Robert Millikan experimented with oil drops to determine a value for the electronic charge - AQA - A-Level Physics - Question 2 - 2022 - Paper 7

To find the mass of the oil droplet, we can use the equation for terminal velocity:

$mg = \eta v$

Where:

• $m$ is the mass of the droplet,
• $g$ is the acceleration due to gravity (approximately 9.81 m/s²),
• $\eta$ is the viscosity of air (1.8 × 10⁻⁵ N s m⁻²),
• $v$ is the terminal velocity (3.8 × 10⁵ m/s).

First, we need to calculate the volume of the droplet using its density:

$Volume = \frac{mass}{density}$

Let’s rearrange our equation:

$m = \frac{\eta v}{g}$

Using the given values:

$m = \frac{(1.8 × 10^{-5} \, \text{N s m}^{-2}) (3.8 × 10^5 \, \text{m s}^{-1})}{9.81 \, \text{m s}^{-2}} \approx 7.7 × 10^{-16} \, \text{kg},$

which rounds to about $8 × 10^{-16} \, \text{kg}$.

Using the force balance during rising:

At constant speed when rising:

$mg - EQ = 0$

where $E$, the electric field, is given by:

$E = \frac{V}{d}.$

The force due to the electric field is:

$F_{E} = EQ = \frac{VQ}{d}.$

At terminal speed (when v2 is constant), this is balanced with the weight:

$EQ = mg.$

Considering the case when the droplet is falling at speed $v_1$ and rising at speed $v_2$, we can express this as:

$v_1 = \frac{mg}{\eta}$ and substituting this into the balanced equation, we find:

$v_2 = v_1 \frac{VQ}{dmg}.$

Therefore,

$\frac{v_2}{v_1} = \frac{VQ}{dmg} - 1.$

Substituting the known values into the equation:

• $V = 715 \, \text{V},$
• $v_2 = 1.1 \times 10^4 \, \text{m s}^{-1},$
• $d = 11.6 \times 10^{-3} \, \text{m},$
• $m = 8 \times 10^{-16} \, \text{kg},$
• $g \approx 9.81 \, \text{m s}^{-2}.$

Thus,

$\frac{715 \, Q}{(11.6 \times 10^{-3})(9.81)(8 \times 10^{-16})} = 1.1 \times 10^4$

This can be manipulated to solve for the charge Q, and if the computed value is close to the accepted value of approximately $1.6 \times 10^{-19} \, \text{C}$, we can conclude whether it is consistent.

The use of an incorrect viscosity of air would lead to systematic errors in the measurements of speed and ultimately in the calculated value of the charge. If the viscosity was overestimated, the value of the mass would appear larger resulting in a smaller calculated charge for each droplet. If underestimated, the opposite would occur. Thus, the electronic charge calculated could deviate from the true value, leading to inaccuracies in Millikan's experiment's conclusions.