Figure 3 shows an arrangement used to investigate the repulsive forces between two identical charged conducting spheres - AQA - A-Level Physics - Question 4 - 2019 - Paper 2

1. Draw the force of repulsion $F_{r}$ directed away from sphere A.
2. Draw the gravitational force $F_{g}$ acting downwards.
3. Label the angle of deviation $\theta$ from the vertical.

One problem is ensuring accurate measurement of distance $d$, especially with any movement of the spheres post-charging. A solution could involve using a ruler with a non-conductive material to minimize interference during measurement.

The electrostatic force can be calculated using Coulomb's law: $F = k \frac{Q^2}{d^2}$ where $k = 8.99 \times 10^{9} \text{ N m}^2/\text{C}^2$, $Q = 52 \times 10^{-9} \text{ C}$, and $d = 40 \times 10^{-3} \text{ m}$. Substituting: $F = (8.99 \times 10^{9}) \frac{(52 \times 10^{-9})^2}{(40 \times 10^{-3})^2} \approx 4 \times 10^{-3} \text{ N}$

The angle $\theta = 7^{\circ}$ indicates that there is a balance between gravitational and electrostatic forces. The calculation of the gravitational force can be used to determine if it significantly affects the dynamics. The tension in the threads must also be taken into account to confirm that the system remains in equilibrium.

The gravitational pull on each sphere owing to its mass $m = 3.2 \times 10^{-3} \text{ kg}$ is given by: $F_{g} = mg = (3.2 \times 10^{-3}) (9.81) \approx 0.0314 \text{ N}$ Comparing this with the calculated electrostatic force $F_{e} \approx 4 \times 10^{-3} \text{ N}$, we find that: $F_{g} = 0.0314 \text{ N} >> F_{e}$ Therefore, the student's statement is not valid; the gravitational force does have a significant impact on the equilibrium state.