WJEC Physics for A2: Study and Rev Guide

How to use this book As examiners and former examiners we have written this new study guide to help you be aware of what is required for – and structured the content to guide you towards – success in the year 13 part of the WJEC A level examination in Physics. There are three main sections to the book: Knowledge and Understanding This rst section covers the key knowledge required for the examination. You’ll nd notes on the content of the examination units: ■ Unit 3 Oscillations and nuclei ■ Unit 4 Fields and options including the practical and data-handling skills which you will need to develop. In addition there are a number of features throughout this section that will give you additional help and advice as you develop your work: Unit introduction The key sub-sections are listed with their page references and their corresponding exam questions. Each then has a short summary giving you an essential overview of the topic, plus a revision checklist as you work through your revision process. ■ Key terms : many of the terms in the WJEC speci cation can be used as the basis of a question, so we have highlighted those terms and offered de nitions. ■ Quick re questions : are designed to test your knowledge and understanding of the material. ■ Pointer and Grade Boost : offer extra examination advice based on experience of what candidates need to do to attain the highest grades. ■ Extra questions : feature at the end of each topic or area of study providing you with further practice at answering questions with a range of dif culty. q F Fig. 4.2.1 Testing for an electric field The electricfield strength atapoint ( P ) is the forceper unitcharge experiencedby a small ‘test’chargeplaced at P . E = F q (not inDatabooklet) KeyTerm Pointer You’ll soon see the sense in using Vm −1 as a unit for E (Section 4.2.3 (d)), but, for now, note that Vm −1 = JC − 1 m − 1 =NmC −1 m −1 =NC −1 Pointer It’s perfectly possible to use a negative test charge. E is then in the opposite direction to the force on the test charge.  Typically, there is anaturallyarising downward electricfield of strength 120Vm −1 , just above theEarth’s surface. If there isawaterdroplet here,witha chargeof −2.0nC ,find: a) theelectric force itexperiences b) thedroplet’s initial acceleration if itsmass is 8.0×10 −9 kg . (Don’t forget the droplet’sweight.) QUICKFIRE QUICKFIRE QUICKFIRE 4.2 Electrostatic and gravitational fields of force These twofieldsarequitedifferent inorigin,but theyhave important features in common: the inverse square law and theapplicabilityof the conceptsof field strength and potential .Sections4.2.1–4.2.5dealwithelectric fields.Sections4.2.6–4.2.10dealwithgravitationalfields,butmorebriefly where the ideasarevery similar to thosemetearlier.Section4.2.11 isauseful summary table from the specification. 4.2.1 Electric field strength We can test for thepresenceofanelectricfieldusingapositive charge, q . We shall call thisa ‘test charge’. If q experiencesa forceproportional to its charge, q ,we say it is inanelectricfield. Forexample, the test chargewill revealan electricfieldnearaworkingVandeGraaff generator (seediagram). The electricfield strength or electric intensity , E ,atapoint isdefined inwords under Key terms ,orasanequation: E = forceon (positive) test charge test charge that is E = F q ■ units : NC −1 ( =Vm −1 )See Pointer . ■ Electricfield strength isavector (because force isavector). ■ Wedefine E as F / q because twice the test chargewill feel twice the force, so F / q doesn’tdependon q ,butonlyon thefield that q is in. ■ Weoftenneed theequation re-arrangedas… F = qE Example A test chargeof 5.0nC experiencesa forceof 0.40mN atapointnearaVan deGraaff sphere.Calculate theelectricfield strength. Answer E = F q = 0.4× 10 −3 N 5.0× 10 −9 C = 80kNC −1 (that is80kVm −1 ) 75 4.2 Electrostatic andgravitational fields of force The mole is theSIunitof amountof substance . It is theamount containing asmanyparticles (e.g. molecules) as thereare atoms in 12g of carbon-12. The Avagadro constant , N A , is thenumberof particlespermole. N A = 6.02×10 23 mol −1 KeyTerms Pointer N A is not arbitrary, but is chosen to give the simple equation M = M r 1000 kgmol −1 Pointer Remember n = Totalmass molarmass Grade boost Try to rememberwhat n , N , M , M r stand for: they’re fairly standard. QUICKFIRE QUICKFIRE QUICKFIRE  Calculate themass (in kg ) ofanoxygenmolecule ( M r =32.0 ). QUICKFIRE QUICKFIRE QUICKFIRE  What is themass of 1.00×10 25 helium molecules? ( M r =4.00) 3.3 Kinetic theory of gases Althoughwe can’t seewhat’sgoingon insideagas,usingeitheroptical ornon-opticalmicroscopes,a ‘model’ofagasasparticles (molecules) in randommotion successfullyexplains thepropertiesweobserve. Inparticular, applyingNewton’s laws to themoleculesaccounts forgas pressure andhow it dependson the containervolumeandon the temperature. 3.3.1 Molecules and moles Gases consistofparticles called molecules . In so-called monatomic gases (suchasheliumandneon) themoleculesare singleatoms. Inhydrogen, oxygen,nitrogenandmanyothergases,eachmolecule consistsofmore than oneatom ‘bonded’ together. The relativemolecularmass , M r ,ofamolecule isdefinedas M r = massofmolecule 1 12 massof 12 6 C atom M r examples:hydrogen: 1.01 ,helium: 4.00 ,oxygen: 32.0 ,nitrogen: 28.0 . (Youdon’thave to remember these!) Moles Amole (abbreviates to ‘mol’)ofmolecules isabatchof 6.02×10 23 molecules. Thenumberofmoleculespermole is called the Avogadro constant , N A . Thus N A = 6.02× 10 23 mol −1 . It follows that, ifwehave N molecules, the amount , n inmoles, is: n = N N A so N = nN A The molarmass , M , is themasspermoleof thegas… There isaneasy relationshipbetween the M r ofamoleculeand themolar mass, M : M /kgmol −1 = M r 1000 Example Howmanymoleculesare there in 20.0kg ofoxygengas? Answer molarmass= M r 1000 kgmol −1 = 32.0 1000 kgmol −1 = 0.0320kgmol −1 So,amount inmoles, n = massof gas molarmass = 20.0kg 0.0320kgmol −1 = 625mol andnumberofmolecules, N = nN A = 625mol×6.02× 10 23 mol −1 =3.76× 10 26 25 3.3Kinetic theory ofgases 4