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?

Question

Neveah Kirk

Physics

21 July, 09:02

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?

+5

Answers (1)

Know the Answer?

Kate Lester · Answer 1

q = 4.04*10^-12C

n = 2.53*10^7

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