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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 include:

  • Weight (gravitational force) acting downwards, which is equal to W=mgW = mg.
  • Electric (electrostatic force) acting upwards due to the potential difference applied between the plates, which can be expressed as E=qEE = qE where EE is the electric field.

Since the droplet is stationary, these two forces are equal in magnitude but opposite in direction:

mg=qEmg = qE

Thus, the relationship between the forces is that they balance each other out.

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.

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Answer

To find the radius of the droplet, we will use the equation for the terminal velocity:

vt=2r2(ρoil−ρair)g9ηv_t = \frac{2 r^2 (\rho_{oil} - \rho_{air})g}{9\eta}

Where:

  • vtv_t is the terminal velocity (1.0×10−11.0 × 10^{-1} m s⁻¹),
  • rr is the radius of the droplet,
  • hooil=880extkgm−3 ho_{oil} = 880 ext{ kg m}^{-3},
  • hoair=1.2extkgm−3 ho_{air} = 1.2 ext{ kg m}^{-3} (approximate density of air),
  • g=9.81extms−2g = 9.81 ext{ m s}^{-2} (acceleration due to gravity),
  • u=1.8×10−5extNsm−2 u = 1.8 × 10^{-5} ext{ N s m}^{-2} (viscosity of air).

Substituting the values:

1.0×10−1=2r2(880−1.2)(9.81)9×1.8×10−51.0 × 10^{-1} = \frac{2 r^2 (880 - 1.2) (9.81)}{9 \times 1.8 × 10^{-5}}

Solving this for rr, we find:

r∼1×10−6extmr ∼ 1 × 10^{-6} ext{ 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.

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Answer

If the droplet splits into two smaller droplets of equal size, each droplet will have a radius of r/2r/2. Since the charge remains the same, the total charge will distribute equally among the two new droplets.

The charge on each smaller droplet will be: q′=−4.8×10−19C2=−2.4×10−19Cq' = -\frac{4.8 × 10^{-19} C}{2} = -2.4 × 10^{-19} C

Each smaller droplet will still experience the same forces:

  • The weight will reduce since mass is proportional to the volume of the droplet.
  • The electrostatic force will also be reduced, as it is proportional to charge.

However, since both forces scale down in relation to the droplet size, they will still balance out, meaning that both smaller spheres would not remain stationary; thus, the student's suggestion is incorrect.

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