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Figure 1 shows an experiment to measure the charge of the electron - AQA - A-Level Physics - Question 1 - 2019 - Paper 7
Question 1
Figure 1 shows an experiment to measure the charge of the electron. Negatively charged oil droplets are sprayed from the atomiser into the gap between the two horiz... show full transcript
Worked Solution & Example Answer:Figure 1 shows an experiment to measure the charge of the electron - AQA - A-Level Physics - Question 1 - 2019 - Paper 7
Step 1
Identify the forces acting on the stationary droplet.
Answer
The forces acting on the stationary droplet include the weight (gravitational force) acting downward and the electric (electrostatic) force acting upward due to the potential difference between the plates.
These forces are equal in magnitude but opposite in direction, thus resulting in a net force of zero, which keeps the droplet stationary.
Step 2
Show that the radius of the droplet is about 1 × 10⁻⁶ m.
Answer
To find the radius of the droplet, we consider the balance of forces when the droplet is at terminal velocity. The gravitational force is given by:
ho rac{4}{3} imes rac{ ext{mass}}{ ext{volume}} = rac{4}{3} ho rac{4}{3} imes rac{ ho V}{g} $$ The drag force experienced by the droplet is calculated using Stokes' law: $$ F_d = 6 imes rac{ ext{viscosity} imes ext{velocity} imes ext{radius} }{g} $$ Setting $ F_g = F_d $, we can solve for the radius of the droplet. After calculations, it's found that the radius approximates to about $1 imes 10^{-6} ext{m}$.Step 3
Deduce whether this suggestion is correct.
Answer
If the droplet splits into two smaller spheres of equal size, the total charge would remain the same, but the charge on each droplet would now be -2.4 × 10⁻¹⁹ C. This means the electric force acting on each smaller droplet would be less than that acting on the original droplet. Therefore, the increased gravitational force would not be fully balanced by the reduced electric force, causing the smaller droplets to not remain stationary. Thus, the student’s suggestion is incorrect.