Figure 1 shows an experiment to measure the charge of the electron - AQA - A-Level Physics - Question 1 - 2019 - Paper 7

To find the radius of the droplet, we apply the following equations:

1. The terminal velocity formula is given as: $v_t = \frac{2r^2(\rho_p - \rho_f)g}{9\eta}$ Where:

   • $v_t$ = terminal velocity = 1.0 × 10⁻¹ m s⁻¹
   • $\rho_p$ = density of the oil = 880 kg m⁻³
   • $\rho_f$ = density of air (approximately 1.225 kg m⁻³ at 20 °C)
   • $g$ = acceleration due to gravity = 9.81 m s⁻²
   • $r$ = radius of the droplet
   • $\eta$ = viscosity of air = 1.8 × 10⁻² N s m⁻²

2. Rearranging the formula to solve for $r$ involves substituting values and solving: $r^2 = \frac{9\eta v_t}{2(\rho_p - \rho_f)g}$ By substituting the values found above, we can determine the radius, showing it to be around 1 × 10⁻⁶ m.

To determine whether the suggestion that two equal droplet spheres would remain stationary is correct, we evaluate the forces on each droplet after splitting:

1. The charge on the original droplet is -4.8 × 10⁻¹⁹ C.
   Consequently, if this charge is split equally, each droplet will possess a charge of -2.4 × 10⁻¹⁹ C.

2. With each sphere having half the charge, their interactions with the electric field will change. The forces experienced by both (due to electrostatic repulsion) would now be different when compared to a single droplet. Therefore even though both will feel a force, with reduced charge, it may not be sufficient to maintain stationarity due to lack of balance with gravitational force.

In essence, while the electrostatic forces may allow for temporary stability, upon further analysis, the droplets cannot remain stationary as they would still have gravitational pull without equal countering electric force.