Grade 12 — Term 3
10 study notes · 139 practice questions · Aligned with NIE syllabus
Counted in the Idasara Academy app, which you open with a free account. The free notes, revision, practice and library pages for each lesson are linked below.
This term in the Idasara Academy app has 10 lessons (NIE Units 5, 6). For each lesson, this page gives you the points you must know and one exam-style question with its answer. None of it needs an account.
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Lessons in This Term
Newton’s Law of Gravitation
What you must know
- Gravitational force is strictly attractive, central (acts along the line joining centers of masses), conservative, and obeys inverse square law.
- Independent of intervening medium: Gravitational force between two masses is completely unaffected by the presence of any intervening physical medium or shielding.
- Earth's mass determination: Mass of Earth M_E = (g0 R_E²) / G ≈ 5.97 × 10²⁴ kg (the 'weighing of the Earth').
Try one question
Two point masses m and 4m are fixed at a distance d apart in a vacuum. At what distance from the mass m along the line joining them is the net gravitational force on a third mass M equal to zero?
- A. d / 3
- B. d / 2
- C. 2d / 3
- D. d / 4
Show the answer
Answer: A. d / 3. Let x be the distance from m. Gravitational forces from both masses must balance: G m M / x² = G (4m) M / (d - x)² => 1 / x² = 4 / (d - x)². Taking square roots: 1 / x = 2 / (d - x) => d - x = 2x => 3x = d => x = d / 3.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Gravitational Field & Field Intensity
What you must know
- Variation of g with Depth (d below surface): Inside uniform Earth, only mass within radius r = R - d contributes to gravity: g_d = g0 (1 - d/R); at Earth's center (d = R), g = 0.
- Variation of g with Latitude (Earth's rotation): Due to centrifugal acceleration: g_apparent = g - ω² R cos² λ (maximum g at poles λ = 90°, minimum at equator λ = 0°).
- Gravitational field lines: Directed radially inward toward attracting mass; density of lines indicates field strength; lines NEVER cross.
Try one question
If g is the gravitational acceleration on the surface of a spherical planet of radius R, what is the gravitational field intensity at an altitude h = R above the planet's surface?
- A. g / 2
- B. g / √2
- C. g / 4
- D. g / 8
Show the answer
Answer: C. g / 4. Gravitational field intensity at distance r from the center of mass is g(r) = GM / r². On the surface, r = R, so g = GM / R². At altitude h = R, the distance from the center is r = R + h = 2R. Thus, g' = GM / (2R)² = GM / (4R²) = g / 4.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Gravitational Potential & Potential Energy
What you must know
- Reference Point at Infinity: Gravitational potential is conventionally zero at infinity (V_∞ = 0); because gravity is attractive, potential everywhere in real space is strictly NEGATIVE.
- Potential difference near Earth's surface: For small height change h << R: ΔU = U(R+h) - U(R) = -GMm/(R+h) - (-GMm/R) ≈ (GMm/R²)*h = mgh.
- Energy Conservation in gravitational fields: (1/2)m v1² - (GMm / r1) = (1/2)m v2² - (GMm / r2).
Try one question
What is the gravitational potential V at a distance r from the center of a spherical planet of mass M, taking infinity as the zero-potential reference point?
- A. V = - GM / r²
- B. V = - GM / r
- C. V = - ½ GM / r
- D. V = + GM / r
Show the answer
Answer: B. V = - GM / r. Gravitational potential V at a point is defined as the work done by an external agent per unit mass in bringing a test mass from infinity to that point without acceleration. Since gravity is an attractive central force, V = - ∫_∞^r (F/m) dr = - GM / r.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Orbital Motion & Satellites
What you must know
- Apparent Weightlessness in Orbit: Astronauts in an orbiting spacecraft feel weightless not because gravity is zero (g at 400 km is ~90% of surface value), but because the astronaut and spacecraft are in continuous free-fall together toward Earth.
- Close-orbit velocity around Earth (h ≈ 0): v_orbit = √(g0 R_E) ≈ √(9.81 × 6.37 × 10⁶) ≈ 7.9 km/s.
- Orbital decay: Residual atmospheric friction at low altitudes dissipates satellite mechanical energy, causing orbital radius to shrink and velocity to paradoxically increase before burning up.
Try one question
A communications satellite moves in a stable circular orbit at radius r around Earth. If its orbital radius is increased to 4r, by what factor does its orbital period T change according to Kepler's Third Law?
- A. 8 times
- B. 4 times
- C. 2 times
- D. 16 times
Show the answer
Answer: A. 8 times. According to Kepler's Third Law for circular orbits, T² ∝ r³ => T ∝ r^(3/2). Increasing radius by a factor of 4 yields: T_new / T_old = (4)^(3/2) = (√4)³ = 2³ = 8.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Escape Velocity & Energy of Orbits
What you must know
- Escape velocity is independent of projectile mass and launch angle (as long as it clears terrain and atmosphere); depends solely on planetary mass M and radius R.
- Planetary Atmospheres: Why Moon has no atmosphere: Moon's escape velocity (v_e ≈ 2.4 km/s) is comparable to thermal rms speed c_rms of gas molecules, allowing gases to leak into space; Earth's v_e (11.2 km/s) comfortably retains O2, N2, and CO2.
- Black Hole Event Horizon: When escape velocity reaches speed of light c: Radius R_s = 2GM / c² (Schwarzschild Radius).
Try one question
What is the relationship between the escape velocity v_e from the surface of Earth (radius R, surface gravity g) and the orbital velocity v_o of a satellite in low Earth orbit skimming the surface?
- A. v_e = 2 · v_o
- B. v_e = 4 · v_o
- C. v_e = √2 · v_o
- D. v_e = v_o / √2
Show the answer
Answer: C. v_e = √2 · v_o. Orbital velocity in low circular orbit is v_o = √(GM / R) = √(gR). Escape velocity from the surface is obtained by equating total mechanical energy to zero: ½ m v_e² - GMm/R = 0 => v_e = √(2GM / R) = √(2gR). Therefore, v_e = √2 · v_o ≈ 1.414 v_o.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Electric Charge & Coulomb’s Law
What you must know
- Point Charge Limitation: Coulomb's inverse square law applies strictly to point charges or non-overlapping spherically symmetric charge distributions (by Shell Theorem).
- Electrostatic vs Gravitational Force: For two electrons: F_e / F_g ≈ 10⁴²; electrostatic force is enormously stronger than gravity at atomic scales, but gravity dominates cosmically due to net electrical neutrality.
- Vector form of Coulomb's Law: F_12 = (1 / (4πε0)) * (q1 q2 / r²₁) * r̂_21.
Try one question
Two point charges +q and +4q are held fixed at a distance r apart in vacuum, repelling each other with an electrostatic force F. If the distance between them is doubled to 2r, what is the new electrostatic force between them?
- A. F / 4
- B. 2F
- C. 4F
- D. F / 2
Show the answer
Answer: A. F / 4. According to Coulomb's Law, the electrostatic force between two point charges is inversely proportional to the square of their separation distance: F ∝ 1 / r². When the separation r becomes 2r, F' ∝ 1 / (2r)² = 1 / (4r²) = F / 4.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Electric Field & Electric Field Intensity
What you must know
- Motion of Charged Particle in Uniform Electric Field: Particle of mass m and charge q experiences constant acceleration a = q E / m; traverses parabolic trajectory analogous to projectile motion under gravity: x = v0 t, y = (1/2)(q E / m) t².
- Conductor in Electrostatic Equilibrium: 1) E = 0 everywhere INSIDE the conductor bulk; 2) Excess charge resides strictly on external surface; 3) E at surface is PERPENDICULAR to surface with magnitude E = σ / ε0; 4) Entire conductor is an equipotential volume.
- Action of Points (Corona Discharge): Charge density σ is highest at sharp points of small radius of curvature (σ ∝ 1/r); intense local electric field ionizes surrounding air molecules, producing electric wind; exploited in lightning conductors.
Try one question
Two large parallel conducting plates separated by a distance d = 2.0 cm in a vacuum are maintained at a potential difference V = 200 V. What is the magnitude of the electrostatic acceleration experienced by an electron (charge e = 1.6 × 10⁻¹⁹ C, mass m = 9.1 × 10⁻³¹ kg) placed between the plates?
- A. 8.80 × 10¹⁴ m s⁻²
- B. 1.76 × 10¹³ m s⁻²
- C. 3.52 × 10¹⁵ m s⁻²
- D. 1.76 × 10¹⁵ m s⁻²
Show the answer
Answer: D. 1.76 × 10¹⁵ m s⁻². Uniform electric field between plates: E = V / d = 200 V / (0.02 m) = 1.0 × 10⁴ V m⁻¹. Electrostatic force on electron: F = e E = (1.6 × 10⁻¹⁹ C) × (1.0 × 10⁴ V m⁻¹) = 1.6 × 10⁻¹⁵ N. Acceleration a = F / m = (1.6 × 10⁻¹⁵ N) / (9.1 × 10⁻³¹ kg) ≈ 1.76 × 10¹⁵ m s⁻².
A model question written by Idasara on the exam pattern. It is not from a past paper.
Electric Potential & Potential Energy
What you must know
- Potential of a Charged Conducting Sphere of radius R carrying charge Q: At external point (r ≥ R): V = Q / (4πε0 r); At surface (r = R): V_surface = Q / (4πε0 R); At ANY internal point (r < R): V_inside = V_surface = Q / (4πε0 R) (constant throughout cavity).
- Electron-Volt (eV): Microscopic energy unit defined as the kinetic energy gained by an electron accelerating through a potential difference of 1 Volt: 1 eV = 1.602 × 10⁻¹⁹ J.
- Zero potential reference: Conventional zero potential is taken at infinity (V_∞ = 0) or Earth (Ground potential V_earth = 0).
Try one question
What is the work done by an external agent in moving a point charge q = +2.0 μC from a point A at electric potential V_A = +50 V to a point B at electric potential V_B = +150 V at constant velocity?
- A. +4.0 × 10⁻⁴ J
- B. +1.0 × 10⁻⁴ J
- C. -2.0 × 10⁻⁴ J
- D. +2.0 × 10⁻⁴ J
Show the answer
Answer: D. +2.0 × 10⁻⁴ J. Work done by an external agent in moving a charge q between two points is W = q ΔV = q (V_B - V_A). Substituting: W = (2.0 × 10⁻⁶ C) × (150 V - 50 V) = 2.0 × 10⁻⁶ C × 100 V = +2.0 × 10⁻⁴ J.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Electric Flux & Gauss’s Law
What you must know
- Choice of Gaussian Surface: Must possess spatial symmetry matching the charge distribution (spherical, cylindrical, or planar pillbox) such that E is either constant and normal to the surface, or parallel to the surface (zero flux).
- Charge Outside Gaussian Surface: Charges located outside the closed surface contribute zero net flux through the surface (field lines entering must exit).
- Field between two oppositely charged parallel conducting plates: Charges migrate to facing internal surfaces; inside the gap E = σ / ε0, outside the plates E = 0.
Try one question
According to Gauss's Law of electrostatics, what is the net electric flux Φ_E emerging through any closed Gaussian surface enclosing a net electric charge Q_enc in a medium of permittivity ε₀?
- A. Φ_E = Q_enc · ε₀
- B. Φ_E = Q_enc / (4π ε₀)
- C. Φ_E = Q_enc / (2 ε₀)
- D. Φ_E = Q_enc / ε₀
Show the answer
Answer: D. Φ_E = Q_enc / ε₀. Gauss's law states that the total outward electric flux through any closed surface in vacuum equals the net enclosed charge divided by the permittivity of free space: ∮ E · dA = Q_enc / ε₀.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Capacitors & Capacitance
What you must know
- Dielectric insertion under Battery Connected vs Disconnected: Battery Connected (V constant): C increases (×ε_r), Q increases (×ε_r), E remains constant (V/d), U increases (×ε_r); Battery Disconnected (Q constant): C increases (×ε_r), V drops (/ε_r), E drops (/ε_r), U decreases (/ε_r, mechanical work done by field pulling dielectric in).
- Energy loss during capacitor sharing: When charged capacitor C1 at V1 is connected to uncharged C2, common voltage V = (C1 V1)/(C1 + C2); Energy lost as heat/radiation: ΔU = (1/2) [ (C1 C2) / (C1 + C2) ] V1².
- Dielectric Breakdown: Maximum electric field a dielectric material can withstand without electrical discharge / sparking is its dielectric strength (Air ≈ 3 × 10⁶ V/m).
Try one question
A parallel-plate capacitor has capacitance C = 10 μF with air between its plates and is connected across a 12 V battery. If a dielectric slab of relative permittivity ε_r = 5 is inserted to completely fill the space between the plates while the battery remains connected, what is the new electrostatic energy stored in the capacitor?
- A. 0.72 mJ
- B. 1.8 mJ
- C. 7.2 mJ
- D. 3.6 mJ
Show the answer
Answer: D. 3.6 mJ. When a dielectric slab of dielectric constant ε_r is inserted, capacitance increases: C' = ε_r C = 5 × 10 μF = 50 μF. Because the battery remains connected, the voltage V remains constant at 12 V. New stored energy U' = ½ C' V² = ½ (50 × 10⁻⁶ F) × (12 V)² = 25 × 10⁻⁶ × 144 = 3600 × 10⁻⁶ J = 3.6 × 10⁻³ J = 3.6 mJ.
A model question written by Idasara on the exam pattern. It is not from a past paper.
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