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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 stationary droplet experiences two main forces: the weight (gravitational force) acting downwards and the electric (electrostatic) force acting upwards due to the electric field between the plates. These forces are equal in magnitude but opposite in direction, which leads to the droplet being in equilibrium.
Step 2
The potential difference between the plates is changed to zero and the droplet falls at a terminal velocity of 1.0 x 10^-1 m s^-1. Show that the radius of the droplet is about 1 x 10^-6 m.
Answer
When the droplet reaches terminal velocity, the net force acting on it is zero. The downward gravitational force equals the upward viscous drag force. The weight of the droplet is given by:
ho V g = ho \left( \frac{4}{3} \pi r^3 \right) g$$ The viscous drag force can be described by Stoke's law as: $$F_d = 6 \pi \eta r v$$ Setting the two forces equal: $$\rho \left( \frac{4}{3} \pi r^3 \right) g = 6 \pi \eta r v$$ Substituting values, we have: - Density of oil, $\rho = 880 \text{ kg m}^{-3}$ - Viscosity of air, $\eta = 1.8 \times 10^{-5} \text{ N s m}^{-2}$ - Terminal velocity, $v = 1.0 \times 10^{-1} \text{ m s}^{-1}$ - Gravitational acceleration, $g \approx 9.81 \text{ m s}^{-2}$. Thus, rearranging and solving this equation will yield an approximate radius $r \approx 1 \times 10^{-6} \text{ m}$.Step 3
The charge on the droplet is -4.8 x 10^-19 C. Deduce whether this suggestion is correct.
Answer
The suggestion is incorrect. When the droplet splits into two equal spheres, each sphere would have half the charge of the original droplet. The charge on a single new droplet would be -2.4 x 10^-19 C. Since the droplet’s charge causes it to behave in an electric field, both new droplets would still experience electrostatic forces in opposite directions, and therefore would be influenced differently by any electric field present, preventing them from remaining stationary under the same conditions.