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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:
- Weight (gravitational force): This acts downwards due to gravity.
- Electric (electrostatic) force: This acts upwards due to the electric field created by the potential difference between the plates.
These two forces are equal in magnitude but opposite in direction, thus resulting in a net force of zero, allowing the droplet to remain 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 must consider the forces acting on it when it reaches terminal velocity.
At terminal velocity, the weight of the droplet is equal to the drag force:
-
Weight: [ F_g = mg = \rho V g = \rho \left( \frac{4}{3} \pi r^3 \right) g ]
where ( \rho ) is the density of the oil, and ( g ) is the acceleration due to gravity. -
Drag Force: [ F_d = 6 \pi \eta r v ]
where ( \eta ) is the viscosity of air, ( r ) is the radius, and ( v ) is the terminal velocity.
Setting these equal gives:
[ \rho \left( \frac{4}{3} \pi r^3 \right) g = 6 \pi \eta r v ]
Canceling ( r ) from both sides (assuming ( r \neq 0 )) and rearranging gives:
[ r^2 = \frac{9 \eta v}{2 \rho g} ]
Substituting in values:
- ( \rho = 880 , \text{kg m}^{-3} )
- ( \eta = 1.8 \times 10^{-5} , \text{Ns m}^{-2} )
- ( v = 1.0 \times 10^{-1} , \text{m s}^{-1} )
- ( g = 9.81 , \text{m s}^{-2} )
We find:
[ r^2 = \frac{9 \times (1.8 \times 10^{-5}) \times (1.0 \times 10^{-1})}{2 \times (880) \times (9.81)} \approx 1.0 \times 10^{-12} , \text{m}^2 ]
Thus, [ r \approx 1.0 \times 10^{-6} , \text{m} ]
Step 3
Deduce whether this suggestion is correct.
Answer
The suggestion that both smaller droplets would remain stationary is incorrect.
When the droplet splits into two equal-sized droplets, each droplet would have half the charge and half the radius. The weight of each smaller droplet would still balance any electrostatic force acting on it, but because of the reduced radius, the drag force acting on the smaller droplets will also change.
Using the formula for drag, since drag force is proportional to the velocity and radius of the droplet, the smaller droplets may not experience the same balance of forces as the larger droplet did, which could lead to them falling instead of remaining stationary.