Secondary 6 Physics Practice Paper: Electrostatic Fields, Capacitors, Electromagnetic Induction, Alternating Current and Transformers, Photoelectric Effect and Nuclear Decay

2026-11-13

Physics – Secondary 6 – Practice Paper

Secondary 6 physics practice paper with full answers: Coulomb law, electric field, capacitors, dielectrics, induction, transformers, photoelectric effect and half-life.

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Paper details　Secondary 6 physics practice paper with answers: Coulomb force between two point charges, electric field strength of a point charge, electron in a field, capacitor charge and energy with a dielectric slab, Faraday law induced emf and Lenz law, transformer turns ratio, rms and peak voltage, photoelectric work function and stopping voltage, half-life and decay constant.

Section A: Coulomb Law and Electric Field (18 marks)

Data 1: two point charges in air, Q1 = +6.0 uC and Q2 = -4.0 uC, are 0.30 m apart. Take the Coulomb constant k as 9.0 x 10^9 N m^2 C^-1 and the electron charge as 1.6 x 10^-19 C.

#	Question	Answer and marking notes
1	(a) Find the magnitude of the electrostatic force between the two charges.	Answer: F = kQ1Q2 / r^2 = (9.0 x 10^9 x 6.0 x 10^-6 x 4.0 x 10^-6) / 0.30^2 = 2.16 x 10^-1 / 0.090 = 2.4 N. (Marking: formula 2 marks, r squared 2 marks, answer with unit 1 mark)
2	(b) State the direction of that electrostatic force.	Answer: the charges have opposite signs, so the force is attractive and acts along the line joining them, towards each other (Q1 is pulled towards Q2 and Q2 towards Q1). (Marking: attractive 2 marks, direction 1 mark)
3	(c) Find the electric field strength produced by Q1 at a point 0.15 m from it.	Answer: E = kQ1 / r^2 = (9.0 x 10^9 x 6.0 x 10^-6) / 0.15^2 = 5.4 x 10^4 / 0.0225 = 2.4 x 10^6 N C^-1, directed radially outwards. (Marking: formula 2 marks, value 2 marks, direction 1 mark)
4	(d) An electron is placed at the point in (c). Find the magnitude and direction of the electric force on it.	Answer: F = eE = 1.6 x 10^-19 x 2.4 x 10^6 = 3.84 x 10^-13 N; the electron is negative, so the force is opposite to the field and points towards Q1. (Marking: F = eE 2 marks, value 1 mark, opposite direction 2 marks)

Section B: Capacitors and Dielectrics (18 marks)

Data 2: a 20 uF parallel plate capacitor is connected to a 60 V d.c. supply. After it is fully charged the supply is disconnected, and a dielectric of relative permittivity 3.0 is then inserted between the plates.

#	Question	Answer and marking notes
5	(a) Find the charge stored on the capacitor.	Answer: Q = CV = 20 x 10^-6 x 60 = 1.2 x 10^-3 C (1.2 mC). (Marking: formula 2 marks, value with unit 2 marks)
6	(b) Find the energy stored before the supply is disconnected.	Answer: E = 0.5 x C x V^2 = 0.5 x 20 x 10^-6 x 60^2 = 0.036 J. (Marking: formula 2 marks, value with unit 2 marks)
7	(c) After inserting the dielectric, find the potential difference across the plates and the new capacitance.	Answer: with the supply disconnected the charge stays the same, so C = 3.0 x 20 = 60 uF and V = Q / C = 1.2 x 10^-3 / 60 x 10^-6 = 20 V. (Marking: charge constant 2 marks, new C 1 mark, new V 2 marks)
8	(d) Find the energy stored after the dielectric is inserted and explain the change.	Answer: E = 0.5 x Q^2 / C = 0.5 x (1.2 x 10^-3)^2 / (60 x 10^-6) = 0.012 J; the energy falls from 0.036 J to 0.012 J, a decrease of 0.024 J, released as the mechanical work done in pulling the dielectric into the gap. (Marking: new energy 2 marks, difference 1 mark, explanation 2 marks)

Section C: Electromagnetic Induction (17 marks)

Data 3: a coil of 200 turns and cross sectional area 0.015 m^2 lies with its plane perpendicular to a uniform magnetic field. The flux density falls uniformly from 0.60 T to 0.20 T in 0.040 s. The resistance of the coil is 6.0 ohm.

#	Question	Answer and marking notes
9	(a) Find the magnitude of the induced emf in the coil.	Answer: change in flux = A x change in B = 0.015 x (0.60 – 0.20) = 6.0 x 10^-3 Wb; emf = N x flux change / time = 200 x 6.0 x 10^-3 / 0.040 = 30 V. (Marking: flux change 2 marks, formula 2 marks, value with unit 1 mark)
10	(b) Find the induced current in the coil.	Answer: I = emf / R = 30 / 6.0 = 5.0 A. (Marking: formula 2 marks, value with unit 2 marks)
11	(c) Use Lenz law to explain how the direction of the induced current is determined.	Answer: the induced current creates a field that opposes the decrease in flux, so its field points in the same direction as the original field to make up for the lost flux; Lenz law fixes the current direction so that it always opposes the change in flux. (Marking: opposes the change 2 marks, same field direction 1 mark)
12	(d) Find the power dissipated in the coil during this time and state where the energy comes from.	Answer: P = I^2 R = 5.0^2 x 6.0 = 150 W, or P = emf x I = 30 x 5.0; the energy comes from the mechanical work done in changing the magnetic field (moving the magnet or the coil). (Marking: power 3 marks, energy source 2 marks)

Section D: Alternating Current and Transformers (16 marks)

Data 4: an ideal transformer has 1100 turns in its primary, connected to a 220 V rms, 50 Hz mains supply. Its secondary has 55 turns and is connected to a 5.5 ohm resistor.

#	Question	Answer and marking notes
13	(a) Find the rms voltage across the secondary coil.	Answer: Vs / Vp = Ns / Np, so Vs = 220 x 55 / 1100 = 11 V. (Marking: turns ratio 2 marks, value with unit 2 marks)
14	(b) Find the peak value of the secondary voltage.	Answer: peak = rms x square root of 2 = 11 x 1.414 = 15.6 V. (Marking: factor 2 marks, value with unit 2 marks)
15	(c) Find the current in the secondary circuit and the power dissipated in the resistor.	Answer: I = Vs / R = 11 / 5.5 = 2.0 A; P = Vs x I = 11 x 2.0 = 22 W. (Marking: current 2 marks, power 2 marks)
16	(d) Assuming the transformer is 100 percent efficient, find the current in the primary coil.	Answer: input power equals output power, 22 W; Ip = P / Vp = 22 / 220 = 0.10 A. (Marking: power conservation 2 marks, value with unit 2 marks)

Section E: Photoelectric Effect (16 marks)

Data 5: the work function of a metal is 2.4 eV. Light of wavelength 400 nm falls on its surface. Take the Planck constant as 6.63 x 10^-34 J s, the speed of light as 3.0 x 10^8 m s^-1 and 1 eV as 1.6 x 10^-19 J.

#	Question	Answer and marking notes
17	(a) Find the energy of an incident photon in eV.	Answer: E = hc / wavelength = (6.63 x 10^-34 x 3.0 x 10^8) / 400 x 10^-9 = 4.97 x 10^-19 J; in eV this is 4.97 x 10^-19 / 1.6 x 10^-19 = 3.11 eV. (Marking: hc/lambda 2 marks, joule value 1 mark, eV value 2 marks)
18	(b) Find the maximum kinetic energy of the emitted photoelectrons, in eV and in J.	Answer: Kmax = E – W = 3.11 – 2.4 = 0.71 eV, which is 0.71 x 1.6 x 10^-19 = 1.13 x 10^-19 J. (Marking: Einstein equation 2 marks, eV value 1 mark, joule value 2 marks)
19	(c) Find the stopping voltage.	Answer: stopping voltage = Kmax / e = 0.71 eV / e = 0.71 V. (Marking: relation 2 marks, value with unit 1 mark)
20	(d) If light of wavelength 550 nm is used instead, will photoelectrons be emitted? Justify with a calculation.	Answer: photon energy = hc / wavelength = 1.989 x 10^-25 / 550 x 10^-9 = 3.62 x 10^-19 J = 2.26 eV; this is less than the 2.4 eV work function, so no photoelectrons are emitted because the wavelength is longer than the threshold wavelength. (Marking: calculation 2 marks, comparison and conclusion 1 mark)

Section F: Radioactive Decay and Half-life (15 marks)

Data 6: a radioisotope has a half-life of 8.0 days. A sample initially contains 6.4 x 10^20 atoms of the isotope.

#	Question	Answer and marking notes
21	(a) Find the percentage of undecayed atoms after 16 days.	Answer: 16 days is 2 half-lives, so the remaining fraction is (1/2)^2 = 1/4, that is 25 percent. (Marking: number of half-lives 2 marks, percentage 2 marks)
22	(b) Find the decay constant in day^-1.	Answer: decay constant = ln2 / half-life = 0.693 / 8.0 = 0.0866 per day, about 8.7 x 10^-2 per day. (Marking: formula 2 marks, value with unit 2 marks)
23	(c) Find the number of undecayed atoms remaining after 24 days.	Answer: 24 days is 3 half-lives, so N = 6.4 x 10^20 / 2^3 = 6.4 x 10^20 / 8 = 8.0 x 10^19 atoms. (Marking: number of half-lives 2 marks, value 2 marks)
24	(d) Find the number of atoms that have decayed within 24 days.	Answer: number decayed = 6.4 x 10^20 – 8.0 x 10^19 = 5.6 x 10^20 atoms. (Marking: subtraction 2 marks, value 1 mark)

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