6 Two positively charged identical metal spheres A and B have their centres separated by a distance of 24 cm, as shown in Fig. 6.1. [Figure 6.1] The variation with distance x from the centre of A of the electric field strength E due to the two spheres, along the line joining their centres, is represented in Fig. 6.2. [Figure 6.2]
- (a)State the radius of the two spheres. radius = ................................................................ cm [1][1]
- (b)The charge on sphere A is 3.6 × 10⁻⁹C. Determine the charge Q_B on sphere B. Assume that spheres A and B can be treated as point charges at their centres. Explain your working. Q_B = ........................................................................ C [3][3]
- (c)(i)Sphere B is removed. Use information from (b) to determine the electric potential on the surface of sphere A. electric potential = ................................................................ V [2][2]
- (c)(ii)Calculate the capacitance of sphere A. capacitance = ................................................................ F [2][2]
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Mark scheme
Marking points from the official mark scheme (Feb/Mar 2020). The mark or mark type is shown in brackets.
- [1] 2.0 cm
- [1] At 16 (cm) from A the electric fields are equal or E_A = E_B
- [1] E = Q / 4πε₀r²
- [1] Q_B = 9.0 × 10⁻¹⁰ C
- [1] V = Q / 4πε₀r_A
- [1] V = 3.6 × 10⁻⁹ / (4 × π × 8.85 × 10⁻¹² × 0.020) = 1600 V
- [1] C = Q / V
- [1] = 3.6 × 10⁻⁹ / 1600 = 2.3 × 10⁻¹² F
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