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

The forces acting on sphere B are as follows:

1. Repulsive force $F_{e}$ due to sphere A, directed horizontally away from sphere A.
2. Tension in the thread $T$, directed diagonally upwards towards the support.
3. Weight $W = mg$, directed vertically downwards.

One problem in measuring $d$ is the difficulty in ensuring that the spheres remain at rest while taking the measurement.
A possible solution would be to use a protractor mounted on a rigid stand to measure the angle consistently, ensuring the instruments do not influence the position of the spheres.

The electrostatic force $F$ between the spheres can be calculated using Coulomb’s law:
$F = k \frac{Q^2}{d^2}$
Where $k \approx 8.99 \times 10^{9} \: N m^{2}/C^{2}$, $Q = 52 \times 10^{-9} \: C$, and $d = 40 \: mm = 0.040 \: m$.
Substituting accordingly, we find:
$F = 8.99 \times 10^{9} \cdot \frac{(52 \times 10^{-9})^2}{(0.040)^2} \approx 4 \times 10^{-3} \: N.$

Given that the angle $\theta = 7^{\circ}$, we can find the horizontal force components acting on the spheres.
The calculated electrostatic force suggests a significant separation of the spheres; thus, a $7^{\circ}$ inclination indicates a matching equilibrium state with the forces acting on both spheres.
This is consistent given the repulsive interaction and the force values derived, confirming the interplay between gravitational and electrostatic forces.

To assess the student’s statement, we can calculate the gravitational force $F_g$ as:
$F_g = mg = (3.2 \times 10^{-3} \: kg)(9.81 \: m/s^{2}) \approx 3.14 \times 10^{-2} \: N.$
Comparing $F_g$ with the electrostatic force found earlier ($4 \times 10^{-3} \: N$), we see that $F_g$ is significantly larger.
Thus, the gravitational force does have a notable effect on the forces acting on the spheres, which contradicts the student's assertion.