Unit 4 High-Yield Topics | Flowxiom

Unit 4 High-Yield Topics

Q: Q1. \(C = 500\ \mu\text{F}\), \(R = 10\ \text{k}\Omega\), \(V_0 = 12\ \text{V}\). Find the voltage after one time constant.

\(\tau = RC = 10000 \times 500\times10^{-6} = 5\ \text{s}\) \(V = 12 \times e^{-1} = 12 \times 0.37 = \mathbf{4.4\ \text{V}}\)

Q: Q2. A satellite of mass \(m\) has orbital radius \(r\) and period \(T\). State the source of centripetal force and write the expression.

Source: gravitational force \(F = \dfrac{GMm}{r^2} = \dfrac{4\pi^2 mr}{T^2}\)

Q: Q3. \(C_1 = 4\ \mu\text{F}\) and \(C_2 = 12\ \mu\text{F}\) in series, connected to \(9\ \text{V}\). Find: (a) equivalent capacitance; (b) total energy stored.

(a) \(\dfrac{1}{C} = \dfrac{1}{4} + \dfrac{1}{12} = \dfrac{4}{12}\), \(C = \mathbf{3\ \mu\text{F}}\) (b) \(E = \dfrac{1}{2}CV^2 = \dfrac{1}{2} \times 3\times10^{-6} \times 81 = \mathbf{1.22\times10^{-4}\ \text{J}}\)

Q: Q4. An electron is accelerated through \(3.0\ \text{kV}\), then enters a magnetic field \(B = 0.050\ \text{T}\) perpendicular to its velocity. Find: (a) electron speed; (b) radius of circular path.

(a) \(v = \sqrt{\dfrac{2eV}{m_e}} = \sqrt{\dfrac{2\times1.6\times10^{-19}\times3000}{9.11\times10^{-31}}} = \mathbf{3.25\times10^7\ \text{m/s}}\) (b) \(r = \dfrac{m_ev}{eB} = \dfrac{9.11\times10^{-31}\times3.25\times10^7}{1.6\times10^{-19}\times0.050} = \mathbf{3.70\times10^{-3}\ \text{m}}\)

Free resource by Flowxiom — Edexcel A-level Physics

Not everything. Just what’s on the paper. High-frequency topics only — covering ~80% of exam marks.

Edexcel A-level Physics | Fields, Capacitors & Particles | WPH14 & WPH15

Topic 1: Capacitor Discharge (Exponential Decay)

Question types: Exponential equation calculation, ln graph linearisation, reading time constant.

Key Formulae

\[Q = Q_0 e^{-t/RC} \qquad V = V_0 e^{-t/RC} \qquad I = I_0 e^{-t/RC}\]

Time constant: \(\tau = RC\)

After one time constant \(\tau\), charge/voltage/current falls to 37% of its initial value (63% lost).

Logarithmic Linearisation (essential for practical questions)

Taking \(\ln\) of \(V = V_0 e^{-t/RC}\):

\[\ln V = \ln V_0 – \frac{1}{RC} \cdot t\]

Plot \(\ln V\) against \(t\):

Gradient \(= -1/RC\) (use to find \(C\) or \(R\))
y-intercept \(= \ln V_0\)

Common Mistakes

❌ Time constant: value falls TO 37%, not BY 37%
❌ Gradient is negative; \(RC = -1/\text{gradient}\)
❌ Charging and discharging curves go in opposite directions

Topic 2: Electromagnetic Induction

Question types: Determine direction of induced current; calculate induced e.m.f.; Lenz’s law explanation (frequent 6-mark).

Faraday’s Law

\[\varepsilon = -N\frac{\Delta\Phi}{\Delta t}\]

Magnitude of induced e.m.f. ∝ rate of change of flux linkage.

Magnetic Flux

\[\Phi = BA\cos\theta\]

\(\theta\) = angle between B field and the normal to the coil plane
Coil parallel to B: \(\theta = 90°\), \(\Phi = 0\)
Coil perpendicular to B: \(\theta = 0°\), \(\Phi = BA\) (maximum)

⚠️ If the question gives the angle between the coil plane and B, use \(\sin\alpha\) not \(\cos\alpha\).

Lenz’s Law Answer Chain (6-mark question)

The induced current flows in a direction such that its magnetic field opposes the change in flux that caused it.

Steps:

State whether flux is increasing or decreasing

By Lenz’s law, the induced field must oppose the change

Use right-hand rule to find current direction

State that this creates a force opposing the motion

Common Mistakes

❌ Confusing angle between B and normal (use cos) with angle between B and plane (use sin)
❌ Must state WHY — the induced current opposes the change in flux
❌ Maximum \(\varepsilon\) occurs when \(\Phi\) is zero (90° phase difference)

Topic 3: Circular Motion

Question types: Identify source of centripetal force; calculate centripetal force, speed, period.

Centripetal Force is Not a New Force

Centripetal force = the resultant force (or component of resultant) directed towards the centre.

Scenario	Source of centripetal force
Satellite orbiting Earth	Gravitational force
Car cornering	Static friction
Conical pendulum	Horizontal component of tension
Top of vertical circle	Weight + normal force (or tension)

Formulae

\[F = \frac{mv^2}{r} = mr\omega^2 = \frac{4\pi^2 mr}{T^2}\]

⚠️ Do NOT draw centripetal force on a free-body diagram — only draw real forces.

Common Mistakes

❌ Centripetal force is not a separate force — do not include it on a free-body diagram
❌ At the top of a vertical circle: centripetal force = weight − normal force

Topic 4: Electric Fields

Question types: Point charge fields, work done in uniform field, calculations involving \(E\), \(V\), \(W\).

Key Formulae

\[E = \frac{F}{q} \qquad E = \frac{V}{d} \text{ (uniform)} \qquad E = \frac{Q}{4\pi\varepsilon_0 r^2} \text{ (point charge)}\]

\[W = qV \qquad V = \frac{Q}{4\pi\varepsilon_0 r}\]

Electric Field vs Gravitational Field

	Electric field	Gravitational field
Source	Charge \(Q\)	Mass \(M\)
Field strength	\(E = Q/4\pi\varepsilon_0 r^2\)	\(g = GM/r^2\)
Potential	\(V = Q/4\pi\varepsilon_0 r\)	\(\varphi = -GM/r\)
Sign	Positive or negative	Always negative

Common Mistakes

❌ \(d\) = separation between the plates, not distance from one plate
❌ Gravitational potential is always negative; electric potential can be positive or negative

Topic 5: Magnetic Force and Charged Particles

Question types: Force direction (Fleming’s left-hand rule), velocity selector, charged particle trajectory.

Force on a Conductor

\[F = BIL\sin\theta\]

Magnetic Force on a Moving Charge

\[F = Bqv\sin\theta\]

Direction: use Fleming’s left-hand rule

(first finger = B field; second finger = conventional current direction; thumb = force)

Velocity Selector (straight-line condition)

\[qE = qvB \implies v = \frac{E}{B}\]

Only particles with speed \(v = E/B\) pass through undeflected — independent of mass and charge.

Common Mistakes

❌ Neutrons are uncharged — no magnetic force, trajectory is a straight line
❌ Negative charges (e.g. electrons) experience force in the opposite direction to positive charges

Topic 6: Capacitors

Question types: Calculate stored energy, series/parallel equivalent capacitance, effect of changing parameters.

Key Formulae

\[C = \frac{Q}{V} \qquad E_{stored} = \frac{1}{2}QV = \frac{1}{2}CV^2 = \frac{Q^2}{2C}\]

Series vs Parallel (opposite to resistors)

	Capacitors in series	Capacitors in parallel
Formula	\(\dfrac{1}{C_{total}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}\)	\(C_{total} = C_1 + C_2\)
Result	Total less than smallest	Total greater than largest

Capacitor combinations are the reverse of resistor combinations.

Changing Parameters

After disconnecting supply: \(Q\) is fixed. Change \(d\) → \(C\) changes → \(V = Q/C\) changes
While connected to supply: \(V\) is fixed. Change \(d\) → \(C\) changes → \(Q = CV\) changes

Common Mistakes

❌ Capacitor series/parallel rules are the reverse of resistors
❌ Choose the correct energy formula based on which quantities are known
❌ After disconnecting: \(Q\) is constant, not \(V\)

Topic 7: Electron Acceleration and Deflection in a Magnetic Field

Question types: Calculate electron speed after acceleration; radius of circular path in B field; mass spectrometer.

Electron Accelerated Through p.d. V

\[eV = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2eV}{m_e}}\]

\(e = 1.6\times10^{-19}\ \text{C}\), \(m_e = 9.11\times10^{-31}\ \text{kg}\)

Circular Motion in Magnetic Field

Lorentz force provides centripetal force:

\[Bqv = \frac{mv^2}{r} \implies r = \frac{mv}{Bq}\]

Larger mass or speed → larger radius; stronger \(B\) or larger charge → smaller radius.

Common Mistakes

❌ Proton mass ≈ 1836 × electron mass — do not use the same value
❌ Negative particles deflect in the opposite direction to positive charges
❌ Magnetic force does no work — it only changes direction, not speed

Topic 8: Particle Physics Fundamentals

Question types: Verify quark composition; use conservation laws to judge reactions; quark explanation of β decay.

Particle Classification

Class	Subclass	Examples	Composition
Hadron	Baryon	Proton, neutron	Three quarks
Hadron	Meson	Pion	Quark + antiquark
Lepton	—	Electron, neutrino, muon	Fundamental — no substructure

Quark Charges

Up quark: charge \(+\frac{2}{3}e\); Down quark: charge \(-\frac{1}{3}e\)
Proton \(= uud\): \(+\frac{2}{3}+\frac{2}{3}-\frac{1}{3} = +1e\) ✓
Neutron \(= udd\): \(+\frac{2}{3}-\frac{1}{3}-\frac{1}{3} = 0\) ✓

β⁻ Decay — Quark Explanation

\[d \to u + e^- + \bar{\nu}_e\]

One down quark in a neutron changes to an up quark → neutron becomes a proton + electron + electron antineutrino.

Conservation Laws

Conserved quantity	Must be conserved
Charge	Equal on both sides
Baryon number	Equal on both sides
Lepton number	Equal on both sides

Antiparticles have negative baryon/lepton numbers (e.g. antiproton: baryon number = −1).

Common Mistakes

❌ Photons are gauge bosons — neither hadrons nor leptons
❌ Electron antineutrino has lepton number −1, not +1
❌ Check all three conservation laws — charge alone is not sufficient

Topic 9: Alternating Current and Transformers

Question types: Convert between RMS and peak values; transformer turns ratio; power loss in transmission lines.

Peak and RMS Values

\[V_{rms} = \frac{V_0}{\sqrt{2}} \qquad I_{rms} = \frac{I_0}{\sqrt{2}} \qquad P_{mean} = V_{rms} I_{rms} = \frac{1}{2}V_0 I_0\]

RMS value = equivalent d.c. value that produces the same heating effect.

Transformer Equations (ideal)

\[\frac{V_s}{V_p} = \frac{N_s}{N_p} \qquad \frac{I_s}{I_p} = \frac{N_p}{N_s} \qquad V_p I_p = V_s I_s\]

Step-up: \(N_s > N_p\) → voltage increases, current decreases
Step-down: \(N_s < N_p\) → voltage decreases, current increases

Power Loss in Transmission

\[P_{loss} = I^2 R_{line}\]

Reason for stepping up voltage: voltage ×\(n\) → current ÷\(n\) → power loss ÷\(n^2\).

Common Mistakes

❌ Divide by \(\sqrt{2}\), not by 2
❌ Transformers only work with a.c. — not d.c.
❌ Step-up transformer: voltage increases, current decreases — power is conserved
❌ Use the current in the transmission line, not the current at the load

Practice Questions

Q1. \(C = 500\ \mu\text{F}\), \(R = 10\ \text{k}\Omega\), \(V_0 = 12\ \text{V}\). Find the voltage after one time constant.

Answer

\(\tau = RC = 10000 \times 500\times10^{-6} = 5\ \text{s}\)

\(V = 12 \times e^{-1} = 12 \times 0.37 = \mathbf{4.4\ \text{V}}\)

Q2. A satellite of mass \(m\) has orbital radius \(r\) and period \(T\). State the source of centripetal force and write the expression.

Answer

Source: gravitational force

\(F = \dfrac{GMm}{r^2} = \dfrac{4\pi^2 mr}{T^2}\)

Q3. \(C_1 = 4\ \mu\text{F}\) and \(C_2 = 12\ \mu\text{F}\) in series, connected to \(9\ \text{V}\). Find: (a) equivalent capacitance; (b) total energy stored.

Answer

(a) \(\dfrac{1}{C} = \dfrac{1}{4} + \dfrac{1}{12} = \dfrac{4}{12}\), \(C = \mathbf{3\ \mu\text{F}}\)

(b) \(E = \dfrac{1}{2}CV^2 = \dfrac{1}{2} \times 3\times10^{-6} \times 81 = \mathbf{1.22\times10^{-4}\ \text{J}}\)

Q4. An electron is accelerated through \(3.0\ \text{kV}\), then enters a magnetic field \(B = 0.050\ \text{T}\) perpendicular to its velocity. Find: (a) electron speed; (b) radius of circular path.

Answer

(a) \(v = \sqrt{\dfrac{2eV}{m_e}} = \sqrt{\dfrac{2\times1.6\times10^{-19}\times3000}{9.11\times10^{-31}}} = \mathbf{3.25\times10^7\ \text{m/s}}\)

(b) \(r = \dfrac{m_ev}{eB} = \dfrac{9.11\times10^{-31}\times3.25\times10^7}{1.6\times10^{-19}\times0.050} = \mathbf{3.70\times10^{-3}\ \text{m}}\)

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