1. 1
   In a simple model of the hydrogen atom, the electron moves around the proton at a constant separation of 5.3 × 10⁻¹¹ m. In this model the kinetic energy of the electron is 2.17 × 10⁻¹⁸ J. Treat the proton and electron as point charges.

   (a)

   What is the electric potential due to the proton at the position of the electron?

   [1 mark]

   • A+27 V
   • B−27 V
   • C+5.1 × 10¹¹ V
   • D+4.3 × 10⁻¹⁸ V

   (b)

   What is the electric potential energy of the proton–electron system?

   [1 mark]

   • A+4.3 × 10⁻¹⁸ J
   • B−2.3 × 10⁻²⁸ J
   • C−8.2 × 10⁻⁸ J
   • D−4.3 × 10⁻¹⁸ J

   (c)

   Determine the minimum energy needed to remove the electron from the atom to infinity.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   An uncharged gold-leaf electroscope has a metal cap joined by a metal rod to a thin gold leaf, all insulated from the case. The leaf hangs flat against the rod when the electroscope is uncharged. A negatively charged polythene rod is brought close to the cap without touching it, and the leaf rises.

   (a)

   Why does the leaf rise when the rod is brought near the cap?

   [1 mark]

   • AElectrons jump through the air from the rod to the cap
   • BFree electrons in the cap are repelled down to the metal rod and leaf, which both become negative and repel
   • CPositive charges move up from the leaf to the cap, leaving the leaf uncharged
   • DThe leaf becomes positive and is attracted towards the rod

   (b)

   With the polythene rod still held near, the cap is touched briefly with a finger. The finger is removed, and then the polythene rod is taken away. What is the final state of the electroscope?

   [1 mark]

   • AIt is uncharged and the leaf hangs flat
   • BIt has a net negative charge and the leaf rises
   • CIt has a net positive charge and the leaf rises
   • DIt has a net positive charge, but the leaf hangs flat because the charge stays on the cap

   (c)

   Explain why, while the finger is touching the cap and the polythene rod is still held near, the leaf falls.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   Two parallel metal plates, A and B, are 5.0 cm apart in a vacuum. Plate A is at 0 V and plate B is at +250 V. A probe measures the electric potential at distances x from plate A: at x = 1.0 cm, V = 50 V; at x = 2.0 cm, V = 100 V; at x = 3.0 cm, V = 150 V; at x = 4.0 cm, V = 200 V. An electron is released from rest at a small hole in plate A.

   (a)

   Use the data to determine the magnitude and direction of the electric field between the plates, and state what the data show about the field.

   [3 marks]

   (b)

   Determine the speed of the electron when it reaches plate B, and explain why the speed would be the same if the electron followed a curved path between the plates.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   Two small spheres, A and B, each carrying a charge of +4.0 nC, are fixed 0.20 m apart in air. M is the midpoint of the line AB. A small test charge of +1.0 nC can be placed at different points near A and B. Treat all charges as point charges.

   (a)

   Determine the electric field strength and the electric potential at M. Explain, with reference to the relationship between field strength and potential, why one of these is zero and the other is not, and calculate the work done in bringing the test charge from infinity to M.

   [6 marks]

   (b)

   Describe the pattern of the electric field lines and equipotential surfaces for this arrangement, and evaluate whether the test charge placed at M is in stable equilibrium.

   [6 marks]

   Total for question 4: 12 marks

End of questions