Grade 12 Science: Physics

In this unit, 'Electric charges, fields and potential', you learn to calculate fields and potential and use Gauss's law. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Observe: see, touch, measure or watch a safe demonstration linked to 'Electric charges, fields and potential' and write what you notice.
• Explain: use the key words of the unit to say what is happening and why, in a few lines.
• Use it: calculate fields and potential and use Gauss's law. Use a diagram, table or short calculation where it fits.
• Test and conclude: say what evidence supports your answer and what you would change.

Step by step.

1. Step 1, Apply Coulomb's law. Two charges of 2 microcoulomb each are 0.3 m apart in air. F = k q1 q2 / r^2 = 9e9 x 4e-12 / 0.09 = 0.4 N, repulsive.
2. Step 2, Find the electric field. A charge of 4 microcoulomb at 0.6 m produces E = kq/r^2 = 9e9 x 4e-6 / 0.36 = 1.0 x 10^5 N/C. The force on a 1 microcoulomb test charge there is 0.1 N.
3. Step 3, Calculate potential. V = kq/r = 9e9 x 4e-6 / 0.6 = 6.0 x 10^4 V. Work to bring 1 microcoulomb from infinity is qV = 0.06 J.
4. Step 4, Use Gauss's law for an infinite sheet with surface charge density sigma = 2 x 10^-6 C/m2: E = sigma / (2 epsilon0) = 2e-6 / (2 x 8.85e-12) = 1.13 x 10^5 N/C. For a sphere, flux = q / epsilon0 gives the same field as a point charge outside.
5. Step 5, Capacitors. A parallel plate capacitor with plate area 0.01 m2 and separation 1 mm has C = epsilon0 A / d = 8.85e-12 x 0.01 / 0.001 = 8.85e-11 F. At 100 V, energy = 0.5 C V^2 = 4.4e-7 J. A dipole has p = q x 2a.
6. Step 6, Your turn: find the field 0.3 m from a 3 microcoulomb point charge. Answer: 9e9 x 3e-6 / 0.09 = 3.0 x 10^5 N/C.

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Calculate fields and potential and use Gauss's law: at least 70 percent on an exam-style task, with the method shown.

In this unit, 'Current electricity and magnetism', you learn to apply Kirchhoff's laws and the Biot-Savart law. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Observe: see, touch, measure or watch a safe demonstration linked to 'Current electricity and magnetism' and write what you notice.
• Explain: use the key words of the unit to say what is happening and why, in a few lines.
• Use it: apply Kirchhoff's laws and the Biot-Savart law. Use a diagram, table or short calculation where it fits.
• Test and conclude: say what evidence supports your answer and what you would change.

Step by step.

1. Step 1, Apply Kirchhoff's laws to a circuit. A 12 V battery with a 2 ohm series resistor feeds two parallel branches of 6 ohm and 3 ohm. Parallel resistance is 2 ohm, total 4 ohm. I = 12 / 4 = 3 A. Voltage across the parallel part is 3 x 2 = 6 V.
2. Step 2, Branch currents. 6 / 6 = 1 A and 6 / 3 = 2 A, which sum to 3 A, consistent with the junction rule. The loop rule: 12 - 3 x 2 - 6 = 0.
3. Step 3, Use a Wheatstone bridge. Balance condition P/Q = R/S. If P = 10, Q = 20 and R = 15, then S = 30 ohm. A potentiometer compares emfs by balance lengths: if E1 balances at 60 cm and E2 at 40 cm, E1/E2 = 1.5.
4. Step 4, Biot-Savart. A coil of 100 turns, radius 0.1 m, carrying 2 A has centre field B = mu0 N I / (2R) = 4e-7 pi x 100 x 2 / 0.2 = 1.26 x 10^-3 T. A long wire carrying 10 A gives B = mu0 I / (2 pi r) = 4e-5 T at 0.05 m.
5. Step 5, Force on a conductor and moving charge: F = B I L sin theta. 0.5 m of wire carrying 4 A in 0.2 T at 90 degrees feels 0.4 N. A proton at 10^6 m/s in 0.1 T moves in a circle of radius r = mv/(qB) = 1.67e-27 x 1e6 / (1.6e-19 x 0.1) = 0.104 m.
6. Step 6, Your turn: find B at 0.1 m from a straight wire carrying 5 A. Answer: 2e-7 x 5 / 0.1 = 1 x 10^-5 T.

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Apply Kirchhoff's laws and the Biot-Savart law: at least 70 percent on an exam-style task, with the method shown.

In this unit, 'Electromagnetic induction and alternating current', you learn to apply Faraday's law and AC circuits. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Observe: see, touch, measure or watch a safe demonstration linked to 'Electromagnetic induction and alternating current' and write what you notice.
• Explain: use the key words of the unit to say what is happening and why, in a few lines.
• Use it: apply Faraday's law and AC circuits. Use a diagram, table or short calculation where it fits.
• Test and conclude: say what evidence supports your answer and what you would change.

Step by step.

1. Step 1, State Faraday's law: the induced emf equals the rate of change of flux linkage, emf = N x change in flux / time. A coil of 100 turns sees flux fall from 0.05 Wb to 0.01 Wb in 0.2 s. The size is 100 x 0.04/0.2 = 20 V, and Lenz's law says it opposes the change.
2. Step 2, Find a motional emf: emf = B l v. A 0.5 m rod moves at 4 m/s at right angles to a 0.2 T field, so emf = 0.2 x 0.5 x 4 = 0.4 V. With a 2 ohm loop the current is 0.4/2 = 0.2 A. Safety: in a demonstration with a magnet and coil, use a small magnet and a sensitive low-voltage meter, never mains.
3. Step 3, Describe AC with V = V0 sin(omega t). The rms value is V0/root 2. Suppose mains of 230 V rms (the value differs by country) has a peak of 230 x 1.414 = 325 V. At f = 50 Hz, omega = 2 pi f = 314 rad/s. Meters read rms values, not peaks.
4. Step 4, Find reactances. For an inductor X_L = omega L, so L = 0.1 H at 50 Hz gives 31.4 ohm. For a capacitor X_C = 1/(omega C), so C = 100 microfarad gives 1/(314 x 0.0001) = 31.8 ohm. Higher frequency raises X_L and lowers X_C.
5. Step 5, Solve a series LCR circuit. With R = 30 ohm, X_L = 60 ohm and X_C = 20 ohm, Z = root(30^2 + 40^2) = 50 ohm. On 100 V rms the current is 2 A. Resonance occurs when X_L = X_C, at f = 1/(2 pi root(LC)), which is about 50.3 Hz for 0.1 H and 100 microfarad.
6. Step 6, Your turn: an ideal transformer has 200 primary turns and 20 secondary turns on 230 V. Find the secondary voltage and say whether it is step-up or step-down. Answer: V_s = 230 x 20/200 = 23 V, a step-down transformer.

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Apply Faraday's law and AC circuits: at least 70 percent on an exam-style task, with the method shown.

In this unit, 'Electromagnetic waves and optics', you learn to use ray and wave optics. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Observe: see, touch, measure or watch a safe demonstration linked to 'Electromagnetic waves and optics' and write what you notice.
• Explain: use the key words of the unit to say what is happening and why, in a few lines.
• Use it: use ray and wave optics. Use a diagram, table or short calculation where it fits.
• Test and conclude: say what evidence supports your answer and what you would change.

Step by step.

1. Step 1, Use electromagnetic waves. c = f lambda. A 100 MHz FM wave has lambda = 3e8 / 1e8 = 3 m. Visible light of 500 nm has f = 6 x 10^14 Hz. Order: radio, microwave, infrared, visible, UV, X-ray, gamma.
2. Step 2, Apply the mirror formula 1/v + 1/u = 1/f. A concave mirror has f = -10 cm and an object at u = -30 cm. 1/v = -1/10 + 1/30 = -2/30, so v = -15 cm, a real inverted image in front of the mirror, with magnification m = -v/u = -0.5.
3. Step 3, Use a lens. A convex lens of f = 20 cm with an object at 60 cm gives 1/v = 1/20 - 1/60 = 1/30, so v = 30 cm and m = v/u = -0.5, a real inverted image. Lens maker's formula 1/f = (n - 1)(1/R1 - 1/R2).
4. Step 4, Use Snell's law. Light going from air to glass n = 1.5 at 30 degrees: sin r = sin 30 / 1.5 = 0.333, so r = 19.5 degrees. Critical angle in glass is sin^-1(1/1.5) = 41.8 degrees, which allows total internal reflection.
5. Step 5, Young's double slit. d = 0.5 mm, D = 1 m, lambda = 600 nm. Fringe width = lambda D / d = 600e-9 x 1 / 0.5e-3 = 1.2 mm. Single slit first minimum at sin theta = lambda / a.
6. Step 6, Your turn: find the image distance for an object 40 cm from a convex lens of f = 10 cm. Answer: 1/v = 1/10 - 1/40 = 3/40, so v = 13.3 cm.

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Use ray and wave optics: at least 70 percent on an exam-style task, with the method shown.

In this unit, 'Dual nature of matter and atoms', you learn to explain the photoelectric effect and atomic spectra. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Observe: see, touch, measure or watch a safe demonstration linked to 'Dual nature of matter and atoms' and write what you notice.
• Explain: use the key words of the unit to say what is happening and why, in a few lines.
• Use it: explain the photoelectric effect and atomic spectra. Use a diagram, table or short calculation where it fits.
• Test and conclude: say what evidence supports your answer and what you would change.

Step by step.

1. Step 1, Apply the photoelectric equation. Light of 400 nm has photon energy E = hc / lambda = 1240 eV nm / 400 nm = 3.10 eV. The metal work function is 2.3 eV. Maximum KE = 3.10 - 2.3 = 0.80 eV, so stopping potential is 0.80 V.
2. Step 2, Find the threshold wavelength for the same metal. lambda0 = 1240 / 2.3 = 539 nm. Light above 539 nm, like red at 700 nm, causes no emission, however intense, since each photon has too little energy.
3. Step 3, De Broglie wavelength. An electron accelerated through 100 V has lambda = 1.227 / sqrt(V) nm = 0.1227 nm. A 0.1 kg ball at 10 m/s has lambda = 6.63e-34 / 1 = 6.6 x 10^-34 m, not observable.
4. Step 4, Bohr model of hydrogen. E_n = -13.6 / n^2 eV. For n = 3 to n = 2: E = 13.6 (1/4 - 1/9) = 1.89 eV, giving lambda = 1240 / 1.89 = 656 nm, the red H-alpha line in the Balmer series.
5. Step 5, Compute orbit radius. r_n = 0.529 n^2 angstrom, so n = 2 gives 2.12 angstrom. Lyman series ends in n = 1, Balmer in n = 2. Shortest Lyman wavelength is 1240/13.6 = 91 nm.
6. Step 6, Your turn: find the photon energy of 200 nm light and say whether it ejects electrons from a metal with work function 4.5 eV. Answer: 1240/200 = 6.2 eV, which exceeds 4.5 eV, so yes.

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Explain the photoelectric effect and atomic spectra: at least 70 percent on an exam-style task, with the method shown.

In this unit, 'Nuclei and semiconductor electronics', you learn to describe nuclear reactions and diodes and transistors. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Observe: see, touch, measure or watch a safe demonstration linked to 'Nuclei and semiconductor electronics' and write what you notice.
• Explain: use the key words of the unit to say what is happening and why, in a few lines.
• Use it: describe nuclear reactions and diodes and transistors. Use a diagram, table or short calculation where it fits.
• Test and conclude: say what evidence supports your answer and what you would change.

Step by step.

1. Step 1, Describe the nucleus. Carbon-14 has Z = 6 and A = 14, so 8 neutrons. Nuclear radius R = R0 A^(1/3) with R0 = 1.2 fm gives R = 1.2 x 14^(1/3) = 2.89 fm. Density is about 2.3 x 10^17 kg/m3.
2. Step 2, Calculate binding energy using atomic masses. Helium-4 has mass 4.0026 u against 2 hydrogen atoms (1.00783 u) and 2 neutrons (1.00866 u) summing to 4.03298 u. Mass defect 0.0304 u x 931.5 MeV/u = 28.3 MeV, so about 7.1 MeV per nucleon.
3. Step 3, Radioactive decay. N = N0 (1/2)^(t/T). A sample with half-life 5 days is 800 mg. After 15 days, N = 800 / 8 = 100 mg. Decay constant lambda = 0.693 / T = 0.139 per day.
4. Step 4, Write nuclear reactions. Alpha decay: U-238 gives Th-234 + He-4. Beta decay: C-14 gives N-14 + electron + antineutrino. Fission: U-235 + n gives Ba-141 + Kr-92 + 3n + about 200 MeV. Fusion: 4 H gives He.
5. Step 5, Semiconductors. A p-n junction diode conducts in forward bias; in reverse bias there is a small leakage. A half-wave rectifier passes one half cycle. Silicon has a band gap of 1.1 eV. N-type doping with phosphorus; p-type with boron.
6. Step 6, Your turn: a 400 mg sample has half-life 2 h. How much remains after 6 h? Answer: 400 / 8 = 50 mg.

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Describe nuclear reactions and diodes and transistors: at least 70 percent on an exam-style task, with the method shown.

On the slide: L1 Basic · Pushes and pulls

Notes: Name forces as pushes or pulls and say how they change movement. Notes: roll a toy car down a ramp onto the floor and mark where it stops; learners push it again with a bigger start and compare. Keep the floor clear so nobody trips.

On the slide: L2 Awareness · Energy and simple circuits

Notes: Describe energy stores and build a simple circuit that lights a bulb. Notes: use one 1.5 V cell, one small bulb and two leads; learners draw the loop and say what happens when it is broken. Use only cells or batteries at low voltage, never mains sockets.

On the slide: L3 Foundation · Speed, distance and time

Notes: Calculate speed from distance and time and read a distance-time graph. Notes: practice case is 100 m in 12.5 s, so speed is 8 m/s. Learners time a walked 20 m with a phone stopwatch, repeat three times, find the mean, and then work out the speed.

On the slide: L4 Practitioner · Force, mass and acceleration

Notes: Apply F = ma to find an unknown force, mass or acceleration. Notes: a 1200 kg car with a net force of 3000 N has acceleration 2.5 m/s2. Learners then find the net force when the engine gives 3500 N and drag is 500 N, and check that the answer matches.

On the slide: L5 Specialist · Series and parallel circuits

Notes: Analyse resistor networks using Ohm law and the rules for current and voltage. Notes: 24 ohm and 12 ohm in parallel give 8 ohm, since 1/24 plus 1/12 is 1/8. On 6 V, current is 0.75 A. Check power ratings and keep currents low before building; use a meter to verify.

This course: Grade 12 Science Physics is at this level. Its units include Electric charges, fields and potential; Current electricity and magnetism; Electromagnetic induction and alternating current; Electromagnetic waves and optics and more.

On the slide: L6 Advanced · Oscillations and waves

Notes: Apply the simple harmonic motion model and solve for period, frequency and energy. Notes: a 0.50 kg mass on a spring of constant 200 N/m has period 2 pi times the square root of 0.0025, about 0.314 s. Learners measure ten swings with a stopwatch and compare, keeping the mass low so it cannot hurt a hand.

On the slide: L7 Super expert · Uncertainty and fitting

Notes: Evaluate measurements and models with uncertainty propagation, fitting and goodness of fit. Notes: a pendulum with L of 1.000 m and T of 2.007 s gives g equal to 4 pi squared L over T squared, 9.80 m/s2. Learners propagate a 0.002 s timing error, plan repeats, and decide if a model is supported. Check clamps and stand stability.