PHY 132 Electricity, Magnetism and Modern Physics Course Guide PDF

PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS PHY 132: ELECTRICITY, MAGNETISM AND MODERN PHYSICS COURSE GUIDE NATIONAL OPEN UNIVERSITY OF NIGERIA NOUN 2 PHY 12...

PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS PHY 132: ELECTRICITY, MAGNETISM AND MODERN PHYSICS COURSE GUIDE NATIONAL OPEN UNIVERSITY OF NIGERIA NOUN 2 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS 1.0 Introduction PHY132 electricity, magnetism and modern physics is a one semester 2 credits, foundation level course. It will be available to all students to take towards the core module of their B.Sc. Education, and other programmes B.Sc computer science, environmental studies and The course comprises 20 study units (4 modules), which involve basic principles of Electricity, Magnetism and Modern Physics. The material ha been developed in such a way that students with at least a credit pass at the ordinary level of equivalent will follow quite easily. There are no compulsory pre requisites for the course. However, you are strongly advised to have adequate grasp of Further Mathematics or Applied Mathematics. This course guide tells you briefly what the course is about, what course materials you will be using and how to work your way through these materials. Is suggests some general guidelines for the TIME to complete it successfully. It also gives you some guidance on your tutor-marked assignments. There are regular tutorial classes that are linked to the course. You are advised to attend these sessions regularly. Details of time and locations of tutorials will be given to you at the point of registration for the course. 2.0 What You Will Learn In This Course The overall aim of PHY132 is to introduce the basic principle and application of Electrical Energy and its association with Magnetism. During the course you will learn that an electric field is always associated with a magnetic field and vice versa. You would see that this bond between Electricity and Magnetism is the basis of many scientific and technology developments during the last century. Towards the end of the course you will be introduced into some aspects of Modern Physics where we have introduced some new concepts to explain sub-atomic phenomena. These include quantum theory, and energy levels in atoms. NOUN 3 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS 3.0 Course Aims The aim of this course can be summarized as follows: this course aims to give you an understanding of Electricity, Magnetism and Modern Physics and their applications in everyday life. This will be achieved by ι Introd ucing you to the funda mental s of Electri city, Magne tism and Moder n Physic s as subject s on their own right. ι Demo nstrati ng how the variou s theorie s can be applie d to real life situati ons. NOUN 4 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS ι Explai ning some funda mental concep ts in Electri city, Magne tism and Moder n Physic s. ι Explai ning the transis tion from Newto nian Mecha nics to Quant um Mecha nics. ι Giving you some insight into possibl e future develo pment in NOUN 5 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS these areas. 4.0 Course Objectives The course sets overall objectives, to achieve the aims set out above. In addition, each unit also has specific objectives. The unit objectives are always included at the beginning of a unit; you should read then before you start working through the unit. You may want to refer to them during your study of the Unit to check your progress. You should always look at the Unit objectives after completing a unit. In this way you can be sure that you have done what was required of you for the unit. Set out below are the objectives of the Course as a whole. By meeting these objectives you should have achieved the aims of the Course as a whole. On successful completion of the Course, you should be able to: 1. Describe the theory of electricity, magnetism and electromagnetic radiation 2. Explain the concepts of electric and magnetic fields. 3. Measure and compute electric current in d.c and a.c. circuit. 4. Illustrate the principles of electromagnetic induction as they apply to both d.c. and a.c. generators. 5. Demonstrate how circuit elements are connected. 6. Describe the principles of cathode ray oscilloscope, ammeters, voltmeters, x-ray tubes and dry cells as well as accumulators. 7. Identifying the advantages and disadvantages of x-rays 8. Describe the structure of the nuclear atom. 9. Distinguish between geographic and geomagnetic meridians. 10. Describe the terrestrial magnetic field. 11. Distinguish between nuclear fusion and nuclear fission. NOUN 6 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS 12. Describe the generation and distribution of electric power. 5. 0 Working Through This Course To complete this course you are required to read the study units, read set books and read other materials provided by NOUN. You will also need to do some practical exercise which will be arranged by your Course Tutor. Each unit contains self-assessment exercises, and at points in the course you are required to submit assignments for assessment purposes. At the end of the course, there is a final examination. The course shall take you about 45 weeks in total to complete. Below you will find listed all the components of the course, what you have to do and how you should allocate your time to each unit in order to complete the course successfully and on time. 6.0 Course Materials 1. Course guide 2. Study units 3. Assignment file 4. Presentation schedule 7.0 Study Units There are 20 Study Units in this Course, as follows: Unit 1 Electric charge, Force and Field Unit 2 Gauss’s Law Unit 3 Electric Potential Unit 4 Potential for Continuous Charge Distribution And Energy Unit 5 Dielectrics and Capacitors Unit 6 Electric Current Unit 7 Direct-Current Circuits and Instruments NOUN 7 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS Unit 8 The Magnetic Field Unit 9 Motion of Charge Particles in Electric and Magnetic Field Unit 10 Electrolysis and Cells Unit 11 Thermal Effects of Electric Currents And Electric Power Unit 12 Magnetic Properties of Matter Unit 13 Terrestrial Magnetism Unit 14 Electromagnetic Induction I Unit 15 Electromagnetic Induction 11 Unit 16 Alternating Current Theory 1 Unit 17 Alternating Current Theory 11 Unit 18 Thermoelectric, Photoelectric Thermionic Effects Unit 19 Modern Physics 1 Unit 20 Modern Physics 11 Each study unit consists of two to three weeks’ work, and includes specific objectives. Each unit contains a number of self-tests. In general, these self- tests, question you on the material you have just covered or require you to apply it in some way and, thereby, help you to gauge your progress and reinforce your understanding of the material. Together with tutor-marked assignments, these exercises will assist you in achieving the stated learning objectives of the individual units and of the course. 8.0 Set Textbooks Duncan Tom (1982) Physics. A Textbook for Advanced Level Students John Murray (Publishers) Ltd. London. S.M. Geddes (1981) Advanced Physics. Macmillan Education Ltd. London McKenzie A.E.E (1973) A Second Course of Electricity. The University Press, Cambridge 9.0 Assignment File NOUN 8 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS The assignment file will be supplied by NOUN. In this file you will find all the details of the work you must submit to your tutor for marking. The marks you obtain for these assignments will count towards the final mark you obtain for this course. Further information on assignments will be found in the assignment file itself and later in this course guide in the section on assessment. 10.0 Presentation Schedule The presentation schedule included in your course materials may show the important dates for the completion of tutor-marked assignments. Remember, you are required to submit all your assignments by the due date as dictated by your facilitator. You should guide against falling behind in your work. 11.0 Assessment There are two aspects to the assessment of the course. First are the tutor- marked assignment; second, there is a written examination. In doing the assignment, you are expected to apply information, knowledge and techniques gathered during the course. The assignments must be submitted to your tutor for formal assessment in accordance with the dead- lines stated in the presentation schedule and the assignment file. The work you submit to your tutor for assessment will count for 40% of your total course work. At the end of the course you will need to sit for a final written examination of three hours ‘duration’. This examination will also count for 60% of your course mark. 12.0 Tutor-Marked Assignments (TMA) The TMAs are listed as item 6.0 in each unit. Generally, you will be able to complete your assignments from the information and martial contained in the study units, set books and other reading. However, it is desirable in all degree level education to demonstrate that you have read and researched more widely than the required minimum. Using other references will give you a broader viewpoint and may provide a deeper understanding of the subject. When you have completed each assignment, send it, together with a TMA form, to your tutor. Make sure that each assignment reaches your tutor on or before the deadline given in the presentation schedule and assignment file. If, for any reason you cannot complete your work on time contact your tutor NOUN 9 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS before the assignment is due to discuss the possibility of an extension. Extensions will not be granted after the due date unless there are exceptional circumstances. 13.0 Final Examination and Grading The final examination for PHY 132 will be of three hours duration and have a value of 60% of the total course grade. The examination will consist of quantities which reflect the types of self-testing practice exercises and tutor- marked problems you have previously encountered. All areas of the course will be assessed. You are advised to use the time between finishing the last unit and sitting the examination to revise the entire course. You might find it useful to review your self-tests, tutor-marked assignments and comments on them before the examination. 14.0 Course Marking Scheme The following table shows how the actual course marking is broken down. Assessment Marks Assignments 40% of course marks Final examination 60% of overall course marks Total 100% of course marks Table 1 course marking scheme NOUN 10 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS ////……………… NOUN 11 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS PHY 132: ELECTRICITY, MAGNETISM AND MODERN PHYSICS COURSE DEVELOPMENT Course Developer Fred Ebunu Unit Writer Fred Ebunu Programme Leader Course Coordinator Dr. (Mrs.) C. A. Okonkwo NATIONAL OPEN UNIVERSITY OF NIGERIA NOUN 12 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS UNIT 1 ELECTRIC CHARGE, FORCE AND FIELD Table of Contents 1.0 Introduction 2.0 Objectives 3.1 Electric charge 3.2 Coulomb’s law 3.3 Principle of superposition 3.4 Electric field 4.0 Conclusion 5.0 Summary 6.0 Tutor Marked Assignment (TMAs) 7.0 References 1.0 Introduction Lightning and thunder are two common phenomena in our hot and humid atmosphere in Nigeria. Have you ever given a thought to what is responsible for the occurrence of the phenomena? A physicist, Benjamin Franklin demonstrated as long ago as 1752 that thunder clouds are charged with electricity. These charged clouds, when discharged in the atmosphere, give rise to a great spark, which is referred to as lightening. It will interest you to know that the amount of electric current during the discharge is about 20KA. The electric discharge which gives rise to lightning also produces a great amount of heat. In a fraction of a second, temperature rises to about 15000 0C. The lightening develops in a small area which is about 20cm in width. However, as a result of the heat amount of great produced in that small area the air molecules move fast and cause the intense sound which we call thunder. When the sound is reflected by clouds, hill or any other obstacle, we hear the roaring of clouds. A very important thing about electric charges is that the forces between them are very large. The force is known as electrostatic force (or electric force) and is responsible for holding electrons to nuclei to form atoms and for holding the groups of the atoms together to form molecules, solids and liquids. NOUN 13 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS The study of these static charges is known as electrostatic. Indeed, electrostatic was the first branch of electricity to be investigated and, for some time, it was regarded as a subject which had no practical value. However, it is now known to have practical industrial applications. For example, we shall see later in this course that a knowledge of electrostatic is important in the design of cathode ray tubes for television, in electrical prospecting for minerals. Electrostatic loudspeakers and microphone are in common use as well as electrostatic photocopying machines. 2.0 Objectives After studying this unit, you will be able to: * Distinguish between the two types of electric charge * Show that the total electric charge in an isolated system is conserved * State Coulomb’s law and use it to find the electrostatic force between two charges * State the superposition principle * Calculate the vector sum of the electric field strength due to a number of point charges. * Sketch the field lines for some simple distribution of charge. 3.1 Electric charge 3.1.1 Types of charges The ancient Greeks discovered that amber when rubbed with silk acquired the property of attracting light objects such as pieces of chaff. William Gilbert discovered that other substances exhibit the same effect, and that the magnitude of the effect is roughly proportional to the area of the surface rubbed. He was then led to the idea of a charge of electricity. Du Fay (1745) discovered that there are two kinds of electricity. Two ebonite rods when rubbed with fur exert a force of repulsion on each other. Two glass rod rubbed with silk also repel one another. However, an ebonite rod which has been rubbed with fur attracts a glass rod which has been rubbed with silk. Any substance rubbed with a different substance acquires a charge of electricity, and is found either to repel charged ebonite and attract charged glass, or vice versa. Since the two kinds of electricity can neutralize each NOUN 14 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS others effect, one is called positive and the other negative. Note that the choice as to which is positive was purely arbitrary. Glass rubbed with silk is said to have a positive charge and ebonite rubbed with fur a negative charge. The origin of the +ve and –ve charge of proton and electron The law of force between charges may be stated as follows: like charges repel, unlike charges attract Let us now understand clearly the origin of the two types of charges; we remember that an atom consists of a positively charged nucleus with negatively charged electrons around it. The nucleus is made up of proton and electron. The neutron is neutral (no charge) which the proton and electron have equal but opposite charges (positive and negative). The proton and neutron in the nucleus are held together very tightly by a nuclear force. So strong is the nuclear force that the protons are un able to move away from the nucleus. On the other hand, the force holding electron to the atomic nucleus is much weaker than the nuclear force. Hence the electrons may move away from the atom. When two different materials are rubbed together, electrons get transferred fairly easily from one material to the other. Since some materials tend to hold their electrons more strongly than others, the direction of transfer of electrons depends on the materials. For example, when a plastic ruler is rubbed with a woolen cloth, electrons flow from wool to plastic, so that it carries net negative charges whereas the wool, with a deficit of electrons, carries a positive charge of equal magnitude. This process of charging the bodies by means of rubbing them together is called charging by friction. In any case, we should note that friction actually has nothing to do with the charging process. It would appear that friction is only borrowed to describe the rubbing process. Question There are two charged bodies, x and y which attract each other. X repels a third charged body Z. Do you think z will attract or repel Y? 3.1.2 Unit of Charge In the System International (SI), electric charge is measured in coulombs (C), which is defined in terms of ampere. A coulomb is the quantity of charge NOUN 15 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS flowing per second through a conductor in which there is a steady current of 1A. Note: The definition of ampere involves force between currents. We shall see this is Modules 3. A Coulombs is the amount of charge that flow through a cross-section of a wire in one second if there is a steady current of one ampere (IA) in the wire. In symbols, q= It …………………………………………..3.1 Where q is in coulombs, if I is in ampere a.d t is in seconds. The main reason for defining the coulomb in terms of ampere is that it is easy to maintain, control and measure a current through a conductor rather than the amount of charge. 3.1.3 Conservation of Charge In the method of charging by friction (rubbing) which is discussed in section 3.1, no new charges are created. The algebraic sum of the individual charges, that is the net charge, always remains constant. Let us see how this is the case. Before the process of rubbing, the two bodies were electrically neutral (having no charge). Therefore, the total charge is zero. After rubbing, one body becomes negatively charged while the other acquires a positive charge of equal magnitude. In effect, the algebraic sum of the equal and opposite charges on the two bodies is zero. This shows that electric charge is a conserved quantity. In other words, conservation of charge implies that the total charge in an isolated system does not change. You should note that this does not mean that the total amount of positive or negative charge in a system is fixed. What we are saying is that for every additional positive charge created, there is always an equal amount of negative charge created. The charge conservation law may be stated as follows: The total electric charge in an isolated system, that is, the algebraic sum of the positive and negative charge present at any time, does not change NOUN 16 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS 3.1.4 Quantization of charge The smallest charge that is possible to obtain is that of an electron or proton. The magnitude of this charge is denoted by e. A charge smaller than e has not been found. If one determines the amount of charge on any charged body (e.g. a charge sphere) or any charged particle (e.g. α-particle) or any ion, its charge is always found to be an integral multiple of e, that is e, 2e, 3e, and so on. No charge will be a fractional multiple of e like 0.7e or 2.5e. This is true for both negative and positive charges and is expressed as q = ne ………………………………..3.2 where n is a positive or negative integer. You have now learnt that charge exist in discrete packets rather than in continuous amount. Whenever a physical quantity possesses discrete values instead of continuous values, then the quantity is said to be quantized. Therefore, we say that charge is quantized. Question A conductor possesses a positive charge of 3.2x 10-19 C. How many electron does it have in excess or deficit (e = 1.60 x 10-19C) 3.2 Coulomb’s law A knowledge of the forces that exist between charge particles is necessary for a good understanding of the structure of the atom and of matter. The magnitude of the forces between charged spheres was first investigated quantitatively in 1785 by Charles Coulomb, a French scientist. He observed that the electrostatic force between the two sphere is proportional to the product of the charges and is inversely proportional to the square of their distance apart. Coulomb’s law may be stated in mathematical terms as F α Q1 Q2 r2 Where F is the electric force between the two charges Q1 Q2, distance r apart. We can turn the above expression of proportionality to as equation by writing F = Q1 Q2 ………………………………….3.3 r2 Where K is a constant. NOUN 17 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS K = 1 ………………………………….3.4 4πε Where the constant ε depends on the material surrounding the charges, and is called permittivity. Note We shall see later that it is advantageous to have the additional constant 4π in any system having “spherical symmetry” i.e. any system in which effects are the same anywhere on the surface of a sphere. The permittivity of a vacuum is denoted by εo (pronounced as epsilon nought) and is called the permittivity of free space εo = 8.85 x 10 -12 C2 N -1 m -2 We can also write 1 = 8.98 x 109 Nm2 C-2 (4π εo ) and also F = 9 x 109 Q1 Q2 /r2 …………………………………3.5 The permittivity of air at standard temperature and pressure (s.t.p) is 1.0005εo. Therefore, we can usually take Eo as the value for air. We shall see in Module 2 that a more widely used unit for permittivity is the Farad per metre (Fm-1). You should note that Coulomb’s law applies to point charges. Sub-atomic particles such as electrons and protons may be regarded as approximating to point charge. In practice two small spheres will only approximate to point charges when they are far apart and there must not be any charge nearby to disturb the uniform distribution of charge on each of them. Example A charge q1, = 5.0μc is placed 30cm to the west of another charge q2=12μc. What is the force exerted by the positive charge on the negative charge? Solution q1 = + 5.0μc q2 = -12μc NOUN 18 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS 30cm Coulomb’s law gives force on negative charge due to the positive charge as follows: F = 1__ q1 q2 4πεo r2 F = (9 x 109 N m2 C-2) (5 x 10-6C) (-12 x 10-6C) (0.30m)2 = -6 N The minus sign shows that the force is in the negative x-direction that is towards west. Therefore, it is a force of attraction. 3.3 Principle of superposition In the last section, we considered the electrostatic forces between two charges. The question is how do we calculate the electrostatic force on a charge q1 due to the presence of other two, three or more charges? This situation is shown in figure 3.1 q2 F13 q3 q1 Figure 3.1 F1 F12 We can still calculate the force between different pair of charges using Coulomb’s law. The total force on q1 will be the vector sum of forces on q1 due to q2 and q3 independently. This is the principle of Superposition. In the other words, the fact that electric forces add vectorially is known as the priciple of superposition. To illustrate the principle, let us go through the following NOUN 19 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS Worked example Example q2 = + 2.0μc F1 10 cm F12 F13 q3 = +4.0μc q1 = -1.0uc 20cm figure 3.2 In figure 3., q1 = 1.0μc, q2.0μc and q3 = 4.0μc. Find the electrostatic force on q1 to the two other charges. You should express your result as a magnitude and direction. Solution The three charges are located at the corners of a right angle triangle. The problem can be solved using the superposition triangle. The problem can be solved using the superposition principle. The force on q1 due to the charge q2 is given by F12 = 1 q1 q2 4πεo r 2 = (9 x 10 9 Nm2 C-2 ) (-1 x 10-6 C) (2x 10-6C) (0.10m)2 = 1.8N (attractive) in the +ve x-direction Similarly, the force on q1 due to q2 is F13 = (9 x9 Nm2 C-2) (-1 x 10-6C) (4x 10-6C) (0.20m)3 = 0.90N (attractive) in the +ve x-direction NOUN 20 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS According to the superposition principle, the resultant force F1 acting on q, is the vector sum of the forces due to q2 and q3. The magnitude of F1 is √(0.90)2 + (1.8)2 2.01N and it make an angle θ = tan -1 (1.8/0.9) = tan -12 = 63.5° with the positive x- axis. 3.4 The Electric Field An electric field is a region where an electric charge experiences a force, just as a football field is an area where the game is played. If a very small, positive point charge q is place at any point in an electric field and it experiences a force F, then the field strength E 9also called the E-field) at that point is defined by the equation. ε = f/q or F = qE ……………………………………..3.5 The magnitude of E is the force per unit charge and its direction is that of F (i.e the direction of the force which acts on a positive charge). Thus E is a vector. 3.4.1 Calculation of the Electric Field In order to measure the electric field in a given region, we introduce a test charge and measure the force on it. However, we should realize that the test charge q exerts forces on the charge that produce the field, so it may change the configuration of the charges. In principle, the test charge should be so small as to have no significant effect on the charge configuration that produces the field. Equation 3.5 shows that the electric field is measured in newtons Coulomb-1 (NC-1). Since F is a vector quantity, E will also be a vector. If q is positive, the electric field E has the same direction as the force acting on the charge. If q is negative, the direction of E is opposite to that of the force F. Let us consider the electric field of a point charge. We already know from coulomb’s law that if we place a point charge q1 at a distance r from another point charge q1 the force on q1 will be. F = 1 qq1 ………………..3.6 4πεo r2 NOUN 21 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS Since the electric field is force per unit charge, we divide the force is equation 3.6 by the charge q1 to obtain the field due to q at the location of q1. That is E = F = 1 = q ………………..3.7 q 4πεo r 2 Equation 3.7 gives the field arising due to the charge q at any location which is at a distance r from q. What is the situation when the electric field is due to two or more charges? The answer is simple. Since the electric force obeys the superposition principle, so does the electric field (force per unit charge). Therefore, the field at a given point due to two or more charges is the vector sum of the fields of individual charges. Example An electric field is set up by two point charges q1 and q2 such that q1 = -q2 = 12 x 10-9 C and separated by a distance of 0.1m as shown in figure 3.3. Find the electric field at the points A and B. E1 B 60° EB O 0.1m E2 0.1m -q2 C +q1 A 0.04m 0.05 0.1m Fig 3.3 example 3. Solution (i) At point A1 the electric field E1 due to q1 is NOUN 22 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS E1 = (9x109 Nm2 C-2) 12 x 10-9 C (0.06)2 = (9 x 109 Nm2 C-2) x 12 x 10-9C 36 x 10-4 m2 = 3 x 104 NC-1 in the +ve x-direction At A , the electric field E2, due to q2 is E2 = (9 x 109) x (-12 x 10 -9 C) (0.04)2 = 9 x 109 Nm2 C-2 x 12 x 10-9C 4 x 10-4m2 = 6.75 x 10 NC-1 in the +ve x-direction 4 Therefore, the net electric field, E1 at point A is given by EA = E1 + E2 = (3 + 6.75) x 104 NC-1 = 9.75 x 104 NC-1 At point B, the electric field E, due to q, is E1 = (9x109 Nm2C -2) 12 x 10-9C (0.1m)2 4 -1 = 1.08 x 10 NC in the direction shown in Fx, 3. Similarly the field E2 due to q2 is E2 = (9x109 Nm2 C-2)2 (-122 x 10-9C) (0.1m)2 = -1.08x 104 NC-1 (The minus sign shows that the electric field points diagonally downward to the right) START Now we have to add the two forces vertically. If we resolve E1 and E2 into components along the x-axis and y-axis, it is clear from Fig. 3.3 that the y – components of vectors E1 and E2 cancel out and those along x-axis, i.e. BO add. You will observe that the angel between either vector and the x-direction is 600 because the triangle formed by B1 q1 and –q2 is an equilateral triangle. The direction of the resultant field is, therefore, along BO and its magnitude is given by NOUN 23 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS EB = (1.08 x 104 cos 600 + 1.8 x 104 cos 600) = 1.08 x 104 NC-1 3.4.2 Field lines An electric field can be represented by electric field lines or lines of force. The lines are drawn so that (a) The field line at a point (or the tangent to it if it is curved) gives the direction of E at the point. This is the direction in which a positive charge would accelerate. (b) The number of lines per unit cross-section area is proportional to E. You should note that the field line is imaginary; the representation serves only the useful purpose of allowing us to know the general features of the electric field in the entire region at a glance. The tangent to the filed at point A in figure 3.4 shows the direction of electric field at that point. The field lines are continuous and extends throughout space depicting the electric field. EA EB A B Field line Figure 3.4 An electric field line Since a field lines is also defined as a path along which a free, positive, point charge would travel in an electric field, it is always drawn with an arrowhead indicating the direction of travel of the position charge. NOUN 24 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS Figure 3.5 Field lines due to a positive point charge Since the direction of an electric field is taken to be that of the direction of the forces on a positive charge, the field surrounding a point positive charge is radially outward, as shown in fig. 3.5 Figure 3.6 (a) + + + + Figure 3.6(b) Figure 3.6: The nature of field lines (a) Two unlike charges (b) Two like charges 4.0 Conclusion NOUN 25 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS You have now gone through the first unit of the course – electricity and magnetism. The most central point pertains to the electric charge. As you would see the concept of electric charge cuts across virtually all aspect of the course. You have also leant about the electric fields due to a single charge and those due to a system of charges. Do not forget that the electric force can be attractive or repulsive. 5.0 Summary * There are only two types of electric charge and they are arbitrarily called positive and negative. Like charges repel and unlike charges attract each other. * The unit of charge is the coulomb (C) * Charge is always conserved. That is, the algebraic sum of the charges in a closed system does not change * Electric charge is quantized, occurring only in discrete amounts. * The force between two charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. The force acts along the line joining the two charges. F = __1_ (q1 q2) 4πεo r2 The value of 1/4πεo is 9x109 Nm2C-2 * The electric field at a point in space is defined as the electric force exerted on a test charge placed at that point. E = E q * The electric field of a point charge q is given by E= 1 q r 4πεo r2 Where r is a unit vector pointing from the point charge q to the location at which the electric field is being calculated. * The electric field due to a distribution of charges, according to the superposition principle, is the vector sum of the fields of the individual charges making up the distribution NOUN 26 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS E = ∑ 1 qn rn n 4πεo 2 rn * Electric lines of force are only a visual way of representing an electric field. The tangent to a line of force at any point shows the direction of the electric field at that point. 6.0 Tutor Marked Assignments 6.1 Two charges +4e and +e are fixed at a distance a. A third charge q is placed on the straight line joining the two charges so that q is in equilibrium. Find the position of q. Under what circumstances will the equilibrium be stable or unstable? 6.2 ABCD is a square of 0.04m side. Charges 16 x 10-9, -16 x 10-9 and 32 x 10-9 coulomb are placed at the points A, C and D respectively. Find the electric field strength at point B. 6.3 A small object carrying a charge of -5x10-9 C experiences a force of 20 x 10-9 N in the negative x-direction when placed at a certain point in an electric field. (a) What is the electric field at the point? (b) What is the magnitude and direction of the force acting on a proton placed at the point ? 7.0 References And Other Resources Physics. A Textbook for Advanced Level Students. Tom Duncan John Murray (Publishers ) Ltd, London 1982 Electrostatics in Free Spare – PHE-07. India Ghandi National Open University. October 2001 A Second Course of Electricity. A. E. E. Mckenzie University Press, Cambridge 1973 NOUN 27 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS UNIT 2 GAUSS’S LAW Table Of Content 1.0 Introduction 2.0 Objectives 3.1 Electric Flux 3.2 Gauss’s Law 3.3 Applications Of Gauss’s Law NOUN 28 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS 3.3.1 Spherical Symmetry 3.3.2 The Electric Field Of A Spherical Charge Distribution 3.3.3 Line Symmetry 3.3.4 Plane Symmetry 4.0 Conclusions 5.0 Summary 6.0 Tutor Marked Assignments(TMAs) 7.0 References And Other Resources 1.0 Introduction Gauss’s law expresses the relation between an electric charge and the electric field that it sets up. It is a consequence of Coulomb’s law. Although it contains no additional information, its mathematical form enables us to solve many problems of electric field calculation far more conveniently than through the use of Coulomb’s law. In unit 1, you learnt that electric field at any point is given by the force experience by a unit positive charge placed at that point. In this unit, we will develop the concept of flux of an electric field and then arrive at the Gauss’s law. We will also see how the gauss’s law allows us to calculate the electric field far more easily than we could using Coulomb’s law. 2.0 Objectives After studying this unit, you will be able to: * relate the electric flux through any surface to: (i) the field strength (ii) the surface area (iii) the orientation of the surface relative to the Field * write relation between the electric flax and the charge enclosed within the surface. * compute the electric flux through any closed surface placed in the electric field. * use Gauss’s law to compute electric fields in the case of spherical, linear and planar symmetry. NOUN 29 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS 3.1 Electric flux The number of the lines of force crossing any surface depends on three factors – the field strength, E, the surface area, S and the orientation of the surface relative to the electric field. S θ E Figure 3.1 To specify the orientation of the surface, S we draw a perpendicular to the surface. If θ is the angle between the electric field and the perpendicular as shown in figure 3.1, then the number of lines of force passing through the surface ranges from maximum to minimum depending on θ. That is When θ = 0; the number of lines of force crossing the surface is maximum When θ= 900, the number of lines of force crossing the surface is zero. We can now see that the number of lines of force crossing a surface is proportional to the projection of the field on to the perpendicular to the surface, i.e cos θ -(Note cos00 = 1, cos 900 = 0). Putting together quantities on which the number of lines of force depends gives the relationship. Number of lines of force crossing a surface α ES cos θ = E. S – 3.1. Where E is the electric field vector and S is a vector whose magnitude is equal to the area of the surface and whose direction is that of the perpendicular to the surface. The quantity on the left side of Eq. is an indefinite number because we can draw as many lines of force as we like. But the quantity on the right hand side of the equation has a definite value. It is called electric flux. Let us denote it by φ. Hence NOUN 30 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS Φ = E. S ………………………3.2 Eq. 3.2 shows that the electric flux is a Scalar, being the scalar product of two vectors. Since E is measured in NC-1, the unit of flax is Nm2C-1 Example 1 ds E E ds E S ds S1 S2 Figure 3.2 Figure 3.2 shows a closed surface S in the form of a cylinder of radius R situated in a uniform electric field F, the axis of the cylinder being parallel to the field. What is the flux φ of the electric field through this closed surface? Solution We can write the total electric flux through the surface S as the sum of three terms, an integral over the surface, S1 i.e the left cylinder cap. S2, the cylinder surface, and S3, the right cap. Thus from Eq. 3.2, we have Φ = ∫E. ds S Φ = ∫ E. ds + ∫ E.ds + ∫ E. ds S1 S2 S3 For the left cap, angle θ for all points is 18000, E is constant, and all the vectors ds are parallel. Therefore , ∫ E. ds = ∫ E (cos 180 ) ds 0 ∫ = -E ds = - πER2 S1 Since the area of the cap is πR2. Similarly, for the right hand cap NOUN 31 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS ∫ E. ds = + π ER2 S3 Since the angle θ = 0 for all points on the cap. Finally, for the surface S2. ∫ E.ds = 0 S2 Since the angle θ = 900 for all points on the cylindrical surface. We can now sum up the total flux through the cylindrical surface S as Φ = - πER2 + 0 + πER2 = 0 Therefore, the total outward flux of the electric field through the closed surface of figure 3.2 is zero. 3.2 Gauss’s Law In section 3.1, we found that: (i) The number of lines of force crossing any closed surface is proportional to the net charge enclosed by the surface (ii) The electric flux through any closed surface is proportional to the total charge enclosed by the surface. We can (ii) Mathematically as follows Φ α q enclosed ………………………………3.3 Or φ = ∫ E. ds α q enclosed……………………………..3.4 E ds q R Figure 3.3 NOUN 32 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS To evaluate the proportionality constant in equation 3.3 or 3.4, let us consider a positive point charge q in free space and a spherical surface of radius R centred on q as shown in fig. 3.3. The flax through any surface is given by equation, i.e Φ = ∫ E.ds = ∫ E.ds cos θ ………………….3.5 Where θ is the angle between the direction of the electric field and the outward normal to the surface. Now, the magnitude of the electric field at a distance R due to point charge q is given by E = 1 q 4πεo R2 The field point radially outward so that the electric field is everywhere parallel to the outward normal to the surface. Then θ = 0 and cos θ = 1. Putting the values of E and cosθ in equation 3.5, the flux through the spherical surface of radius R becomes Φ = ∫ 1 q ds = 1 q ∫ ds sphere 4πεo R 2 4πεo R2 Note that the expression for the magnitude of the electric field has been taken outside the integral sign because it has the same value (it is constant) everywhere on the spherical surface. The remaining integral is just the areas of all infinitesimal elements, ds, on the sphere. In other word, the remaining integral is the surface area of the sphere, that is ∫ ds = 4πR2 Then the flax becomes Φ = 1__ q 4πR2 = q/Eo ……….3.6 4πεo R2 Comparing eqns 3.4 and 3.6, we observe that the proportionality constant is 1/εo. Hence we have Φ = ∫ E. ds = q enclosed ……………………3.7 εo NOUN 33 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS Equation 3.7 is known as Gauss’s law. It tells us that the electric flax through the sphere is proportional to the charge and is independent of the radius of the surface. 3.3 Application of Gauss’s Law Gauss’s law applies to any hypothetical closed surface (called a Gaussian surface ) and enclosing a charge distribution. However, the evaluation of the surface integral becomes simple only when the charge distribution has sufficient symmetry. In such situation, Gauss’s law allows us to calculate the electric field far more easily than we could using Coulomb’s law. Since gauss’s law is valid for an arbitrary closed surface, we will use this freedom to choose a surface having the same symmetry as that of the charge distribution to evaluate the surface integral. We shall now illustrate the use of Gauss’s law for some important symmetries. 3.3.1 Spherical Symmetry A charge distribution is spherically symmetric if the charge density (that is, the charge per unit volume) at any point depends only on the distance of the point from a central point (called centre of symmetry) and not on the direction. Figure 3.4 Figure 3.4 represents a spherically symmetric distribution of charge such that the charge density is high at the centre and zero beyond r. Spherical symmetry of charge distribution implies that the magnitude of electric field NOUN 34 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS also depends on the distance r from the centre of symmetry. In such a situation, the only possible direction of the field consistent with the symmetry is the redial direction – outward for a positive charge (fig 3.4) and inward for a negative charge. Example of spherically symmetric charge distributions are: (i) a point charge (ii) a uniformly charged sphere, and (iii) a uniformly charged thin spherical shell Example 1. Use Gauss’s law to derive the expression for the electric field of a point charge. Solution r ds q Figure 3.5 Figure 3.5 shows a positive point charge q. Using Gauss’s law, let us find out the electric field at a distance r from the charge q. Draw a concentric spherical Guass’s surface of radius r. we know from symmetry that E points radially outward. If we divide the Gaussian surface into differential areas ds, then both E and ds will be at right angles to the surface, the angle θ between them being zero. Thus the quantity E.ds becomes simply Eds and Gauss’s law becomes εo φ E.ds = Eo φ Eds = q Since E has the same magnitude for all points on the Gaussian surface , we can write εo φ E. ds = εo φ Eds = εo φds = q …….3.8 However, the integral in equation 3.8 is simply the area of the spherical surface, i.e 4πr2 Hence NOUN 35 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS εo E (4πr2) = q Or E = q/4πεor2 ……………………3.9 Equation 3.9 is coulomb’s law in the form we derived it in unit 1. We can now see that Gauss’s law and Coulomb’s law are not two independent physical laws but the same law expressed in different ways. 3.3.2 The Electric Field of a Spherical Charge Distribution E1 P1 P2 E2 Figure 3.6 Let us consider a total charge Q which s spread uniformly throughout a sphere of radius R as shown is figure 3.6. We want to find the electric field at some point such as P1 outside the distribution and at point P2 inside it. (a) For points outside the charge distribution, let us draw a Gaussian surface S1of radius r, through the point P1. How do we now take advantage of the spherical symmetry? Answer: Because of the spherical symmetry, the electric field is the same at all points on the Gaussian surface What is the direction of the electric field? Answer: At any point on the Gaussian surface, the field is radically directed, i.e perpendicular to the surface, so that the angle between the normal to the surface and the electric field direction is zero. That is cos θ = cos 0 = 1 (Here it is assumed that the sphere has a positive charge, if there is a net negative charge, the field will point radially (inward whereby θ = 180° and cosθ = -1). NOUN 36 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS Then the flux through the Gaussian sphere S, becomes Φ = ∫E1. ds1, = ∫E1 cos θ ds1, = E1 ∫ds1 (since cos θ = 1) = 4πr21 E1 ………………………….3.10 Since ∫ds, is just the surface area of the sphere S1, i.e. 4πr2 Applying Gauss’s law 4πr12E1 = Q/Eo Since the charge enclosed within the sphere S1 is Q E1 = 1 Q 4πEo r12 …………………………3.11 Equation 3.11 shows that the field at all points on surface S1 is the same as if all the charges within the surface S1 were concentrated at the centre. (b) For points inside the charge distribution, the electric field depends on how the charge is distributed. This is because any Gaussian sphere with r< R1 such as surface S2 in figure 3.6 does not enclose the entire charge Q. The charge enclosed depends on the charge distribution. Suppose a Gaussian sphere S2 of radius r2 is drawn passing through the point P2 where we wish to find the electric field. Let the field b denoted by E2. Inside the sphere S2, eqn. 3.7 for the flux still holds. However, the charge enclosed is some fraction of Q. The volume of the charge sphere is 4πR3 and it 3 contains a total charge Q. Since the charge is spreading uniformly throughout the sphere, the volume charge density ρ is constant and is given by. ρ = Q 4π R3 3 NOUN 37 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS Therefore, the charge enclosed by the sphere S2 will be just the volume of that sphere multiplied by the volume charge density, that is q enclosed = 4π r23 x Q = Q r23 3 4π R3 R3 Applying the Gauss’s law (eqn. 3.7), we have 4π r22 E2 = Q r22/R3 So that E2 = 1 Q r2 4πEo R3 …………………3.12 Illustrate with a sketch the variation of the electric fields both inside and outside a spherical charge distribution. Answer: fig. 3.7 E R 0 R P The electric field inside the distribution increases generally with distance from the centre (E α r). On the other hand, outside the charge distribution, the electric field falls off as 1/r2 figure 3.7 is a sketch of the fields inside and outside the sphere. NOUN 38 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS 3.3.3 Line Symmetry A charge distribution has cylindrical symmetry when it is infinity long and has a charge density that depends only on the perpendicular distance from a line called symmetry axis (see fig. 3.8 r ℓ Gussian Surface E Figure 3.8 By symmetry, the electric field will point radially outward from the axis and its magnitude will depend only on the perpendicular distance from the axis. (We assume positive charge, for negative charge the field points inward). Let us find expression for the electric field, E at a distance r from the line charge (say a wire). We draw a Gaussian surface which is a circular cylindrical of radius r and length l closed at each end by plane caps normal to the axis as shown in figure 3.8 To calculate the flux, you should note that the electric field also has cylindrical symmetry, which implies that its magnitude at a point depends only on the perpendicular distance of the point form the symmetry axis, and its direction has to be radially outwards. The flax through the cylindrical surface is Φ = ∫ E.ds = ∫Eds = E∫ ds = 2πrlE ………..3.13 Where 2π rℓ is the area of the curved surface. NOUN 39 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS The flax through the end of the cylinder is zero because the field lines are parallel to the plane caps of the Gaussian surface. Mathematically, E and ds are perpendicular, so that cos θ = 0 in the scalar (dot) product E. ds. Therefore, the only flax is through the curved part of the cylinder. Gauss’s law tells us that the flax is proportional to the charge enclosed within the cylindrical Gaussian surface, i.e. 2π rι E = q enclosed εo So that E = q enclosed ……………………3.14 2πεorl If the line charge density is λ, then the charge enclosed by the Gaussian cylinder of length λl. Hence eqn 3.14 becomes E = q enclosed = λl = λ …………..3.15 2 πεo rℓ 2πεorl 2πεor With regard to the electric field inside the wire, we shall consider two cases: (i) Suppose the charge is distributed uniformly within the wire with charge density ρ. ℓ P E r R Figure 3.9 An enlarged view of the wire Let the radius of the wire be r. to find E at the inner point P 1 a distance r apart from the axis of the wire, we draw a Gaussian surface (i.e a cylinder) of radius r and length ℓ passing through P as shown in figure 3.9. As we saw earlier, the flux is due to the curved surface only. Hence from Gauss’s law. NOUN 40 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS ∫ E. ds = 2π rℓ E = q1/Eo The charge q1 inside this Gaussian surface = π p r2 ι E 2 π r ℓ= π ρ r2ℓ εo Or E = rρ ……………………….3.16 2εo Thus the electric field at a point inside an infinite uniformly charged wire is radially directed and varies as the distance from its axis. (ii) When the charge is on the surface of the wire or cylinder only, the electric field at any point inside it is zero because the net charge in the Gaussian surface through this point is zero. How does the electric field due to a charged wire or cylinder depend on its radius? Answer: Equations 3.15 and 3.16 show that electric field due to a charged wire or cylinder does not depend upon its radius. In effect, the field behaves as if the charge on the wire or cylinder were concentrated in a line along its axis. 3.3.4 Plane Symmetry When the charge density depends only on the perpendicular distance from a plane, the charge distribution is said to have plane symmetry. The electric field is everywhere normal to the plane sheet as shown in figure 3.10, pointing outward, if positively charged and inward, if negatively charged. s r s r NOUN 41 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS Fig 3.10 A charged distribution with plane Symmetry showing electric field To find the electric field at a distance in front of the plane sheet, it is required to construct a Gaussian surface. A convenient Gaussian surface is a closed cylinder of cross-sectional area S and length 2r. The sides of the Gaussian surface are perpendicular to the symmetry plane and the ends of the surface are parallel to it. Since no lines of force cross the sides, the flux through the sides is zero. But the lines of force cross perpendicular to the ends, so that E and the area element vector ds on the ends are parallel. The cos θ in the product E.ds is 1 over both ends (-1 if charge is negative). Since the flux through the sides is zero, the total flux through our Gaussian surface then becomes Φ = ∫ Eds = 2ES both ends The factor, 2 arises because there are two ends. The Gauss’s law gives 2 ES = q enclosed Eo If σ is the surface charge density, then the charge enclosed is σs. Hence E = σ ……………………3.17 2εo 4.0 Conclusion This unit is a follow-up of the previous unit. In particular, you have now learnt to apply the Gauss’s law for the solution of problems involving electric charges with linear spherical and plane symmetry. You appreciate that the solution of problems with the Gauss’s law is not as tedious as the use of coulomb’s law. Always remember that Gauss’s law applies to a closed surface, usually referred to as a Gaussian surface) and enclosing any charge distribution. For problems involving the application of Gauss’s law, choose a surface having the same symmetry as that of the charge distribution to evaluate the surface integral. However we note that unlike Coulomb’s law its not sufficient to determine the electric fields in all cases. 5.0 Summary NOUN 42 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS * the number of lines of force crossing a closed surface is proportional to the total charge enclosed by the surface. * the concept of electric flux quantifies the notion “number of lines of force crossing a surface” the electric flux, φ is defined as the surface integral of the electric field E over a surface as follows: φ = ∫ E.ds where ds is an infinitesimal vector whose direction at any point is towards the outward drawn normal to the surface at that point, its magnitude being the area of the surface. * Gauss’s law is ∫ E.ds = q/εo In which q is the net charge inside an imaginary closed surface (called a Gaussian surface) and εo is permittivity of free space. Gauss’s law expresses an important property of the electric field. * The electric field outside a spherically symmetrical shell with radius r and total charge q is directed radially and has magnitude E = 1 q ( r > R) 4 πεo r 2 The charge behaves as if it were all concentrated at the centre of the sphere. * The electric field due to an infinite line of charge with uniform charge per unit length, λ, is a direction perpendicular to the line of charge and has magnitude E = λ 2 πεor * The electric field due to an infinite sheet of charge is perpendicular to the plane of the sheet and has magnitude. E = σ / 2εo Where σ is the surface charge density. 6.0 Tutor Marked Assignments NOUN 43 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS 1. The electric field in a certain space is given by E = 200 r. How much flux passes through an area A if it is a portion of (a) The xy – plane (b) The xz – plane (c) The yz – plane 2. A flat sheet of area 50cm2 carries a uniform surface charge density σ. An electron 1.5cm a point near the center of the sheet experience a force of 1.8x10-12N directed away from the sheet. Find the total charge on the sheet. 3. Suppose that a positive charge is uniformly distributed throughout a spherical volume of radius R, the charge per unit volume being e. (a) Use Gauss’s law to prove that the electric field inside the volume and at a distance r from the centre, is E = er 3εo (b) What is the electric field at a point outside the spherical volume at a distance r from the centre ? Express your answer in terms of the total charge q within the spherical volume. (c) Compare your answer to (a) and (b) when r = R. (d) Sketch a graph of the magnitude of E as a function of r, from r = o to r = 3R. 7.0 Reference And Other Resources College Physics. Sears F.W, Zemansky M.W and Young H.D. Addison-Wesley publishing Company, London. 1975 Electrostatic in Free Space. Indira Ghandi Open University. PHE -07, 2001 Physics. A textbook for Advanced Level Students. Tom Duncan John Murray (Publishers) Ltd. London. 1982 NOUN 44 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS UNIT 3 ELECTRIC POTENTIAL Table Of Content NOUN 45 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS 1.0 Introduction 2.0 Objectives 3.1 Equipotential Surface 3.2 Potential due to a Point Charge 3.3 Potential due to a System of Charge 3.4 Potential Difference 3.5 Relation Between Electric Field and Electric Potential 3.6 Electric Field and Potential of an Electric Dipole 4.0 Conclusions 5.0 Summary 6.0 Tutor Marked Assignment 7.0 References And Other Resources 1.0 Introduction You will recall from your study of PHY 101. Elementary mechanics that work isI done when the point of application of a force undergoes a displacement in its own direction. If a body A exerts a force on another body B and work is done, a transfer of energy occurs which is measured by the work done. For example, if we lift a mass, m through a vertical height, h, the work done, W by the force we apply is W = mgh. The energy transfer is mgh and we consider that the system gains and store that amount of gravitational potential energy which is obtained from the conversion of chemical energy by our muscular activity. When the mass falls, the system loses gravitational potential energy, and, neglecting air resistance, there is a transfer of kinetic energy to the mass equal to the work done by gravity. The meaning of electric potential can be illustrated by the above analogy from basic mechanics. A body when raised above the earth’s surface is said to acquire potential energy because it can do work in falling. If it is free to move in a gravitational field it will fall to the position in which its potential energy is zero. Similarly, a charged body in an electric field has potential energy, and it will tend to move to those parts of the field where its potential energy is smaller. When a positive charge is repelled by another positively charged body and moves away, its potential energy decreases. It will be zero when it is completely away from the influence of the charged body, that is at infinity. We select as the zero of electric potential the potential at an infinite distance from any electric charges. Definition of Electric Potential NOUN 46 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS The electric potential at a point in a field can be defined as the work done per unit charge moving from infinity to the point. You should note that we are always assuming that the charge does not affect the field in any way. The choice of the zero of potential is purely arbitrary and whilst infinity may be a few hundred metres in some cases, in atomic physics where distances of 10-10m are involved, it need only be a very small distance away from the charge responsible for the field. Electric potential is a property of a point in a field and is a scalar since it deals with a quantity of work done or potential energy per unit charge. The symbol for potential is V and the unit a joule per coulomb (JC-1) or volts (V). 2.0 Objectives After studying this unit, you will be able to: * Compute the work done in taking a charge q from one point to another in an electric field. * Compute the electric potential at a point due to a single charge * Relate the electric potential and electric field, and thereby compute the electric field at a point knowing the electric potential. * Compute the electric potential at a point due to a dipole and a quadrupole 3.1 Equipotential Surfaces All point in a field which have the same potential can be imagined as lying on a surface, called an equipotential surface. When a charge moves on such a surface no energy change occurs and no work is done. The force due the field must therefore act at right angles to the equipotential surface at any point. Therefore, equipotential surfaces and field lines always interst at right angles. NOUN 47 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS Field line Equipeteateal Surface Figure 3.1 Equipotential surface for a point charge We can now see that as electric field can therefore be represented pictorially by field lines and by equipotential surface (or lines in two dimensional diagrams). Figure 3.1 shows the equipotential surfaces for a point charge. These are concentric spheres (circles in two dimensions). If equipotential are drawn so that the change of potential from one to the next is constant, then the spacing will be closer where the field is stronger. This follows from the fact that in order to perform a certain amount of work in such regions a shorter distance need be travelled. 3.2 Potential due to a point charge +Qo +Q A B C r δx x fig. 3.2 We wish to find the potential at point A in the field of an isolated point charge +Q situated at point 0, such that 0 A = r as shown in fig. 3.2. Let us imagine a very small point charge +Qo is moved by an external agent from C distance x from A, through a very small distance δx to B without affecting the field due to +Q. Assuming the force F on Qo due to the field remains constant over, δx, the work δW by the external agent over δx against the force of the field is NOUN 48 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS δW = F (-δx) ……………………………….3.1 Do you know why we have to insert the negative sign for δx? This is to take account of the fact that the displacement δx is in the opposite direction to that in which F acts. Applying coulomb’s law. F = Q Qo 1 4πεo x2 δW = Q Qo (-δx) 4πεo x2 The total work done, W in bringing Qo from infinite Qo from infinity to A is W = -Q Qo ∫r∞ dx = - Q Qo -1/x r∞ 4πεo x2 4πεo = Q Qo …………………………………3.2 4πε ϒ 3.3 Potential due to a system of charges Let us now consider a system of charges. Like in the previous case of electric field we shall find the superposition principle very useful. That is, the total potential, Vp at a point P due to a system of charges q1, q2, ……..,qN is equal to the sum of the potentials due to the individual charges at that point. If ϒ1, ϒ2, ….ϒN are the distances of the charges q1, q2, ………. qN respectively from the point P, the potential at that point is p = q1 + q2 + ……. + qN ………………3.4 4πεoϒ1 4πεoϒ2 4πεoϒN You will note in eq. 3.4 that each charge is acting as if no other charge is present. The potential at point P. may be written in a summation form as p = 1 N qi/ri ………………………3.5 4πεo ∑ i = 1 Example 1 NOUN 49 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS The following point charges are placed on the x - axis: 2 μc at x = 20cm, -3μc at x = 30cm, -4μc at x = 40cm. Find the potential on the x – axis at the origin. Solution We know that potential is a scalar quantity. Using the superposition principle, the potential at the origin (x = 0) is given by o= 1 3 qi/ri 4πεo ∑ I=1 On substituting the numerical values of qi and ϒi, we obtain Vo = 9 x 109 Nm2 C-2 2 x 10-6 C - 3 x 10 -6 C - 4 x 10-6C 0.20m 0.30m 0.40m = 9 x 109 Nm2 C-2 105 Cm-1 – 10-5 Cm -1 – 10-5 Cm-1 = -9 x 104 Nm C-1 = -9 x 104V 3.4 Potential Difference Let us write down the amount of work done in bringing a unit positive charge from infinity, first to point A and then to point B, shown in figure 3.3. Remember that point A and B are within the field of charge q. YA A YB B NOUN 50 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS Fig. 3.3 Using equation 3.3, the potentials at point A and B are VA = q 4πεoϒA And VB = q 4πεoϒB The difference of these two potentials (i.e VB – VA) is the work done in taking a unit charge from A to B, and it is called the potential difference the two points B and A. it is written as VBA = VB –VA = q 1 - 1 …………..3.6 4πε ϒB ϒA We observe that the work done is carrying the charge in an electric field is independent of path. It is just this path independence that enables us to define the concept of potential. By the way, you should note that if, instead of the unit positive charge, we transport a charge q1 between A and B, then the work W done is given by: W = q1 VBA = q1 ( VB – VA ) ……….3.7 On a final note, I want you to bear in mind that the potential difference (p.d. for short) is a very importance concept in the field of electrostatics and current electricity. Its knowledge helps us to determine the exact value of the current which flows between any two points in an electric circuit, provided the resistance between the two points is known. We shall see this later in this course (module 3) 3.5 Relation Between Electric Field and Electric Potential A B E E 0 charge +Q x δx Fig 3.4 NOUN 51 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS Let us consider a charge +Q at a point A in an electric field where the field strength is E. This configuration is illustrated in figure 3.4 The force, F on Q is given by F = EQ If Q moves a very short distance δx from A to B in the direction of E, then the work done δW by the electric force on q is δW = force x distance = Fδx = EQδx (assuming E is constant over AB) If the p.d between B and A is δV, we have by the definition of p.d. δV = work done per unit charge = δW = - EQ δx Q Q That is δV = -Eδx The negative sign is inserted to show that if displacements in the direction of E are taken to be positive, then when δx is positive, δV is negative, i.e the potential decreases. In the limit, as δx ----01 E becomes the field strength at a point (A). In calculus notation E = Lim δV = dv δx---0 δx dx dv/dx is called the potential gradient in the direction and so the field strength at a point equals the negative of the potential gradient there. Question: Is potential gradient a vector or a scalar? Answer: You notice that potential gradient involves displacement. Therefore, it is a vector. It is measured in volts per metre (Vm-1). The Vm-1 and NC-1 are both units of E, but the Vm-1 is the one that is commonly used. NOUN 52 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS If the electric field E is uniform, that is it is constant in magnitude and direction at all points, it follows that dv is constant i.e dv = -E ………………….3.8 dx dx ∴ V = Ex ∴ E = -v …………………….3.9 x in other words, the potential changes steadily with distance (Recall the case of ds/dt = constant, for uniform velocity) Example 2 Question: Two large horizontal, parallel metal plates are 2.0cm apart in air and the upper is maintained at a positive potential relative to the lower so that field strength between them is 2.5 x 10 5 Vm-1. (a) What is the p.d. between the plates? (b) If an electron of charge 1.6 x 10-19C and mass 9.1 x 10-31kg is liberated from rest at the lower plates, what is its speed on reaching the upper plate? Solution (a) If E is the field strength (assumed uniform) and V is the p.d. between the plates which are at a distance d apart, we have (from equation 3.9) V = Ed = (2.5 x 105 Vm-1) x (2.0 x 10-2m) = 5.0 x 103 volts (b) The energy change (i.e work done ) W which occurs when a charge Q moves through a p.d of V volts in an electric field is given by W = QV There is a transfer of electrical potential energy from the field to k.e. of the electron, Hence we have QV = ½ mv2 where v is the required speed and m is the mass of the electron NOUN 53 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS Therefore, ½ mv2 = QV V = 2qv m i.e v = 2x16x10-19C) x (5.0 x 103 V) (9.1 x 10-31kg) = 2 x 1.6 x 10-19 x 5.0 x 103 C JC-1 9.1 x 10-31 kg (IV = 1 JC-1) = 16 x 1015m2 5-2 9.1 (I J = INm = 1 kgm S-2m) = 4.2 x 101ms-1 3.6 Electric Field and Potential of an Electric Dipole A pair of equal and opposite charge, ±q, separated by a vector distance a is called a dipole. (see figure 3.5). The vector a, which is also along the axis of the dipole, is drawn from the negative to the positive charge. A molecule consisting of a positive and negative ion is an example of an electric dipole in nature. An atom consists of equal amount of positive and negative charges whose centre coincide, hence an atom is neutral for all points outside it. However, in the presence of an external electric field, the centres of positive and negative charges get separated. The atom then becomes a dipole. As we shall see in module 2, the electric field and potential in the vicinity of a dipole forms the first step in understanding the behaviour of dielectrics under the influence of an external electric field. 3.6.1 Electric Field at a point P along the axis of the dipole -q +q a c Ŷ P A B Y figure 3.5: Electric dipole AB with centre C and axis a. NOUN 54 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS The point P is along the axis. Let the distance between the mid-point of the dipole and the point which is along axis be equal to r. We shall evaluate the electric field at P. The electric field at P due to +q. is given by q r 4πεo (r-a/2)2 ϒ is a unit vector in the direction Cp. And that due to –q is E = -q r 4πεo (r + a/2)2 The resultant field at P is E = E+ + E- = qr 1 - 1 4πε r–a 2 r+a 2 2 2 = q (2a r) r 4πεo r2-a2 2 4 E≏ 2Pr for r >> a ……………………3.10 3 4πεor Where p = qa r ………………..3.11 You will notice that, in the denominator, we have neglected a2/4 as compared to r2, since a a. 3.6.2 Potential due to a dipole Let us evaluate the potential Vp at P, a distance r from the mid-point C of the dipole. (see figure 3.6). the line joining P to C makes an angle θ with the dipole axis, a. NOUN 55 PHY 121 ELECTRICITY, MAGNETISM AND MODERN PHYSICS The distances of P from –q and +q are AP and BP respectively. From the geometry, you will notice that BP = SP = PC – CS = r – a 2 cosθ And AP = TP = TC + CP = r + a 2 cosθ R α r s -q a θ +q A C B T Hence the potential at P is equal to Vp = q 1 1 4πEo ( r – /2 cos θ) (r + a/2 cos θ)