In Millikan's experiment, an oil drop of radius 2.342 μm and density 0.816 g/cm3 is suspended in chamber C when a downward-pointing electric field of 1.92 ✕ 105 N/C is applied. Find the charge on the drop, in terms of e. (Include the sign of the value in your answer.)

-14 e

Explanation:

Assuming oil drop is spherical then V = 4pi.r^3/3

density = 0.816 g/cm^3= 0.816 x 10^-3kg/cm^3= 0.816 x 10^-3/10^-6 kg/m^3= 0.816 x 10^3 kg/m^3

mass =(4 x 3.142 x (2.342 x 10^-6)^3 x 0.816 x 10^3)/3 kg

mass = (12.568 x 12.85 x 10^-18 x 0.816 x 10^3)/3 kg

m= 4.39 x 10^-14 kg

F=qE…. Eqn. 1

F=mg… Eqn. 2

Compare both eqn.1 and eqn. 2

qE=mg

q=mg/E

q = (4.39 x 10^-14 x 9.81)/1.92 x 10^5

q = 2.24 x 10^-18 C

Number of charges, n = 2.24 x 10^-18/1.62 x 10^-19

n=13.827

Charges come in integer values, nearest is 14

Number of charges is 14.

The direction of the field (downwards) is the direction a positive test charge would move. This means a positive charge would move downwards and repelled by the top plate which must be positive.

Since mg acts downwards then the other force must act up if the oil drop is suspended. This means that the charge is being attracted to the top plate which we have already determined is positive. for the charge to be attracted to the positive top plate then the charge must be negative.

Charge is electron hence the sign is minus.

-14 e

i think the answer would be b