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An oil droplet is sprayed into a uniform electric field of adjustable magnitude. The 0.11 g droplet hovers
motionless (gravity force equalling electrostatic force) when the electric field is set to 370 N/C and directed
downward.
a) Determine the sign and magnitude of the charge on the oil droplet.
b) Determine the approximate number of excess electrons that are on the oil droplet.

Respuesta :

Answer:

The direction of the field is downward, and negatively charged particles will experience an upwards force due to the field.

F = N e E     where E is the value of the field and N e the charge Q

M g = N e E      and M g is the weight of the drop

N = M g / (e E)

N = 1.1E-4 * 9.8 / (1.6E-19 * 370) = 1.1 * 9.8 / (1.6 * 370) * E15 = 1.82E13

.00011 kg is a very large drop

Q = N e = M g / E = .00011 * 9.8 / 370 = 2.91E-6 Coulombs

Check:     N = Q / e = 2.91E-6 / 1.6E-19 = 1.82E13   electrons