Question

In Millikan's experiment, an oil drop of radius 2.012 µm and density 0.828 g/cm3 is suspended in chamber C (Figure 22-14) when a downward-pointing electric field of 1.92* 10^5 N/C is applied. Find the charge on the drop, in terms of e. (Include the sign.)

Oil spray Insulating chamber wall drop Microscope Fig. 22-14

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Answer 1

For equilibrium in vertical direction on drop,

gravitational force on drop = electric force on drop

m*g = q*E

here, m = mass of drop = rho*V

where, rho = density of oil = 0.828 gm/cm^3

V = volume of drop = (4/3)*pi*r^3

where r = radius of drop = 2.012 $\mu$m = 2.012*10^-4 cm

So, V = (4/3)*pi*(2.012*10^-4)^3

V = 3.41*10^-11 cm^3

So, m = 0.828*(3.41*10^-11)

m = 2.82*10^-11 gm = 2.82*10^-11 kg

q = charge on drop = ??

E = electric field = 1.92*10^5 N/C

now by bold equation,

q = m*g/E

q = (2.82*10^-14*9.81)/(1.92*10^5)

q = 1.44*10^-18 C

Since electric field is downward

So, for equilibrium

q = -1.44*10^-18 C

charge on electron(e) = -1.6*10^-19 C

So, charge on drop = [(-1.44*10^-18)/(-1.60*10^-19)]*e

charge on drop = 9e

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