12. -1 points SerCP11 15.P.042 My Notes Ask Your Teacher In the Millikan oil-dro

Question

12. -1 points SerCP11 15.P.042 My Notes Ask Your Teacher In the Millikan oil-drop experiment illustrated in the figure below, an atomizer (a sprayer with a fine nozzle) is used to introduce many tiny droplets of oil between two oppositely charged parallel metal plates. Some of the droplets pick up one or more excess electrons. The charge on the plates is adjusted so that the electric force on the excess electrons exactly balances the weight of the droplet. The idea is to look for a droplet that has the smallest electric force and assume it has only one excess electron. Suppose we are using an electric field of 8.85 × 104 N/C. The charge on one electron is 1.60 × 10-19 C. Calculate the radius of an oil drop of density 906 kg/m3 for whih its weight could be balanced by the electric force of this field on one electron. Oil droplets Pinhole 0 Telescope with scale in eyepiece Light Need Help? Read it

Explanation / Answer

We know that electric force is given as

F = qE

where q is charge and E is electric field

F = 1.6e-19*8.85e4

F = 1.416e-14 N

Now, this electric force will balance the weight of the oil drop

F = Fg

F = mg

F = density*volume*g

F = 4/3*pi*r3*density*g

r3 = 3F/4*pi*density*g

Now, put in the values

r3 = 3*1.416e-14 / 4*pi*906*9.8

r = 7.247e-7 m

we need r in um

Therefore,

r = 0.7247 um

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