Answer 8_2_5

In an experiment to determine the charge on a charged oil droplet, the droplet was held stationary in a vertical electric field of strength 57 kV m-1.  After the field was switched off, the droplet fell at a steady speed, taking 18.3 s to fall through a vertical distance of 2.0 mm

 

Viscosity of air = 1.8 × 10-5 N s m-2

Density of oil = 970 kg m-3

g = 10 m/s-2

 

(a)    Calculate the speed of the droplet as it falls.

(b)   Show that the droplet’s radius is 9.7 ´ 10-7 m

(c)    Calculate the charge of the droplet.

(d)   Compare this to the electronic charge.  What does it suggest?                   

 

(a) Speed = 2.0 ´ 10-3 ¸ 18 = 1.11 ´ 10-4 m/s (P)

 

(b) r2 = 9hv  (PP) = 9 ´ 1.8 ´ 10-5 ´ 1.1 ´ 10-4 = 9.37 ´ 10-13 m2 (P)

            2rg              2 ´ 970 ´ 9.8

 

 

r = 9.7 ´ 10-7 m

 

(c)  Find out the mass

 

Mass = 4/3 ´ p ´ (9.7 ´ 10-7m)3 ´ 970 kg m-3 = 3.69 ´ 10-15 kg (P)

 

Weight = 3.69 ´ 10-15 kg ´ 9.8 m/s2 = 3.61 ´ 10-14 N

 

F = Eq = weight

 

57000 N C-1 ´ q = 3.61 ´ 10-14 (P)

 

q = 6.37 ´ 10-19 C (P)

 

(d)  This is about 4 times the electronic charge.  (P)

 

It tells us the charge is quantised (goes up in whole number multiples) (P)

 

 

 

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