given is correct)

Explanation:

Weight of the drop = mg

Electric force = Eq

Both the forces are acting in equal & opposite directions to prevent falling of drop

Therefore,

mg = Eq

q = mg / E

= (1.6 × 10 g × 10 m/s²) / (100 V/m)

= 1.6 C

Number of electrons = q/e

= (1.6 C) / (1.6 × 10^(-19))

= 10^(19)

P.S: Please check the mass of drop which you've provided, there must be a typo in it.