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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

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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.

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Answer

The forces acting on the stationary droplet are the gravitational force (weight) pulling it downwards and the electrostatic force acting upwards due to the electric field between the plates. These two forces must be equal in magnitude for the droplet to remain stationary, which can be expressed as:

Fextgravity=FextelectrostaticF_{ ext{gravity}} = F_{ ext{electrostatic}}

Thus, the relationship is that both forces are equal in magnitude and opposite in direction.

Step 2

Show that the radius of the droplet is about 1 × 10⁻⁶ m.

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Answer

The terminal velocity can be analyzed using the equation derived from Stokes' law for a sphere falling through a viscous fluid:

vt=2r2(ρ−σ)g9ηv_t = \frac{2r^2 (\rho - \sigma) g}{9\eta}

Where:

  • vt=1.0×10−1 m s−1v_t = 1.0 \times 10^{-1} \text{ m s}^{-1} is the terminal velocity.
  • ρ=880 kg m−3\rho = 880 \text{ kg m}^{-3} is the density of the oil.
  • σ=1.2 kg m−3\sigma = 1.2 \text{ kg m}^{-3} is the density of air (given in standard conditions).
  • g=9.81 m s−2g = 9.81 \text{ m s}^{-2} is the acceleration due to gravity.
  • η=1.8×10−5 N s m−2\eta = 1.8 \times 10^{-5} \text{ N s m}^{-2} is the viscosity of air.

Rearranging the formula to solve for radius rr:

r=9ηvt2(ρ−σ)gr = \sqrt{\frac{9\eta v_t}{2 (\rho - \sigma) g}}

Substituting the values, we find:

r≈1×10−6 mr \approx 1 \times 10^{-6} \text{ m}

Step 3

Deduce whether this suggestion is correct.

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Answer

The student's suggestion is incorrect. If the droplet splits into two smaller droplets of equal size, the charge on each droplet will be half of the original droplet's charge. Since the charge is proportional to the electrostatic force acting on a droplet in the electric field, a smaller charge will create less upward electrostatic force. Thus, without additional forces acting on these smaller droplets, they would not be able to balance their weight and would fall due to gravity, rather than remaining stationary.

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