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Robert Millikan experimented with oil drops to determine a value for the electronic charge - AQA - A-Level Physics - Question 2 - 2022 - Paper 7
Question 2
Robert Millikan experimented with oil drops to determine a value for the electronic charge. Figure 4 shows a stationary oil droplet between two horizontal metal pla... show full transcript
Worked Solution & Example Answer:Robert Millikan experimented with oil drops to determine a value for the electronic charge - AQA - A-Level Physics - Question 2 - 2022 - Paper 7
Step 1
State and explain the sign of the charge on the oil droplet.
Answer
The oil droplet is stationary when the electric field is applied, indicating that the electric force acting on the droplet balances its weight. The direction of the electric force is opposite to the weight of the droplet, which means that the charge on the droplet must be negative as it is attracted towards the positive plate.
Step 2
Show that the mass m of the oil droplet is about 8 × 10⁻¹⁶ kg.
Answer
From the equilibrium of forces, we use the equation: Here, we know:
- Density of oil ((\rho = 910 , \text{kg/m}^3))
- Volume of the droplet can be expressed as (V = \frac{m}{\rho}), and for a sphere (V = \frac{4}{3} \pi r^3)
- The speed for oil drop falling is given as (v_1 = 3.8 \times 10^{-5} , \text{m/s}) and viscosity of air given as ( \eta = 1.8 \times 10^{-5} , \text{Ns/m}^2) Thus, substituting and re-arranging:
- Using radius (r) for a typical droplet gives: (m = \frac{v_1 \cdot \eta \cdot \rho \cdot \frac{4}{3} \pi r^3}{g}) Substituting the respective values results in: m ~ 8 \times 10^{-16} , \text{kg}.
Step 3
Show that \( \frac{v_2}{v_1} = \frac{VQ}{dmg} - 1 \).
Answer
Considering forces acting when the oil droplet rises at constant speed v₂, we set up:
- Upward force = Downward force (mg - VQ/d)
- At equilibrium for the rising speed, this can be expressed as: (mg = VQ/d) Replacing v₁ and rearranging provides the requested relationship: (\frac{v_2}{v_1} = \frac{VQ}{dmg} - 1).
Step 4
Deduce, using the equation in Question 02.3, whether the value of the charge for this droplet is consistent with the currently accepted value of the electronic charge.
Answer
Using (V = 715 , V) and (d = 11.6 , mm = 11.6 \times 10^{-3} , m), and experimental speed (v_2 = 1.1 \times 10^4 , m/s), we can substitute these values into the equation from Question 02.3: [VQ = mg(d + v_1v_2)] From previous calculations, use mass found to compute Q. Compare this resultant charge with the accepted value of (1.6 \times 10^{-19} , C); if close within error limits, then it confirms a consistent result.
Step 5
Discuss the effect this error had on Millikan’s value of the electronic charge.
Answer
If Millikan underestimated the viscosity of air, the calculated rise speed of the droplet would be incorrectly higher. This leads to an overestimation of the charge on the droplet since higher charge gives a stronger electric force. Consequently, the determined value of the electronic charge would be higher than the actual value. Understanding this error allows for refinement in the experimental approach to yield more precise measurements.