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

One problem in measuring the distance (d) is the possibility of the spheres swinging or moving due to external disturbances. A solution could be to use a rigid support to hold the spheres steady while measuring the distance.

The electrostatic force (F) between two charges is given by Coulomb's law:

$F = k \frac{Q_1 Q_2}{r^2}$

Where:

• $k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2$ (Coulomb's constant)
• $Q_1 = Q_2 = 52 \times 10^{-9} \text{ C}$
• $r = 0.04 \text{ m}$ (Distance between centers)

Substituting these values:

$F = (8.99 \times 10^9) \frac{(52 \times 10^{-9})(52 \times 10^{-9})}{(0.04)^2} \approx 4 \times 10^{-3} \text{ N}$

The measured angle θ = 7° indicates a relatively small angle, suggesting that the forces involved (electrostatic and gravitational) are in a comparable range. However, given that the gravitational force is significantly weaker than the electrostatic force in this scenario, this measurement aligns well with the observed effects and the equilibrium state described in the investigation.

To validate the student's claim about the gravitational force's insignificance, we compare the gravitational force (F_g) and the electrostatic force (F_e).

The gravitational force is given by:

$F_g = m g$

Where:

• Mass (m) = $3.2 \times 10^{-3} \text{ kg}$
• Gravitational acceleration (g) = $9.81 \text{ m/s}^2$

Calculating:

$F_g = (3.2 \times 10^{-3}) (9.81) \approx 3.14 \times 10^{-2} \text{ N}$

Comparatively, since the electrostatic force (F_e) is about $4 \times 10^{-3} \text{ N}$, we see that: