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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 are the weight (gravitational force) acting downward and the electric (electrostatic) force acting upward due to the electric field between the plates. Since the droplet is stationary, these forces must be equal in magnitude and opposite in direction, hence:
- Weight (W) = mg, where m is mass and g is the acceleration due to gravity.
- Electrostatic force (F) = qE, where q is the charge on the droplet and E is the electric field strength.
Thus, we can state:
This reflects that the net force is zero when the droplet is stationary.
Step 2
The potential difference between the plates is changed to zero and the droplet falls at a terminal velocity of 1.0 × 10^{−1} m s^{−1}. Show that the radius of the droplet is about 1 × 10^{−6} m.
Answer
At terminal velocity, the forces acting on the droplet are balanced:
- The weight of the droplet:
ho g $$
where V is the volume of the droplet, and is the density of the oil.
- The drag force acting against the motion of the droplet as it falls:
u rac{r^2
ho_{air} v}{h} $$
where is the viscosity of air, is the radius of the droplet, is the density of air, and is the terminal velocity.
Setting these forces equal:
ho g = 6 u r^2 ho_{air} v $$ Substituting the variables and solving for the radius: 1. Compute volume from radius: $$ V = rac{4}{3} ext{π} r^3 $$ 2. Then inserting and rearranging for $r$ gives an approximate radius of: $$r ext{ is approximately } 1 imes 10^{−6} m.$$Step 3
The potential difference between the plates is restored to its initial value and the droplet becomes stationary. Deduce whether this suggestion is correct.
Answer
The student's claim that splitting the droplet into two equal-sized spheres would allow both to remain stationary is incorrect. When a droplet splits, the total charge is redistributed between the two smaller droplets. Each smaller droplet will have half the original charge:
q' = rac{q}{2}
Given that the force due to electric field is proportional to charge, each new droplet experiences a reduced upward electrostatic force. At the same time, the weight remains the same for each droplet. As a result, the balance of forces is disrupted, causing them not to remain stationary under the same electric field. Therefore, both droplets would not balance out like the original droplet did.