PREFACE ii STPM Scheme of Assessment v Term of Study Paper Code and Name Theme/Title Type of Test Mark (Weighting) Duration Administration First Term 960/1 Physics Paper 1 Mechanics and Thermodynamics Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 1 to 11. 60 (26.67%) 15 Second Term 960/2 Physics Paper 2 Electricity and Magnetism Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 12 to 18. 30 2 assessment Third Term 960/3 Physics Paper 3 Oscillations and Waves, Optics and Modern Physics Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 19 to 25. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment 960/5 Physics Paper 5 Written Physics Practical Written practical test 3 compulsory structured questions to be answered. 45 (20%) 1—1 2 hours Central First, assessment Second and Third Terms 960/4 Physics Paper 4 Physics Practical School-based Assessment of Practical 13 compulsory experiments and one project to be carried out. 225 To be scaled to 45 (20%) Throughout the three terms School-based assessment 00 PRELIMS PHYSICS T2.indd 5 10/12/ 12 1 ELECTROSTATICS Bilingual Keywords *OHYNPUN! 4LUNLJHZ ,SLJ[YPJ WV[LU[PHS! 2L\WH`HHU LSLR[YPR ,SLJ[YVU]VS[ L=! ,SLR[YVU]VS[ L= ,X\PWV[LU[PHS Z\YMHJLZ! 7LYT\RHHU ZHTH RL\WH`HHU .H\ZZ» SH^! /\R\T .H\ZZ 0UZ\SH[VY! 7LULIH[ 7VPU[ JOHYNL! *HZ [P[PR 7V[LU[PHS KPMMLYLUJL! )LaH RL\WH`HHU 1 Concept Map &RXORPE¶V/DZ Q1Q2 F = ——— 4SH0r 2 (OHFWULF)LHOG E = F q Motion of point charge in ngb_hkf^e^\mkb\Û^e]E. *DXVV¶V/DZ ΣQ Φ = ——ε0 Φ = EA Ngb_hkf\aZk`^]ieZm^3 E = σ 2ε0 Q Ihbgm\aZk`^3E = ——— 4Sε0 r 2 <aZk`^]lia^k^3 For r R, E = 0 Q For r R, E = ——— 4Sε0r 2 (OHFWURVWDWLFV 10/12/18 3:27 PM Physics Term 2 STPM Chapter 14 Electric Current 14 87 Quick Check 3 1. (a) The density of copper is 8940 kg m-3 and mass of one mole of copper is 0.0635 kg. A current of 50 A fl ows in a copper cube of sides 1.0 cm. Assuming that each copper atom contributes one free electron, determine the number of free electron in the cube. (b) (i) The free electrons in copper drift when a current fl ows, what causes the free electrons to drift? (ii) Find the drift velocity of free electrons in the copper cube. 2. (a) A copper wire has a cross sectional area A. The number of free electrons per unit volume in copper is n, and the electrons have a drift velocity v. Derive an expression for the current in the wire in terms of A, n, v and the electronic charge, e. (b) A current of 4.5 A fl ows in a copper wire of cross sectional area 3.2 u m2 . If n = 8.5 ufl m-3, fi nd (i) the current density, (ii) the drift velocity of the free electrons. 14.4 Electric Conductivity and Resistivity Students should be able to: t EFSJWFBOEVTFUIFFRVBUJPOσ = ne2 t m t EFmOFSFTJTUJWJUZ BOEVTFUIFGPSNVMBρ = RA l t TIPXUIFFRVJWBMFODFCFUXFFO0INTMBXBOEUIFSFMBUJPOTIJQJ = σE t FYQMBJOUIFEFQFOEFODFPGSFTJTUJWJUZPOUFNQFSBUVSFGPSNFUBMTBOETFNJDPOEVDUPSTCZVTJOHUIFFRVBUJPO t EJTDVTTUIFFGGFDUTPGUFNQFSBUVSFDIBOHFPOUIFSFTJTUJWJUZPGDPOEVDUPST TFNJDPOEVDUPSTBOETVQFSDPOEVDUPST Learning Outcomes Resistivity 1. The resistance R of a piece of wire is s DIRECTLY PROPORTIONAL TO ITS LENGTH l s INVERSELY PROPORTIONAL TO ITS CROSS SECTIONAL AREA A, s DEPENDENT ON THE MATERIAL 2. Hence the resistance of a piece of wire R = l —A where is a constant which depends on the material of the wire, it is known as the resistivity of the material. 3. Resistivity, = RA —– l The resistivity of a material is the resistance of a unit length of the material which has a unit cross-sectional area. 4. The unit for resistivity is : m2 ——– m = : m. 5. The table below shows the resistivity of three metals. Metal Resistivity/: m Copper 1.69 u 10–8 Aluminium 3.21 u 10–8 Eureka (Constantan) 49 u 10–8 2011/P1/Q27, 2012/P1/Q26, 2013/P2/Q6,Q19, 2015/P2/ Q18(a), 2016/P2/Q6, 2017/P2/Q5,Q6 3 29 PM Summary of Key, Quantities and Units QUANTITY KUANTITI BASE QUANTITY KUANTITI ASAS SYMBOL SIMBOL UNIT UNIT Mass Jisim m kg Length Panjang l m Time Masa t s Electric current Arus elektrik l A Thermodynamic temperature Suhu termodinamik T K Amount of substance Amaun bahan n mol Other Quantities Kuantiti Lain Distance Jarak d m Displacement Sesaran s, x m Area Luas A m2 Volume Isi padu V m3 Density Ketumpatan ρ kg m–3 Speed Laju u, v m s–1 Velocity Halaju u, v m s–1 Acceleration Pecutan a m s–2 Force Daya F N Weight Berat W N Work Kerja W J Energy Tenaga E, U J Potential energy Tenaga keupayaan U J Heat Haba Q J Power Kuasa P W Pressure Tekanan p Pa Period Kala T s Frequency Frekuensi f, v Hz Wavelength Panjang gelombang λ m Electric charge Cas elektrik Q, q C Current density Ketumpatan arus J A m–2 Electric potential Keupayaan elektrik V V Electric potential difference Beza keupayaan elektrik V V Electromotive force Daya gerak elektrik ε, E V Resistance Rintangan R : Resistivity Kerintangan ρ : m Conductance Konduktans G S = :–1 Conductivity Kekonduksian σ S m–1 = :–1 m–1 Electric fi eld strength Kekuatan medan elektrik E N C–1 Capacitance Kapasitans C F Magnetic fl ux Fluks magnet φ Wb Magnetic fl ux density Ketumpatan fl uks magnet B T Self inductance Swainduktans L H Mutual inductance Induktans saling M H Reactance Reaktans X : Impedance Impedans Z : Force constant Pemalar daya k N m–1 Temperature Suhu T, θ °C 273 273 00 Summary PHYSICS T2.indd 273 Pre-U STPM Text Physics Term 2 is written based on the new syllabus prepared by the Malaysian Examinations Council (MEC) that has been implemented since 2012. The book is well designed and organised with the following features to help students understand the concepts taught.
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 198 Example 1 The diagram shows two coils P and Q arranged coaxially. The coil Q is connected in series with a battery, rheostat and switch K. Initially the switch K is closed. For each of the following changes, show on a diagram the direction of the induced current in coil P. (a) Coil P is moved closer to coil Q. (b) Coil Q is moved closer to coil P. (c) Switch K is opened. (d) The slider of the rheostat is moved to the right. Exam Tips When applying Lenz’s law, identify 1. the change 2. what is the opposite of the change? 3. what is the direction of the induced current to produce the desired effect in (2)? Solution: (a) The change is: Coil P moving closer to coil Q. Opposite of this change is: Coil P is moved away from Q. In order for coil P to be repelled from Q, the direction of the induced current in P must be opposite to that of Q. P Q Current from battery Induced current (b) The change: Coil Q moving towards coil P. Opposite to this change is: Coil Q repelled from coil P. In order for coil P to be repelled, the induced current in P must be opposite to the direction of the current in Q. P Q Current from battery Induced current (c) The change: Switch K is opened. Magnetic fl ux produced by current in Q decreases to zero. Opposite to this change: Maintain magnetic fl To achieve this, the induced current in P must be in the s ux in the same direction. ame direction as that in Q. P B Initial flux linkage Q P Flux linkage produced by induced current Q Induced current P Q Physics Term 2 STPM Chapter 14 Electric Current 14 99 mean time between collisions and m is the mass of electron. (iii) Use the expression to explain the effect of temperature increase on the conductivity of semiconductors. (b) At room temperature, the resistivity of silicon is 2.1 u 10–3 : m, and number density of charge carriers is 3.0 u 1022 m–3. Calculate the mean free time between collisions of the charge carriers. 27. (a) Electric conduction in a metal can be explained in terms of motion of free Give an estimated value at room (i) the mean speed of free electrons, (ii) the drift velocity of free electrons in (b) Show that the expression for current where σ is electrical conductivity, and E the electric fi eld, is equivalent to Ohm’s law. Hence, deduce the relation between the resistance R of a wire with its crosssectional area A. (c) A wire of length 1.50 m and crosssectional area 1.2 mm2 carries a current of 5.0 A when a potential difference of 0.24 V is applied across the wire. (i) Calculate the drift velocity of the free electrons, if the number of free electrons is 1.5 u 1029 m–3. (ii) What is the force on each electron? (iii) Calculate the resistivity of the material of the wire. 1 1. (a) Q = It = (0.20)(5 u 60) = 60 C (b) n(1.60 u 10–19) = 0.20 A n = 1.25 u 1018 s–1 2 1. A 2. C: I = nAve, n is large. 3. B 4. C 3 1. (a) Mass = Vρ M Number of free e- = number of Cu atoms = (Vρ M )(6.02 u1023 = 8.48 u (b) (i) Drift velocity is caused by the electric force on the free electrons due to the potential GLͿHUHQFHDSSOLHGDFURVVWKHFRSSHUFXEH (ii) I = nAve v = 50 _ 8.48 u (0.010)3 +(0.010)2 (1.60 u 10–19) m s-1 = 3.69 u m s-1 2. (a) Number of free e- that passes a cross section of nAve = 1.41 u A m-2 m s-1 = 100 m 4.00 u 10–3 ۙ 1.50 u 10–2ۙP 1. D 2. D 3. C 4. B 5. C 6. B 7. A 8. C 9. B: J = I A = σE E = I Aσ ANSWERS 15 103 5. of a conductor is the ratio of the potential difference V across the conductor to the Resistance, R = V I 6. Resistance is measured in ohms ( ). The resistance ) if the current in the conductor is 1 A when a potential difference of 1 V is Internal Resistance 1. The e.m.f. of a cell can be measured approximately by connecting a high resistance voltmeter across the cell as shown in Figure 15.1, with the switch S open. The voltmeter reading is approximately the e.m.f. E of the cell. 2. When the switch S in the circuit shown is closed so that the cell is connected to a resistor of resistance R, the voltmeter shows a lower reading, V. 3. The loss in potential difference = (E – V). This loss of potential difference represents the energy required to drive one coulomb of charge through the materials of the cell. 4. If I is the current in the circuit, then E – V ——– I = r, the internal resistance of the cell. 5. The internal resistance r of a cell is the resistance due to the chemical materials in the cell. 6. From E – V ——– I = r E = V + Ir Also, V = IR .............................. { Hence, E = I(R + r) .............................. | { —|: E —V = R + r ——– R E —V = 1 + r —R 7. An experimental method to determine the internal resistance of a cell is to use the circuit shown in Figure 15.1. With the switch S closed, the voltmeter reading V is noted for difference values of the resistance R. A graph of 1 —V against 1 —R is plotted as shown in Figure 15.2. 1 R – Figure 15.1 V Cell S R Info Physics The most common type of resistor used in electronic circuits is the carbon resistor. VIDEO *VUK\J[VYZ HUK 0UZ\SH[VYZ Physics Term 2 STPM Chapter 15 Direct Current Circuits iii 262 STPM Model Paper (960/2) Paper 2 Kertas 2 (1 hour 30 minutes) (1 jam 30 minit) Section A [15 marks] Bahagian A [15 markah] Instructions There are fi fteen questions in this section. Each question is followed by four choices of answer. Select the best answer. Answer all questions. Marks will not be deducted for wrong answers. Arahan Ada lima belas soalan dalam bahagian ini. Setiap soalan diikuti dengan empat pilihan jawapan. Pilih jawapan yang terbaik. Jawab semua soalan. Markah tidak akan ditolak bagi jawapan yang salah. 1. The electric potential at a point distance r from a point charge is V. The electric fi eld strength E at that point is given by Keupayaan elektrik di suatu titik pada jarak r dari suatu cas titik ialah V. Kekuatan medan elektrik E di titik itu diberi oleh A E = rV B E = r2 V C E = V r D E = V r2 2. Two point charges of +2Q and -Q are separated by a distance of r. Jarak di antara dua cas titik yang bercas +2Q dan -Q ialah r. +2Q –Q The electric fi eld potential at the point P between the charges is zero. What is the distance of the point P from the charge of +2 Keupayaan elektrik di suatu titik P yang terletak di antara dua c Q? as itu ialah sifar. Berapakah jarak titik P dari cas titik +2Q? A r 4 B r 3 C r 2 D 3r 3 3. A parallel-plate capacitor is charged to a potential difference Q V and the charge in the capacitor is . The capacitor is disconnected from the charging voltage and the dielectric is inserted between the plates of the capacitor. The new potential difference is V1 and the charge is Q1. Which of the following is correct? Suatu kapasitor plat selari dicas sehingga beza keupayaan merentasinya ialah V dan cas pada kapasitor ialah Q. Kapasitor itu ditanggalkan daripada voltan mengecas dan sekeping dielektrik diselitkan di antara plat-plat kapasitor. Keupayaan elektrik merentasi kapasitor menjadi V1 dan cas padanya Q1 A V . Antara berikut, yang manakah betul? 1 > V, Q1 > Q C V1 = V, Q1 B V1 > V, Q > Q 1 = Q D V1 < V, Q1 = Q 15 140 Important Formulae 1. Current, I = dQ—–– dt 2. Resistance, R = V —– I , V = IR 3. In series: R = R1 + R2 + R3 —–—–– 2 )V and internal resistance, r 15 STPM PRACTICE 1. A bulb which is rated 60 W, 240 V is connected to a 110 V supply. What is the power dissipated? A 12.6 W C 27.5 W B 15.0 W D 60.0 W The variation of potential difference V across a conductor with current I is as shown in the 0 0.2 1.0 0.4 0.6 0.8 1.0 I / A What is the power dissipated from the conductor when the current is 0.5 A? A 0.156 W B 0.312 W C 0.625 W D 1.25 W 3. Three resistors of resistance 2.0 :, 3.0 : and 5.0 :are connected to two cells of negligible internal resistance. The current from the cells are 3.2 A and 1.4 A as shown in the diagram. 2.0 1.4 A 5.0 3.0 3.2 A What is the potential difference across the 3.0 :resistor? A 5.4 V C 7.0 V B 6.4 V D 13.8 V 4. A potentiometer is used to determine the resistance R of a resistor as shown in the circuit. The slide wire XY is 100.0 cm long. The centre-zero galvanometer G is connected to the point P. It is balanced when the slider is at the 35.0 cm mark. When connected to the point Q, the galvanometer G is balanced when the slider is at the 63.0 cm mark. 20 X 63.0 cm Y P Q R 0 35.0 cm G G What is the value of the resistance R? A 16 : C 37 : B 36 : D 65 : Physics Term 2 STPM Chapter 15 Direct Current Circuits 10/12/18 3:44
iv Analysis of STPM Papers (2015 - 2018) Chapter 2015 2016 2017 2018 ABCABCABCABC 12 Electrostatics 2 2 12 2 1 13 Capacitors 21 2 12 12 14 Electric Current 2 121 21 3 15 Direct Current Circuits 21 3 3 2 1 16 Magnetic Fields 2 2 2 121 17 Electromagnetic Induction 3 1 21 2 2 1 18 Alternating Current Circuits 2 1 2 1 2 1 121 Total 15 2 3 15 2 3 15 2 3 15 2 3
STPM Scheme of Assessment v Term of Study Paper Code and Name Theme/Title Type of Test Mark (Weighting) Duration Administration First Term 960/1 Physics Paper 1 Mechanics and Thermodynamics Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 1 to 11. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment Second Term 960/2 Physics Paper 2 Electricity and Magnetism Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 12 to 18. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment Third Term 960/3 Physics Paper 3 Oscillations and Waves, Optics and Modern Physics Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 19 to 25. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment 960/5 Physics Paper 5 Written Physics Practical Written practical test 3 compulsory structured questions to be answered. 45 (20%) 1—1 2 hours Central assessment First, Second and Third Terms 960/4 Physics Paper 4 Physics Practical School-based Assessment of Practical 13 compulsory experiments and one project to be carried out. 225 To be scaled to 45 (20%) Throughout the three terms School-based assessment
Chapter ttttttttttttttttttttttttttttttttttttttttttttt 12 ELECTROSTATICS 1 12.1 Coulomb’s Law 2 12.2 Electric Field 6 12.3 Gauss’s Law 9 12.4 Electric Potential 15 STPM PRACTICE 12 32 QQ1 38 Chapter ttttttttttttttttttttttttttttttttttttttttttttt 13 CAPACITORS 43 13.1 Capacitance 44 13.2 Parallel – Plate Capacitors 45 13.3 Dielectrics 47 13.4 Capacitors in Series and in Parallel 49 13.5 Energy Stored in a Charged Capacitor 56 13.6 Charging and Discharging of a Capacitor 63 STPM PRACTICE 13 71 QQ2 77 Chapter ttttttttttttttttttttttttttttttttttttttttttttt 14 ELECTRIC CURRENT 81 14.1 Conduction of Electricity 82 14.2 Drift Velocity 84 14.3 Current Density 85 14.4 Electric Conductivity and Resistivity 87 STPM PRACTICE 14 96 QQ3 99 Chapter ttttttttttttttttttttttttttttttttttttttttttttt 15 DIRECT CURRENT CIRCUITS 101 15.1 Internal Resistance 102 15.2 Kirchhoff ’s Laws 108 15.3 Potential Divider 119 15.4 Potentiometer and Wheatstone Bridge 127 STPM PRACTICE 15 140 QQ4 147 vi Chapter ttttttttttttttttttttttttttttttttttttttttttttt 16 MAGNETIC FIELDS 150 16.1 Concept of a Magnetic Field 151 16.2 Force on a Moving Charge 152 16.3 Force on a Current-Carrying Conductor 156 16.4 Magnetic Fields due to Currents 159 16.5 Force between Two Current- Carrying Conductors 167 16.6 Determination of the Ratio e m 172 STPM PRACTICE 16 185 QQ5 191 Chapter ttttttttttttttttttttttttttttttttttttttttttttt 17 ELECTROMAGNETIC INDUCTION 195 17.1 Magnetic Flux 196 17.2 Faraday’s Law and Lenz’s Law 197 17.3 Self-Induction 212 17.4 Energy Stored in an Inductor 215 17.5 Mutual Induction 216 STPM PRACTICE 17 221 QQ6 228 Chapter ttttttttttttttttttttttttttttttttttttttttttttt 18 ALTERNATING CURRENT CIRCUITS 232 18.1 Alternating Current Th rough a Resistor 233 18.2 Alternating Current Th rough an Inductor 239 18.3 Alternating Current Th rough a Capacitor 243 18.4 R-C and R-L Circuits in Series 246 STPM PRACTICE 18 253 QQ7 258 STPM Model Paper 262 Summary of Key Quantities and Units 273 CONTENTS
12 1 CHAPTER ELECTROSTATICS 12 Bilingual Keywords *OHYNPUN! 4LUNLJHZ *VUK\J[VY! 2VUK\R[VY *V\SVTI»Z 3H^! /\R\T *V\SVTI +PWVSL! +^PR\[\I +PZJOHYNPUN! 4LU`HOJHZ ,SLJ[YPJ ÄLSK! 4LKHU LSLR[YPR ,SLJ[YPJ Å\_! -S\RZ LSLR[YPR ,SLJ[YPJ WV[LU[PHS! 2L\WH`HHU LSLR[YPR ,SLJ[YVU]VS[ L=! ,SLR[YVU]VS[ L= ,X\PWV[LU[PHS Z\YMHJLZ! 7LYT\RHHU ZHTH RL\WH`HHU .H\ZZ» SH^! /\R\T .H\ZZ 0UZ\SH[VY! 7LULIH[ 7VPU[ JOHYNL! *HZ [P[PR 7V[LU[PHS KPMMLYLUJL! )LaH RL\WH`HHU 1 Concept Map &RXORPE¶V/DZ Q1Q2 F = ————— 4SH0r 2 (OHFWULF)LHOG E = F q Motion of point charge in ngb_hkf^e^\mkb\ Û^e] E. (OHFWULF3RWHQWLDO 9 V = –∫ r ∞E ]x E = – ]V ]x U = qV *DXVV¶V/DZ ΣQ Φ = ——ε0 Φ = EA Ngb_hkf\aZk`^]ieZm^3 E = σ 2ε0 Q Ihbgm\aZk`^3 E = ——— 4Sε0 r 2 <aZk`^]lia^k^3 For r R, E = 0 Q For r R, E = ——— 4Sε0r2 (OHFWURVWDWLFV
12 Physics Term 2 STPM Chapter 12 Electrostatics 2 INTRODUCTION 1. What causes lightning? Why do the hairs on your hand stand when you put your hand into a new plastic bag? 2. This phenomenon is due to static charges or electrostatics. 3. In gravitation, you learned that the gravitational fi eld around a mass M is an example of inverse square fi eld because the gravitational fi eld strength is inversely proportional to the square of the distance as given by GM g = – —— r2 where G = universal gravitational constant, r = distance from the mass M. 4. In this chapter, you will fi nd that the electric fi eld of a point charge is also an example of inverse square fi eld. 12.1 Coulomb’s Law Students should be able to: VWDWH&RXORPEҋVODZDQGXVHWKHIRUPXODF = Qq 4Sε0 r2 Learning Outcome 1. There are two types of electric charge, positive (+) charge, and negative (–) charge. Like charges repel, and unlike charges attract. (a) Repulsion between two like charges (b) Attraction between two unlike charges r F F +Q1 +Q2 r F +Q1 –Q2 Figure 12.1 2. The magnitude of the force F between two point charges Q1 and Q2 is given by Coulomb’s law. 3. The force F between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between the charges. Q1 Q2 Force, F ∝ ——— .............................. { r2 4. The forces between two point charges form an action-reaction pair that obeys Newton‘s third law of motion. From { above, F = k——– Q1 Q2 r2 where k = constant of proportionality In SI, k = ——– 1 4Sε0 2010/P1/Q24, 2011/P1/Q23, 2013/P2/Q1, 2014/P2/Q1
12 Physics Term 2 STPM Chapter 12 Electrostatics 3 where ε0 is known as the permittivity of free space for charges in free space, and ε0 = 8.85 u 10–12 F m–1 Hence, Q1 Q2 ——— 4Sε0 r 2 F = 5. For charges in an insulating medium, the force F = ——— Q1 Q2 4Sεr 2 where ε is the permittivity of the medium. ε = εr ε0 where εr is a constant, known as the relative permittivity of the medium. The insulating medium is known as a dielectric. εr is also known as the dielectric constant of the medium. Table 12.1 Dielectric constant, εr Material εr Material εr Vacuum 1 Water 80 Air 1.0006 Paraffi n 2.5 Mica 5 Pyrex glass 5.6 Rubber 3 Paper 3.7 6. The dielectric constant of vacuum is exactly 1 whereas that of air is 1.0006. Hence, the difference of the electric force between two point charges in air compared to the force between the charges in vacuum is negligible. Example 1 The diagram shows two small conducting spheres hanging from fi ne nylon threads of length l. The mass of each sphere is m and the charges on the spheres are Q1 and Q2 respectively. Assume that the angle θ is small such that tan θ < sin θ, deduce an expression for x, the separation between the spheres in terms of Q1 , Q2 , l, m, g and ε0 . Solution: Let T = tension in each string ? T sin θ = F = ——— Q1 Q2 4Sε0 x2 ............................. { and T cos θ = mg ..............................| { | : tan θ = ———— Q1 Q2 4Sε0 x2 mg tan θ = sin θ (θ is small). = — x 2 yl ?—– x 2l = ———— Q1 Q2 4Sε0 x2 mg x = Q1 Q2 l (———– 2Sε ) 0 mg 1 —3 x Q1 Q2 l l θ θ x Q1 Q2 l θ θ l T T mg mg F F
12 Physics Term 2 STPM Chapter 12 Electrostatics 4 Example 2 Assume that a hydrogen atom consists of an electron of charge e– orbiting a proton of charge e+. Calculate the ratio of the electrostatic force between the electron and proton to the gravitational force between them. (Mass of electron me = 9.1 u 10–31 kg, mass of proton mp = 1.67 u 10–27 kg) Solution: Electrostatic force, Fe = ——— e u e 4Sε0 r2 where r = radius of electron orbits around the proton. Gravitational force, Fg = G——– me mp r2 F —e Fg = e2 ——–– 4Sε0 r2 r 2 (———– Gm ) e mp = (1.6 u 10–19)2 ————————————————————————————– 4S u (8.85 u 10–12) u (6.67 u 10–11) u(9.1 u 10–31) u (1.67 u 10–27) = 2.27 u 1039 Exam Tips 7KHDERYHFDOFXODWLRQVKRZVWKDWWKHIRUFHVEHWZHHQWKHHOHFWURQDQGQXFOHXVLVPDLQO\HOHFWULFLQQDWXUH Example 3 Three charged particles A, B, and C are fi xed along a straight line as shown in the diagram. Deduce an expression for the force on the particle B. Solution: Force on B due to A, FA = Q2 ——— 4Sε0 d2 in the direction BC Force on B due to C, FC = Q(2Q) ——— 4Sε0 d2 in the direction BA = 2Q2 ——— 4Sε0 d2 Resultant force on B, F = FC – FA in the direction BA = 2Q2 ——— 4Sε0 d2 – Q2 ——— 4Sε0 d2 = Q2 ——— 4Sε0 d2 d A B C +Q +Q +2Q d A B C +Q +Q +2Q FC FA
12 Physics Term 2 STPM Chapter 12 Electrostatics 5 Example 4 In the diagram shown, ABC is an isosceles triangle such that angle BAC is 90°, AB = AC = 5.0 cm. The charge q = 1.0 u 10–7 C. Calculate the force on the charge at A. Solution: The force on the charge at A due to the charge at B, FB = q2 ——— 4Sε0 r2 (r = AB) in the direction BA, since like charges repel. The force in the charge at A due to the charge at C, FC = q2 ——— 4Sε0 r2 (r = AB = AC) in the direction AC, since unlike charges attract. By using the vector sum of FB and FC , the resultant force F on the charge at A is along the x-direction, and of magnitude F = F B 2 + FC 2 FB = FC = 2FB = 2 q2 ——— 4Sε0 r2 = 2 (1.0 u 10–7)2 ——————————————— 4S u (8.85 u 10–12) u (5.0 u 10–2)2 = 5.09 u 10–2 N y x A B C +q –q +q y x A B C +q –q +q 45° F FB FC Quick Check 1 1. Two small conducting spheres S1 and S2 of the same mass are hung from points P1 and P2 on the same horizontal level using light nylon threads of the same length. The charge on S1 is Q, and that on S2 is 2Q. Repulsion between the spheres causes the threads to incline at angles of θ1 and θ2 respectively from the vertical axis. What is the value of the ratio θ1 θ2 ? A 4.0 B 2.0 C 1.0 D 0.5 2. Two small charged spheres P and Q of small mass are hung from a fi xed point X using fi ne nylon threads. When the spheres are in static equilibrium, the angle a is greater than the angle b. Which of the following statements must be correct? A The charge on P is numerically greater than that on Q. B The mass of P is less than that of Q. C The charge on P and Q are both positive. D The charge on P and Q are both negative. a b P Q X
12 Physics Term 2 STPM Chapter 12 Electrostatics 6 3. The charge on the uranium nucleus is 1.5 u 10–17 C and the charge on a proton is 1.6 u 10–19 C. What is the electrostatic force between a uranium nucleus and a proton when separated by a distance of 1.0 u 10–13 m? A 2.2 u 10–20 N C 2.2 N B 2.2 u 10–13 N D 2.2 u 1010 N 4. The diagram shows two identically charged spheres with charge +Q in a static equilibrium at the ends of threads each of length l. The weight of each sphere is W and the spheres are separated by a distance x. The angle θ is not small. Deduce an expression for Q2 in terms of x, l, W and ε0 , the permittivity of free space. 5. In the diagram shown, ABC is a right-angled triangle. If Q = 2 u 10–7 C, and a = 10.0 cm, A B C a a +Q +Q –Q (a) calculate the resultant force on the charge at B, (b) what is the direction of the resultant force? 12.2 Electric Field Students should be able to: H[SODLQWKHPHDQLQJRIHOHFWULFÀHOGDQGVNHWFKWKHÀHOGSDWWHUQIRUDQLVRODWHGSRLQWFKDUJHDQHOHFWULFGLSROHDQGDXQLIRUPO\ FKDUJHGVXUIDFH GHÀQHWKHHOHFWULFÀHOGVWUHQJWKDQGXVHWKHIRUPXODE = F q GHVFULEHWKHPRWLRQRIDSRLQWFKDUJHLQDXQLIRUPHOHFWULFÀHOG Learning Outcomes 1. An electric fi eld exists if there is an electrostatic force on a charged particle. (a) Electric fi eld around a positive point charge (c) Electric fi eld for a uniform positively charged surface (b) Electric fi eld around a dipole (d) Electric fi eld for a uniform negatively charged surface + + – Figure 12.2 x +Q +Q l l e 2017/P2/Q1
12 Physics Term 2 STPM Chapter 12 Electrostatics 7 2. An electric fi eld can be represented in a fi gure by lines of forces. Each line of force represents the path of a free charge in the fi eld. 3. The direction of an electric fi eld is the direction of motion of a free positive test charge. 4. Regions where the lines of forces are close to each other represent stronger electric fi eld. Figure 12.2 shows some electric fi eld patterns. 5. The electric fi eld strength (or electric fi eld intensity) E at a point in an electric fi eld is defi ned as the force per unit charge on a positive charge placed at that point. That is electric fi eld strength, F E = —q 6. Electric fi eld strength E is a vector quantity. Its direction is the direction of the force on a free positive charge. The unit for E is N C–1 or V m–1. Motion of a Point Charge in a Uniform Electric Field 1. Figure 12.3(a) shows a particle of charge +q moving with a velocity v in the direction of the electric fi eld E. The force F = qE which is in the direction of v causes the particle of charge +q to accelerate along the direction of the electric fi eld E. Acceleration, a = F m = qE m (m = mass of charged particle) v q F eE v F qE E v F +q E Parabolic path (a) (b) Figure 12.3 2. A negative point charge such as an electron moving with a velocity v in the direction of E experiences a force F = eE in the direction opposite to that of E. This force causes the electron to decelerate. 3. Figure 12.3(b) shows a particle of charge +q entering a uniform electric fi eld E with a velocity v which is perpendicular to E. The force F = qE on the particle is constant and in the direction of E. This is identical to the motion of a projectile under gravity. s 4HE HORIZONTAL COMPONENT OF VELOCITY OF THE PARTICLE REMAINS CONSTANT v. s 4HE VERTICAL COMPONENT OF VELOCITY INCREASES WITH AN ACCELERATION a = F m = qE m s 4HE SUBSEQUENT PATH OF THE PARTICLE IN THE ELECTRIC l ELD IS A PARABOLA 2014/P2/Q2, 2016/P2/Q3 VIDEO ,SLJ[YPJ-PLSKZ HUK,SLJ[YPJ-PLSK 3PULZ
12 Physics Term 2 STPM Chapter 12 Electrostatics 8 Example 5 The diagram shows three ink drops in a bubble-jet printer that have the same mass but different charge entering a uniform electric fi eld E with the same velocity v perpendicular to the electric fi eld. Sketch the paths of the particles in the electric fi eld. Explain for the differences in the paths. Solution: Differences in the paths: s 4HE CHARGE nq is defl ected in the direction opposite to that of the positive charges. Electric forces on the charge +q and +2q are in the direction of E, but the force on –q is in the opposite direction. s 4HE CHARGE q is defl ected more than the charge +q. The force on +2q is (2q)E, but the force on +q is qE. Quick Check 2 1. Draw the electric fi eld pattern around (a) a positive point charge, (b) a negative point charge, (c) a dipole, (d) two positive point charges separated by a distance. 2. The diagram shows the electric fi eld between two point charges P and Q. P Q (a) Identify the type of charge on (i) P and (ii) Q. (b) Which of the charges has the greater magnitude? E v v v –q 2q q E v v q v q 2q
12 Physics Term 2 STPM Chapter 12 Electrostatics 9 3. (a) In diagram (a) the velocity of a positive point charge and negative point charge in the uniform electric fi eld at a particular instant is v. E v v E v v (a) (b) Draw the paths of the point charges in the electric fi eld. State the difference in the subsequent motion of the charges. (b) Diagram (b) shows a positive point charge and negative point charge entering a uniform electric fi eld E with the same velocity. Sketch the paths of the charges in the fi eld. 12.3 Gauss’s Law Students should be able to: VWDWH*DXVVҋVODZDQGDSSO\LWWRGHULYHWKHHOHFWULFÀHOGVWUHQJWKIRUDQLVRODWHGSRLQW FKDUJHDQLVRODWHGFKDUJHGFRQGXFWLQJVSKHUHDQGDXQLIRUPO\FKDUJHGSODWH Learning Outcome 1. Figure12.4 shows an electric fi eld E at an angle of θ from the normal to a small area ('A). The electric fl ux through the area ('A) is Φ = (component of E normal to the surface) u (area) = (E cos θ )('A) If the electric fi eld E is normal to the area ('A), then electric fl ux ) = E('A) 2. Gauss‘s law states that for a closed surface, ε0 ) = ¦Q Where ¦Q is the algebraic sum of the charges inside the closed surface, and ) is the electric fl ux through the closed surface. 3. Gauss‘s law is used to derive the formula for the electric fi eld strength of different electric fi elds. Electric Field due to a Point Charge 2007/P1/Q25, 2010/P1/Q24 1. In section 12.2, the formula for the electric fi eld strength E due to a point charge Q was derived using Coulomb’s law. This formula can also be derived using Gauss‘s law. 2. To derive the formula for the electric fi eld strength E at a point P, distance r from the positive point charge +Q, a suitable close surface known as Gaussian surface has to be chosen. Normal E Area e 6A E Area 6A (a) Electric fl ux (b) Electric fl ux ) = (E cos)('A) ) = E('A) Figure 12.4 2016/P2/Q1, 2018/P2/Q1, Q2, Q18
12 Physics Term 2 STPM Chapter 12 Electrostatics 10 3. The electric field pattern due to a positive point charge +Q is radial in nature as shown in Figure 12.5. +Q r P E Figure 12.5 4. A suitable Gaussian surface is a sphere of radius r and centred at the charge. The electric field E is normal to the surface of the sphere anywhere on the sphere surface. 5. Hence, the electric flux, ) = E(surface area of sphere) = E(4Sr2 ) 6. Using Gauss‘s law, ε0 ) = ¦Q (¦Q = Q) ε0 (4Sr2 E) = Q E = Q ——— 4Sε0 r2 7. A test charge +q at the point P experiences an electrostatic force, F = qE = qQ ——— 4Sε0 r2 which is Coulomb’s law. Hence, Gauss‘s law is equivalent to Coulomb’s law. 8. Conversely, the expression for E can be derived using Coulomb’s law. Force on a charge q at a distance r from a positive point charge Q is F = Qq ——— 4Sε0 r2 From the definition of E = F q , electric field at a distance from point charge Q is E = 1 Qq ——— 4Sε0 r2 2 1 q = Q ——— 4Sε0 r2 Technique of Using Gauss’s Law 1. Make a sketch of the electric field pattern. 2. The field pattern helps us to decide a suitable Gaussian surface to be chosen. 3. Try to select a Gaussian surface where E is always normal to the surface. 4. Apply Gauss‘s law, ε0 ) = ¦Q Try to identify the steps mentioned above in the following derivations. Exam Tips 7RGHWHUPLQHWKHGLUHFWLRQRIDQHOHFWULFðHOG,, place a positive charge at the point. Direction of , = direction of force on the positive charge.
12 Physics Term 2 STPM Chapter 12 Electrostatics 11 Charged Sphere 1. When a conductor is charged, the charge stays only on the surface. This is because like charges repel, and since charges are able to move in a conductor, the force of repulsion between like charges pushes the charges to the surface of the conductor. 2. Figure 12.6 shows the distribution of charges on a solid sphere (Figure 12.6(a)), and on a hollow sphere (Figure 12.6(b)). Since the radius of the curvature of sphere is uniform, the charges are distributed uniformly on the surface of the sphere. 3. The electric field radiates normally from the surface of the sphere. 4. To derive the formula for the electric field strength E at a point P, distance r from the centre O of the sphere, a suitable Gaussian surface is a sphere S1 of radius r centred at O. 5. Using Gauss‘s law, ε0 ) = ¦Q ¦Q = Q for r R, radius of sphere ε0 (E u 4Sr2 ) = Q E = Q ——— 4 Sε0 r2 for r R .............................. 6. For a point D inside the sphere, r < R, consider the Gaussian surface S2 , a sphere centred at O and of radius r (r < R). Using Gauss‘s law, ε0 ) = ¦Q ¦Q = 0 as there is no charge in the Gaussian surface S2 . ε0 (E u 4Sr2 ) = 0 E = 0 for r < R .............................. E +Q r 0 R E = –––––– , (r R) Q 4πε0r 2 Figure 12.7 7. Formulae and above are also true for a hollow sphere. Figure 12.7 shows the variation of the magnitude of E with distance r from the centre of a charged sphere. O D S2 S1 P Q + + + + + + + + Q + + + + + + + + O (a) Solid sphere (b) Hollow sphere Figure 12.6
12 Physics Term 2 STPM Chapter 12 Electrostatics 12 Uniformly Charged Plate 1. Figure 12.8 shows a uniformly charged large non-conducting plate. A free positive charge placed near the plate would be repelled and moved in a direction perpendicular to the plate. Hence the electric fi eld E on both sides of the plate is uniform and perpendicular to the plate. E E A A Q A E E A A x (a) (b) Figure 12.8 2. The surface charge density of the charged plate is σ. To fi nd the electric fi eld E at a distance x from the plate, the Gaussian surface used is a right-cylinder of cross-sectional area A and length 2x. 3. No electric lines of force pass through the curved surface of the cylinder. The electric fl ux through the two ends of the right-cylinder is ) = (EA + EA) = 2EA The charge enclosed by the right-cylinder ¦Q = σA 4. Applying Gauss’s law ε0 ) = ¦Q ε0 (2EA) = σA Electric fi eld strength, E = σ ––—2ε0 5. The electric fi eld strength E is independent of the distance x from the charged plate. Example 6 A gold nucleus has a radius of 5 u 10–15 m and contains 79 protons. Assuming that the protons are uniformly distributed in the nucleus, fi nd the electric fi eld strength at the surface of the nucleus. (Charge of proton = 1.6 u 10–19 C) Solution: The electric fi eld strength at the surface of a charged sphere, E = Q ——— 4Sε0 R2 Hence, at the surface of gold nucleus E = 79 u (1.6 u 10–19) ————––——————— 4S(8.85 u 10–12)(5 u 10–15)2 = 4.55 u 1021 V m–1 Info Physics ,QVXODWLRQ RI WKH DLU EUHDNV GRZQ DQG HOHFWULF FXUUHQW ÁRZV ZKHQ WKH HOHFWULF ÀHOG VWUHQJWK H[FHHGV09P–17KLVLVZKDWKDSSHQVZKHQ OLJKWQLQJ RFFXUV 'R QRW VHHN VKHOWHU XQGHU D WDOOWUHHGXULQJDVWRUP7KHGLVWDQFHLVVPDOOHVW EHWZHHQWKHFKDUJHGFORXGVDQGWKHWUHHWRS,W LVPRVWOLNHO\WKDWWKHHOHFWULFÀHOGVWUHQJWKDERYH WKHWUHHZRXOGH[FHHG09P–1 2014/P2/Q16
12 Physics Term 2 STPM Chapter 12 Electrostatics 13 Example 7 A charge Q is fi xed at the centre of a hollow, conducting sphere that is initially uncharged. What is the net charge on (a) the inner surface, (b) the outer surface of the sphere? +Q + + + + + + + – – – – – – – – Solution: The charge +Q induces (a) a charge of –Q on the inner surface, (b) a charge of +Q on the outer surface. Example 8 (a) Deduce an expression for the electric fi eld strength on the surface of a charged conducting sphere in terms of the charge density V. (b) The electric fi eld strength on the surface of the Earth is 130 V m–1 vertically downwards. What is the mean charge density on the surface of the Earth? What is the sign of the charge on the Earth‘s surface? Solution: (a) The electric fi eld strength on the surface of sphere with charge Q and radius R is E = Q ——— 4Sε0 R2 = Q ——– 4SR2 — 1 ε0 Q ——– 4SR2 = Q —A = σ = —σ ε0 (b) Using E = —σ ε0 , charge density on the surface of the Earth σ = ε0 E = (8.85 u 10–12)(130) = 1.15 u 10–9 C m–2 Since the direction of the electric fi eld is vertically downwards, the charge on the Earth‘s surface is negative. 2018/P2/Q2
12 Physics Term 2 STPM Chapter 12 Electrostatics 14 Example 9 A capacitor is constructed from two concentric spheres. The inner sphere is charged positively and the outer sphere is being earthed as shown in the diagram. Sketch three graphs, using the same axes, to show how the electric fi eld E due to (a) the charges on the inner sphere alone, (b) the charges on the outer sphere alone, (c) the combination of all charges, varies with distance r from the centre. Solution: Explanation (a) E1 = Q ——— 4Sε0 r2 r a = 0 r < a (b) E2 = –Q ——— 4Sε0 r2 r b (charge of –Q is induced = 0 r < b on the outer sphere) (c) E = E1 + E2 For r b, E = 0 because E1 and E2 are of the same magnitude but in opposite directions. Quick Check 3 + + + + + + + + + + + + a b E(r) –b b r –a a (a) and (c) 0 (b) (a) (c) 1. The diagram shows the electric fi eld round an electric dipole. Which of the following A, B, C and D has no net electric fl ux? A C B D 2. Each charge of magnitude Q is at the corners X, Y and Z of a square as shown in the figure. Calculate the magnitude and direction of the electric fi eld strength at the point P. –Q +Q +Q Y Z X P a a a a 3. (a) Write an expression for the electric fi eld intensity E on the surface of a conducting sphere of radius r and charge Q. (b) Assuming that the uranium nucleus 238 92U is a sphere of radius 7 u 10–15 m, fi nd the electric fi eld intensity on its surface. 4. Two point charges of opposite sign and same magnitude 2.0 u 10–7 C are separated by a distance of 15 cm. (a) Find the magnitude and direction of the electric fi eld strength E at the mid-point between the charges. (b) What is the magnitude and direction of the force on an electron placed at the mid-point? 5. Two small point charges +Q1 and +Q2 are placed at X and Y respectively as shown in the fi gure. r X Y +Q1 +Q2
12 Physics Term 2 STPM Chapter 12 Electrostatics 15 (a) Draw the direction of the electric fi eld at Y which is due to the charge +Q1 . (b) Write the expression of the electric fi eld strength at Y due to the charge +Q1 . (c) What is the force caused by +Q1 on +Q2 ? (d) Sketch the pattern of the electric fi eld. 6. The diagram shows a charge Q = 1.0 u 10–7 C at the centre of a spherical cavity of radius 3.0 cm inside a metal block. Q A B Metal Cavity Use Gauss‘s law to calculate the electric fi eld strength E (a) at the point A, distance 1.5 cm from the charge Q, (b) at the point B, distance 4.0 cm from the charge Q. 7. The diagram shows three point charges at the corners of a triangle. Find the electric fi eld strength, magnitude and direction at the point P in terms of q and a. A P C a B a +q +q +2q 8. The diagram shows a sphere S of mass 1.0 u 10–3 g and charge 2.0 u 10–8 C hanging from a thread which makes an angle θ = 30° with the charged non-conducting plane. + + + + + + θ S Charged plane Thread Calculate the charge density of the plane. 12.4 Electric Potential Students should be able to: GHÀQHHOHFWULFSRWHQWLDO XVHWKHIRUPXODV = Q 4Sε0 r H[SODLQWKHPHDQLQJRIHTXLSRWHQWLDOVXUIDFHV XVHWKHUHODWLRQVKLSE = – dV dr XVHWKHIRUPXODU = qV Learning Outcomes 1. The electric potential V at a point in an electric fi eld is the work done to bring a unit positive charge from infi nity to the point. 2. The electric potential at infi nity is zero. Electric potential is a scalar quantity. Unit for electric potential is the volt (V). 2007/P2/Q4, 2012/P1/Q22
12 Physics Term 2 STPM Chapter 12 Electrostatics 16 Electric Potential due to a Point Charge 2013/P2/Q4, 2015/P2/Q2, 2017/P2/Q18 1. Figure 12.9 shows a test charge +q at a point A in the electric field produced by a point charge +Q. The distance of the test charge +q from the point charge +Q is x. Figure 12.9 F x Dx +Q B +q A Point charge Test charge Direction of E Positive direction 2. The force on the test charge +q is F = qQ ——— 4Sε0 x 2 in the same direction as the electric field E. 3. The work done to bring the test charge +q from A to B through a small distance δx is δW = (F)(– δx) = – Fδx where – δx is the displacement which is opposite to the positive direction. 4. Hence, the total work done to bring the test charge +q from infinity to a point distance r from the point charge +Q is W = –qQ ——— 4Sε0 x ∫ 2 dx r ∞ = [ ] r ∞ qQ ——— 4Sε0 x = qQ ——— 4Sε0 r From the definition, the electric potential at a point distance r from a point charge Q, V = —W q V = Q ——— 4Sε0 r 5. All points at the same distance from the point charge Q have the same electric potential V. These points form a surface known as the equipotential surface. 6. An equipotential surface is a surface where all points on the surface have the same electric potential. 7. Figure 12.10 shows the equipotential surfaces around a point charge. These equipotential surfaces consist of concentric spheres. Figure 12.10
12 Physics Term 2 STPM Chapter 12 Electrostatics 17 Electric Potential due to a Charged Sphere 2013/P2/Q2 1. Figure 12.11 shows the electric field around a charged sphere, and around a point charge. The electric field outside the charged sphere is identical to the electric field due to a point charge. Figure 12.11 (a) Charged sphere (b) Point charge Q R + + + + + + + + +Q 2. Hence, the electric potential V at a point distance r R, (R = radius of sphere) from the centre of the charged sphere, V = Q ——— 4Sε0 r for r R For points inside the charged sphere, E = 0 Hence, the electric potential V is constant and given by V = Q ——— 4Sε0 R for r R 3. Figure 12.12(a) shows the variation of the electric field strength E with distance r from the centre of the charged sphere. Figure 12.12(b) shows the variation of the electric potential V with distance r. Effect of a Neighbouring Conductor 1. Figure 12.13 shows the variation of the electric potential V with distance r for the electric field around a charged sphere (a) before and (b) after an uncharged conductor is brought close to the charged sphere. 2. Electrostatic induction causes the end B of the conductor to be charged negative (–Q), and the end C to be charged positive (+Q). The conductor remains uncharged as (–Q) + (+Q) = 0. 3. The electric potential of the uncharged conductor is constant. 4. The effect is that the electric potential on the charged sphere decreases. Figure 12.12 –R R +Q V E r r 0 –R 0 R (a) (b) Q –––––– Q –––––– Q E = –––––– Q V = –––––– Figure 12.13 V r A B C 0 + – + (b) After Charged sphere Uncharged conductor (a) Before
12 Physics Term 2 STPM Chapter 12 Electrostatics 18 5. Figure 12.14 shows the effect on V, the electric potential when an earthed conductor is brought close to the charged sphere A. 6. The positive charge at C is discharged by electron from the Earth. Hence, the earthed conductor is negatively charged, but its electric potential is equal to the Earth‘s electric potential, which is zero. 7. The decrease in the electric potential on the charged sphere A is much bigger compared to when the neighbouring uncharged conductor is not earthed. Electric Potential Energy, U 2009/P1/Q24, 2016/P2/Q2 1. The electric potential energy U of a charge q at a point in an electric fi eld is the work done to bring the charge q from infi nity to the point. 2. The electric potential V at a point had been defi ned as the work done per unit positive charge. Hence, the electric potential energy U of a charge q is U = qV 3. At a point distance r from a point charge Q, the electric potential V = Q ——— 4Sε0 r Hence, the electric potential energy, U of a charge q at a point distance r from point charge Q is U = qV U = qQ ——— 4 Sε0 r 4. The electric potential energy of a system of charges is the total work done to bring the charges from infi nity to form the system. Example 10 The diagram shows a system of three point charges q1 , q2 and q3 . The distance between the charges are as shown. (a) Derive the expression for the electrical potential energy of the system. (b) If q1 = +2.0 u 10–7 C, q2 = –1.0 u 10–7 C, q3 = +3.0 u 10–7 C, and r12 = r13 = r23 = 10.0 cm, calculate the electric potential energy of the system. Solution: (a) Work done to bring the fi rst charge q1 from infi nity to the position shown, U1 = 0. Reason: The charge q1 does not experience any attractive of repulsive force as there were no other charges present. Figure 12.14 V 0 + r B C A – After Before Earthed conductor Charged sphere q1 q3 q2 r13 r23 r12
12 Physics Term 2 STPM Chapter 12 Electrostatics 19 Work done to bring the second charge q2 to the point distance r12 from q1 is U2 = q1 q ———– 2 4Sε0 r12 Work done to bring the third charge q3 to the point distance r13 from q1 and r23 from q2 is U3 = q1 q ———– 3 4Sε0 r13 + q2 q ———– 3 4Sε0 r23 Hence, electric potential energy of the system U = U1 + U2 + U3 = 0 + q1 q ———– 2 4Sε0 r12 + ( q1 q ———– 3 4Sε0 r13 + q2 q ———– 3 4Sε0 r23 ) = q1 q ———– 2 4Sε0 r12 + q1 q ———– 3 4Sε0 r13 + q2 q ———– 3 4Sε0 r23 (b) q1 = +2.0 u 10–7 C, q2 = –1.0 u 10–7 C, q3 = +3.0 u 10–7 C, and r12 = r13 = r23 = 10.0 cm Electrical potential energy of the system U1 = q1 q ———– 2 4Sε0 r12 + q1 q ———– 3 4Sε0 r13 + q2 q ———– 3 4Sε0 r23 = (+2.0 u 10–7)(–1.0 u 10–7) ——–————————— 4S(8.85 u 10–12)(0.100) + (+2.0 u 10–7)(+3.0 u 10–7) ——–————————— 4S(8.85 u 10–12)(0.100) + (–1.0 u 10–7)(+3.0 u 10–7) ——–————————— 4S(8.85 u 10–12)(0.100) J = +8.99 u 10–4 J Example 11 An insulated metal sphere of radius 0.10 m is given a charge. The work done to bring a small charge from a distant point to a point 0.50 m from the centre of the charged sphere is W, and at its fi nal position, the force on the small charge is F. If the small charge is only brought to distance 1.00 m from the centre of the charged sphere, what is the magnitude of the work done, and the force on the small charge at its fi nal position? Solution: At a point P distance r > R from the sphere of charge Q, V = Q ——— 4Sε0 r R = radius of sphere Work done to bring a small charge q from infi nity to P W = qV = qQ ——— 4Sε0 r When r = 0.50 m, W = Q ———–— 4Sε0 (0.5) when r = 1.00 m, W1 = qQ ———–— 4Sε0 (1.0) = W —2
12 Physics Term 2 STPM Chapter 12 Electrostatics 20 Force on charge q at P, F = qQ ——— 4Sε0 r2 when r = 0.50 m, F = qQ ———––— 4Sε0 (0.5)2 when r = 1.00 m, F1 = qQ ———––—– 4Sε0 (1.00)2 = F —4 Example 12 The diagram shows three points X, Y and Z forming an equilateral triangle of sides s in a uniform electric fi eld of strength E. A unit positive charge is moved from X to Y, Y to Z, and back to X. What is the work done against the electric forces in moving the charge (a) from X to Y, (b) Y to Z, (c) Z back to X? Solution: Work done against electrical force W = (force on charge) u (displacement in the direction against the electric fi eld) = (qE)(displacement) (q = 1) (a) From X to Y, W = E u (0) (since displacement is perpendicular to E) = 0 (b) From Y to Z, W = E u (–s sin 60°) = – Es sin 60° (c) From Z to X, W = E u (s sin 60°) = Es sin 60° Example 13 The diagram shows two charges of –q and +2q at the point B and C. If q = 1.0 u 10–7 C, calculate (a) the electric potential at D, (b) the electric potential at A, (c) the work done to move a charge of 3.0 u 10–7 C from A to D. Solution: (a) The electric potential at D is the sum of the electric potential due to the charge –q at B and the charge +2q at C. VD = –q ———— 4Sε0 (BD) + 2q ———— 4Sε0 (DC) = q ———–– 4Sε0 (BD) = 1.0 u 10–7 ——————–——————— 4S u 8.85 u 10–12 u (5.0 u 10–2) = 1.8 u 104 V s Z E X Y A B C –q D +2q 10 cm 10 cm 10 cm Exam Tips Electric potential is a scalar quantity.
12 Physics Term 2 STPM Chapter 12 Electrostatics 21 (b) Electric potential at A, VA = –q ———— 4Sε0 (BA) + 2q ———— 4Sε0 (CA) = q ———–– 4Sε0 (BA) = 1.0 u 10–7 ——————–————– 4S u 8.85 u 10–12 u (0.10) = 9.00 u 103 V (c) Work done to move a charge q from A to D, W = q(VD – VA) = (3.0 u 10–7)(1.80 u 104 – 9.00 u 103 ) = 2.7 u 10–3 J Quick Check 4 1. The work done to move a charge of 3 C from infi nity to a point X in an electric fi eld is 15 J. The electric potential at X is A 45 V B 15 V C 5 V D 0.2 V 2. Four point charges are arranged at the corner of a square as shown. +Q –Q O –Q +Q Which of the following is true about the electric field strength E, and the electric potential V at the point O in the midde of the square? A B C D E Not zero Not zero Zero Zero V Zero Not zero Not zero Zero 3. In the diagram, the point charge Q1 causes an electric potential of 50 V and an electric fi eld strength of 30 V m–1 at P, and the point charge Q2 separately causes an electric potential of 110 V and a fi eld strength of 40 V m–1 at P. P Q1 Q2 Which of the following gives the possible values of potential and fi eld strength at P due to the combined action of Q1 and Q2 ? Potential/ V Field strength/ V m–1 A 160 70 B 160 50 C 60 50 D 60 10 4. An electron enters an electric fi eld E = 200 V m-1 at the point X and moves to Y as shown in the diagram. E = 200 V m–1 Y X Electron 4.0 m 3.0 m What is the change in kinetic energy of the electron? A – 1.28 u–J C + 9.60 u–J B – 9.60 u–J D +1.28 u–J
12 Physics Term 2 STPM Chapter 12 Electrostatics 22 5. A and B are two identical conducting spheres, each carrying a charge +Q. They are placed in vacuum with their centres distance d apart as shown in the figure. d A B Explain why force F between them is not given by the expression F = Q2 ——— 4Sε0 d2 6. In a helium nucleus, the distance between the two protons is 1.5 u 10–15 m. (a) Find the electrostatic force between the protons. (b) How much work is done against this force to bring two protons from infinity to this separation? 7. A water droplet is spherical and carries a charge of 3.0 u 10–11 C. The electric potential on its surface is 500 V. (a) Calculate the radius of the droplet. (b) If two water droplets of the same radius and carrying the same charge combine to form a bigger spherical droplet, what is the electric potential on the surface of the big droplet? 8. The charge density on the surface of the Earth is 2.0 u 10–19 C m–2. (a) What is the electric potential on the Earth‘s surface? (b) What is the electric field strength at a point close to the Earth‘s surface? (Radius of the Earth = 6.4 u 106 m) 9. Two points A and B are at a distance of 4.0 m and 3.0 m respectively from a point charge Q = 1.0 u 10–6 C as shown in the figure. B Q A 3.0 m 4.0 m (a) What is the electric potential at (i) A, (ii) B? (b) Calculate the work done to move a charge of 2.0 u 10–6 C from B to A. 10. (a) (i) Give the expression for the electric field strength E at a distance d from a point charge Q. (ii) The electric field due to an isolated charged sphere is the same as that due to a point charge. Explain how far this statement is true. (b) A helium nucleus (4 2 He) travelling at a speed of 7.0 u 106 m s–1 is used to bombard a gold foil ( 79 Au) 197 . By taking into account the electrostatic repulsion, calculate the closest distance between the helium nucleus to the gold nucleus. Discuss the effect of the gravitational force in your answer. Potential Difference 2017/P2/Q2 1. The potential difference V between two points in an electric field is the work done per unit charge to bring a positive charge from one point to the other against the electric field. 2. The unit for potential difference is the volt, V. 3. If the potential difference between two points is V, that means the work done to move charge q from one point to the other is W = qV
12 Physics Term 2 STPM Chapter 12 Electrostatics 23 4. From the principle of conservation of energy, the work done is equal to the change in the electric potential energy. Change in electric potential energy, 'U = W = qV. Relationship between E and V 1. Figure 12.15 shows a charge q at the point A in an electric field of intensity E. Force on the charge, F = qE Dx x E B A q F = qE 0 Positive direction Figure 12.15 2. When the charge, q is moved a small distance Gx against the electric field, the work done GW = Force u Displacement = F(–Gx) = – qEGx If GV = Potential difference between the points A and B, then, GV = Work done per unit charge = —– GW q = – qEGx ——– q = – EGx E = – —– dV dx E = –(Potential gradient) Hence, the unit for the electric field intensity can also be expressed as V m–1. 3. Alternatively, from E = – —– dV dx dV = – E dx V = – ∫ r ∞ E dx = –(area under E – x graph) 4. Figure 12.16 shows the variation of the electric potential V with distance x for two different electric fields. If the magnitude of the gradient of graph is bigger, it denotes a stronger field. Figure 12.16 x V 0 Stronger field Weaker field Exam Tips , = – dV–—dx Do not leave out the negative sign.
12 Physics Term 2 STPM Chapter 12 Electrostatics 24 5. The formula E = – —– dV dx is used to obtain the electric fi eld strength between two charged conducting plates. Figure 12.17 + – V d If V = Potential difference between the plates d = Plate separation Then, E = ——————————— Potential difference Distance between plates = — V d Example 13 The electric potentials V at distances x from point P along a line PQ are given below. V/V 13 15 18 21 23 x/m 0.020 0.030 0.040 0.050 0.060 Find the component of the electric fi eld along PQ for x = 0.04 m. Solution: Using E = – —– dV dx , E = – (21 – 18) ——————— (0.050 – 0.040) = –300 V m–1 The negative sign shows that the direction of E is towards P. Example 14 The diagram shows a graph of the electric potential V against distance x from a fi xed point. Sketch a graph to show the variation of the electric fi eld intensity E against distance x. V/V x/m 3 4 5 3 2 1 2 1 0 –1 –1 –2 –5 –4 –3 –2
12 Physics Term 2 STPM Chapter 12 Electrostatics 25 Solution: Using E = – —– dV dx , –4 x < –2, E = – —3 2 = – 1.5 V m–1 –2 x < 1, E = 0 1 x < 3, E = –(–5 —2 ) = 2.5 V m–1 3 x < 4, E = –(—2 1 ) = –2 V m–1 Example 15 Electrons ’leak‘ from the surface of many stars so that such stars acquire positive charges. The charging stops when the charge on the star is so large that protons on the surface also begin to be repelled. This occurs when the sum of the gravitational potential energy and the electric potential energy of a proton near the surface is zero. (a) Write down the equation relating these two energies. (b) Show that, in the steady state, the maximum charge carried by a star of given mass is independent of its radius. (c) Calculate the maximum charge on the Sun whose mass is 2.0 u1030 kg. Solution: (a) Gravitational potential energy + Electric potential energy = 0 – GmM ——– r + qQ ——– 4Sε0 r = 0 .............................. where m = mass of proton q = charge of proton M = mass of star Q = charge of star r = radius of star (b) From : qQ ——– 4Sε0 r = GmM ——– r Charge on star, Q = 4Sε0 GmM ———–— q (c) Q = 4Sε0 GmM ———–— q = 4S u (8.85 u 10–12) u (6.67 u 10–11) u (1.67 u 10–27) u (2 u 1030) ———–———————————————————————— 1.6 u 10–19 = 155 C E/V m–1 –5 –4 –3 –2 –1 –1 0 –1.5 –2 2.5 2 1 3 1 2 3 4 5 x/m
12 Physics Term 2 STPM Chapter 12 Electrostatics 26 Example 16 A light conducting sphere X of radius R and charge Q is hung between two parallel metal plates which are connected to a cell B. The separation between the plates is 2.0 cm and the sphere experiences an electric force of 4 u 10–4 N. B X (a) What is the charge density (i) inside the sphere, (ii) on the surface of the sphere in terms of Q and R? (b) The sphere X then touches an uncharged sphere Y of twice the radius. The sphere Y is then removed. What is the fi nal electric force on sphere X? (c) The sphere X is recharged to a charge Q again and the separation between the plates is increased to 3.0 cm. What is the fi nal electric force on sphere X? (d) Another two cells which are identical to cell B are connected in series with cell B. The charge on sphere X remains as Q, and the separation between the plates is 3.0 cm. What is the fi nal electric force on the sphere X? Solution: (a) (i) Inside the sphere X, there is no charge, as all the charges are on its surface. Hence, charge density = 0. (ii) Charge on surface of sphere = Q Charge density = Charge ——— Area = Q——4SR2 (b) When spheres X and Y touch each other, charge fl ow between X and Y until the electric potential of both spheres are equal. If Q1 and Q2 are the fi nal charge on X and Y respectively, ———Q1 4Sε0 R = ———— Q2 4Sε0 (2R) Q2 = 2Q1 Since charge is conserved, Q = Q1 + Q2 = Q1 + 2Q1 Final charge on X, Q1 = Q —3 Electric force on sphere X when the charge is Q F = 4.0 u 10–4 N = QE Final electric force on sphere X, F1 = ( Q —3 )E = —1 3 u 4.0 u 10–4 = 1.33 u 10–4 N
12 Physics Term 2 STPM Chapter 12 Electrostatics 27 (c) When the separation between the plates is 2.0 cm, the electric fi eld strength E = ——–—— V 2.0 u 10–2 V m–1 V = Voltage of cell When the plate separation is 3.0 cm, E1 = ——–—— V 3.0 u 10–3 V m–1 = —2 3 E Electric force on sphere X, F2 = QE1 = — 2 3 QE (QE = 4.0 u 10–4 N) = —2 3 u (4.0 u 10–4) = 2.67 u 10–4 N (d) When two extra cells are added, the potential difference between the plates = 3V Hence, E3 = (3V) ——–— 3 u 10–2 V m–1 = 2E Electric force on sphere, F3 = QE3 = 2QE = 2 u (4 u 10–4) = 8.0 u 10–4 N Example 17 A beam of electrons travelling at a speed of 1.50 u 107 m s–1 in a vacuum enters the space between two parallel plates of a cathode-ray oscilloscope as shown in the diagram. The length of each plate is 40 mm and they are separated by a distance of 20 mm. The top plate is at a positive potential V relative to the bottom plate. The electron beam leaves the plates at R with a velocity of 1.60 u 107 m s–1 at an angle of 20.0° to the original direction. (a) Find the potential difference required to accelerate the electron from rest to a speed of 1.50 u 107 m s–1. (b) Use a suitable scale to draw a vector diagram showing the change of velocity 'v of an electron during its motion from the point Q to R. Use your vector diagram to fi nd the magnitude and direction of 'v. (c) What is the time taken for electron to move from Q to R? Hence, calculate the acceleration of the electron from Q to R. (d) Neglecting the effect of gravity, fi nd the value of V. (e) You are told to neglect the effect of gravity. Show that the assumption is correct. P Q R +V 20 mm 40 mm 0 V 20.0° v = 1.60 107 m s–1 u = 1.50 107 m s–1
12 Physics Term 2 STPM Chapter 12 Electrostatics 28 Solution: (a) Kinetic energy gained by electron = loss of electric potential energy —1 2 mv2 = eVacc Accelerating potential difference, Vacc = mv2 —– 2e = (9.11 u 10–31) u(1.50 u 107 )2 ————————————– 2(1.60 u 10–19) V = 640 V (b) Using a scale of 1 cm : 0.1 u 107 m s–1, draw accurately the vector diagram shown below. 20° u = 1.50 107 m s–1 v = 1.60 107 m s–1 Δv From measurements made from the vector diagram, magnitude of 'v = 5.6 u 106 m s–1 at an angle of 90° to the original direction of motion. (c) Time taken to move from Q to R, t = 40 u 10–3 ———— 1.5 u 107 = 2.67 u 10–9 s (d) Acceleration = —– 'v t = 5.6 u 106 ————— 2.67 u 10–9 = 2.10 u 1015 m s–2 E = — V d a = —– eE m = —e m (—V d ) V = mad —–e = (9.11 u 10–31)u(2.10 u 1015) u (20 u 10–3) ————————————————–––— (1.6 u 10–19) = 239 V (e) Gravitational force on electron = mg W = (9.11 u 10–31) u 9.81 = 8.93 u 10–30 N
12 Physics Term 2 STPM Chapter 12 Electrostatics 29 1. The graph shows the variation of the electric fi eld intensity E with the distance d from a point charge. E P 0 Q d The shaded area between P and Q represents A the potential difference between P and Q. B the electric fi eld intensity between P and Q. C the work done to move a unit charge from P to Q. D the power dissipated when a unit charge moves from P to Q. 2. A charge q is at a point P in an electric fi eld of strength E which acts in the x-direction. The electric potential at P is V, the force on q in the x-direction is F and the electric potential energy of the charge is U. Which one of the following sets of equations correctly expresses the relationship between E, F, q, U and V? A F = qE; U = qV; V = – —– dE dx ; U = – —– dF dx B U = qF; V = qE; U = – —– dV dx ; F = – —– dE dx C V = qE; U = qF; F = —– dE dx ; U = —– dF dx D F = qE; U = qV; E = – —– dV dx ; F = – —– dU dx 3. Which of the following statements about the electric fi eld strength, electric potential and potential gradient is correct? A The electric potential at a point is the force on unit positive charge at that point. B Unit potential difference exists between two points if one joule of work is done to transfer one coulomb of charge between the points. C The electric potential and potential gradient are both scalar quantities. D Electric fi eld strength at a point is the work done to bring unit positive charge from infi nity to the point. 4. The diagram shows the electric fi eld between a positive charge and a negative charge. + – 1 2 3 4 Which of the following points show the greatest potential difference? A 1 and 2 B 1 and 3 C 1 and 4 D 2 and 4 Electrostatic force on electron FE = eE = e(—V d ) = (1.60 u 10–19)(———— 239 20 u 10–3 ) = 1.91 u 10–15 N —– W FE = 8.93 u 10–30 ——–——— 1.91 u 10–15 = 4.68 u 10–15 Hence, the gravitational force W is negligible compared to the electrostatic force FE. Quick Check 5
12 Physics Term 2 STPM Chapter 12 Electrostatics 30 5. The electric field lines around two point charges P and Q are shown below. The field strength at the point X is zero. X P Q Which of the following statements is true? A The potential at X is zero. B Both the charges P and Q are negative. C P has smaller charge than Q. D The electric potential at P is higher. 6. The diagram shows equipotential lines spaced 2.0 cm apart in a uniform electric field. 2.0 cm Electric field 100 V 50 V 0 V –50 V What is the force on a charge of +6.0 PC in the electric field? A 5.0 u 10–6 N B 1.5 u 10–2 N C 6.0 u 103 N D 3.0 u 108 N 7. Electrons in a cathode-ray tube are accelerated from the cathode to the anode by a potential difference of 500 V. If this potential difference is increased to 2 000 V, the electrons will arrive at the anode with A twice the kinetic energy and four times the velocity. B four times the kinetic energy and twice the velocity. C four times the kinetic energy and sixteen times the velocity. D sixteen times the kinetic energy and four times the velocity. 8. Which of the following graphs shows the correct relationship between the force F on an electron in the electric field between two opposite charged, large, parallel plates and the perpendicular distance x of the electron from the positive plate? A C F 0 x F 0 x B D F 0 x F 0 x 9. In an evacuated enclosure, a metal plate PQ is maintained at a negative potential V relative to a second plate MN. Electron of velocity v enters the space between the plates as shown. N Q d P M v Q Evacuated Given that the electron charge is e and the electron mass is me . Electrons just reach the plate PQ if A — l 2 me v 2 = e—V d B —l 2 me (v cos θ)2 = eV C —l 2 me (v sin θ)2 = e—V d D —l 2 me (v sin θ)2 = eV
12 Physics Term 2 STPM Chapter 12 Electrostatics 31 10. An electron beam enters a region in an evacuated tube where there is a uniform electric field as shown in the diagram. Electron beam X P Q R Which of the following is a possible path for the electron beam? A A straight line XQ B A curve line from X to P C A curve line from X to R D A line curving out of the plane of the figure. 11. The graph shows the variation of the electric potential V with the distance x from a fixed point along a straight line. V/V x/m 4 2 0 –6 2 –4 –2 4 6 –2 –4 Sketch the graph of electric intensity Ex against x. 12. (a) A thunder cloud at a potential of 1 u 107 V with respect to the Earth delivers a lightning stroke of 50 C of charge to the Earth. Calculate the energy dissipated if the potential difference remains constant. (b) Calculate the speed of an electron accelerated from rest in a vacuum by a potential difference of 1 u 107 V. (Specific charge of electron, —e me = – 1.76 u 1011 C kg–1) (c) Is the result possible? 13. An oil drop of mass m and carrying a charge q is maintained stationary between two horizontal parallel metal plates by applying a suitable electric field between the plates. (a) Write an expression for the electric field E in terms of m, q and g, the acceleration of free fall. What is the direction of E? (b) State what would happen to the oil drop if (i) the drop acquires additional charge of the same sign, (ii) the plates move closer together, the potential difference between them remains constant. 14. The diagram shows two parallel metal plates PQ and RS in an evacuated enclosure. The separation of the plates is 15 mm and PQ is at a potential +100 V relative to RS. A and B are two slits in the plate PQ separated by 25 mm. A collimated beam containing electrons of different kinetic energy is directed at A at an angle of 60° to the plate, as shown. R P S Q A B 60n 25 mm 15 mm 0 V +100 V (a) Find the initial kinetic energy of the electrons that just reach the plate RS. (b) Find the initial velocity of the electrons that emerge from B. (Mass of electron, me = 9.11 u 10–31 kg, electron charge, e = –1.6 u 10–19 C) 15. The deflection system of a cathode-ray oscilloscope consists of two parallel horizontal metal plates, each of length 2.0 cm, with a separation of 0.50 cm as shown in the diagram. 15 cm 0.50 cm 2.0 cm + – Screen Electron 3.1 107 m s–1
12 Physics Term 2 STPM Chapter 12 Electrostatics 32 The centre of the plates is situated 15 cm from the screen. A potential difference of 80 V between the plates provides a uniform electric fi eld in the region between the plates. The electrons are accelerated by the potential difference between the cathode and anode to a speed of 3.1 u 107 m s–1 before entering the electric fi eld at right angles. Calculate (a) the accelerating potential difference between the cathode and the anode, (b) the time taken for an electron to pass between the plates, (c) the electric fi eld strength between the plates, (d) the force on an electron due to the electric fi eld, (e) the acceleration of the electron along the direction of the electric fi eld, (f) the speed of the electron at right angles to its original direction of motion as it leaves the region between the plates, (g) the defl ection of the electron on the screen. Important Formulae 1. Coulomb’s law: F = ——— Q1 Q2 4Sε0 r2 2. Electric fi eld strength, E = —F q , hence F = qE 3. Electric fi eld strength due to a point charge, E = Q ——— 4Sε0 r2 4. Electric fi eld strength due to a charged sphere, For r R, E = 0 For r R, E = Q ——— 4Sε0 r2 5. Gauss’s law: electric fl ux, ) = ΣQ ——ε0 , also ) = EA 6. Electric potential, V = Work done ———––— Charge 7. Relation between E and V V = – ∫ r ∞E dx and E = – —– dV dx 8. Electric fi eld between two parallel charged plates, E = —V d 9. Electric potential energy, U = qV STPM PRACTICE 12 1. Two point charges q1 and q2 are separated by a distance of 2.59 m. The force of attraction between them is 0.630 N. If q1 = +40 μC, what is q2 ? A +27 μC C –11 μC B +11 μC D –27 μC 2. An oil drop is charged negative by eleven electrons and it has a weight of 4.5 u 10–15 N. The oil drop is in the uniform electric fi eld between two horizontal charged metal plates. The top plate is charged positive and the electric fi eld strength is 550 N C–1. What is the resultant force on the oil drop? A 3.53 u 10–15 N downwards B 3.53 u 10–15 N upwards C 5.47 u 10–15 N downwards D 5.47 u 10–15 N upwards 3. A D-shaped metal disc is placed in the electric fi eld between two charged parallel plates. Which is the electric fi eld between the plates? A + – C + –
12 Physics Term 2 STPM Chapter 12 Electrostatics 33 B + – D + – 4. Three point charges q are fixed at the corners of an equilateral triangle of sides a. What is the electric potential energy of the system of charges? A q —–— 4πε0 r C q2 —–— 4πε0 r B 3q —–— 4πε0 r D 3q2 —–— 4πε0 r 5. Three point charges Q1 , Q2 and Q3 are arranged in a straight line as shown in the diagram below. Q1 = –30 PC Q2 = +5 PC Q3 = –60 PC 0.20 m 0.40 m The electric force on the charge Q2 is A 16.8 N in the positive-x direction. B 16.8 N in the negative-x direction. C 50.6 N in the positive-x direction. D 50.6 N in the negative-x direction. 6. A sphere of radius R is positively charged. Which graph shows the variation of electric field strength E with distance r from the centre of the sphere? A E r R 0 C E r R 0 B E r R 0 D E r R 0 7. Three equal charges Q form an equilateral triangle of sides r are as shown in the diagram below. Q r Q r r Q What is the work done to bring one of the charges to infinity? A Q2 —–—πε0 r C Q2 —–— 4πε0 r B Q2 —–— 2πε0 r B Q2 —–— 8πε0 r 8. Three point charges are at the vertices of an equilateral triangle as shown in the diagram. q q q O T S Z Y R X What is the direction of the electric field at O? A OS C OY B OZ B OX 9. What is the work done to bring two point charges of +Q and –Q from infinity to a separation of d? A – Q2 ——— 2Sε0 d B – Q2 ——— 4Sε0 d C Q2 ——— 4Sε0 d D Q2 ——— 2Sε0 d
12 Physics Term 2 STPM Chapter 12 Electrostatics 34 10. The electric field lines around an electric dipole are shown below. The points P, Q, R and S are equidistant from the positive charge. Q S R P The electric field is strongest at A P B Q C R D S 11. A charge of + 5.0 C is at the centre of a sphere of radius 10.0 cm. 10.0 cm 5.0 C The electric flux through the sphere is A 9.95 N m2 C–1 B 39.8 N m2 C–1 C 5.65 u 1011 N m2 C–1 D 7.51 u 1011 N m2 C–1 12. The work done to move a charge q from point X to point Y in an electric field is W. What is the potential difference between X and Y? A qW B q2 W C — W q D — W q2 13. An alpha-particle of charge +2e approaches a stationary point charge of +20e head-on. The closest distance between the alphaparticle and the stationary point charge is d. If the same alpha-particle makes a head-on approach another point charge of +40e, what is the closest distance between the alphaparticle and the stationary point charge of +40e? A 0.25d C 2.0d B 0.50d D 4.0d 14. Four equal point charges of +Q are fixed in a square as shown in the diagram below. W +Q Y O Z X +Q +Q +Q What is the direction of the resultant electric force on the charge +Q at the centre O of the square? A OW B OX C OY D OZ 15. At a point X distance r from a point charge, the electric potential is V and the electric field is E. A point Y is at a distance of 2r from the point charge. What is the electric potential and electric field? Electric potential Electric field A —1 4 V —1 4 E B —1 2 V —1 4 E C —1 4 V —1 2 E D —1 2 V —1 2 E 16. A point charge +q is in equilibrium between two stationary point charges +2Q and +Q. 2Q q x y Q Which of the following is the correct relationship between x and y? A x = 2 y B x = 2y C x = 4y D x = 2 2 y
12 Physics Term 2 STPM Chapter 12 Electrostatics 35 17. A point charge q is at the centre of a cube of sides L. L q The electric flux through the surface of the cube is A q —ε0 C 6qL2 —–– ε0 B qL2 —– ε0 D ——– q 4Sε0 L2 18. The points X, Y and Z are in an electric field E = 200 V m–1. E = 200 V m–1 Z X Y Proton 4.0 m 3.0 m The work done by the electric field to move a proton from the point X to Y is WXY and from Y to Z is WYZ. Which row correctly gives the values of WXY and WYZ? A WXY = 0, WYZ = 9.60 u 10–17 J B WXY = 9.60 u 10–17 J, WYZ = 0 C WXY = 9.60 u 10–17 J, WYZ = 1.28 u 10–16 J D WXY = 1.28 u 10–16 J, WYZ = 9.60 u 10–17 J 19. Two parallel metal plates each of area 0.40 m2 , have opposite charges of the same magnitude. The electric field strength between the plates is 55 N C–1. What is the magnitude of the charge on each plate? A 7.80 u 10–10 C C 1.95 u 10–10 C B 3.90 u 10–10 C D 1.22 u 10–9 C 20. Point charges, each of magnitude Q, are placed at three corners of a square as shown. What is the direction of the resultant force on an electron at the fourth corner? +Q +Q –Q e– B A C D 21. A helium nucleus of radius r is represented by the symbol 4 2 He. If e is the elementary charge, the electric field strength at the surface of an isolated helium nucleus is A ——– 2e 4Sε0 r C ——– 4e 4Sε0 r B ——– 2e 4Sε0 r2 D ——– 4e 4Sε0 r2 22. The diagram shows two points P and Q in front of a large charged plane. The electric field strength at P and Q are EP and EQ respectively. r 2r P Q Charged plane Which of the following is correct? A EP = EQ C EP = 4EQ B EP = 2EQ D EP = —l 2 EQ 23. P and Q are two charged spheres initially at a different potential. Then, the spheres are connected by a very thin wire. P Q Insulated stand Which of the following is correct? A Charge is conserved and the electric potential of the spheres is the same. B There is a net loss of charge, the electric potential of the spheres are the same. C There is a net loss of charge, the electric potential of the spheres remains unchanged. D Charge is conserved, the electric potential of the spheres remains unchanged.
12 Physics Term 2 STPM Chapter 12 Electrostatics 36 24. The graphs below which have the same scale show the electric potential V between two points P and Q. Which graph shows that the electric field strength between P and Q is the strongest? A C 0 V P Q 0 V P Q B D 0 V P Q 0 V P Q 25. A point charge Q at the centre of a square of sides x is surrounded by four identical charges Q at the corners of the square. x x x x Q Q Q Q Q How much work is done by the force of electrostatic repulsion when the charge at the centre is moved to infinity? A Zero C Q2 ——–– 2Sε0 x B 2Q2 ——Sε0 x D Q2 ——Sε0 x 26. What is the potential difference required to accelerate an electron in a vacuum from rest to a velocity of 5.0 u 106 m s–1? A 14 V C 142 V B 71 V D 712 V 27. The figure shows two isolated conducting spheres X and Y initially uncharged placed on insulated stands. A positively charged rod is then brought close to each of the spheres. X + + + + Y + + + + The electric potentials of the spheres X and Y are A B C D Sphere X Positive Positive Negative Negative Sphere Y Positive Zero Positive Zero 28. The diagram shows two concentric spheres, X of radius a and charge Q and Y of radius b is earthed. b a X Y Which of the following is the graph of electric intensity E against distance r from the centre of the spheres? A C E r a b 0 E r a b 0 B D E r a b 0 E r a b 0
12 Physics Term 2 STPM Chapter 12 Electrostatics 37 29. The diagram below shows equipotential lines in an electric field. 4.0 V 8.0 V 12.0 V 16.0 V (a) Copy the figure and draw a line to show the direction of the electric field. (b) An electron moves from the equipotential line of –4.0 V to –16.0 V. Calculate the change in the kinetic energy of the electron. Is the changes increasing or decreasing? 30. The initial velocity of an electron in an electric field E is 5.00 u 106 m s–1 in the direction of the electric field. The electron stops momentarily after travelling through a distance of 3.00 cm. (a) Find the strength of the electric field. (b) State and explain the motion of the electron before and after it stops momentarily. 31. (a) Draw a diagram to illustrate the electric field, and three equipotential surfaces round a negative point charge. (b) Three point charges Q1 = – 3.0 μC, Q2 = + 4.0 μC and Q3 = + 3.0 μC are fixed at the vertices of an isosceles triangle as shown in the diagram. 0.50 m 0.50 m Q2 Q1 Q3 P (i) Find the magnitude and direction of the electric force on the charge Q1. (ii) Determine the electric potential at the point P which is midway between Q2 and Q3. 32. ABC is a triangle in which CA = CB = r. In Figure(a) charges –Q and +Q are placed at A and B respectively and in diagram (b) equal point masses m are placed at A and B. –Q +Q y x A B C y x A B C m m Figure (a) Figure (b) (a) In Figure (a), find the direction of the resultant electric field at C relative to the x and y directions. (b) In Figure (b), find the direction of the resultant gravitational field at C relative to the x and y direction. (c) Find the electrostatic and gravitational potential at C. 33. An oil drop of mass 2.30 u 10–14 kg is stationary between two parallel conducting plates separated by a distance of 1.80 cm. The potential difference between the plates is 2.80 kV. Estimate the number of electrons on the oil drop. 34. P and Q are two charged metal spheres of radius r1 and r2. The spheres are then connected by a thin wire. Calculate the ratio of the final charge density σ —1 σ2 , of the spheres in terms of r1 and r2 . 35. Two identical small charged conducting spheres each of mass 5.0 u 10–4 kg,are suspended by nylon threads, each of length 0.20 m from the same point. The threads are separated with the angle between them is 30°. Calculate the charge on each sphere. 36. (a) Define the electric potential, and the electric potential energy. (b) The diagram shows a system which consists of four charges at the corners of a square of sides 2a in a vacuum. 2a 2a A +Q +Q –Q –Q B D C
12 Physics Term 2 STPM Chapter 12 Electrostatics 38 Find (i) the electric potential at the centre of the square, (ii) the electric potential energy of the system, (iii) the work done to move the charges closer to form a square of sides —3 4 a. State whether work is done by the system or on the system and show that the deduction can be obtained by considering the electrostatic force on the system of charges. 37. (a) By referring to Coulomb’s law in electrostatic and Newton’s law of gravitation, explain what is meant by inverse square fi eld. (b) With the aid of labelled diagrams, explain how a conductor (i) can be at Earth’s potential, yet positively charged, (ii) can be at positive potential relative to the Earth, yet uncharged. (c) A metal sphere is of radius 0.20 m. Insulation of the air breaks down when the electric fi eld strength is 3.0 u 106 V m–1. If the sphere is charged, calculate (i) the maximum charge that the sphere can hold, (ii) the maximum electric potential of the sphere. Discuss the effect of the maximum electric potential if the surface of the sphere is rough. 38. (a) (i) State Coulomb’s law. (ii) Define electric field strength E, and use Coulomb’s law to derive an expression for the electric fi eld intensity E at a point distance x from a point charge Q. (b) In the diagram below, a charge of +9Q is fi xed at A and a charge of –Q is fi xed at B at a distance of y from A. y A C B D H –Q +9Q At the points C and D, the magnitude of the electric fi eld strength due to the charge +9Q is equal to that due to the charge –Q. (i) Find in terms of y, the distance of the point C, and the point D from A. (ii) Discuss which of the points C and D is the resultant electric fi eld strength due to both the charges zero. (iii) Sketch diagrams to show the electric fi eld due to the charges +9Q and –Q. (iv) State the direction of motion of an electron which is placed at different points along the line between B and H. (v) Find the work done to bring a unit positive charge from infi nity to the point D. 39. (a) State Gauss’s law and show that it is consistent with Coulomb’s law. (b) Derive a formula for the electric potential at the surface of a charged isolated conducting sphere. (c) An ion beam which contains a mixture of charged sodium ions (Na+) and potassium ions (K+) is accelerated from rest in a vacuum by a potential difference of 5 kV. The beam current is 20 PA. (i) What is the total number of ions passing any fi xed point in the beam every second? (ii) The accelerating potential difference is now applied only for a short time so as to produce a ‘bunch’ of accelerated ions which then travel 20 m in a vacuum to a detector. Pulses due to the arrival of the sodium ions and potassium ions occur at slightly different times. Calculate this time difference. (Ionic masses: Na+ : 3.81 u 10–26 kg K+ : 6.49 u 10–26 kg)
12 Physics Term 2 STPM Chapter 12 Electrostatics 39 1 1. C 2. B 3. C 4. Q2 = 4 Sε0 x3 W ———– 4l 2 – x2 5. (a) 0.050 N Use F = ——– Q1 Q2 4Sε0 r2 (b) Parallel to AC 2 1. (a) (b) – (c) (d) 2. (a) (i) P: Positive charge (ii) Q: Negative charge (b) Magnitude of Q charge is greater. 3. D 'LͿHUHQFHffl3RVLWLYHFKDUJHDFFHOHUDWHVQHJDWLYH charge decelerates. (a) v v E (b) v v E 3 1. A: Apply Gauss' law to surface A (OHFWULFÁX[ ¦Q ε0 = (q) + (–q) ε0 = 0 2. E = Q ——– 4Sε0 a2 (2 – 1 –– 2) in the direction parallel to YP. 3. (a) E = Q ——– 4Sε0 r2 (b) 2.7 u1021 V m–1 (Q = 92 u1.6 u10–19 C) 4. (a) 6.4 u 106 V m–1 towards –Q (b) 1.0 u 10–13 N towards +Q 5. (a) X Y +Q1 +Q2 E (b) E = ——– Q1 4Sε0 r2 (c) F = Q2 E = ——– Q1 Q2 4Sε0 r2 +Q1 +Q2 6. (a) 4.00 u 106 V m–1 (b) E = 0 ¦Q = (+Q – Q) = 0 7. E = q ——Sε0 a2 parallel to BP 8. σ = 5.0 u 10–9 C m–2 Using T cos 30° = mg T sin 30° = qE = q 1 σ––– 2ε0 2 ANSWERS
12 Physics Term 2 STPM Chapter 12 Electrostatics 40 4 1. C 2. D 3. B 4. A: Change in K.E. = work done againstHOHFWULFÀHOG = (eE)x = (– 1.60 u 10-19)(200)(4.0) J = - 1.28 u 10-16 J 5. Charged spheres are not point charges. 6. (a) 102 N (b) 1.53 u 10–13 J Work = QV V = —–—Q 4Sε0 r 7. (a) 5.40 u 10–4 m V = Q —–— 4Sε0 R 1 — (b) 794 V Q = 2QR = 2 3 R 8. (a) 0.145 V V = Q —–— 4Sε0 R (b) 2.26 u 10–8 V m–1 E = —–—Q 4Sε0 R2 9. (a) (i) 2.25 u 103 V V = —–—Q 4Sε0 R (ii) 3.00 u 103 V (b) 1.50 u 10–3 J W = q(VB – VA) 10. (a) (i) E = Q —–— 4Sε0 d2 (ii) True for r RUDGLXVRIVSKHUH (b) 2.24 u 10–13 m 1 –– 2 mv2 = qQ —–— 4Sε0 r Negligible because masses of 4 2 He and gold nucleus are small. 5 1. A 2. D 3. B 4. B 5. C 6. B 7. B 8. B 9. B 10. C 11. –6 –4 –2 0 2 468 2 4 6 –2 E/V m–1 x/m 12. (a) 5.0 u 108 J W = QV (b) 1.88 u 109 m s–1 (c) Not possible because 1.88 u 109 m s–1 3.0 u 108 m s–1VSHHGRIOLJKW 13. (a) E = mg —–q upwards. (b) (i) Accelerates upwards because F = qE increases. (ii) Accelerates upwards because E increases. 14. (a) 2.13 u 10–17 J 1 –– 2 m(v sin 60°)2 = eV K.E = 1 –– 2 mv 2 (b) 5.81 u 106 m s–1 In time t = —–—— 25 mm v cos 60° , vertical displacement = 0 s = ut + 1 –– 2 at2 15. (a) 2.74 u 103 V 1 –– 2 mv2 = eV (b) 6.45 u 10–10 s l = vt (c) 1.60 u 104 V m–1 E = V–– d (d) 2.56 u 10–15 N F = eE (e) 2.81 u 1015 m s–2 a = eE––m (f) 1.81 u 106 m s–1 v = u + at (g) 8.76 u 10–3 m tan θ = vy –——–– 3.1 u 107 = y –—– 0.15 STPM PRACTICE 12 1. C: F = –0.630 N = (+40 u 10–6)q ———–——–2 4Sε 0 (2.50)2 q2 = –11 μC 2. A: FE = (ne)E upwards = 9.68 u 10-16 N F = (4.5 u 10-15 – 9.68 u 10-16) N = 3.53 u 10-15 N downwards 3. C: Electric field is perpendicular to surface of conductor. 4. D: U = work done to bring the three charges from LQÀQLW\ = (q)(q) ——— 4Sε0 r + 3 (q)(q) ——— 4Sε0 r + (q)(q) ——— 4Sε0 r 4 = 3q2 ——– 4Sε0 r 5. B F12 F32 Q2 F12 = ——— Q1 Q2 4Sε0 r2 = (30 u 10–6)(5 u 10–6) ——————–———— 4S(8.85 u 10–12)(0.20)2 = 33.7 N F32 = ——— Q2 Q3 4Sε0 r2 = (5 u 10–6)(60 u 10–6) —————––———— 4S(8.85 u 10–12)(0.40)2 = 16.9 N Resultant force F = (33.7 – 16.9) N = 16.8 N in the negative-x direction.
12 Physics Term 2 STPM Chapter 12 Electrostatics 41 6. A: r R, E =r R, E = Q ——— 4Sε0 r2 7. %ffl(OHFWULFSRWHQWLDODWRQHRIWKHDSH[GXHWRWKH FKDUJHVDWWKHWZRRWKHUDSH[HVV = 2 ( Q ——— 4Sε0 r ) Work done = electric potential energy of charge QDWWKHDSH[ = 2( Q2 ——— 4Sε0 r ) = Q2 ——— 2Sε0 r 8. A: Resultant E DORQJ26 VHHÀJXUH q q q O T S Z Y R E X E Z E Z X 9. B: Work done = (–Q) 1 Q —–— 4Sε0 d 2 10. $ffl)LHOGLVVWURQJHVWZKHUHWKHÀHOGOLQHVDUHFORVH together. 11. &ffl8VLQJ*DXVV·VODZÁX[ 6Q—–ε0 12. C: V is work per coulomb 13. C: 1 —2 mu2 = (2e)(20e) —–––– 4Sε0 d 1 —2 mu2 = (2e)(40e) —–––– 4Sε0 d1 d1 = 2d 14. D: Directions of the forces F on the charge at O are as shown in the diagram. Resultant force in the direction OZ. W +Q Y O F F F Z X +Q +Q +Q 15. B: V = Q —–— 4Sε0 r E = Q —–— 4Sε0 r2 16. A: F[ = Fy (2Q)q —–––– 4Sε0 x2 = (Q)q —–––– 4Sε0 y2 Hence, x = 2 y 17. $ffl8VLQJ*DXVV·VODZÁX[ 6Q—–ε0 18. B: WXY = (qE)(x) = 1.28 u 10–16 J $ORQJ<=F = qE is normal to E. WXY = 0 19. C 20. D 21. B 22. A 23. A 24. C 25. B 26. B 27. B 28. D 29. (a) 4.0 V 8.0 V 12.0 V 16.0 V E (b) Change in electric potential energy = (–e)[(–16.0) – (–4.0)] = (–1.60 u 10–19)(–12.0) = + 1.92 u 10–18 J (increase) Decrease in kinetic energy = 1.92 u 10–18 J 30. (a) Gain in electrical P.E = loss in K.E. eV = —1 2 mu2 E = —V d = mu2 ——2ed = (9.11 × 10–31)(5.00 × 106 )2 ——————–————— 2(1.60 × 10–19)(3.00 × 10–2) V m–1 = 2.37 kV m-1 Alternative method: 'HFHOHUDWLRQRIHOHFWURQa Use v2 = 0 = u2 – 2as 2 (—– eE m )d = u2 gives E = mu2 —–– 2ed (b) Electron retards uniformly because the electron force F = eE is in the opposite direction of its motion. $IWHULW VWRSV WKH HOHFWULF IRUFH FDXVHV WKH electron to accelerate backwards. 31. (a) Equipotential surface E (b) (i) F12 = Q1 Q2 4S ε0 x2 F13 = Q1 Q3 4S ε0 x2 Q2 Q1 Q3 P F12 F13 F e F = F 2 12 + F 2 13 = 0.4322 + 0.3242 N = 0.540 N θ = tan-1( F12 F13 ) = 53o
12 Physics Term 2 STPM Chapter 12 Electrostatics 42 (ii) y2 + y2 = 0.502 gives y = 0.354 m VP = Q1 4S ε0 y + Q2 4S ε0 y + Q3 4S ε0 y = 102 kV 32. (a) Negative x–direction (b) Negative y–direction F =HURV = –—–– 2Gm r 33. 9 electrons (ne)E = mgE = V–– d 34. σ —1 σ2 = r —2 r1 VP = VQσ = Q——4Sr2 ——– Q1 4Sε0 r1 = ——– Q2 4Sε0 r2 35. 3.95 u10–8 C 30° F mg mg F = ——— 4πε0r 2 Q2 l 36. (a) Refer page 15 and 18 (b) (i) 0 (ii) U2a = (—– 1 2 – 2) Q2 ——– 4Sε0 a U2a = –1.293 Q2 ——– 4Sε0 a (iii) U—3 4 a = –1.724 Q2 ——– 4Sε0 a W = U—3 4 a – U2a = – 0.431 Q2 ——– 4Sε0 a Work done by system. Attraction between charges. 37. (a) Force F v 1 –– r2 (b) (i) – – – – – + + + + + Conductor (ii) + + + + + + + + + + + + + + + – – – – – Conductor Positive charged sphere (c) (i) 1.33 u 10–5 C E = Q ——– 4Sε0 R2 (ii) 6.0 u 105 V V = ER VPD[LPXPLVORZHU EHFDXVH RI VKDUS points on rough surface. 38. (a) (i) Refer page 2 (ii) Refer page 10 (b) (i) AC = 0.75y E = Q ——– 4Sε0 r2 AD = 1.50y (ii) ED = 0 EqQ = –E–Q (iii) D Neutral point (iv) Between B and D : Towards the right At D : No force (stationary) Between D and H : Towards the left (v) W = Q——Sε0 y W = VD 39. (a) Refer page 10 (b) Refer page 11 (c) (i) 1.25 u 1014 ion s–1I = nq (q = 1.6 uйй–19 C) (ii) 2.98 u 10–5 s t = ———— Distance Velocity 1 –– 2 mv2 = qV
43 13 CHAPTER CAPACITORS 13 &DSDFLWDQFH Q C = —V (QHUJ\6WRUHGLQ &DSDFLWRU U = 1 2 CV2 = 1 2 QV = 1 2 Q2 V 'LHOHFWULF +Q –Q V d C = εr ε0A d &DSDFLWRUV 3DUDOOHO3ODWH&DSDFLWRU C = ε0A d &KDUJLQJD&DSDFLWRU (a) V = E(1 – e t – —–CR ) (b) Q = Q0(1 – e t – —–CR ) (c) I = I0e t – —–CR 0 E V t 0 Q0 Q t 0 I0 I t 'LVFKDUJLQJD&DSDFLWRU (a) V = V0e t –—–CR (b)Q = Q0e t –—–CR (c) I = I0e t –—–CR 0 V0 V t 0 Q0 Q t 0 I 0 I t 43 Bilingual Keywords *HWHJP[HUJL!2HWHZP[HUZ *HWHJP[VY!2HWHZP[VY *OHYNPUN!4LUNLJHZ *VUJLU[YPJZWOLYL!:MLYHZMLYHZLW\ZH[ +PLSLJ[YPJZ!+PLSLR[YPR +PZJOHYNPUN!4LU`HOJHZ ,ULYN`Z[VYLK!;LUHNHKPZPTWHU 0UWHYHSSLS!:LSHYP 0UZLYPLZ!:LZPYP 7HYHSSLSWSH[L!7SH[ZLSHYP Concept Map
Physics Term 2 STPM Chapter 13 Capacitors 44 13 INTRODUCTION 1. Capacitors are widely used in electronic circuits. Capacitors store electric charge. 2. Electrons fl ow in and out but not through a capacitor. A capacitor blocks direct current but alternating current is able to pass through it. 3. Basically, a capacitor consists of two parallel metal plates with an insulator known as dielectric in between. 4. Example of dielectrics are air, paper, wax and mica. Figure 13.1 Figure 13.2 5. Figure 13.1 shows the circuit symbol of a capacitor. 6. Figure 13.2 shows capacitors of various sizes used in electrical and electronic circuits. 13.1 Capacitance Students should be able to: GHÀQHFDSDFLWDQFH Learning Outcome 1. Figure 13.3 shows what happens when a capacitor is connected to a battery. Electrons from the battery charges the plate X of the capacitor with a charge of –Q and a charge of +Q is induced on the opposite plate Y. Hence, the charge on both plates of the capacitor is the same magnitude but opposite in sign. 2. As the charges on the plates of the capacitor increase, the potential difference across the capacitor increases until it is equal to the e.m.f. of the battery. 3. The charges remain in the capacitor even after the battery is disconnected. 4. The quantity of charge a capacitor is able to store depends on its capacitance. 5. The capacitance C of a capacitor is the ratio of the charge on a plate of the capacitor to the potential difference between the plates. Capacitance, C = Charge on either plate of capacitor —————————————————– Potential difference between the plates C = Q —V Figure 13.3 _ _ _ _ + + _ + Battery + X Capacitor Y + Figure 13.4 –Q +Q V 2016/P2/Q3, 2018/P2/Q4