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

For sphere B, the forces need to be represented by arrows:

• The electrostatic repulsion force ( $F_{e}$) acting horizontally away from sphere A, labeled as 'Electrostatic Force'.
• The weight force ( $F_{g}$) acting vertically downward due to gravity, labeled as 'Weight'.
• The tension in the supporting thread ( $T$) acting at an angle towards the support, labeled as 'Tension'.

The arrows should clearly indicate directions with respect to sphere B.

One problem in measuring the distance d is that the spheres may swing slightly due to air currents or external vibrations. A suggested solution is to use a measuring device that is fixed and does not disturb the setup, such as a digital caliper, to measure the distance between the spheres while ensuring minimal movement.

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

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

Where:

• $k ≈ 8.99 × 10⁹ N m²/C²$ (Coulomb's constant)
• $Q_1 = Q_2 = 52 × 10^{-9} C$ (charges on the spheres)
• $r = d = 40 × 10^{-3} m$

Substituting the values:

$F = (8.99 × 10^9) \frac{(52 × 10^{-9})^2}{(40 × 10^{-3})^2} ≈ 4.028 × 10^{-3} N$

Thus, the magnitude of the electrostatic force on each sphere is approximately 4 × 10⁻³ N.

The measurement of the angle θ can provide insight into the forces acting on the spheres. Since θ = 7°, we can analyze the balance of forces. The sine of the angle can be correlated to the ratio of forces acting on the spheres:

$tan(θ) = \frac{F_{e}}{F_{g}}$

If the ratio aligns with our previously determined forces, this indicates consistency between the angle measurement and the electrostatic force calculated. Thus, measuring the angle is valid in assessing the equilibrium state.

To test the statement regarding the gravitational force, we calculate:

$F_{g} = m imes g$

Where:

• m = 3.2 × 10⁻³ kg (mass of the sphere)
• g = 9.81 m/s² (acceleration due to gravity)

Thus,

$F_{g} = 3.2 × 10^{-3} kg imes 9.81 m/s² ≈ 0.0314 N$

Comparing the gravitational force to the electrostatic force ( $F_{e} ≈ 4 × 10^{-3} N$), we find:

$\frac{F_{g}}{F_{e}} ≈ \frac{0.0314}{0.004} = 7.85$

This shows that the gravitational force is significant compared to the electrostatic force, suggesting the student's assertion is not valid.