Question

In a particular Millikan oil-drop apparatus, the plates are 2.25 cm apart. The oil used has a density of 0.815 g/cm3 , and the atomizer that sprays the oil drops produces drops of diameter 1.00×10⁻³ mm.  

A) What strength of electric field is needed to hold such a drop stationary against gravity if the drop contains five extra electrons?

B) What should be the potential difference across the plates to produce this electric field?

C) If another drop of the same oil requires a plate potential of 65.3 V to hold it stationary, how many excess electrons did it contain?

Answer 1

In Milikan's experiment

Using force balance on the oil drop

Since we know that gravitational force will balance the electrostatic force on oil drop. And gravitational force is always downward, which means electrostatic force must be upward.

Now given that Electric force is upward, and charge on oil drop is negative, So that electric field will be downwards.

Now

Fnet = 0 = mg - qE

E = mg/q

g = 9.81 m/sec^2

m = mass of drops = density*Volume

density = 0.815 gm/cm^3 = 815 kg/m^3

Volume = 4*pi*r^3/3

r = radius of drop = diameter/2 = 1.00*10^-3 mm = 5.0*10^-7 m

m = 815*4*pi*(5.0*10^-7)^3/3 = 4.267*10^-16 kg

Now

q = 5e = 5*(1.60*10^-19) = -8.0*10^-19 C

Which gives

|E| = 4.267*10^-16*9.81/(8.0*10^-19) = 5232.4 N/C = Strength of electric field

Part B.

Potential difference between plates will be given by:

V = E*d

d = separation distance between plates = 2.25 cm = 2.25*10^-2 m

V = 5232.4*2.25*10^-2 = 117.7 V

Part C.

Now on 2nd drop V = 65.3 V, then

E = V/d = 65.3/(2.25*10^-2) = 2902.22 N/C

Now for this electric field charge on drop will be:

q = mg/E

q = 4.267*10^-16*9.81/(2902.22) = 1.44*10^-18 C

q = 14.4*10^-19 C = (14.4*10^-19)/(-1.6*10^-19) = -9

excess electrons in 2nd drop = 9 electrons

Let me know if you've any query.