Question

A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 8480 N/C. The mass of the water drop is3.50×10−9kg(a) Is the excess charge on the water drop positive or negative? Why? (b) How many excess electrons or protons reside on the drop?

Answer 1

q = 4.04×10^-12C

n = 2.53×10^7

Step-by-step explanation:

1. Electrostatic force on charge due to electrical field E is F = qE

Gravitational force due to weight of drop of mass M is F = mg, when net force is zero we have to find the charges on the drop.

2. Charge on drop: since gravitational force is acting downward, electric field is acting upward and we equate both equations.

Therefore, Fg = Fe

qe = mg

q = mg/E

q = 3.50×10^-9kg×9.8m/s²/8480N/c

q = 4.04 × 10^-12

3. Number of protons by quantitization law

n = q/e

n = 4.04× 10^-12/8480

n = 2.53× 10^7