Question 6 The diagram below shows two small identical spheres, \( P \) and \( Q \), on insulated stands. The charge on sphere \( P \) is \( -3 \times 10^{-6} \mathrm{C} \) and the charge on sphere \( Q \) is unknown. \( -3 \times 10^{-6} \mathrm{C} \) 6.1.1. State the principle of charge quantization 6.1.2. Calculate the number of electrons in excess on sphere \( P \). The two spheres are brought into contact and are then returned to their original positions. Each sphere now carries a charge of \( -1 \times 10^{-6} \mathrm{C} \). 6.2. Calculate the original charge on sphere \( Q \) before the spheres were brought into contact. 6.3. Were electrons transferred from \( P \) to \( Q \) or from \( Q \) to \( P \) during contact? (1) [9]

Below is a worked solution for each part: ────────────────────────────── 6.1.1. Principle of Charge Quantization The principle of charge quantization states that any charge must be an integer multiple of the elementary charge (e). In other words, the net charge Q on any object can only have values of Q = n · e, where n is an integer and e = 1.6 × 10⁻¹⁹ C. ────────────────────────────── 6.1.2. Number of Excess Electrons on Sphere P Sphere P has a charge of –3 × 10⁻⁶ C. Since each electron carries a charge of –1.6 × 10⁻¹⁹ C, the number of excess electrons (n) is given by   n = (Magnitude of total charge)/(Magnitude of one electron’s charge)     = (3 × 10⁻⁶ C)/(1.6 × 10⁻¹⁹ C)     ≈ 1.875 × 10¹³ Thus, there are approximately 1.9 × 10¹³ excess electrons on sphere P. ────────────────────────────── 6.2. Original Charge on Sphere Q When the spheres are brought into contact, charge redistributes equally between the two identical spheres. After separation, each sphere carries –1 × 10⁻⁶ C, so the total combined charge is   Total charge = (–1 × 10⁻⁶ C) + (–1 × 10⁻⁶ C) = –2 × 10⁻⁶ C. Before contact, the total charge was the sum of the charges on P and Q. Since sphere P was initially –3 × 10⁻⁶ C, let the original charge on Q be Q₀. Thus,   (–3 × 10⁻⁶ C) + Q₀ = –2 × 10⁻⁶ C Solve for Q₀:   Q₀ = (–2 × 10⁻⁶ C) – (–3 × 10⁻⁶ C)     = +1 × 10⁻⁶ C. So, sphere Q originally had a charge of +1 × 10⁻⁶ C. ────────────────────────────── 6.3. Direction of Electron Transfer Electrons are negatively charged. Initially, sphere P had a more negative charge (–3 × 10⁻⁶ C) and sphere Q had a positive charge (+1 × 10⁻⁶ C). When they were brought into contact, the charges equalized to –1 × 10⁻⁶ C on each sphere. Notice that: • Sphere P went from –3 × 10⁻⁶ C to –1 × 10⁻⁶ C, meaning it lost some negative charge (or lost electrons). • Sphere Q went from +1 × 10⁻⁶ C to –1 × 10⁻⁶ C, meaning it gained negative charge (gained electrons). Therefore, electrons were transferred from sphere P to sphere Q. ────────────────────────────── Summary of Answers: 6.1.1. Charge quantization means that the charge on any object is an integer multiple of the elementary charge (e = 1.6 × 10⁻¹⁹ C). 6.1.2. The number of excess electrons on sphere P is approximately 1.9 × 10¹³. 6.2. The original charge on sphere Q was +1 × 10⁻⁶ C. 6.3. Electrons were transferred from sphere P to sphere Q during contact.