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

Arrows should be drawn on sphere B indicating the following forces:

• The gravitational force acting downwards (F_gravity).
• The electrostatic repulsive force acting horizontally away from sphere A (F_electrostatic).
• The tension in the thread (T) acting upwards and at an angle.

One problem in measuring d is that the apparatus may not be stable during measurement due to the forces involved. A solution could be to use a ruler with a non-conductive material to prevent interference. Additionally, conducting the measurement far from any conductive surfaces would reduce errors.

The electrostatic force can be calculated using Coulomb's law: $F = k \frac{Q_1 Q_2}{d^2}$ Where:

• k = 8.99 \times 10^9 , \text{N m}^2 / \text{C}^2
• Q1 = Q2 = 52 \times 10^{-9} C
• d = 0.040 m (the distance between the centers)

Thus: $F = 8.99 \times 10^9 \frac{(52 \times 10^{-9})^2}{(0.040)^2} \approx 4 \times 10^{-3} N$

The angle θ recorded (7°) should be reasonable given the forces acting on each sphere. If the electrostatic force is significant compared to the gravitational force, it would influence the angle. If the calculated force significantly differs, this could suggest an inconsistency or measurement error regarding the gravitational force's impact.

To assess the student's statement, we can compare the gravitational force to the electrostatic force. The gravitational force (F_gravity) can be calculated using: $F_{gravity} = mg = (3.2 \times 10^{-3} \text{ kg})(9.81 \text{ m/s}^2) \approx 0.0314 \text{ N}$ This value is much larger than the calculated electrostatic force (4 × 10^-3 N), suggesting that the gravitational force has a significant effect on the arrangement and thus influences the angle. Therefore, the student's statement is not valid.