Chapter 1. Electric Charges and Fields PDF

Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Chapter One ELECTRIC CHARGES AND FIELDS om.c i ya un D...

Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Chapter One ELECTRIC CHARGES AND FIELDS om.c i ya un D ls 1.1 INTRODUCTION ia All of us have the experience of seeing a spark or hearing a crackle when we take off our synthetic clothes or sweater, particularly in dry weather. or Have you ever tried to find any explanation for this phenomenon? Another common example of electric discharge is the lightning that we see in the sky during thunderstorms. We also experience a sensation of an electric t shock either while opening the door of a car or holding the iron bar of a Tu bus after sliding from our seat. The reason for these experiences is discharge of electric charges through our body, which were accumulated due to rubbing of insulating surfaces. You might have also heard that this is due to generation of static electricity. This is precisely the topic we are going to discuss in this and the next chapter. Static means anything that does not move or change with time. Electrostatics deals with the study of forces, fields and potentials arising from static charges. 1.2 ELECTRIC CHARGE Historically the credit of discovery of the fact that amber rubbed with wool or silk cloth attracts light objects goes to Thales of Miletus, Greece, around 600 BC. The name electricity is coined from the Greek word Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Physics om FIGURE 1.1 Rods: like charges repel and unlike charges attract each other. elektron meaning amber. Many such pairs of materials were known which on rubbing could attract light objects like straw, pith balls and bits of.c papers. It was observed that if two glass rods rubbed with wool or silk cloth ya are brought close to each other, they repel each other [Fig. 1.1(a)]. The two strands of wool or two pieces of silk cloth, with which the rods were rubbed, also repel each other. However, the glass rod and wool attracted i each other. Similarly, two plastic rods rubbed with cat’s fur repelled each un other [Fig. 1.1(b)] but attracted the fur. On the other hand, the plastic rod attracts the glass rod [Fig. 1.1(c)] and repel the silk or wool with which the glass rod is rubbed. The glass rod repels the fur. D These seemingly simple facts were established from years of efforts and careful experiments and their analyses. It was concluded, after many ls careful studies by different scientists, that there were only two kinds of an entry which is called the electric charge. We say that the bodies like glass or plastic rods, silk, fur and pith balls are electrified. They acquire ia an electric charge on rubbing. There are two kinds of electrification and we find that (i) like charges repel and (ii) unlike charges attract each or other. The property which differentiates the two kinds of charges is called the polarity of charge. When a glass rod is rubbed with silk, the rod acquires one kind of t charge and the silk acquires the second kind of charge. This is true for Tu any pair of objects that are rubbed to be electrified. Now if the electrified glass rod is brought in contact with silk, with which it was rubbed, they no longer attract each other. They also do not attract or repel other light objects as they did on being electrified. Thus, the charges acquired after rubbing are lost when the charged bodies are brought in contact. What can you conclude from these observations? It just tells us that unlike charges acquired by the objects neutralise or nullify each other’s effect. Therefore, the charges were named as positive and negative by the American scientist Benjamin Franklin. By convention, the charge on glass rod or cat’s fur is called positive and that on plastic rod or silk is termed negative. If an object possesses an electric charge, it is said to be electrified or charged. When it has no charge it is said to be electrically neutral. 2 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com  Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Electric Charges and Fields A simple apparatus to detect charge on a body is the gold-leaf electroscope [Fig. 1.2(a)]. It consists of a vertical metal rod housed in a box, with two thin gold leaves attached to its bottom end. When a charged object touches the metal knob at the top of the rod, charge flows on to the leaves and they diverge. The degree of divergance is an indicator of the amount of charge. Try to understand why material bodies acquire charge. You know that all matter is made up of atoms and/or molecules. Although normally the materials are electrically neutral, they do contain charges; but their charges are exactly balanced. Forces that hold the molecules together, forces that om hold atoms together in a solid, the adhesive force of glue, forces associated with surface tension, all are basically electrical in nature, arising from the forces between charged particles. Thus the electric force is all pervasive and it encompasses almost each and every field associated with our life. It is.c therefore essential that we learn more about such a force. To electrify a neutral body, we need to add or remove one kind of ya charge. When we say that a body is charged, we always refer to this excess charge or deficit of charge. In solids, some of the electrons, being less tightly bound in the atom, are the charges which are transferred from one body to the other. A body can thus be charged positively by i un losing some of its electrons. Similarly, a body can be charged negatively by gaining electrons. When we rub a glass rod with silk, some of the electrons from the rod are transferred to the silk cloth. Thus the rod gets positively charged and the silk gets negatively charged. No new charge is D created in the process of rubbing. Also the number of electrons, that are transferred, is a very small fraction of the total number of electrons in the ls material body. ia 1.3 CONDUCTORS AND INSULATORS Some substances readily allow passage of electricity through them, others or do not. Those which allow electricity to pass through them easily are called conductors. They have electric charges (electrons) that are comparatively free to move inside the material. Metals, human and animal t bodies and earth are conductors. Most of the non-metals like glass, Tu porcelain, plastic, nylon, wood offer high resistance to the passage of electricity through them. They are called insulators. Most substances fall into one of the two classes stated above*. When some charge is transferred to a conductor, it readily gets distributed over the entire surface of the conductor. In contrast, if some charge is put on an insulator, it stays at the same place. You will learn why this happens in the next chapter. This property of the materials tells you why a nylon or plastic comb gets electrified on combing dry hair or on rubbing, but a metal article * There is a third category called semiconductors, which offer resistance to the movement of charges which is intermediate between the conductors and insulators. 3 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Physics like spoon does not. The charges on metal leak through our body to the ground as both are conductors of electricity. However, if a metal rod with a wooden or plastic handle is rubbed without touching its metal part, it shows signs of charging. 1.4 BASIC PROPERTIES OF ELECTRIC CHARGE We have seen that there are two types of charges, namely om positive and negative and their effects tend to cancel each other. Here, we shall now describe some other properties of the electric charge. If the sizes of charged bodies are very small as.c compared to the distances between them, we treat them as point charges. All the charge content of the body is assumed to be concentrated at one point in space. FIGURE 1.2 Electroscopes: (a) ya 1.4.1 Additivity of charges We have not as yet given a quantitative definition of a i un The gold leaf electroscope, (b) charge; we shall follow it up in the next section. We shall Schematics of a simple tentatively assume that this can be done and proceed. If electroscope. a system contains two point charges q1 and q2, the total charge of the system is obtained simply by adding D algebraically q 1 and q2 , i.e., charges add up like real numbers or they are scalars like the mass of a body. If a system contains n charges q1, ls q2, q3, …, qn, then the total charge of the system is q1 + q2 + q3 + … + qn. Charge has magnitude but no direction, similar to mass. However, ia there is one difference between mass and charge. Mass of a body is always positive whereas a charge can be either positive or negative. or Proper signs have to be used while adding the charges in a system. For example, the total charge of a system containing five charges +1, +2, –3, +4 and –5, in some arbitrary unit, is (+1) + (+2) + (–3) + (+4) + (–5) = –1 in t the same unit. Tu 1.4.2 Charge is conserved We have already hinted to the fact that when bodies are charged by rubbing, there is transfer of electrons from one body to the other; no new charges are either created or destroyed. A picture of particles of electric charge enables us to understand the idea of conservation of charge. When we rub two bodies, what one body gains in charge the other body loses. Within an isolated system consisting of many charged bodies, due to interactions among the bodies, charges may get redistributed but it is found that the total charge of the isolated system is always conserved. Conservation of charge has been established experimentally. It is not possible to create or destroy net charge carried by any isolated system although the charge carrying particles may be created or destroyed 4 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com  Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Electric Charges and Fields in a process. Sometimes nature creates charged particles: a neutron turns into a proton and an electron. The proton and electron thus created have equal and opposite charges and the total charge is zero before and after the creation. 1.4.3 Quantisation of charge Experimentally it is established that all free charges are integral multiples of a basic unit of charge denoted by e. Thus charge q on a body is always given by om q = ne where n is any integer, positive or negative. This basic unit of charge is the charge that an electron or proton carries. By convention, the charge on an electron is taken to be negative; therefore charge on an electron is written as –e and that on a proton as +e..c The fact that electric charge is always an integral multiple of e is termed as quantisation of charge. There are a large number of situations in physics ya where certain physical quantities are quantised. The quantisation of charge was first suggested by the experimental laws of electrolysis discovered by English experimentalist Faraday. It was experimentally demonstrated by Millikan in 1912. i un In the International System (SI) of Units, a unit of charge is called a coulomb and is denoted by the symbol C. A coulomb is defined in terms the unit of the electric current which you are going to learn in a D subsequent chapter. In terms of this definition, one coulomb is the charge flowing through a wire in 1 s if the current is 1 A (ampere), (see Chapter 1 ls of Class XI, Physics Textbook , Part I). In this system, the value of the basic unit of charge is ia e = 1.602192 × 10–19 C Thus, there are about 6 × 1018 electrons in a charge of –1C. In or electrostatics, charges of this large magnitude are seldom encountered and hence we use smaller units 1 mC (micro coulomb) = 10–6 C or 1 mC (milli coulomb) = 10–3 C. t If the protons and electrons are the only basic charges in the Tu universe, all the observable charges have to be integral multiples of e. Thus, if a body contains n1 electrons and n2 protons, the total amount of charge on the body is n2 × e + n1 × (–e) = (n2 – n1) e. Since n1 and n2 are integers, their difference is also an integer. Thus the charge on any body is always an integral multiple of e and can be increased or decreased also in steps of e. The step size e is, however, very small because at the macroscopic level, we deal with charges of a few mC. At this scale the fact that charge of a body can increase or decrease in units of e is not visible. In this respect, the grainy nature of the charge is lost and it appears to be continuous. This situation can be compared with the geometrical concepts of points and lines. A dotted line viewed from a distance appears continuous to us but is not continuous in reality. As many points very close to 5 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Physics each other normally give an impression of a continuous line, many small charges taken together appear as a continuous charge distribution. At the macroscopic level, one deals with charges that are enormous compared to the magnitude of charge e. Since e = 1.6 × 10–19 C, a charge of magnituOde, say 1 mC, contains something like 1013 times the electronic charge. At this scale, the fact that charge can increase or decrease only in units of e is not very different from saying that charge can take continuous values. Thus, at the macroscopic level, the quantisation of charge has no practical consequence and can be ignored. However, at the microscopic level, where the charges involved are of the order of a few tens or hundreds om of e, i.e., they can be counted, they appear in discrete lumps and quantisation of charge cannot be ignored. It is the magnitude of scale involved that is very important..c Example 1.1 If 109 electrons move out of a body to another body every second, how much time is required to get a total charge of 1 C ya on the other body? Solution In one second 109 electrons move out of the body. Therefore the charge given out in one second is 1.6 × 10–19 × 109 C = 1.6 × 10–10 C. The time required to accumulate a charge of 1 C can then be estimated i un to be 1 C ÷ (1.6 × 10–10 C/s) = 6.25 × 109 s = 6.25 × 109 ÷ (365 × 24 × 3600) years = 198 years. Thus to collect a charge of one coulomb, from a body from which 109 electrons move out every second, we will need approximately 200 years. One coulomb is, therefore, a very large EXAMPLE 1.1 D unit for many practical purposes. It is, however, also important to know what is roughly the number of ls electrons contained in a piece of one cubic centimetre of a material. A cubic piece of copper of side 1 cm contains about 2.5 × 10 24 electrons. ia Example 1.2 How much positive and negative charge is there in a or cup of water? Solution Let us assume that the mass of one cup of water is 250 g. The molecular mass of water is 18g. Thus, one mole t (= 6.02 × 1023 molecules) of water is 18 g. Therefore the number of EXAMPLE 1.2 Tu molecules in one cup of water is (250/18) × 6.02 × 1023. Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons. Hence the total positive and total negative charge has the same magnitude. It is equal to (250/18) × 6.02 × 1023 × 10 × 1.6 × 10–19 C = 1.34 × 107 C. 1.5 COULOMB’S LAW Coulomb’s law is a quantitative statement about the force between two point charges. When the linear size of charged bodies are much smaller than the distance separating them, the size may be ignored and the charged bodies are treated as point charges. Coulomb measured the force between two point charges and found that it varied inversely as the square of the distance between the charges and was directly 6 proportional to the product of the magnitude of the two charges and Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com  Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Electric Charges and Fields acted along the line joining the two charges. Thus, if two point charges q1, q2 are separated by a distance r in vacuum, the magnitude of the force (F) between them is given by q1 q 2 F =k (1.1) r2 How did Coulomb arrive at this law from his experiments? Coulomb used a torsion balance* for measuring the force between two charged metallic spheres. When the separation between two spheres is much larger than the radius of each om sphere, the charged spheres may be regarded as point charges. However, the charges on the spheres were unknown, to begin with. How then could he discover a relation like Eq. (1.1)? Coulomb thought of the following simple way: Suppose the Charles Augustin de.c Coulomb (1736 – 1806) charge on a metallic sphere is q. If the sphere is put in contact Coulomb, a French with an identical uncharged sphere, the charge will spread over physicist, began his ya the two spheres. By symmetry, the charge on each sphere will CHARLES AUGUSTIN DE COULOMB (1736 –1806) career as a military be q/2*. Repeating this process, we can get charges q/2, q/4, engineer in the West etc. Coulomb varied the distance for a fixed pair of charges and Indies. In 1776, he measured the force for different separations. He then varied the i returned to Paris and un charges in pairs, keeping the distance fixed for each pair. retired to a small estate Comparing forces for different pairs of charges at different to do his scientific distances, Coulomb arrived at the relation, Eq. (1.1). research. He invented a Coulomb’s law, a simple mathematical statement, was torsion balance to D initially experimentally arrived at in the manner described measure the quantity of above. While the original experiments established it at a a force and used it for ls determination of forces macroscopic scale, it has also been established down to of electric attraction or subatomic level (r ~ 10–10 m). repulsion between small ia Coulomb discovered his law without knowing the explicit charged spheres. He magnitude of the charge. In fact, it is the other way round: thus arrived in 1785 at Coulomb’s law can now be employed to furnish a definition the inverse square law or for a unit of charge. In the relation, Eq. (1.1), k is so far relation, now known as arbitrary. We can choose any positive value of k. The choice Coulomb’s law. The law of k determines the size of the unit of charge. In SI units, the had been anticipated by t Nm 2 Priestley and also by Tu value of k is about 9 × 109. The unit of charge that Cavendish earlier, C2 though Cavendish results from this choice is called a coulomb which we defined never published his earlier in Section 1.4. Putting this value of k in Eq. (1.1), we results. Coulomb also see that for q1 = q2 = 1 C, r = 1 m found the inverse F = 9 × 109 N square law of force That is, 1 C is the charge that when placed at a distance between unlike and like of 1 m from another charge of the same magnitude in vacuum magnetic poles. experiences an electrical force of repulsion of magnitude * A torsion balance is a sensitive device to measure force. It was also used later by Cavendish to measure the very feeble gravitational force between two objects, to verify Newton’s Law of Gravitation. * Implicit in this is the assumption of additivity of charges and conservation: two charges (q/2 each) add up to make a total charge q. 7 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Physics 9 × 109 N. One coulomb is evidently too big a unit to be used. In practice, in electrostatics, one uses smaller units like 1 mC or 1 mC. The constant k in Eq. (1.1) is usually put as k = 1/4pe0 for later convenience, so that Coulomb’s law is written as 1 q1 q2 F = (1.2) 4 π ε0 r2 om e0 is called the permittivity of free space. The value of e0 in SI units is  0 = 8.854 × 10–12 C2 N–1m–2.c Since force is a vector, it is better to write Coulomb’s law in the vector notation. Let the position ya vectors of charges q1 and q2 be r1 and r2 respectively [see Fig.1.3(a)]. We denote force on q1 due to q2 by FIGURE 1.3 (a) Geometry and F12 and force on q2 due to q1 by F21. The two point i (b) Forces between charges. charges q1 and q2 have been numbered 1 and 2 for un convenience and the vector leading from 1 to 2 is denoted by r21: r21 = r2 – r1 D In the same way, the vector leading from 2 to 1 is denoted by r12: ls r12 = r1 – r2 = – r21 The magnitude of the vectors r21 and r12 is denoted by r21 and r12 , ia respectively (r12 = r21). The direction of a vector is specified by a unit vector along the vector. To denote the direction from 1 to 2 (or from 2 to or 1), we define the unit vectors: r r rɵ 21 = 21 , rɵ 12 = 12 , rɵ 21 − rɵ 12 t r21 r12 Tu Coulomb’s force law between two point charges q1 and q2 located at r1 and r2, respectively is then expressed as 1 q1 q 2 ɵ F21 = r 21 (1.3) 4 π εo 2 r21 Some remarks on Eq. (1.3) are relevant: · Equation (1.3) is valid for any sign of q1 and q2 whether positive or negative. If q1 and q2 are of the same sign (either both positive or both negative), F21 is along r̂ 21, which denotes repulsion, as it should be for like charges. If q1 and q2 are of opposite signs, F21 is along – rɵ 21(= rɵ 12), which denotes attraction, as expected for unlike charges. Thus, we do not have to write separate equations for the cases of like and unlike charges. Equation (1.3) takes care of both cases correctly [Fig. 1.3(b)]. 8 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com  Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Electric Charges and Fields · The force F12 on charge q1 due to charge q2, is obtained from Eq. (1.3), by simply interchanging 1 and 2, i.e., 1 q1 q 2 F12 = rˆ12 = −F21 4 π ε0 2 r12 Thus, Coulomb’s law agrees with the Newton’s third law. · Coulomb’s law [Eq. (1.3)] gives the force between two charges q1 and q2 in vacuum. If the charges are placed in matter or the intervening space has matter, the situation gets complicated due to the presence om http://webphysics.davidson.edu/physlet_resources/bu_semester2/menu_semester2.html Interactive animation on Coulomb’s law: of charged constituents of matter. We shall consider electrostatics in matter in the next chapter..c Example 1.3 Coulomb’s law for electrostatic force between two point charges and Newton’s law for gravitational force between two stationary point masses, both have inverse-square dependence on the distance ya between the charges and masses respectively. (a) Compare the strength of these forces by determining the ratio of their magnitudes (i) for an electron and a proton and (ii) for two protons. (b) Estimate the accelerations of electron and proton due to the electrical force of their i mutual attraction when they are 1 Å (= 10-10 m) apart? (mp = 1.67 × un 10–27 kg, me = 9.11 × 10–31 kg) Solution (a) (i) The electric force between an electron and a proton at a distance D r apart is: 1 e2 Fe = − ls 4 πε 0 r 2 where the negative sign indicates that the force is attractive. The ia corresponding gravitational force (always attractive) is: m p me FG = −G or r2 where mp and me are the masses of a proton and an electron respectively. t Fe e2 = = 2.4 × 1039 Tu FG 4 πε 0Gm pm e (ii) On similar lines, the ratio of the magnitudes of electric force to the gravitational force between two protons at a distance r apart is: Fe e2 = = 1.3 × 1036 FG 4πε 0Gm p m p However, it may be mentioned here that the signs of the two forces are different. For two protons, the gravitational force is attractive in nature and the Coulomb force is repulsive. The actual values EXAMPLE 1.3 of these forces between two protons inside a nucleus (distance between two protons is ~ 10-15 m inside a nucleus) are Fe ~ 230 N, whereas, FG ~ 1.9 × 10–34 N. The (dimensionless) ratio of the two forces shows that electrical forces are enormously stronger than the gravitational forces. 9 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Physics (b) The electric force F exerted by a proton on an electron is same in magnitude to the force exerted by an electron on a proton; however, the masses of an electron and a proton are different. Thus, the magnitude of force is 1 e2 |F| = = 8.987 × 109 Nm2/C2 × (1.6 ×10–19C)2 / (10–10m)2 4 πε 0 r 2 = 2.3 × 10–8 N Using Newton’s second law of motion, F = ma, the acceleration that an electron will undergo is om a = 2.3×10–8 N / 9.11 ×10–31 kg = 2.5 × 1022 m/s2 Comparing this with the value of acceleration due to gravity, we can conclude that the effect of gravitational field is negligible on EXAMPLE 1.3 the motion of electron and it undergoes very large accelerations under the action of Coulomb force due to a proton..c The value for acceleration of the proton is 2.3 × 10–8 N / 1.67 × 10–27 kg = 1.4 × 1019 m/s2 ya Example 1.4 A charged metallic sphere A is suspended by a nylon i un thread. Another charged metallic sphere B held by an insulating D ls ia t or Tu EXAMPLE 1.4 10 FIGURE 1.4 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com  Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Electric Charges and Fields handle is brought close to A such that the distance between their centres is 10 cm, as shown in Fig. 1.4(a). The resulting repulsion of A is noted (for example, by shining a beam of light and measuring the deflection of its shadow on a screen). Spheres A and B are touched by uncharged spheres C and D respectively, as shown in Fig. 1.4(b). C and D are then removed and B is brought closer to A to a distance of 5.0 cm between their centres, as shown in Fig. 1.4(c). What is the expected repulsion of A on the basis of Coulomb’s law? Spheres A and C and spheres B and D have identical sizes. Ignore the sizes of A and B in comparison to the separation between their om centres. Solution Let the original charge on sphere A be q and that on B be q¢. At a distance r between their centres, the magnitude of the electrostatic force on each is given by 1 qq ′.c F = 4 πε 0 r 2 neglecting the sizes of spheres A and B in comparison to r. When an ya identical but uncharged sphere C touches A, the charges redistribute on A and C and, by symmetry, each sphere carries a charge q/2. Similarly, after D touches B, the redistributed charge on each is i q¢/2. Now, if the separation between A and B is halved, the magnitude EXAMPLE 1.4 un of the electrostatic force on each is 1 (q / 2 )(q ′ / 2) 1 (qq ′ ) F′ = = =F 4 πε 0 (r / 2)2 4 πε 0 r 2 D Thus the electrostatic force on A, due to B, remains unaltered. ls 1.6 FORCES BETWEEN MULTIPLE CHARGES The mutual electric force between two charges is given by ia Coulomb’s law. How to calculate the force on a charge where there are not one but several charges around? Consider a or system of n stationary charges q1, q2, q3,..., qn in vacuum. What is the force on q1 due to q2, q3,..., qn? Coulomb’s law is not enough to answer this question. Recall that forces of t mechanical origin add according to the parallelogram law of Tu addition. Is the same true for forces of electrostatic origin? Experimentally, it is verified that force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges. This is termed as the principle of superposition. To better understand the concept, consider a system of three charges q1, q2 and q3, as shown in Fig. 1.5(a). The force on one charge, say q1, due to two other charges q2, q3 can therefore be obtained by performing a vector addition of the forces due to each one of these charges. Thus, if the force on q1 due to q2 is denoted by F12, F12 is given by Eq. (1.3) even though other charges are present. FIGURE 1.5 A system of 1 q1q 2 (a) three charges Thus, F12 = r̂12 (b) multiple charges. 11 4 πε 0 r12 2 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Physics In the same way, the force on q1 due to q3, denoted by F13, is given by 1 q1q3 F13 = rˆ13 4 πε 0 r13 2 which again is the Coulomb force on q1 due to q3, even though other charge q2 is present. Thus the total force F1 on q1 due to the two charges q2 and q3 is given as 1 q1q 2 1 q1q 3 F1 = F12 + F13 = rˆ12 + rˆ13 (1.4) 4 πε 0 r12 2 4 πε 0 r13 2 om The above calculation of force can be generalised to a system of charges more than three, as shown in Fig. 1.5(b). The principle of superposition says that in a system of charges q1, q2,..., qn, the force on q1 due to q2 is the same as given by Coulomb’s law,.c i.e., it is unaffected by the presence of the other charges q3, q4,..., qn. The total force F1 on the charge q1, due to all other charges, is then given by ya the vector sum of the forces F12, F13,..., F1n: i.e., 1  q1q 2  F1 = F12 + F13 +...+ F1n = i q1q 3 q1qn  2 rˆ12 + 2 rˆ13 +... + 2 rˆ1n  un 4 πε 0  r12 r13 r1n  n q1 q = 4πε 0 ∑ r 2i r̂1i (1.5) D i = 2 1i The vector sum is obtained as usual by the parallelogram law of addition of vectors. All of electrostatics is basically a consequence of ls Coulomb’s law and the superposition principle. ia Example 1.5 Consider three charges q1, q2, q3 each equal to q at the vertices of an equilateral triangle of side l. What is the force on a or charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in Fig. 1.6? t Tu EXAMPLE 1.5 FIGURE 1.6 Solution In the given equilateral triangle ABC of sides of length l, if we draw a perpendicular AD to the side BC, AD = AC cos 30º = ( 3 /2 ) l and the distance AO of the centroid O 12 from A is (2/3) AD = ( 1/ 3 ) l. By symmatry AO = BO = CO. Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Electric Charges and Fields Thus, 3 Qq Force F1 on Q due to charge q at A = along AO 4 πε 0 l2 3 Qq Force F2 on Q due to charge q at B = 4 πε l 2 along BO 0 3 Qq Force F3 on Q due to charge q at C = 4 πε l 2 along CO 0 3 Qq om The resultant of forces F 2 and F 3 is 4 πε l 2 along OA, by the 0 3 Qq parallelogram law. Therefore, the total force on Q = 4 πε l 2 ( rˆ − rˆ ) 0 EXAMPLE 1.5.c = 0, where r̂ is the unit vector along OA. It is clear also by symmetry that the three forces will sum to zero. Suppose that the resultant force was non-zero but in some direction. about O. i ya Consider what would happen if the system was rotated through 60° Example 1.6 Consider the charges q, q, and –q placed at the vertices un of an equilateral triangle, as shown in Fig. 1.7. What is the force on each charge? D ls ia t or Tu FIGURE 1.7 Solution The forces acting on charge q at A due to charges q at B and –q at C are F12 along BA and F13 along AC respectively, as shown in Fig. 1.7. By the parallelogram law, the total force F1 on the charge q at A is given by F1 = F r̂1 where r̂1 is a unit vector along BC. The force of attraction or repulsion for each pair of charges has the EXAMPLE 1.6 q2 same magnitude F = 4 π ε0 l 2 The total force F2 on charge q at B is thus F2 = F r̂ 2, where r̂ 2 is a unit vector along AC. 13 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com  Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Physics Similarly the total force on charge –q at C is F3 = 3 F n̂ , where n̂ is the unit vector along the direction bisecting the ÐBCA. It is interesting to see that the sum of the forces on the three charges EXAMPLE 1.6 is zero, i.e., F1 + F2 + F3 = 0 The result is not at all surprising. It follows straight from the fact that Coulomb’s law is consistent with Newton’s third law. The proof is left to you as an exercise. om 1.7 ELECTRIC FIELD Let us consider a point charge Q placed in vacuum, at the origin O. If we place another point charge q at a point P, where OP = r, then the charge Q will exert a force on q as per Coulomb’s law. We may ask the question: If.c charge q is removed, then what is left in the surrounding? Is there nothing? If there is nothing at the point P, then how does a force act ya when we place the charge q at P. In order to answer such questions, the early scientists introduced the concept of field. According to this, we say that the charge Q produces an electric field everywhere in the surrounding. i When another charge q is brought at some point P, the field there acts on un it and produces a force. The electric field produced by the charge Q at a point r is given as 1 Q 1 Q E ( r) = rˆ = rˆ D 4πε 0 r 2 4πε 0 r 2 (1.6) where rˆ = r/r, is a unit vector from the origin to the point r. Thus, Eq.(1.6) ls specifies the value of the electric field for each value of the position vector r. The word “field” signifies how some distributed quantity (which ia could be a scalar or a vector) varies with position. The effect of the charge has been incorporated in the existence of the electric field. We obtain the or force F exerted by a charge Q on a charge q, as 1 Qq F= rˆ (1.7) 4 πε 0 r 2 t Note that the charge q also exerts an equal and opposite force on the Tu charge Q. The electrostatic force between the charges Q and q can be looked upon as an interaction between charge q and the electric field of Q and vice versa. If we denote the position of charge q by the vector r, it experiences a force F equal to the charge q multiplied by the electric field E at the location of q. Thus, F(r) = q E(r) (1.8) Equation (1.8) defines the SI unit of electric field as N/C*. Some important remarks may be made here: (i) From Eq. (1.8), we can infer that if q is unity, the electric field due to FIGURE 1.8 Electric a charge Q is numerically equal to the force exerted by it. Thus, the field (a) due to a electric field due to a charge Q at a point in space may be defined charge Q, (b) due to a as the force that a unit positive charge would experience if placed charge –Q. 14 * An alternate unit V/m will be introduced in the next chapter. Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com  Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Electric Charges and Fields at that point. The charge Q, which is producing the electric field, is called a source charge and the charge q, which tests the effect of a source charge, is called a test charge. Note that the source charge Q must remain at its original location. However, if a charge q is brought at any point around Q, Q itself is bound to experience an electrical force due to q and will tend to move. A way out of this difficulty is to make q negligibly small. The force F is then negligibly small but the ratio F/q is finite and defines the electric field:  F E = lim   (1.9) om q →0  q  A practical way to get around the problem (of keeping Q undisturbed in the presence of q) is to hold Q to its location by unspecified forces! This may look strange but actually this is what happens in practice..c When we are considering the electric force on a test charge q due to a charged planar sheet (Section 1.14), the charges on the sheet are held to inside the sheet. ya their locations by the forces due to the unspecified charged constituents (ii) Note that the electric field E due to Q, though defined operationally in i un terms of some test charge q, is independent of q. This is because F is proportional to q, so the ratio F/q does not depend on q. The force F on the charge q due to the charge Q depends on the particular location of charge q which may take any value in the space around D the charge Q. Thus, the electric field E due to Q is also dependent on the space coordinate r. For different positions of the charge q all over ls the space, we get different values of electric field E. The field exists at every point in three-dimensional space. ia (iii) For a positive charge, the electric field will be directed radially outwards from the charge. On the other hand, if the source charge is or negative, the electric field vector, at each point, points radially inwards. (iv) Since the magnitude of the force F on charge q due to charge Q depends only on the distance r of the charge q from charge Q, t the magnitude of the electric field E will also depend only on the Tu distance r. Thus at equal distances from the charge Q, the magnitude of its electric field E is same. The magnitude of electric field E due to a point charge is thus same on a sphere with the point charge at its centre; in other words, it has a spherical symmetry. 1.7.1 Electric field due to a system of charges Consider a system of charges q1, q2,..., qn with position vectors r1, r2,..., rn relative to some origin O. Like the electric field at a point in space due to a single charge, electric field at a point in space due to the system of charges is defined to be the force experienced by a unit test charge placed at that point, without disturbing the original positions of charges q1, q2,..., qn. We can use Coulomb’s law and the superposition principle to determine this field at a point P denoted by position vector r. 15 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com  Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Physics Electric field E1 at r due to q1 at r1 is given by 1 q1 E1 = r̂1P 4 πε 0 r12P where r̂1P is a unit vector in the direction from q1 to P, and r1P is the distance between q1 and P. In the same manner, electric field E2 at r due to q2 at r2 is 1 q2 E2 = r̂2P om 4 πε 0 r22P where r̂2P is a unit vector in the direction from q2 to P FIGURE 1.9 Electric field at a point and r 2P is the distance between q 2 and P. Similar expressions hold good for fields E3, E4,..., En due to.c due to a system of charges is the vector sum of the electric fields at charges q3, q4,..., qn. the point due to individual charges. By the superposition principle, the electric field E at r ya due to the system of charges is (as shown in Fig. 1.9) E(r) = E1 (r) + E2 (r) + … + En(r) 1 q1 1 q2 i 1 qn = rˆ + 2 1P rˆ2 P +... + rˆnP un 4 πε 0 r1P 4 πε 0 r2 P 2 4 πε 0 rn2P 1 n q E(r) = 4π ε 0 ∑ r 2i r̂i P (1.10) D i =1 i P E is a vector quantity that varies from one point to another point in space ls and is determined from the positions of the source charges. 1.7.2 Physical significance of electric field ia You may wonder why the notion of electric field has been introduced here at all. After all, for any system of charges, the measurable quantity or is the force on a charge which can be directly determined using Coulomb’s law and the superposition principle [Eq. (1.5)]. Why then introduce this intermediate quantity called the electric field? t For electrostatics, the concept of electric field is convenient, but not Tu really necessary. Electric field is an elegant way of characterising the electrical environment of a system of charges. Electric field at a point in the space around a system of charges tells you the force a unit positive test charge would experience if placed at that point (without disturbing the system). Electric field is a characteristic of the system of charges and is independent of the test charge that you place at a point to determine the field. The term field in physics generally refers to a quantity that is defined at every point in space and may vary from point to point. Electric field is a vector field, since force is a vector quantity. The true physical significance of the concept of electric field, however, emerges only when we go beyond electrostatics and deal with time- dependent electromagnetic phenomena. Suppose we consider the force between two distant charges q1, q2 in accelerated motion. Now the greatest speed with which a signal or information can go from one point to another 16 is c, the speed of light. Thus, the effect of any motion of q1 on q2 cannot Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com  Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Electric Charges and Fields arise instantaneously. There will be some time delay between the effect (force on q2) and the cause (motion of q1). It is precisely here that the notion of electric field (strictly, electromagnetic field) is natural and very useful. The field picture is this: the accelerated motion of charge q1 produces electromagnetic waves, which then propagate with the speed c, reach q2 and cause a force on q2. The notion of field elegantly accounts for the time delay. Thus, even though electric and magnetic fields can be detected only by their effects (forces) on charges, they are regarded as physical entities, not merely mathematical constructs. They have an independent dynamics of their own, i.e., they evolve according to laws om of their own. They can also transport energy. Thus, a source of time- dependent electromagnetic fields, turned on for a short interval of time and then switched off, leaves behind propagating electromagnetic fields transporting energy. The concept of field was first introduced by Faraday.c and is now among the central concepts in physics. Example 1.7 An electron falls through a distance of 1.5 cm in a ya uniform electric field of magnitude 2.0 × 104 N C–1 [Fig. 1.10(a)]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Fig. 1.10(b)]. Compute i the time of fall in each case. Contrast the situation with that of ‘free un fall under gravity’. D ls ia FIGURE 1.10 Solution In Fig. 1.10(a) the field is upward, so the negatively charged or electron experiences a downward force of magnitude eE where E is the magnitude of the electric field. The acceleration of the electron is ae = eE/me t where me is the mass of the electron. Tu Starting from rest, the time required by the electron to fall through a 2h 2h m e distance h is given by t e = = ae eE For e = 1.6 × 10–19C, me = 9.11 × 10–31 kg, E = 2.0 × 104 N C–1, h = 1.5 × 10–2 m, te = 2.9 × 10–9s In Fig. 1.10 (b), the field is downward, and the positively charged proton experiences a downward force of magnitude eE. The EXAMPLE 1.7 acceleration of the proton is ap = eE/mp where mp is the mass of the proton; mp = 1.67 × 10–27 kg. The time of fall for the proton is 17 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Physics 2h 2h m p tp = = = 1.3 × 10 –7 s ap eE Thus, the heavier particle (proton) takes a greater time to fall through the same distance. This is in basic contrast to the situation of ‘free fall under gravity’ where the time of fall is independent of the mass of the body. Note that in this example we have ignored the acceleration due to gravity in calculating the time of fall. To see if this is justified, let us calculate the acceleration of the proton in the given electric field: om eE ap = mp (1.6 × 10−19 C) × (2.0 × 10 4 N C −1 ) = 1.67 × 10 −27 kg.c EXAMPLE 1.7 = 1.9 × 1012 m s –2 ya which is enormous compared to the value of g (9.8 m s –2), the acceleration due to gravity. The acceleration of the electron is even greater. Thus, the effect of acceleration due to gravity can be ignored in this example. i un Example 1.8 Two point charges q1 and q2, of magnitude +10–8 C and –10–8 C, respectively, are placed 0.1 m apart. Calculate the electric fields at points A, B and C shown in Fig. 1.11. D ls ia t or Tu FIGURE 1.11 Solution The electric field vector E1A at A due to the positive charge q1 points towards the right and has a magnitude (9 × 109 Nm 2C-2 ) × (10 −8 C) E1A = = 3.6 × 104 N C–1 (0.05 m)2 The electric field vector E2A at A due to the negative charge q2 points EXAMPLE 1.8 towards the right and has the same magnitude. Hence the magnitude of the total electric field EA at A is EA = E1A + E2A = 7.2 × 104 N C–1 EA is directed toward the right. 18 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com  Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Electric Charges and Fields The electric field vector E1B at B due to the positive charge q1 points towards the left and has a magnitude (9 × 109 Nm2 C –2 ) × (10 −8 C) E1B = = 3.6 × 104 N C–1 (0.05 m)2 The electric field vector E2B at B due to the negative charge q2 points towards the right and has a magnitude (9 × 109 Nm 2 C –2 ) × (10 −8 C) E 2B = = 4 × 103 N C–1 (0.15 m)2 om The magnitude of the total electric field at B is EB = E1B – E2B = 3.2 × 104 N C–1 EB is directed towards the left. The magnitude of each electric field vector at point C, due to charge q1 and q2 is.c (9 × 109 Nm 2C –2 ) × (10−8 C) E1C = E2C = = 9 × 103 N C–1 (0.10 m)2 ya The directions in which these two vectors point are indicated in EXAMPLE 1.8 Fig. 1.11. The resultant of these two vectors is π π i EC = E1c cos + E 2c cos = 9 × 103 N C–1 un 3 3 EC points towards the right. D 1.8 ELECTRIC FIELD LINES We have studied electric field in the last section. It is a vector quantity ls and can be represented as we represent vectors. Let us try to represent E due to a point charge pictorially. Let the point charge be placed at the ia origin. Draw vectors pointing along the direction of the electric field with their lengths proportional to the strength of the field at each point. Since the magnitude of electric or field at a point decreases inversely as the square of the distance of that point from the charge, the vector gets shorter as one goes away from the origin, always pointing t radially outward. Figure 1.12 shows such a picture. In Tu this figure, each arrow indicates the electric field, i.e., the force acting on a unit positive charge, placed at the tail of that arrow. Connect the arrows pointing in one direction and the resulting figure represents a field line. We thus get many field lines, all pointing outwards from the point charge. Have we lost the information about the strength or magnitude of the field now, because it was contained in the length of the arrow? No. Now the magnitude of the field is represented by the density of field lines. E is strong near the charge, so the density of field lines is more near the charge and the lines are closer. Away from the charge, FIGURE 1.12 Field of a point charge. the field gets weaker and the density of field lines is less, resulting in well-separated lines. Another person may draw more lines. But the number of lines is not important. In fact, an infinite number of lines can be drawn in any region. 19 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Physics It is the relative density of lines in different regions which is important. We draw the figure on the plane of paper, i.e., in two- dimensions but we live in three-dimensions. So if one wishes to estimate the density of field lines, one has to consider the number of lines per unit cross-sectional area, perpendicular to the lines. Since the electric field decreases as the square of the distance from a point charge and the area enclosing the charge increases as the square of the distance, the number of field lines crossing the enclosing area remains constant, om whatever may be the distance of the area from the charge. We started by saying that the field lines carry information about the direction of electric field at different points in space. FIGURE 1.13 Dependence of Having drawn a certain set of field lines, the relative density electric field strength on the.c distance and its relation to the (i.e., closeness) of the field lines at different points indicates number of field lines. the relative strength of electric field at those points. The field lines crowd where the field is strong and are spaced apart ya where it is weak. Figure 1.13 shows a set of field lines. We can imagine two equal and small elements of area placed at points R and S normal to the field lines there. The number of field lines in our picture i cutting the area elements is proportional to the magnitude of field at un these points. The picture shows that the field at R is stronger than at S. To understand the dependence of the field lines on the area, or rather the solid angle subtended by an area element, let us try to relate the D area with the solid angle, a generalisation of angle to three dimensions. Recall how a (plane) angle is defined in two-dimensions. Let a small ls transverse line element Dl be placed at a distance r from a point O. Then the angle subtended by Dl at O can be approximated as Dq = Dl/r. ia Likewise, in three-dimensions the solid angle* subtended by a small perpendicular plane area DS, at a distance r, can be written as DW = DS/r2. We know that in a given solid angle the number of radial or field lines is the same. In Fig. 1.13, for two points P1 and P2 at distances r1 and r2 from the charge, the element of area subtending the solid angle DW is r12 DW at P1 and an element of area r22 DW at P2, respectively. The t number of lines (say n) cutting these area elements are the same. The Tu number of field lines, cutting unit area element is therefore n/( r12 DW) at P1 and n/( r22 DW) at P2 , respectively. Since n and DW are common, the strength of the field clearly has a 1/r 2 dependence. The picture of field lines was invented by Faraday to develop an intuitive non-mathematical way of visualising electric fields around charged configurations. Faraday called them lines of force. This term is somewhat misleading, especially in case of magnetic fields. The more appropriate term is field lines (electric or magnetic) that we have adopted in this book. Electric field lines are thus a way of pictorially mapping the electric field around a configuration of charges. An electric field line is, in general, * Solid angle is a measure of a cone. Consider the intersection of the given cone with a sphere of radius R. The solid angle DW of the cone is defined to be equal 20 2 to DS/R , where DS is the area on the sphere cut out by the cone. Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com  Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Electric Charges and Fields a curve drawn in such a way that the tangent to it at each point is in the direction of the net field at that point. An arrow on the curve is obviously necessary to specify the direction of electric field from the two possible directions indicated by a tangent to the curve. A field line is a space curve, i.e., a curve in three dimensions. Figure 1.14 shows the field lines around some simple charge configurations. As mentioned earlier, the field lines are in 3-dimensional space, though the figure shows them only in a plane. The field lines of a single positive charge om are radially outward while those of a single negative charge are radially inward. The field lines around a system of two positive charges (q, q) give a vivid pictorial description of their mutual repulsion, while those around.c the configuration of two equal and opposite charges (q, –q), a dipole, show clearly the mutual attraction ya between the charges. The field lines follow some important general properties: (i) Field lines start from positive charges and end at negative charges. If there is a single charge, they may i un start or end at infinity. (ii) In a charge-free region, electric field lines can be taken to be continuous curves without any breaks. (iii) Two field lines can never cross each other. (If they did, D the field at the point of intersection will not have a unique direction, which is absurd.) ls (iv) Electrostatic field lines do not form any closed loops. This follows from the conservative nature of electric ia field (Chapter 2). 1.9 ELECTRIC FLUX or Consider flow of a liquid with velocity v, through a small flat surface dS, in a direction normal to the surface. The t rate of flow of liquid is given by the volume crossing the Tu area per unit time v dS and represents the flux of liquid flowing across the plane. If the normal to the surface is not parallel to the direction of flow of liquid, i.e., to v, but makes an angle q with it, the projected area in a plane perpendicular to v is δ dS cos q. Therefore, the flux going out of the surface dS is v. n̂ dS. For the case of the electric field, we define an analogous quantity and call it electric flux. We should, however, note that there is no flow of a physically observable quantity unlike the case of liquid flow. In the picture of electric field lines described above, FIGURE 1.14 Field lines due to we saw that the number of field lines crossing a unit area, some simple charge configurations. placed normal to the field at a point is a measure of the strength of electric field at that point. This means that if 21 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Physics we place a small planar element of area DS normal to E at a point, the number of field lines crossing it is proportional* to E DS. Now suppose we tilt the area element by angle q. Clearly, the number of field lines crossing the area element will be smaller. The projection of the area element normal to E is DS cosq. Thus, the number of field lines crossing DS is proportional to E DS cosq. When q = 90°, field lines will be parallel to DS and will not cross it om at all (Fig. 1.15). The orientation of area element and not merely its magnitude is important in many contexts. For example, in a stream, the amount.c of water flowing through a ring will naturally depend on how you hold the ring. If you hold it normal to the flow, maximum water will flow ya FIGURE 1.15 Dependence of flux on the inclination q between E and n̂. through it than if you hold it with some other orientation. This shows that an area element i should be treated as a vector. It has a magnitude and also a direction. How to specify the direction of a planar un area? Clearly, the normal to the plane specifies the orientation of the plane. Thus the direction of a planar area vector is along its normal. How to associate a vector to the area of a curved surface? We imagine D dividing the surface into a large number of very small area elements. Each small area element may be treated as planar and a vector associated ls with it, as explained before. Notice one ambiguity here. The direction of an area element is along its normal. But a normal can point in two directions. Which direction do ia we choose as the direction of the vector associated with the area element? This problem is resolved by some convention appropriate to the given or context. For the case of a closed surface, this convention is very simple. The vector associated with every area element of a closed surface is taken to be in the direction of the outward normal. This is the convention used t in Fig. 1.16. Thus, the area element vector DS at a point on a closed Tu surface equals DS n̂ where DS is the magnitude of the area element and n̂ is a unit vector in the direction of outward normal at that point. We now come to the definition of electric flux. Electric flux Df through an area element DS is defined by Df = E.DS = E DS cosq (1.11) which, as seen before, is proportional to the number of field lines cutting the area element. The angle q here is the angle between E and DS. For a closed surface, with the convention stated already, q is the angle between FIGURE 1.16 E and the outward normal to the area element. Notice we could look at Convention for the expression E DS cosq in two ways: E (DS cosq ) i.e., E times the defining normal n̂ and DS. * It will not be proper to say that the number of field lines is equal to EDS. The number of field lines is after all, a matter of how many field lines we choose to draw. What is physically significant is the relative number of field lines crossing 22 a given area at different points. Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com  Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Electric Charges and Fields projection of area normal to E, or E^ DS, i.e., component of E along the normal to the area element times the magnitude of the area element. The unit of electric flux is N C–1 m2. The basic definition of electric flux given by Eq. (1.11) can be used, in principle, to calculate the total flux through any given surface. All we have to do is to divide the surface into small area elements, calculate the flux at each element and add them up. Thus, the total flux f through a surface S is f ~ S E.DS (1.12) The approximation sign is put because the electric field E is taken to om be constant over the small area element. This is mathematically exact only when you take the limit DS ® 0 and the sum in Eq. (1.12) is written as an integral..c 1.10 ELECTRIC DIPOLE An electric dipole is a pair of equal and opposite point charges q and –q, ya separated by a distance 2a. The line connecting the two charges defines a direction in space. By convention, the direction from –q to q is said to be the direction of the dipole. The mid-point of locations of –q and q is called the centre of the dipole. i un The total charge of the electric dipole is obviously zero. This does not mean that the field of the electric dipole is zero. Since the charge q and –q are separated by some distance, the electric fields due to them, when D added, do not exactly cancel out. However, at distances much larger than the separation of the two charges forming a dipole (r >> 2a), the fields ls due to q and –q nearly cancel out. The electric field due to a dipole therefore falls off, at large distance, faster than like 1/r 2 (the dependence on r of the field due to a single charge q). These qualitative ideas are ia borne out by the explicit calculation as follows: or 1.10.1 The field of an electric dipole The electric field of the pair of charges (–q and q) at any point in space can be found out from Coulomb’s law and the superposition principle. t The results are simple for the following two cases: (i) when the point is on Tu the dipole axis, and (ii) when it is in the equatorial plane of the dipole, i.e., on a plane perpendicular to the dipole axis through its centre. The electric field at any general point P is obtained by adding the electric fields E–q due to the charge –q and E+q due to the charge q, by the parallelogram law of vectors. (i) For points on the axis Let the point P be at distance r from the centre of the dipole on the side of the charge q, as shown in Fig. 1.17(a). Then q E −q = − p [1.13(a)] 4πε0 (r + a )2 where p̂ is the unit vector along the dipole axis (from –q to q). Also q E +q = p [1.13(b)] 23 4 π ε 0 (r − a )2 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com  Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Physics The total field at P is q  1 1  E = E +q + E − q =  − p 4 π ε0  (r − a ) 2 (r + a )2  q 4a r = p (1.14) 4 π εo ( r 2 − a 2 )2 For r >> a 4qa E= ˆ p (r >> a) (1.15) 4 π ε 0r 3 om (ii) For points on the equatorial plane The magnitudes of the electric fields due to the two charges +q and –q are given by.c q 1 E +q = [1.16(a)] 4 πε 0 r + a 2 2 ya q 1 E –q = [1.16(b)] 4 πε 0 r + a 2 2 FIGURE 1.17 Electric field of a dipole i un and are equal. at (a) a point on the axis, (b) a point The directions of E +q and E –q are as shown in on the equatorial plane of the dipole. Fig. 1.17(b). Clearly, the components normal to the dipole p is the dipole moment vector of axis cancel away. The components along the dipole axis D magnitude p = q × 2a and directed from –q to q. add up. The total electric field is opposite to p̂. We have E = – (E +q + E –q ) cosq p̂ ls 2q a =− p (1.17) 4 π ε o (r 2 + a 2 )3 / 2 ia At large distances (r >> a), this reduces to or 2qa E=− ˆ p (r >> a ) (1.18) 4 π εo r 3 From Eqs. (1.15) and (1.18), it is clear that the dipole field at large t distances does not involve q and a separately; it depends on the product Tu qa. This suggests the definition of dipole moment. The dipole moment vector p of an electric dipole is defined by p = q × 2a p̂ (1.19) that is, it is a vector whose magnitude is charge q times the separation 2a (between the pair of charges q, –q) and the direction is along the line from –q to q. In terms of p, the electric field of a dipole at large distances takes simple forms: At a point on the dipole axis 2p E= (r >> a) (1.20) 4 πε o r 3 At a point on the equatorial plane p E=− (r >> a) (1.21) 24 4πε or 3 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com  Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Electric Charges and Fields Notice the important point that the dipole field at large distances falls off not as 1/r 2 but as1/r 3. Further, the magnitude and the direction of the dipole field depends not only on the distance r but also on the angle between the position vector r and the dipole moment p. We can think of the limit when the dipole size 2a approaches zero, the charge q approaches infinity in such a way that the product p = q × 2a is finite. Such a dipole is referred to as a point dipole. For a point dipole, Eqs. (1.20) and (1.21) are exact, true for any r. 1.10.2 Physical significance of dipoles om In most molecules, the centres of positive charges and of negative charges* lie at the same place. Therefore, their dipole moment is zero. CO2 and CH4 are of this type of molecules. However, they develop a dipole moment when an electric field is applied. But in some molecules, the centres of.c negative charges and of positive charges do not coincide. Therefore they have a permanent electric dipole moment, even in the absence of an electric field. Such molecules are called polar molecules. Water molecules, H2O, ya is an example of this type. Various materials give rise to interesting properties and important applications in the presence or absence of electric field. i un Example 1.9 Two charges ±10 mC are placed 5.0 mm apart. Determine the electric field at (a) a point P on the axis of the dipole 15 cm away from its centre O on the side of the positive charge, as shown in Fig. D 1.18(a), and (b) a point Q, 15 cm away from O on a line passing through O and normal to the axis of the dipole, as shown in Fig. 1.18(b). ls ia t or Tu EXAMPLE 1.9 FIGURE 1.18 * Centre of a collection of positive point charges is defined much the same way ∑ qi ri as the centre of mass: rcm = i. ∑ qi 25 i Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Physics Solution (a) Field at P due to charge +10 mC 10 −5 C 1 = × 4 π (8.854 × 10 −12 2 C N −1 m ) −2 (15 − 0.25)2 × 10 −4 m 2 = 4.13 × 106 N C–1 along BP Field at P due to charge –10 mC 10 –5 C 1 = × −12 −1 4 π (8.854 × 10 C N m ) 2 −2 (15 + 0.25)2 × 10 −4 m 2 = 3.86 × 106 N C–1 along P