Answer to Question #165800 in Physics for Juliet K

Question

Millikan’s oil drop experiment was replicated in the laboratory (Figure 7.5). The droplet radius was 7 × 10^-7 m, and the density of the oil was 900 kg m^-3.

The plates were spaced 6 mm apart. When a potential difference of 500 V was applied to the plates, the oil droplet was observed to be stationary.

a) Calculate the charge on the oil drop.

b) Comment on the most likely measurement to be responsible for the discrepancy between the calculated charge and the accepted value of the charge on the electron.

c) If the potential difference is raised above 500 V what will happen to the oil drop?

Answer 1

(a) In order to oil droplet to be stationary, the electrostatic force must balance the force of gravity:

$$F_e=F_g,$$ $$qE=mg,$$ $$q\dfrac{V}{d}=mg,$$ $$q=\dfrac{mgd}{V}.$$

We can find the mass of the oil droplet as follows:

$$m=\rho_{oil}V=\dfrac{4}{3}\pi r^{3}\rho.$$

Substituting $m$ into the previous equation we can find charge on the oil drop:

$$q=\dfrac{\dfrac{4}{3}\pi r^{3}\rho gd}{V},$$ $$q=\dfrac{\dfrac{4}{3}\pi\cdot(7\cdot10^{-7}\ m)^{3}\cdot900\ \dfrac{kg}{m^3}\cdot9.8\ \dfrac{m}{s^2}\cdot6\cdot10^{-3}\ m}{500\ V}=1.52\cdot10^{-19}\ C.$$

(b) The most likely measurement to be responsible for the discrepancy between the calculated charge and the accepted value of the charge on the electron ($1.6\cdot10^{-19}\ C$) is the voltage measurements.

(c) If the potential difference is raised above 500 V, then the electrostatic force (directed upward) would be greater than the gravitational force (directed downward). As a result, the oil drop would start to rise (in the upward direction).