Unit 4 High-Yield Topics | Flowxiom<\/title>\n<meta name=\"description\" content=\"Unit 4 High-Yield Topics \u2014 Edexcel A-level Physics WPH. Question types: Exponential equation calculation, ln graph linearisation, reading time constant. LATE...\">\n<link rel=\"canonical\" href=\"https:\/\/flowxiom.com\/edexcel-physics-unit-4-high-yield-topics\/\">\n<script type=\"application\/ld+json\">\n{\n \"@context\": \"https:\/\/schema.org\",\n \"@type\": \"FAQPage\",\n \"mainEntity\": [\n {\n \"@type\": \"Question\",\n \"name\": \"Q1. \\\\(C = 500\\\\ \\\\mu\\\\text{F}\\\\), \\\\(R = 10\\\\ \\\\text{k}\\\\Omega\\\\), \\\\(V_0 = 12\\\\ \\\\text{V}\\\\). Find the voltage after one time constant.\",\n \"acceptedAnswer\": {\n \"@type\": \"Answer\",\n \"text\": \"\\\\(\\\\tau = RC = 10000 \\\\times 500\\\\times10^{-6} = 5\\\\ \\\\text{s}\\\\) \\n \\\\(V = 12 \\\\times e^{-1} = 12 \\\\times 0.37 = \\\\mathbf{4.4\\\\ \\\\text{V}}\\\\)\"\n }\n },\n {\n \"@type\": \"Question\",\n \"name\": \"Q2. A satellite of mass \\\\(m\\\\) has orbital radius \\\\(r\\\\) and period \\\\(T\\\\). State the source of centripetal force and write the expression.\",\n \"acceptedAnswer\": {\n \"@type\": \"Answer\",\n \"text\": \"Source: gravitational force \\n \\\\(F = \\\\dfrac{GMm}{r^2} = \\\\dfrac{4\\\\pi^2 mr}{T^2}\\\\)\"\n }\n },\n {\n \"@type\": \"Question\",\n \"name\": \"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.\",\n \"acceptedAnswer\": {\n \"@type\": \"Answer\",\n \"text\": \"(a) \\\\(\\\\dfrac{1}{C} = \\\\dfrac{1}{4} + \\\\dfrac{1}{12} = \\\\dfrac{4}{12}\\\\), \\\\(C = \\\\mathbf{3\\\\ \\\\mu\\\\text{F}}\\\\) \\n (b) \\\\(E = \\\\dfrac{1}{2}CV^2 = \\\\dfrac{1}{2} \\\\times 3\\\\times10^{-6} \\\\times 81 = \\\\mathbf{1.22\\\\times10^{-4}\\\\ \\\\text{J}}\\\\)\"\n }\n },\n {\n \"@type\": \"Question\",\n \"name\": \"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.\",\n \"acceptedAnswer\": {\n \"@type\": \"Answer\",\n \"text\": \"(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}}\\\\) \\n (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}}\\\\)\"\n }\n }\n ]\n}\n<\/script>\n\n<script>\nMathJax = {\n tex: { inlineMath: [['\\\\(','\\\\)']], displayMath: [['\\\\[','\\\\]']] },\n svg: { fontCache: 'global' }\n};\n<\/script>\n<script async src=\"https:\/\/cdn.jsdelivr.net\/npm\/mathjax@3\/es5\/tex-svg.js\"><\/script>\n\n<style>\n:root{--accent:#2563eb;--warn-bg:#fef9c3;--warn-border:#ca8a04;--formula-bg:#eff6ff;--formula-border:#2563eb;}\nbody{font-family:system-ui,sans-serif;max-width:860px;margin:0 auto;padding:1.5rem;line-height:1.7;color:#1e293b;}\nh1{font-size:2rem;border-bottom:3px solid var(--accent);padding-bottom:.4rem;}\nh2{font-size:1.4rem;color:var(--accent);margin-top:2rem;}\nh3{font-size:1.1rem;margin-top:1.4rem;}\ntable{border-collapse:collapse;width:100%;margin:1rem 0;}\nth,td{border:1px solid #cbd5e1;padding:.5rem .75rem;text-align:left;}\nth{background:#e2e8f0;}\npre,code{background:#f1f5f9;border-radius:4px;}\npre{padding:1rem;overflow-x:auto;}\ncode{padding:.1rem .3rem;font-size:.9em;}\n.seo-warning-box{background:var(--warn-bg);border-left:4px solid var(--warn-border);padding:.75rem 1rem;margin:1rem 0;border-radius:0 6px 6px 0;}\n.seo-note-box{background:#f0fdf4;border-left:4px solid #16a34a;padding:.75rem 1rem;margin:1rem 0;border-radius:0 6px 6px 0;}\nsection.formula-card{background:var(--formula-bg);border:1px solid var(--formula-border);border-radius:8px;padding:1rem 1.25rem;margin:1.5rem 0;}\ndetails{border:1px solid #e2e8f0;border-radius:6px;padding:.5rem 1rem;margin:.75rem 0;}\nsummary{cursor:pointer;font-weight:600;}\n.answer-block{margin-top:.5rem;padding-top:.5rem;border-top:1px solid #e2e8f0;overflow-x:auto;}\n.answer-block p{margin:.25rem 0;}\n.numbered-step{padding-left:1.5rem;position:relative;}\nhr{border:none;border-top:1px solid #e2e8f0;margin:2rem 0;}\n.toc{background:#f8fafc;border:1px solid #e2e8f0;border-radius:8px;padding:1rem 1.5rem;margin-bottom:2rem;}\n.toc h2{margin-top:0;font-size:1.1rem;}\n.toc ul{margin:0;padding-left:1.2rem;}\n.toc a{color:var(--accent);text-decoration:none;}\n.site-footer{margin-top:3rem;padding-top:1rem;border-top:2px solid var(--accent);font-size:.9rem;color:#64748b;}\n<\/style>\n<\/head>\n<body>\n<nav class=\"toc\"><h2>Contents<\/h2><ul>\n <li><a href=\"#topic-1-capacitor-discharge-exponential-decay\">Topic 1: Capacitor Discharge (Exponential Decay)<\/a><\/li>\n <li><a href=\"#topic-2-electromagnetic-induction\">Topic 2: Electromagnetic Induction<\/a><\/li>\n <li><a href=\"#topic-3-circular-motion\">Topic 3: Circular Motion<\/a><\/li>\n <li><a href=\"#topic-4-electric-fields\">Topic 4: Electric Fields<\/a><\/li>\n <li><a href=\"#topic-5-magnetic-force-and-charged-particles\">Topic 5: Magnetic Force and Charged Particles<\/a><\/li>\n <li><a href=\"#topic-6-capacitors\">Topic 6: Capacitors<\/a><\/li>\n <li><a href=\"#topic-7-electron-acceleration-and-deflection-in-a-magnetic-field\">Topic 7: Electron Acceleration and Deflection in a Magnetic Field<\/a><\/li>\n <li><a href=\"#topic-8-particle-physics-fundamentals\">Topic 8: Particle Physics Fundamentals<\/a><\/li>\n <li><a href=\"#topic-9-alternating-current-and-transformers\">Topic 9: Alternating Current and Transformers<\/a><\/li>\n <li><a href=\"#practice-questions\">Practice Questions<\/a><\/li>\n<\/ul><\/nav>\n<h1>Unit 4 High-Yield Topics<\/h1>\n<p><strong>Free resource by <a href=\"https:\/\/flowxiom.com\">Flowxiom<\/a> \u2014 Edexcel A-level Physics<\/strong><\/p>\n<p><em>Not everything. Just what’s on the paper. High-frequency topics only \u2014 covering ~80% of exam marks.<\/em><\/p>\n<p>Edexcel A-level Physics | Fields, Capacitors & Particles | WPH14 & WPH15<\/p>\n<hr>\n<h2 id=\"topic-1-capacitor-discharge-exponential-decay\">Topic 1: Capacitor Discharge (Exponential Decay)<\/h2>\n<p><strong>Question types:<\/strong> Exponential equation calculation, ln graph linearisation, reading time constant.<\/p>\n<h3 id=\"key-formulae\">Key Formulae<\/h3>\n<p>\\[Q = Q_0 e^{-t\/RC} \\qquad V = V_0 e^{-t\/RC} \\qquad I = I_0 e^{-t\/RC}\\]<\/p>\n<p><strong>Time constant:<\/strong> \\(\\tau = RC\\)<\/p>\n<p>After one time constant \\(\\tau\\), charge\/voltage\/current falls to <strong>37%<\/strong> of its initial value (63% lost).<\/p>\n<h3 id=\"logarithmic-linearisation-essential-for-practical-questions\">Logarithmic Linearisation (essential for practical questions)<\/h3>\n<p>Taking \\(\\ln\\) of \\(V = V_0 e^{-t\/RC}\\):<\/p>\n<p>\\[\\ln V = \\ln V_0 – \\frac{1}{RC} \\cdot t\\]<\/p>\n<p>Plot \\(\\ln V\\) against \\(t\\):<\/p>\n<ul>\n<li>Gradient \\(= -1\/RC\\) (use to find \\(C\\) or \\(R\\))<\/li>\n<li>y-intercept \\(= \\ln V_0\\)<\/li>\n<\/ul>\n<div class=\"seo-warning-box\" id=\"common-mistakes\">\n<h3>Common Mistakes<\/h3>\n<ul>\n<li>\u274c Time constant: value falls <strong>TO<\/strong> 37%, not <strong>BY<\/strong> 37%<\/li>\n<li>\u274c Gradient is negative; \\(RC = -1\/\\text{gradient}\\)<\/li>\n<li>\u274c Charging and discharging curves go in opposite directions<\/li>\n<\/ul>\n<\/div>\n<hr>\n<h2 id=\"topic-2-electromagnetic-induction\">Topic 2: Electromagnetic Induction<\/h2>\n<p><strong>Question types:<\/strong> Determine direction of induced current; calculate induced e.m.f.; Lenz’s law explanation (frequent 6-mark).<\/p>\n<h3 id=\"faradays-law\">Faraday’s Law<\/h3>\n<p>\\[\\varepsilon = -N\\frac{\\Delta\\Phi}{\\Delta t}\\]<\/p>\n<p>Magnitude of induced e.m.f. \u221d rate of change of flux linkage.<\/p>\n<h3 id=\"magnetic-flux\">Magnetic Flux<\/h3>\n<p>\\[\\Phi = BA\\cos\\theta\\]<\/p>\n<ul>\n<li>\\(\\theta\\) = angle between B field and the <strong>normal<\/strong> to the coil plane<\/li>\n<li>Coil parallel to B: \\(\\theta = 90\u00b0\\), \\(\\Phi = 0\\)<\/li>\n<li>Coil perpendicular to B: \\(\\theta = 0\u00b0\\), \\(\\Phi = BA\\) (maximum)<\/li>\n<\/ul>\n<div class=\"seo-warning-box\">\n<p>\u26a0\ufe0f If the question gives the angle between the coil <strong>plane<\/strong> and B, use \\(\\sin\\alpha\\) not \\(\\cos\\alpha\\).<\/p>\n<\/div>\n<h3 id=\"lenzs-law-answer-chain-6mark-question\">Lenz’s Law Answer Chain (6-mark question)<\/h3>\n<p>The induced current flows in a direction such that its magnetic field <strong>opposes<\/strong> the change in flux that caused it.<\/p>\n<p>Steps:<\/p>\n<p class=\"numbered-step\">State whether flux is increasing or decreasing<\/p>\n<p class=\"numbered-step\">By Lenz’s law, the induced field must oppose the change<\/p>\n<p class=\"numbered-step\">Use right-hand rule to find current direction<\/p>\n<p class=\"numbered-step\">State that this creates a force opposing the motion<\/p>\n<div class=\"seo-warning-box\" id=\"common-mistakes\">\n<h3>Common Mistakes<\/h3>\n<ul>\n<li>\u274c Confusing angle between B and normal (use cos) with angle between B and plane (use sin)<\/li>\n<li>\u274c Must state WHY \u2014 the induced current opposes the change in flux<\/li>\n<li>\u274c Maximum \\(\\varepsilon\\) occurs when \\(\\Phi\\) is zero (90\u00b0 phase difference)<\/li>\n<\/ul>\n<\/div>\n<hr>\n<h2 id=\"topic-3-circular-motion\">Topic 3: Circular Motion<\/h2>\n<p><strong>Question types:<\/strong> Identify source of centripetal force; calculate centripetal force, speed, period.<\/p>\n<h3 id=\"centripetal-force-is-not-a-new-force\">Centripetal Force is Not a New Force<\/h3>\n<p>Centripetal force = the resultant force (or component of resultant) directed towards the centre.<\/p>\n<table>\n<thead><tr><th>Scenario<\/th><th>Source of centripetal force<\/th><\/tr><\/thead>\n<tr><td>Satellite orbiting Earth<\/td><td>Gravitational force<\/td><\/tr>\n<tr><td>Car cornering<\/td><td>Static friction<\/td><\/tr>\n<tr><td>Conical pendulum<\/td><td>Horizontal component of tension<\/td><\/tr>\n<tr><td>Top of vertical circle<\/td><td>Weight + normal force (or tension)<\/td><\/tr>\n<\/table>\n\n<h3 id=\"formulae\">Formulae<\/h3>\n<p>\\[F = \\frac{mv^2}{r} = mr\\omega^2 = \\frac{4\\pi^2 mr}{T^2}\\]<\/p>\n<div class=\"seo-warning-box\">\n<p>\u26a0\ufe0f Do <strong>NOT<\/strong> draw centripetal force on a free-body diagram \u2014 only draw real forces.<\/p>\n<\/div>\n<div class=\"seo-warning-box\" id=\"common-mistakes\">\n<h3>Common Mistakes<\/h3>\n<ul>\n<li>\u274c Centripetal force is not a separate force \u2014 do not include it on a free-body diagram<\/li>\n<li>\u274c At the top of a vertical circle: centripetal force = weight \u2212 normal force<\/li>\n<\/ul>\n<\/div>\n<hr>\n<h2 id=\"topic-4-electric-fields\">Topic 4: Electric Fields<\/h2>\n<p><strong>Question types:<\/strong> Point charge fields, work done in uniform field, calculations involving \\(E\\), \\(V\\), \\(W\\).<\/p>\n<h3 id=\"key-formulae\">Key Formulae<\/h3>\n<p>\\[E = \\frac{F}{q} \\qquad E = \\frac{V}{d} \\text{ (uniform)} \\qquad E = \\frac{Q}{4\\pi\\varepsilon_0 r^2} \\text{ (point charge)}\\]<\/p>\n<p>\\[W = qV \\qquad V = \\frac{Q}{4\\pi\\varepsilon_0 r}\\]<\/p>\n<h3 id=\"electric-field-vs-gravitational-field\">Electric Field vs Gravitational Field<\/h3>\n<table>\n<thead><tr><th><\/th><th>Electric field<\/th><th>Gravitational field<\/th><\/tr><\/thead>\n<tr><td>Source<\/td><td>Charge \\(Q\\)<\/td><td>Mass \\(M\\)<\/td><\/tr>\n<tr><td>Field strength<\/td><td>\\(E = Q\/4\\pi\\varepsilon_0 r^2\\)<\/td><td>\\(g = GM\/r^2\\)<\/td><\/tr>\n<tr><td>Potential<\/td><td>\\(V = Q\/4\\pi\\varepsilon_0 r\\)<\/td><td>\\(\\varphi = -GM\/r\\)<\/td><\/tr>\n<tr><td>Sign<\/td><td>Positive or negative<\/td><td>Always negative<\/td><\/tr>\n<\/table>\n\n<div class=\"seo-warning-box\" id=\"common-mistakes\">\n<h3>Common Mistakes<\/h3>\n<ul>\n<li>\u274c \\(d\\) = separation between the plates, not distance from one plate<\/li>\n<li>\u274c Gravitational potential is always negative; electric potential can be positive or negative<\/li>\n<\/ul>\n<\/div>\n<hr>\n<h2 id=\"topic-5-magnetic-force-and-charged-particles\">Topic 5: Magnetic Force and Charged Particles<\/h2>\n<p><strong>Question types:<\/strong> Force direction (Fleming’s left-hand rule), velocity selector, charged particle trajectory.<\/p>\n<h3 id=\"force-on-a-conductor\">Force on a Conductor<\/h3>\n<p>\\[F = BIL\\sin\\theta\\]<\/p>\n<h3 id=\"magnetic-force-on-a-moving-charge\">Magnetic Force on a Moving Charge<\/h3>\n<p>\\[F = Bqv\\sin\\theta\\]<\/p>\n<p>Direction: use <strong>Fleming’s left-hand rule<\/strong><\/p>\n<p><em>(first finger = B field; second finger = conventional current direction; thumb = force)<\/em><\/p>\n<h3 id=\"velocity-selector-straightline-condition\">Velocity Selector (straight-line condition)<\/h3>\n<p>\\[qE = qvB \\implies v = \\frac{E}{B}\\]<\/p>\n<p>Only particles with speed \\(v = E\/B\\) pass through undeflected \u2014 independent of mass and charge.<\/p>\n<div class=\"seo-warning-box\" id=\"common-mistakes\">\n<h3>Common Mistakes<\/h3>\n<ul>\n<li>\u274c Neutrons are uncharged \u2014 no magnetic force, trajectory is a straight line<\/li>\n<li>\u274c Negative charges (e.g. electrons) experience force in the <strong>opposite<\/strong> direction to positive charges<\/li>\n<\/ul>\n<\/div>\n<hr>\n<h2 id=\"topic-6-capacitors\">Topic 6: Capacitors<\/h2>\n<p><strong>Question types:<\/strong> Calculate stored energy, series\/parallel equivalent capacitance, effect of changing parameters.<\/p>\n<h3 id=\"key-formulae\">Key Formulae<\/h3>\n<p>\\[C = \\frac{Q}{V} \\qquad E_{stored} = \\frac{1}{2}QV = \\frac{1}{2}CV^2 = \\frac{Q^2}{2C}\\]<\/p>\n<h3 id=\"series-vs-parallel-opposite-to-resistors\">Series vs Parallel (opposite to resistors)<\/h3>\n<table>\n<thead><tr><th><\/th><th>Capacitors in series<\/th><th>Capacitors in parallel<\/th><\/tr><\/thead>\n<tr><td>Formula<\/td><td>\\(\\dfrac{1}{C_{total}} = \\dfrac{1}{C_1} + \\dfrac{1}{C_2}\\)<\/td><td>\\(C_{total} = C_1 + C_2\\)<\/td><\/tr>\n<tr><td>Result<\/td><td>Total less than smallest<\/td><td>Total greater than largest<\/td><\/tr>\n<\/table>\n\n<div class=\"seo-note-box\">\n<p><strong>Capacitor combinations are the reverse of resistor combinations.<\/strong><\/p>\n<\/div>\n<h3 id=\"changing-parameters\">Changing Parameters<\/h3>\n<ul>\n<li><strong>After disconnecting supply<\/strong>: \\(Q\\) is fixed. Change \\(d\\) \u2192 \\(C\\) changes \u2192 \\(V = Q\/C\\) changes<\/li>\n<li><strong>While connected to supply<\/strong>: \\(V\\) is fixed. Change \\(d\\) \u2192 \\(C\\) changes \u2192 \\(Q = CV\\) changes<\/li>\n<\/ul>\n<div class=\"seo-warning-box\" id=\"common-mistakes\">\n<h3>Common Mistakes<\/h3>\n<ul>\n<li>\u274c Capacitor series\/parallel rules are the <strong>reverse<\/strong> of resistors<\/li>\n<li>\u274c Choose the correct energy formula based on which quantities are known<\/li>\n<li>\u274c After disconnecting: \\(Q\\) is constant, not \\(V\\)<\/li>\n<\/ul>\n<\/div>\n<hr>\n<h2 id=\"topic-7-electron-acceleration-and-deflection-in-a-magnetic-field\">Topic 7: Electron Acceleration and Deflection in a Magnetic Field<\/h2>\n<p><strong>Question types:<\/strong> Calculate electron speed after acceleration; radius of circular path in B field; mass spectrometer.<\/p>\n<h3 id=\"electron-accelerated-through-pd-v\">Electron Accelerated Through p.d. V<\/h3>\n<p>\\[eV = \\frac{1}{2}mv^2 \\implies v = \\sqrt{\\frac{2eV}{m_e}}\\]<\/p>\n<p>\\(e = 1.6\\times10^{-19}\\ \\text{C}\\), \\(m_e = 9.11\\times10^{-31}\\ \\text{kg}\\)<\/p>\n<h3 id=\"circular-motion-in-magnetic-field\">Circular Motion in Magnetic Field<\/h3>\n<p>Lorentz force provides centripetal force:<\/p>\n<p>\\[Bqv = \\frac{mv^2}{r} \\implies r = \\frac{mv}{Bq}\\]<\/p>\n<p>Larger mass or speed \u2192 larger radius; stronger \\(B\\) or larger charge \u2192 smaller radius.<\/p>\n<div class=\"seo-warning-box\" id=\"common-mistakes\">\n<h3>Common Mistakes<\/h3>\n<ul>\n<li>\u274c Proton mass \u2248 1836 \u00d7 electron mass \u2014 do not use the same value<\/li>\n<li>\u274c Negative particles deflect in the <strong>opposite<\/strong> direction to positive charges<\/li>\n<li>\u274c Magnetic force does no work \u2014 it only changes direction, not speed<\/li>\n<\/ul>\n<\/div>\n<hr>\n<h2 id=\"topic-8-particle-physics-fundamentals\">Topic 8: Particle Physics Fundamentals<\/h2>\n<p><strong>Question types:<\/strong> Verify quark composition; use conservation laws to judge reactions; quark explanation of \u03b2 decay.<\/p>\n<h3 id=\"particle-classification\">Particle Classification<\/h3>\n<table>\n<thead><tr><th>Class<\/th><th>Subclass<\/th><th>Examples<\/th><th>Composition<\/th><\/tr><\/thead>\n<tr><td>Hadron<\/td><td>Baryon<\/td><td>Proton, neutron<\/td><td>Three quarks<\/td><\/tr>\n<tr><td>Hadron<\/td><td>Meson<\/td><td>Pion<\/td><td>Quark + antiquark<\/td><\/tr>\n<tr><td>Lepton<\/td><td>\u2014<\/td><td>Electron, neutrino, muon<\/td><td>Fundamental \u2014 no substructure<\/td><\/tr>\n<\/table>\n\n<h3 id=\"quark-charges\">Quark Charges<\/h3>\n<ul>\n<li>Up quark: charge \\(+\\frac{2}{3}e\\); Down quark: charge \\(-\\frac{1}{3}e\\)<\/li>\n<li>Proton \\(= uud\\): \\(+\\frac{2}{3}+\\frac{2}{3}-\\frac{1}{3} = +1e\\) \u2713<\/li>\n<li>Neutron \\(= udd\\): \\(+\\frac{2}{3}-\\frac{1}{3}-\\frac{1}{3} = 0\\) \u2713<\/li>\n<\/ul>\n<h3 id=\"\u03b2-decay-quark-explanation\">\u03b2\u207b Decay \u2014 Quark Explanation<\/h3>\n<p>\\[d \\to u + e^- + \\bar{\\nu}_e\\]<\/p>\n<p>One down quark in a neutron changes to an up quark \u2192 neutron becomes a proton + electron + electron antineutrino.<\/p>\n<h3 id=\"conservation-laws\">Conservation Laws<\/h3>\n<table>\n<thead><tr><th>Conserved quantity<\/th><th>Must be conserved<\/th><\/tr><\/thead>\n<tr><td>Charge<\/td><td>Equal on both sides<\/td><\/tr>\n<tr><td>Baryon number<\/td><td>Equal on both sides<\/td><\/tr>\n<tr><td>Lepton number<\/td><td>Equal on both sides<\/td><\/tr>\n<\/table>\n\n<p>Antiparticles have negative baryon\/lepton numbers (e.g. antiproton: baryon number = \u22121).<\/p>\n<div class=\"seo-warning-box\" id=\"common-mistakes\">\n<h3>Common Mistakes<\/h3>\n<ul>\n<li>\u274c Photons are gauge bosons \u2014 neither hadrons nor leptons<\/li>\n<li>\u274c Electron antineutrino has lepton number <strong>\u22121<\/strong>, not +1<\/li>\n<li>\u274c Check <strong>all three<\/strong> conservation laws \u2014 charge alone is not sufficient<\/li>\n<\/ul>\n<\/div>\n<hr>\n<h2 id=\"topic-9-alternating-current-and-transformers\">Topic 9: Alternating Current and Transformers<\/h2>\n<p><strong>Question types:<\/strong> Convert between RMS and peak values; transformer turns ratio; power loss in transmission lines.<\/p>\n<h3 id=\"peak-and-rms-values\">Peak and RMS Values<\/h3>\n<p>\\[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\\]<\/p>\n<p>RMS value = equivalent d.c. value that produces the same heating effect.<\/p>\n<h3 id=\"transformer-equations-ideal\">Transformer Equations (ideal)<\/h3>\n<p>\\[\\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\\]<\/p>\n<ul>\n<li>Step-up: \\(N_s > N_p\\) \u2192 voltage increases, current decreases<\/li>\n<li>Step-down: \\(N_s < N_p\\) \u2192 voltage decreases, current increases<\/li>\n<\/ul>\n<h3 id=\"power-loss-in-transmission\">Power Loss in Transmission<\/h3>\n<p>\\[P_{loss} = I^2 R_{line}\\]<\/p>\n<p>Reason for stepping up voltage: voltage \u00d7\\(n\\) \u2192 current \u00f7\\(n\\) \u2192 power loss \u00f7\\(n^2\\).<\/p>\n<div class=\"seo-warning-box\" id=\"common-mistakes\">\n<h3>Common Mistakes<\/h3>\n<ul>\n<li>\u274c Divide by \\(\\sqrt{2}\\), not by 2<\/li>\n<li>\u274c Transformers only work with a.c. \u2014 not d.c.<\/li>\n<li>\u274c Step-up transformer: voltage increases, <strong>current decreases<\/strong> \u2014 power is conserved<\/li>\n<li>\u274c Use the current in the transmission line, not the current at the load<\/li>\n<\/ul>\n<\/div>\n<hr>\n<h2 id=\"practice-questions\">Practice Questions<\/h2>\n<p><strong>Q1.<\/strong> \\(C = 500\\ \\mu\\text{F}\\), \\(R = 10\\ \\text{k}\\Omega\\), \\(V_0 = 12\\ \\text{V}\\). Find the voltage after one time constant.<\/p>\n<details><summary>Answer<\/summary><div class=\"answer-block\"><p>\\(\\tau = RC = 10000 \\times 500\\times10^{-6} = 5\\ \\text{s}\\)<\/p>\n<p>\\(V = 12 \\times e^{-1} = 12 \\times 0.37 = \\mathbf{4.4\\ \\text{V}}\\)<\/p><\/div><\/details>\n<hr>\n<p><strong>Q2.<\/strong> A satellite of mass \\(m\\) has orbital radius \\(r\\) and period \\(T\\). State the source of centripetal force and write the expression.<\/p>\n<details><summary>Answer<\/summary><div class=\"answer-block\"><p>Source: <strong>gravitational force<\/strong><\/p>\n<p>\\(F = \\dfrac{GMm}{r^2} = \\dfrac{4\\pi^2 mr}{T^2}\\)<\/p><\/div><\/details>\n<hr>\n<p><strong>Q3.<\/strong> \\(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.<\/p>\n<details><summary>Answer<\/summary><div class=\"answer-block\"><p>(a) \\(\\dfrac{1}{C} = \\dfrac{1}{4} + \\dfrac{1}{12} = \\dfrac{4}{12}\\), \\(C = \\mathbf{3\\ \\mu\\text{F}}\\)<\/p>\n<p>(b) \\(E = \\dfrac{1}{2}CV^2 = \\dfrac{1}{2} \\times 3\\times10^{-6} \\times 81 = \\mathbf{1.22\\times10^{-4}\\ \\text{J}}\\)<\/p><\/div><\/details>\n<hr>\n<p><strong>Q4.<\/strong> 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.<\/p>\n<details><summary>Answer<\/summary><div class=\"answer-block\"><p>(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}}\\)<\/p>\n<p>(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}}\\)<\/p><\/div><\/details>\n<hr>\n<p><em>Want more? Visit <a href=\"https:\/\/flowxiom.com\">flowxiom.com<\/a><\/em><\/p>\n<footer class=\"site-footer\">\n <p>Free resource by <a href=\"https:\/\/flowxiom.com\">Flowxiom<\/a> \u2014 Edexcel A-level Physics<br>\n High-frequency topics only, covering ~80% of exam marks.<\/p>\n<\/footer>\n<\/body>\n<\/html>\n\n","protected":false},"excerpt":{"rendered":"<p>This “Sprint Pack” is an essential guide for Unit 4 Physics, simplifying complex field theories and particle physics into easy-to-digest sections. Key highlights include:<\/p>\n<p>Capacitors & Fields: Step-by-step guides for discharge equations and electric\/magnetic field strengths.<\/p>\n<p>Electromagnetism: Clear “answer chains” for Lenz’s Law and Faraday’s Law questions.<\/p>\n<p>Further Mechanics: Mastering circular motion and centripetal force without common mistakes.<\/p>\n<p>Particle Physics: A straightforward map of quarks, leptons, and conservation laws.<\/p>\n<p>Practical Exam Tips: Specific advice on how to linearize graphs and interpret RMS values in AC circuits.<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"class_list":["post-45","post","type-post","status-publish","format-standard","hentry","category-exam-sprint-pack-physics-exam-sprint-pack"],"_links":{"self":[{"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/posts\/45","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/comments?post=45"}],"version-history":[{"count":2,"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/posts\/45\/revisions"}],"predecessor-version":[{"id":47,"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/posts\/45\/revisions\/47"}],"wp:attachment":[{"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/media?parent=45"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/categories?post=45"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/tags?post=45"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}

{"id":45,"date":"2026-04-17T16:21:16","date_gmt":"2026-04-17T16:21:16","guid":{"rendered":"https:\/\/flowxiom.com\/?p=45"},"modified":"2026-04-17T16:23:47","modified_gmt":"2026-04-17T16:23:47","slug":"unit-4-high-yield-topics","status":"publish","type":"post","link":"https:\/\/flowxiom.com\/index.php\/2026\/04\/17\/unit-4-high-yield-topics\/","title":{"rendered":"Unit 4 High-Yield Topics"},"content":{"rendered":"\n\n\n\n\n\nUnit 4 High-Yield Topics | Flowxiom<\/title>\n<meta name=\"description\" content=\"Unit 4 High-Yield Topics \u2014 Edexcel A-level Physics WPH. Question types: Exponential equation calculation, ln graph linearisation, reading time constant. LATE...\">\n<link rel=\"canonical\" href=\"https:\/\/flowxiom.com\/edexcel-physics-unit-4-high-yield-topics\/\">\n<script type=\"application\/ld+json\">\n{\n \"@context\": \"https:\/\/schema.org\",\n \"@type\": \"FAQPage\",\n \"mainEntity\": [\n {\n \"@type\": \"Question\",\n \"name\": \"Q1. \\\\(C = 500\\\\ \\\\mu\\\\text{F}\\\\), \\\\(R = 10\\\\ \\\\text{k}\\\\Omega\\\\), \\\\(V_0 = 12\\\\ \\\\text{V}\\\\). Find the voltage after one time constant.\",\n \"acceptedAnswer\": {\n \"@type\": \"Answer\",\n \"text\": \"\\\\(\\\\tau = RC = 10000 \\\\times 500\\\\times10^{-6} = 5\\\\ \\\\text{s}\\\\) \\n \\\\(V = 12 \\\\times e^{-1} = 12 \\\\times 0.37 = \\\\mathbf{4.4\\\\ \\\\text{V}}\\\\)\"\n }\n },\n {\n \"@type\": \"Question\",\n \"name\": \"Q2. A satellite of mass \\\\(m\\\\) has orbital radius \\\\(r\\\\) and period \\\\(T\\\\). State the source of centripetal force and write the expression.\",\n \"acceptedAnswer\": {\n \"@type\": \"Answer\",\n \"text\": \"Source: gravitational force \\n \\\\(F = \\\\dfrac{GMm}{r^2} = \\\\dfrac{4\\\\pi^2 mr}{T^2}\\\\)\"\n }\n },\n {\n \"@type\": \"Question\",\n \"name\": \"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.\",\n \"acceptedAnswer\": {\n \"@type\": \"Answer\",\n \"text\": \"(a) \\\\(\\\\dfrac{1}{C} = \\\\dfrac{1}{4} + \\\\dfrac{1}{12} = \\\\dfrac{4}{12}\\\\), \\\\(C = \\\\mathbf{3\\\\ \\\\mu\\\\text{F}}\\\\) \\n (b) \\\\(E = \\\\dfrac{1}{2}CV^2 = \\\\dfrac{1}{2} \\\\times 3\\\\times10^{-6} \\\\times 81 = \\\\mathbf{1.22\\\\times10^{-4}\\\\ \\\\text{J}}\\\\)\"\n }\n },\n {\n \"@type\": \"Question\",\n \"name\": \"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.\",\n \"acceptedAnswer\": {\n \"@type\": \"Answer\",\n \"text\": \"(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}}\\\\) \\n (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}}\\\\)\"\n }\n }\n ]\n}\n<\/script>\n\n<script>\nMathJax = {\n tex: { inlineMath: [['\\\\(','\\\\)']], displayMath: [['\\\\[','\\\\]']] },\n svg: { fontCache: 'global' }\n};\n<\/script>\n<script async src=\"https:\/\/cdn.jsdelivr.net\/npm\/mathjax@3\/es5\/tex-svg.js\"><\/script>\n\n<style>\n:root{--accent:#2563eb;--warn-bg:#fef9c3;--warn-border:#ca8a04;--formula-bg:#eff6ff;--formula-border:#2563eb;}\nbody{font-family:system-ui,sans-serif;max-width:860px;margin:0 auto;padding:1.5rem;line-height:1.7;color:#1e293b;}\nh1{font-size:2rem;border-bottom:3px solid var(--accent);padding-bottom:.4rem;}\nh2{font-size:1.4rem;color:var(--accent);margin-top:2rem;}\nh3{font-size:1.1rem;margin-top:1.4rem;}\ntable{border-collapse:collapse;width:100%;margin:1rem 0;}\nth,td{border:1px solid #cbd5e1;padding:.5rem .75rem;text-align:left;}\nth{background:#e2e8f0;}\npre,code{background:#f1f5f9;border-radius:4px;}\npre{padding:1rem;overflow-x:auto;}\ncode{padding:.1rem .3rem;font-size:.9em;}\n.seo-warning-box{background:var(--warn-bg);border-left:4px solid var(--warn-border);padding:.75rem 1rem;margin:1rem 0;border-radius:0 6px 6px 0;}\n.seo-note-box{background:#f0fdf4;border-left:4px solid #16a34a;padding:.75rem 1rem;margin:1rem 0;border-radius:0 6px 6px 0;}\nsection.formula-card{background:var(--formula-bg);border:1px solid var(--formula-border);border-radius:8px;padding:1rem 1.25rem;margin:1.5rem 0;}\ndetails{border:1px solid #e2e8f0;border-radius:6px;padding:.5rem 1rem;margin:.75rem 0;}\nsummary{cursor:pointer;font-weight:600;}\n.answer-block{margin-top:.5rem;padding-top:.5rem;border-top:1px solid #e2e8f0;overflow-x:auto;}\n.answer-block p{margin:.25rem 0;}\n.numbered-step{padding-left:1.5rem;position:relative;}\nhr{border:none;border-top:1px solid #e2e8f0;margin:2rem 0;}\n.toc{background:#f8fafc;border:1px solid #e2e8f0;border-radius:8px;padding:1rem 1.5rem;margin-bottom:2rem;}\n.toc h2{margin-top:0;font-size:1.1rem;}\n.toc ul{margin:0;padding-left:1.2rem;}\n.toc a{color:var(--accent);text-decoration:none;}\n.site-footer{margin-top:3rem;padding-top:1rem;border-top:2px solid var(--accent);font-size:.9rem;color:#64748b;}\n<\/style>\n<\/head>\n<body>\n<nav class=\"toc\"><h2>Contents<\/h2><ul>\n <li><a href=\"#topic-1-capacitor-discharge-exponential-decay\">Topic 1: Capacitor Discharge (Exponential Decay)<\/a><\/li>\n <li><a href=\"#topic-2-electromagnetic-induction\">Topic 2: Electromagnetic Induction<\/a><\/li>\n <li><a href=\"#topic-3-circular-motion\">Topic 3: Circular Motion<\/a><\/li>\n <li><a href=\"#topic-4-electric-fields\">Topic 4: Electric Fields<\/a><\/li>\n <li><a href=\"#topic-5-magnetic-force-and-charged-particles\">Topic 5: Magnetic Force and Charged Particles<\/a><\/li>\n <li><a href=\"#topic-6-capacitors\">Topic 6: Capacitors<\/a><\/li>\n <li><a href=\"#topic-7-electron-acceleration-and-deflection-in-a-magnetic-field\">Topic 7: Electron Acceleration and Deflection in a Magnetic Field<\/a><\/li>\n <li><a href=\"#topic-8-particle-physics-fundamentals\">Topic 8: Particle Physics Fundamentals<\/a><\/li>\n <li><a href=\"#topic-9-alternating-current-and-transformers\">Topic 9: Alternating Current and Transformers<\/a><\/li>\n <li><a href=\"#practice-questions\">Practice Questions<\/a><\/li>\n<\/ul><\/nav>\n<h1>Unit 4 High-Yield Topics<\/h1>\n<p><strong>Free resource by <a href=\"https:\/\/flowxiom.com\">Flowxiom<\/a> \u2014 Edexcel A-level Physics<\/strong><\/p>\n<p><em>Not everything. Just what’s on the paper. High-frequency topics only \u2014 covering ~80% of exam marks.<\/em><\/p>\n<p>Edexcel A-level Physics | Fields, Capacitors & Particles | WPH14 & WPH15<\/p>\n<hr>\n<h2 id=\"topic-1-capacitor-discharge-exponential-decay\">Topic 1: Capacitor Discharge (Exponential Decay)<\/h2>\n<p><strong>Question types:<\/strong> Exponential equation calculation, ln graph linearisation, reading time constant.<\/p>\n<h3 id=\"key-formulae\">Key Formulae<\/h3>\n<p>\\[Q = Q_0 e^{-t\/RC} \\qquad V = V_0 e^{-t\/RC} \\qquad I = I_0 e^{-t\/RC}\\]<\/p>\n<p><strong>Time constant:<\/strong> \\(\\tau = RC\\)<\/p>\n<p>After one time constant \\(\\tau\\), charge\/voltage\/current falls to <strong>37%<\/strong> of its initial value (63% lost).<\/p>\n<h3 id=\"logarithmic-linearisation-essential-for-practical-questions\">Logarithmic Linearisation (essential for practical questions)<\/h3>\n<p>Taking \\(\\ln\\) of \\(V = V_0 e^{-t\/RC}\\):<\/p>\n<p>\\[\\ln V = \\ln V_0 – \\frac{1}{RC} \\cdot t\\]<\/p>\n<p>Plot \\(\\ln V\\) against \\(t\\):<\/p>\n<ul>\n<li>Gradient \\(= -1\/RC\\) (use to find \\(C\\) or \\(R\\))<\/li>\n<li>y-intercept \\(= \\ln V_0\\)<\/li>\n<\/ul>\n<div class=\"seo-warning-box\" id=\"common-mistakes\">\n<h3>Common Mistakes<\/h3>\n<ul>\n<li>\u274c Time constant: value falls <strong>TO<\/strong> 37%, not <strong>BY<\/strong> 37%<\/li>\n<li>\u274c Gradient is negative; \\(RC = -1\/\\text{gradient}\\)<\/li>\n<li>\u274c Charging and discharging curves go in opposite directions<\/li>\n<\/ul>\n<\/div>\n<hr>\n<h2 id=\"topic-2-electromagnetic-induction\">Topic 2: Electromagnetic Induction<\/h2>\n<p><strong>Question types:<\/strong> Determine direction of induced current; calculate induced e.m.f.; Lenz’s law explanation (frequent 6-mark).<\/p>\n<h3 id=\"faradays-law\">Faraday’s Law<\/h3>\n<p>\\[\\varepsilon = -N\\frac{\\Delta\\Phi}{\\Delta t}\\]<\/p>\n<p>Magnitude of induced e.m.f. \u221d rate of change of flux linkage.<\/p>\n<h3 id=\"magnetic-flux\">Magnetic Flux<\/h3>\n<p>\\[\\Phi = BA\\cos\\theta\\]<\/p>\n<ul>\n<li>\\(\\theta\\) = angle between B field and the <strong>normal<\/strong> to the coil plane<\/li>\n<li>Coil parallel to B: \\(\\theta = 90\u00b0\\), \\(\\Phi = 0\\)<\/li>\n<li>Coil perpendicular to B: \\(\\theta = 0\u00b0\\), \\(\\Phi = BA\\) (maximum)<\/li>\n<\/ul>\n<div class=\"seo-warning-box\">\n<p>\u26a0\ufe0f If the question gives the angle between the coil <strong>plane<\/strong> and B, use \\(\\sin\\alpha\\) not \\(\\cos\\alpha\\).<\/p>\n<\/div>\n<h3 id=\"lenzs-law-answer-chain-6mark-question\">Lenz’s Law Answer Chain (6-mark question)<\/h3>\n<p>The induced current flows in a direction such that its magnetic field <strong>opposes<\/strong> the change in flux that caused it.<\/p>\n<p>Steps:<\/p>\n<p class=\"numbered-step\">State whether flux is increasing or decreasing<\/p>\n<p class=\"numbered-step\">By Lenz’s law, the induced field must oppose the change<\/p>\n<p class=\"numbered-step\">Use right-hand rule to find current direction<\/p>\n<p class=\"numbered-step\">State that this creates a force opposing the motion<\/p>\n<div class=\"seo-warning-box\" id=\"common-mistakes\">\n<h3>Common Mistakes<\/h3>\n<ul>\n<li>\u274c Confusing angle between B and normal (use cos) with angle between B and plane (use sin)<\/li>\n<li>\u274c Must state WHY \u2014 the induced current opposes the change in flux<\/li>\n<li>\u274c Maximum \\(\\varepsilon\\) occurs when \\(\\Phi\\) is zero (90\u00b0 phase difference)<\/li>\n<\/ul>\n<\/div>\n<hr>\n<h2 id=\"topic-3-circular-motion\">Topic 3: Circular Motion<\/h2>\n<p><strong>Question types:<\/strong> Identify source of centripetal force; calculate centripetal force, speed, period.<\/p>\n<h3 id=\"centripetal-force-is-not-a-new-force\">Centripetal Force is Not a New Force<\/h3>\n<p>Centripetal force = the resultant force (or component of resultant) directed towards the centre.<\/p>\n<table>\n<thead><tr><th>Scenario<\/th><th>Source of centripetal force<\/th><\/tr><\/thead>\n<tr><td>Satellite orbiting Earth<\/td><td>Gravitational force<\/td><\/tr>\n<tr><td>Car cornering<\/td><td>Static friction<\/td><\/tr>\n<tr><td>Conical pendulum<\/td><td>Horizontal component of tension<\/td><\/tr>\n<tr><td>Top of vertical circle<\/td><td>Weight + normal force (or tension)<\/td><\/tr>\n<\/table>\n\n<h3 id=\"formulae\">Formulae<\/h3>\n<p>\\[F = \\frac{mv^2}{r} = mr\\omega^2 = \\frac{4\\pi^2 mr}{T^2}\\]<\/p>\n<div class=\"seo-warning-box\">\n<p>\u26a0\ufe0f Do <strong>NOT<\/strong> draw centripetal force on a free-body diagram \u2014 only draw real forces.<\/p>\n<\/div>\n<div class=\"seo-warning-box\" id=\"common-mistakes\">\n<h3>Common Mistakes<\/h3>\n<ul>\n<li>\u274c Centripetal force is not a separate force \u2014 do not include it on a free-body diagram<\/li>\n<li>\u274c At the top of a vertical circle: centripetal force = weight \u2212 normal force<\/li>\n<\/ul>\n<\/div>\n<hr>\n<h2 id=\"topic-4-electric-fields\">Topic 4: Electric Fields<\/h2>\n<p><strong>Question types:<\/strong> Point charge fields, work done in uniform field, calculations involving \\(E\\), \\(V\\), \\(W\\).<\/p>\n<h3 id=\"key-formulae\">Key Formulae<\/h3>\n<p>\\[E = \\frac{F}{q} \\qquad E = \\frac{V}{d} \\text{ (uniform)} \\qquad E = \\frac{Q}{4\\pi\\varepsilon_0 r^2} \\text{ (point charge)}\\]<\/p>\n<p>\\[W = qV \\qquad V = \\frac{Q}{4\\pi\\varepsilon_0 r}\\]<\/p>\n<h3 id=\"electric-field-vs-gravitational-field\">Electric Field vs Gravitational Field<\/h3>\n<table>\n<thead><tr><th><\/th><th>Electric field<\/th><th>Gravitational field<\/th><\/tr><\/thead>\n<tr><td>Source<\/td><td>Charge \\(Q\\)<\/td><td>Mass \\(M\\)<\/td><\/tr>\n<tr><td>Field strength<\/td><td>\\(E = Q\/4\\pi\\varepsilon_0 r^2\\)<\/td><td>\\(g = GM\/r^2\\)<\/td><\/tr>\n<tr><td>Potential<\/td><td>\\(V = Q\/4\\pi\\varepsilon_0 r\\)<\/td><td>\\(\\varphi = -GM\/r\\)<\/td><\/tr>\n<tr><td>Sign<\/td><td>Positive or negative<\/td><td>Always negative<\/td><\/tr>\n<\/table>\n\n<div class=\"seo-warning-box\" id=\"common-mistakes\">\n<h3>Common Mistakes<\/h3>\n<ul>\n<li>\u274c \\(d\\) = separation between the plates, not distance from one plate<\/li>\n<li>\u274c Gravitational potential is always negative; electric potential can be positive or negative<\/li>\n<\/ul>\n<\/div>\n<hr>\n<h2 id=\"topic-5-magnetic-force-and-charged-particles\">Topic 5: Magnetic Force and Charged Particles<\/h2>\n<p><strong>Question types:<\/strong> Force direction (Fleming’s left-hand rule), velocity selector, charged particle trajectory.<\/p>\n<h3 id=\"force-on-a-conductor\">Force on a Conductor<\/h3>\n<p>\\[F = BIL\\sin\\theta\\]<\/p>\n<h3 id=\"magnetic-force-on-a-moving-charge\">Magnetic Force on a Moving Charge<\/h3>\n<p>\\[F = Bqv\\sin\\theta\\]<\/p>\n<p>Direction: use <strong>Fleming’s left-hand rule<\/strong><\/p>\n<p><em>(first finger = B field; second finger = conventional current direction; thumb = force)<\/em><\/p>\n<h3 id=\"velocity-selector-straightline-condition\">Velocity Selector (straight-line condition)<\/h3>\n<p>\\[qE = qvB \\implies v = \\frac{E}{B}\\]<\/p>\n<p>Only particles with speed \\(v = E\/B\\) pass through undeflected \u2014 independent of mass and charge.<\/p>\n<div class=\"seo-warning-box\" id=\"common-mistakes\">\n<h3>Common Mistakes<\/h3>\n<ul>\n<li>\u274c Neutrons are uncharged \u2014 no magnetic force, trajectory is a straight line<\/li>\n<li>\u274c Negative charges (e.g. electrons) experience force in the <strong>opposite<\/strong> direction to positive charges<\/li>\n<\/ul>\n<\/div>\n<hr>\n<h2 id=\"topic-6-capacitors\">Topic 6: Capacitors<\/h2>\n<p><strong>Question types:<\/strong> Calculate stored energy, series\/parallel equivalent capacitance, effect of changing parameters.<\/p>\n<h3 id=\"key-formulae\">Key Formulae<\/h3>\n<p>\\[C = \\frac{Q}{V} \\qquad E_{stored} = \\frac{1}{2}QV = \\frac{1}{2}CV^2 = \\frac{Q^2}{2C}\\]<\/p>\n<h3 id=\"series-vs-parallel-opposite-to-resistors\">Series vs Parallel (opposite to resistors)<\/h3>\n<table>\n<thead><tr><th><\/th><th>Capacitors in series<\/th><th>Capacitors in parallel<\/th><\/tr><\/thead>\n<tr><td>Formula<\/td><td>\\(\\dfrac{1}{C_{total}} = \\dfrac{1}{C_1} + \\dfrac{1}{C_2}\\)<\/td><td>\\(C_{total} = C_1 + C_2\\)<\/td><\/tr>\n<tr><td>Result<\/td><td>Total less than smallest<\/td><td>Total greater than largest<\/td><\/tr>\n<\/table>\n\n<div class=\"seo-note-box\">\n<p><strong>Capacitor combinations are the reverse of resistor combinations.<\/strong><\/p>\n<\/div>\n<h3 id=\"changing-parameters\">Changing Parameters<\/h3>\n<ul>\n<li><strong>After disconnecting supply<\/strong>: \\(Q\\) is fixed. Change \\(d\\) \u2192 \\(C\\) changes \u2192 \\(V = Q\/C\\) changes<\/li>\n<li><strong>While connected to supply<\/strong>: \\(V\\) is fixed. Change \\(d\\) \u2192 \\(C\\) changes \u2192 \\(Q = CV\\) changes<\/li>\n<\/ul>\n<div class=\"seo-warning-box\" id=\"common-mistakes\">\n<h3>Common Mistakes<\/h3>\n<ul>\n<li>\u274c Capacitor series\/parallel rules are the <strong>reverse<\/strong> of resistors<\/li>\n<li>\u274c Choose the correct energy formula based on which quantities are known<\/li>\n<li>\u274c After disconnecting: \\(Q\\) is constant, not \\(V\\)<\/li>\n<\/ul>\n<\/div>\n<hr>\n<h2 id=\"topic-7-electron-acceleration-and-deflection-in-a-magnetic-field\">Topic 7: Electron Acceleration and Deflection in a Magnetic Field<\/h2>\n<p><strong>Question types:<\/strong> Calculate electron speed after acceleration; radius of circular path in B field; mass spectrometer.<\/p>\n<h3 id=\"electron-accelerated-through-pd-v\">Electron Accelerated Through p.d. V<\/h3>\n<p>\\[eV = \\frac{1}{2}mv^2 \\implies v = \\sqrt{\\frac{2eV}{m_e}}\\]<\/p>\n<p>\\(e = 1.6\\times10^{-19}\\ \\text{C}\\), \\(m_e = 9.11\\times10^{-31}\\ \\text{kg}\\)<\/p>\n<h3 id=\"circular-motion-in-magnetic-field\">Circular Motion in Magnetic Field<\/h3>\n<p>Lorentz force provides centripetal force:<\/p>\n<p>\\[Bqv = \\frac{mv^2}{r} \\implies r = \\frac{mv}{Bq}\\]<\/p>\n<p>Larger mass or speed \u2192 larger radius; stronger \\(B\\) or larger charge \u2192 smaller radius.<\/p>\n<div class=\"seo-warning-box\" id=\"common-mistakes\">\n<h3>Common Mistakes<\/h3>\n<ul>\n<li>\u274c Proton mass \u2248 1836 \u00d7 electron mass \u2014 do not use the same value<\/li>\n<li>\u274c Negative particles deflect in the <strong>opposite<\/strong> direction to positive charges<\/li>\n<li>\u274c Magnetic force does no work \u2014 it only changes direction, not speed<\/li>\n<\/ul>\n<\/div>\n<hr>\n<h2 id=\"topic-8-particle-physics-fundamentals\">Topic 8: Particle Physics Fundamentals<\/h2>\n<p><strong>Question types:<\/strong> Verify quark composition; use conservation laws to judge reactions; quark explanation of \u03b2 decay.<\/p>\n<h3 id=\"particle-classification\">Particle Classification<\/h3>\n<table>\n<thead><tr><th>Class<\/th><th>Subclass<\/th><th>Examples<\/th><th>Composition<\/th><\/tr><\/thead>\n<tr><td>Hadron<\/td><td>Baryon<\/td><td>Proton, neutron<\/td><td>Three quarks<\/td><\/tr>\n<tr><td>Hadron<\/td><td>Meson<\/td><td>Pion<\/td><td>Quark + antiquark<\/td><\/tr>\n<tr><td>Lepton<\/td><td>\u2014<\/td><td>Electron, neutrino, muon<\/td><td>Fundamental \u2014 no substructure<\/td><\/tr>\n<\/table>\n\n<h3 id=\"quark-charges\">Quark Charges<\/h3>\n<ul>\n<li>Up quark: charge \\(+\\frac{2}{3}e\\); Down quark: charge \\(-\\frac{1}{3}e\\)<\/li>\n<li>Proton \\(= uud\\): \\(+\\frac{2}{3}+\\frac{2}{3}-\\frac{1}{3} = +1e\\) \u2713<\/li>\n<li>Neutron \\(= udd\\): \\(+\\frac{2}{3}-\\frac{1}{3}-\\frac{1}{3} = 0\\) \u2713<\/li>\n<\/ul>\n<h3 id=\"\u03b2-decay-quark-explanation\">\u03b2\u207b Decay \u2014 Quark Explanation<\/h3>\n<p>\\[d \\to u + e^- + \\bar{\\nu}_e\\]<\/p>\n<p>One down quark in a neutron changes to an up quark \u2192 neutron becomes a proton + electron + electron antineutrino.<\/p>\n<h3 id=\"conservation-laws\">Conservation Laws<\/h3>\n<table>\n<thead><tr><th>Conserved quantity<\/th><th>Must be conserved<\/th><\/tr><\/thead>\n<tr><td>Charge<\/td><td>Equal on both sides<\/td><\/tr>\n<tr><td>Baryon number<\/td><td>Equal on both sides<\/td><\/tr>\n<tr><td>Lepton number<\/td><td>Equal on both sides<\/td><\/tr>\n<\/table>\n\n<p>Antiparticles have negative baryon\/lepton numbers (e.g. antiproton: baryon number = \u22121).<\/p>\n<div class=\"seo-warning-box\" id=\"common-mistakes\">\n<h3>Common Mistakes<\/h3>\n<ul>\n<li>\u274c Photons are gauge bosons \u2014 neither hadrons nor leptons<\/li>\n<li>\u274c Electron antineutrino has lepton number <strong>\u22121<\/strong>, not +1<\/li>\n<li>\u274c Check <strong>all three<\/strong> conservation laws \u2014 charge alone is not sufficient<\/li>\n<\/ul>\n<\/div>\n<hr>\n<h2 id=\"topic-9-alternating-current-and-transformers\">Topic 9: Alternating Current and Transformers<\/h2>\n<p><strong>Question types:<\/strong> Convert between RMS and peak values; transformer turns ratio; power loss in transmission lines.<\/p>\n<h3 id=\"peak-and-rms-values\">Peak and RMS Values<\/h3>\n<p>\\[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\\]<\/p>\n<p>RMS value = equivalent d.c. value that produces the same heating effect.<\/p>\n<h3 id=\"transformer-equations-ideal\">Transformer Equations (ideal)<\/h3>\n<p>\\[\\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\\]<\/p>\n<ul>\n<li>Step-up: \\(N_s > N_p\\) \u2192 voltage increases, current decreases<\/li>\n<li>Step-down: \\(N_s < N_p\\) \u2192 voltage decreases, current increases<\/li>\n<\/ul>\n<h3 id=\"power-loss-in-transmission\">Power Loss in Transmission<\/h3>\n<p>\\[P_{loss} = I^2 R_{line}\\]<\/p>\n<p>Reason for stepping up voltage: voltage \u00d7\\(n\\) \u2192 current \u00f7\\(n\\) \u2192 power loss \u00f7\\(n^2\\).<\/p>\n<div class=\"seo-warning-box\" id=\"common-mistakes\">\n<h3>Common Mistakes<\/h3>\n<ul>\n<li>\u274c Divide by \\(\\sqrt{2}\\), not by 2<\/li>\n<li>\u274c Transformers only work with a.c. \u2014 not d.c.<\/li>\n<li>\u274c Step-up transformer: voltage increases, <strong>current decreases<\/strong> \u2014 power is conserved<\/li>\n<li>\u274c Use the current in the transmission line, not the current at the load<\/li>\n<\/ul>\n<\/div>\n<hr>\n<h2 id=\"practice-questions\">Practice Questions<\/h2>\n<p><strong>Q1.<\/strong> \\(C = 500\\ \\mu\\text{F}\\), \\(R = 10\\ \\text{k}\\Omega\\), \\(V_0 = 12\\ \\text{V}\\). Find the voltage after one time constant.<\/p>\n<details><summary>Answer<\/summary><div class=\"answer-block\"><p>\\(\\tau = RC = 10000 \\times 500\\times10^{-6} = 5\\ \\text{s}\\)<\/p>\n<p>\\(V = 12 \\times e^{-1} = 12 \\times 0.37 = \\mathbf{4.4\\ \\text{V}}\\)<\/p><\/div><\/details>\n<hr>\n<p><strong>Q2.<\/strong> A satellite of mass \\(m\\) has orbital radius \\(r\\) and period \\(T\\). State the source of centripetal force and write the expression.<\/p>\n<details><summary>Answer<\/summary><div class=\"answer-block\"><p>Source: <strong>gravitational force<\/strong><\/p>\n<p>\\(F = \\dfrac{GMm}{r^2} = \\dfrac{4\\pi^2 mr}{T^2}\\)<\/p><\/div><\/details>\n<hr>\n<p><strong>Q3.<\/strong> \\(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.<\/p>\n<details><summary>Answer<\/summary><div class=\"answer-block\"><p>(a) \\(\\dfrac{1}{C} = \\dfrac{1}{4} + \\dfrac{1}{12} = \\dfrac{4}{12}\\), \\(C = \\mathbf{3\\ \\mu\\text{F}}\\)<\/p>\n<p>(b) \\(E = \\dfrac{1}{2}CV^2 = \\dfrac{1}{2} \\times 3\\times10^{-6} \\times 81 = \\mathbf{1.22\\times10^{-4}\\ \\text{J}}\\)<\/p><\/div><\/details>\n<hr>\n<p><strong>Q4.<\/strong> 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.<\/p>\n<details><summary>Answer<\/summary><div class=\"answer-block\"><p>(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}}\\)<\/p>\n<p>(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}}\\)<\/p><\/div><\/details>\n<hr>\n<p><em>Want more? Visit <a href=\"https:\/\/flowxiom.com\">flowxiom.com<\/a><\/em><\/p>\n<footer class=\"site-footer\">\n <p>Free resource by <a href=\"https:\/\/flowxiom.com\">Flowxiom<\/a> \u2014 Edexcel A-level Physics<br>\n High-frequency topics only, covering ~80% of exam marks.<\/p>\n<\/footer>\n<\/body>\n<\/html>\n\n","protected":false},"excerpt":{"rendered":"<p>This “Sprint Pack” is an essential guide for Unit 4 Physics, simplifying complex field theories and particle physics into easy-to-digest sections. Key highlights include:<\/p>\n<p>Capacitors & Fields: Step-by-step guides for discharge equations and electric\/magnetic field strengths.<\/p>\n<p>Electromagnetism: Clear “answer chains” for Lenz’s Law and Faraday’s Law questions.<\/p>\n<p>Further Mechanics: Mastering circular motion and centripetal force without common mistakes.<\/p>\n<p>Particle Physics: A straightforward map of quarks, leptons, and conservation laws.<\/p>\n<p>Practical Exam Tips: Specific advice on how to linearize graphs and interpret RMS values in AC circuits.<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"class_list":["post-45","post","type-post","status-publish","format-standard","hentry","category-exam-sprint-pack-physics-exam-sprint-pack"],"_links":{"self":[{"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/posts\/45","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/comments?post=45"}],"version-history":[{"count":2,"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/posts\/45\/revisions"}],"predecessor-version":[{"id":47,"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/posts\/45\/revisions\/47"}],"wp:attachment":[{"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/media?parent=45"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/categories?post=45"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/flowxiom.com\/index.php\/wp-json\/wp\/v2\/tags?post=45"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}