• Physics FIGURE 1.4 Charging by induction. electric current from the power station and the third is earthed by connecting it to the buried metal plate. Metallic bodies of the electric appliances such as electric iron, refrigerator, TV are connected to the earth wire. When any fault occurs or live wire touches the metallic body, the charge flows to the earth without damaging the appliance and without causing any injury to the humans; this would have otherwise been unavoidable since the human body is a conductor of electricity. 1.4 CHARGINGBY INDUCTION When we touch a pith ball with an electrified plastic rod, some of the negative charges on the rod are transferred to the pith ball and it also gets charged. Thus the pith ball is charged by contact. It is then repelled by the plastic rod but is attracted by a glass rod which is oppositely charged. However, why a electrified rod attracts light objects, is a question we have still left unanswered. Let us try to understand what could be happening by performing the following experiment. (i) Bring two metal spheres, A and B, supported on insulating stands, in contact as shown in Fig. 1.4(a). (ii) Bring a positively charged rod near one of the spheres, say A, taking care that it does not touch the sphere. The free electrons in the spheres are attracted towards the rod. This leaves an excess of positive charge on the rear surface of sphere B. Both kinds of charges are bound in the metal spheres and cannot escape. They, therefore, reside on the surfaces, as shown in Fig. 1.4(b). The left surface of sphere A, has an excess of negative charge and the right surface of sphere B, has an excess of positive charge. However, not all of the electrons in the spheres have accumulated on the left surface of A. As the negative charge starts building up at the left surface of A, other electrons are repelled by these. In a short time, equilibrium is reached under the action of force of attraction of the rod and the force of repulsion due to the accumulated charges. Fig. 1.4(b) shows the equilibrium situation. The process is called induction of charge and happens almost instantly. The accumulated charges remain on the surface, as shown, till the glass rod is held near the sphere. If the rod is removed, the charges are not acted by any outside force and they redistribute to their original neutral state. (iii) Separate the spheres by a small distance while the glass rod is still held near sphere A, as shown in Fig. 1.4(c). The two spheres are found to be oppositely charged and attract each other. (iv) Remove the rod. The charges on spheres rearrange themselves as shown in Fig. 1.4(d). Now, separate the spheres quite apart. The charges on them get uniformly distributed over them, as shown in Fig. 1.4(e). In this process, the metal spheres will each be equal and oppositely charged. This is charging by induction. The positively charged glass rod does not lose any of its charge, contrary to the process of charging by contact. When electrified rods are brought near light objects, a similar effect takes place. The rods induce opposite charges on the near surfaces of the objects and similar charges move to the farther side of the object. Electric Charges and Fields [This happens even when the light object is not a conductor. The mechanism for how this happens is explained later in Sections 1.10 and 2.10.] The centres of the two types of charges are slightly separated. We know that opposite charges attract while similar charges repel. However, the magnitude of force depends on the distance between the charges and in this case the force of attraction overweighs the force of repulsion. As a result the particles like bits of paper or pith balls, being light, are pulled towards the rods. Example 1.1 How can you charge a metal sphere positively without touching it? Solution Figure 1.5(a) shows an uncharged metallic sphere on an insulating metal stand. Bring a negatively charged rod close to the metallic sphere, as shown in Fig. 1.5(b). As the rod is brought close to the sphere, the free electrons in the sphere move away due to repulsion and start piling up at the farther end. The near end becomes positively charged due to deficit of electrons. This process of charge distribution stops when the net force on the free electrons inside the metal is zero. Connect the sphere to the ground by a conducting wire. The electrons will flow to the ground while the positive charges at the near end will remain held there due to the attractive force of the negative charges on the rod, as shown in Fig. 1.5(c). Disconnect the sphere from the ground. The positive charge continues to be held at the near end [Fig. 1.5(d)]. Remove the electrified rod. The positive charge will spread uniformly over the sphere as shown in Fig. 1.5(e). Interactive animation on charging a two-sphere system by induction:EXAMPLE 1.1 FIGURE 1.5 In this experiment, the metal sphere gets charged by the process of induction and the rod does not lose any of its charge. Similar steps are involved in charging a metal sphere negatively by induction, by bringing a positively charged rod near it. In this case the electrons will flow from the ground to the sphere when the sphere is connected to the ground with a wire. Can you explain why? PhysicsExample 1.5 A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by an insulating handle is brought close to A such that the distance between their centres is 10 cm, as shown in Fig. 1.7(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. 1.7(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. 1.7(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 centres. EXAMPLE 1.5 FIGURE 1.7 Physicswhich again is the Coulomb force on q due to q, even though other13charge q2 is present. Thus the total force F1 on q1 due to the two charges q2 and q3 is given as 1 qq 1 qq 12 13F = F + F = rˆ + rˆ1 12 13 212 213 (1.4)4πε r 4πε r012 013 The above calculation of force can be generalised to a system of charges more than three, as shown in Fig. 1.8(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, 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 the vector sum of the forces F12, F13, ..., F1n : i.e., 1 ⎡qq qq 1 n ⎤qq 12 13F = F + F + ...+ F = ˆr + ˆr + ... + rˆ1 12 13 1n ⎢ 212 213 21n ⎥4πε0 ⎣ r12 r13 r1n ⎦ q nq= 1 i rˆ∑ 21i (1.5)4πε r0 i =21i The vector sum is obtained as usual by the parallelogram law of addition of vectors. All of electrostatics is basically a consequence of Coulomb’s law and the superposition principle. Example 1.6 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 charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in Fig. 1.9? EXAMPLE 1.6 FIGURE 1.9 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 from A is (2/3) AD = (1/ 3) l. By symmatry AO = BO = CO. Physics we place a small planar element of area ΔS normal to E at a point, the number of field lines crossing it is proportional* to E ΔS. Now suppose we tilt the area element by angle θ. Clearly, the number of field lines crossing the area element will be smaller. The projection of the area element normal to E is ΔS cosθ. Thus, the number of field lines crossing ΔS is proportional to E ΔS cosθ. When θ = 90°, field lines will be parallel to ΔS and will not cross it at all (Fig. 1.18). The orientation of area element and not merely its magnitude is important in many contexts. For example, in a stream, the amount 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 flowFIGURE 1.18 Dependence of flux on the through it than if you hold it with some otherinclination θ between E and nˆ. orientation. This shows that an area element should be treated as a vector. It has a magnitude and also a direction. How to specify the direction of a planar 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 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 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 we choose as the direction of the vector associated with the area element? This problem is resolved by some convention appropriate to the given 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 in Fig. 1.19. Thus, the area element vector ΔS at a point on a closed surface equals ΔS nˆ where ΔS is the magnitude of the area element and nˆ is a unit vector in the direction of outward normal at that point. We now come to the definition of electric flux. Electric flux Δφ through an area element ΔS is defined by Δφ = E.ΔS = E ΔS cosθ (1.11) which, as seen before, is proportional to the number of field lines cutting the area element. The angle θ here is the angle between E and ΔS. For a closed surface, with the convention stated already, θ is the angle between E and the outward normal to the area element. Notice we could look atFIGURE 1.19 the expression EΔS cosθ in two ways: E (ΔS cosθ) i.e., E times theConvention for defining normal nˆ and ΔS. * It will not be proper to say that the number of field lines is equal to EΔS. 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 a given area at different points. Electric Charges and Fields the density of a liquid, we are referring to its macroscopic density. We regard it as a continuous fluid and ignore its discrete molecular constitution. The field due to a continuous charge distribution can be obtained in much the same way as for a system of discrete charges, Eq. (1.10). Suppose a continuous charge distribution in space has a charge density ρ. Choose any convenient origin O and let the position vector of any point in the charge distribution be r. The charge density ρ may vary from point to point, i.e., it is a function of r. Divide the charge distribution into small volume elements of size ΔV. The charge in a volume element ΔV is ρΔV. Now, consider any general point P (inside or outside the distribution) with position vector R (Fig. 1.24). Electric field due to the charge ρΔV is given by Coulomb’s law: 1 ρΔVΔE = rˆ' 2 (1.26)4πε0 r' where r′ is the distance between the charge element and P, and rˆ′ is a unit vector in the direction from the charge element to P. By the superposition principle, the total electric field due to the charge distribution is obtained by summing over electric fields due to different volume elements: 1 ρΔV E ≅Σ ˆr' 2 (1.27)all ΔV4πε0 r' Note that ρ, r′, ˆr′ all can vary from point to point. In a strict mathematical method, we should let ΔV→0 and the sum then becomes an integral; but we omit that discussion here, for simplicity. In short, using Coulomb’s law and the superposition principle, electric field can be determined for any charge distribution, discrete or continuous or part discrete and part continuous. 1.14 GAUSS’S LAW As a simple application of the notion of electric flux, let us consider the total flux through a sphere of radius r, which encloses a point charge q at its centre. Divide the sphere into small area elements, as shown in Fig. 1.25. The flux through an area element ΔS is qΔφ = Ei Δ S = rˆi ΔS2 (1.28)4πε0 r where we have used Coulomb’s law for the electric field due to a single charge q. The unit vector ˆr is along the radius vector from the centre to the area element. Now, since the normal to a sphere at every point is along the radius vector at that point, the area element ΔS and ˆr have the same direction. Therefore, FIGURE 1.25 Flux qΔφ = 2 ΔS (1.29) through a sphere4πε r0 enclosing a point since the magnitude of a unit vector is 1. charge q at its centre. The total flux through the sphere is obtained by adding up flux through all the different area elements: Electric Charges and Fields 1.15 APPLICATIONSOF GAUSS’S LAW The electric field due to a general charge distribution is, as seen above, given by Eq. (1.27). In practice, except for some special cases, the summation (or integration) involved in this equation cannot be carried out to give electric field at every point in space. For some symmetric charge configurations, however, it is possible to obtain the electric field in a simple way using the Gauss’s law. This is best understood by some examples. 1.15.1 Field due to an infinitely long straight uniformly charged wire Consider an infinitely long thin straight wire with uniform linear charge density λ. The wire is obviously an axis of symmetry. Suppose we take the radial vector from O to P and rotate it around the wire. The points P, P′, P′′ so obtained are completely equivalent with respect to the charged wire. This implies that the electric field must have the same magnitude at these points. The direction of electric field at every point must be radial (outward if λ > 0, inward if λ < 0). This is clear from Fig. 1.29. Consider a pair of line elements P1 and P2 of the wire, as shown. The electric fields produced by the two elements of the pair when summed give a resultant electric field which is radial (the components normal to the radial vector cancel). This is true for any such pair and hence the total field at any point P is radial. Finally, since the wire is infinite, electric field does not depend on the position of P along the length of the wire. In short, the electric field is everywhere radial in the plane cutting the wire normally, and its magnitude depends only on the radial distance r. To calculate the field, imagine a cylindrical Gaussian surface, as shown in the Fig. 1.29(b). Since the field is everywhere radial, flux through the two ends of the cylindrical Gaussian surface is zero. At the cylindrical FIGURE 1.29 (a) Electric field due to anpart of the surface,E is normal to the surface infinitely long thin straight wire is radial,at every point, and its magnitude is constant, (b) The Gaussian surface for a long thin since it depends only on r. The surface area wire of uniform linear charge density. of the curved part is 2πrl, where l is the length of the cylinder. Physics Flux through the Gaussian surface = flux through the curved cylindrical part of the surface = E × 2πrl The surface includes charge equal to λ l. Gauss’s law then gives E × 2πrl = λl/ε0 λ i.e., E = 2πε0r Vectorially, E at any point is given by λ E = nˆ (1.32)2πε0r where nˆ is the radial unit vector in the plane normal to the wire passing through the point. E is directed outward if λ is positive and inward if λ is negative. Note that when we write a vector A as a scalar multiplied by a unit vector, i.e., as A = A aˆ, the scalar A is an algebraic number. It can be negative or positive. The direction ofA will be the same as that of the unit vector aˆif A > 0 and opposite to aˆ if A < 0. When we want to restrict to non-negative values, we use the symbol A and call it the modulus of A. Thus, A ≥ 0. Also note that though only the charge enclosed by the surface (λl) was included above, the electric field E is due to the charge on the entire wire. Further, the assumption that the wire is infinitely long is crucial. Without this assumption, we cannot take E to be normal to the curved part of the cylindrical Gaussian surface. However, Eq. (1.32) is approximately true for electric field around the central portions of a long wire, where the end effects may be ignored. 1.15.2 Field due to a uniformly charged infinite plane sheet Let σ be the uniform surface charge density of an infinite plane sheet (Fig. 1.30). We take the x-axis normal to the given plane. By symmetry, the electric field will not depend on y and z coordinates and its direction at every point must be parallel to the x-direction. We can take the Gaussian surface to be a rectangular parallelepiped of cross sectional area A, as shown. (A cylindrical surface will also do.) As seen from the figure, only the two faces 1 and 2 will contribute to the flux; electric field lines are parallel to the other faces and they, therefore, do not contribute to the total flux. The unit vector normal to surface 1 is in –x direction while the unit vector normal to surface 2 is in the +x direction. Therefore, flux E.ΔS through both the surfaces are equal and add up. ThereforeFIGURE 1.30 Gaussian surface for a the net flux through the Gaussian surface is 2 EA. uniformly charged infinite plane sheet. The charge enclosed by the closed surface is σA. Therefore by Gauss’s law, Physics SUMMARY 1. Electric and magnetic forces determine the properties of atoms, molecules and bulk matter. 2. From simple experiments on frictional electricity, one can infer that there are two types of charges in nature; and that like charges repel and unlike charges attract. By convention, the charge on a glass rod rubbed with silk is positive; that on a plastic rod rubbed with fur is then negative. 3. Conductors allow movement of electric charge through them, insulators do not. In metals, the mobile charges are electrons; in electrolytes both positive and negative ions are mobile. 4. Electric charge has three basic properties: quantisation, additivity and conservation. Quantisation of electric charge means that total charge (q) of a body is always an integral multiple of a basic quantum of charge (e) i.e., q = n e, where n = 0, ±1, ±2, ±3, .... Proton and electron have charges +e, –e,respectively. For macroscopic charges for which n is a very large number, quantisation of charge can be ignored. Additivity of electric charges means that the total charge of a system is the algebraic sum (i.e., the sum taking into account proper signs) of all individual charges in the system. Conservation of electric charges means that the total charge of an isolated system remains unchanged with time. This means that when bodies are charged through friction, there is a transfer of electric charge from one body to another, but no creation or destruction of charge. 5. Coulomb’s Law: The mutual electrostatic force between two point charges q1 and q2 is proportional to the product q1q2 and inversely proportional to the square of the distance r21 separating them. Mathematically, k (qq ) = force on q2 due to q = 12 ˆr1 221 F21r21 1 where rˆ21 is a unit vector in the direction from q1 to q2 and k = 4πε0 is the constant of proportionality. In SI units, the unit of charge is coulomb. The experimental value of the constant ε0 is ε0 = 8.854 × 10–12 C2 N–1 m–2 The approximate value of k is k = 9 × 109 N m2 C–2 6. The ratio of electric force and gravitational force between a proton and an electron is 2ke 24 × 10 39 ≅ . G mm ep 7. Superposition Principle: The principle is based on the property that the forces with which two charges attract or repel each other are not affected by the presence of a third (or more) additional charge(s). For an assembly of charges q1, q2, q3, ..., the force on any charge, say q1, is Physics 11. The electric field due to a charge configuration with total charge zero is not zero; but for distances large compared to the size of the configuration, its field falls off faster than 1/r 2, typical of field due to a single charge. An electric dipole is the simplest example of this fact. EXERCISES 1.1 What is the force between two small charged spheres having charges of 2 × 10–7C and 3 × 10–7C placed 30 cm apart in air? 1.2 The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first? 1.3 Check that the ratio ke2/G memp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify? 1.4 (a) Explain the meaning of the statement ‘electric charge of a body is quantised’. (b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges? 1.5 When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge. 1.6 Four point charges qA = 2 µC, qB = –5 µC, qC = 2 µC, and qD = –5 µC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 µC placed at the centre of the square? 1.7 (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not? (b) Explain why two field lines never cross each other at any point? 1.8 Two point charges qA = 3 µC and qB = –3 µC are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges? (b) If a negative test charge of magnitude 1.5 × 10–9 C is placed at this point, what is the force experienced by the test charge? 1.9 A system has two charges q = 2.5 × 10–7 C and q = –2.5 × 10–7 CA Blocated at points A: (0, 0, –15 cm) and B: (0,0, +15 cm), respectively. What are the total charge and electric dipole moment of the system? 1.10 An electric dipole with dipole moment 4 × 10–9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 104 NC–1. Calculate the magnitude of the torque acting on the dipole. 1.11 A polythene piece rubbed with wool is found to have a negative charge of 3 × 10–7 C. (a) Estimate the number of electrons transferred (from which to which?) (b) Is there a transfer of mass from wool to polythene? 1.12 (a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of Electric Charges and Fields electrostatic repulsion if the charge on each is 6.5 × 10–7 C? The radii of A and B are negligible compared to the distance of separation. (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved? 1.13 Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B? 1.14 Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio? FIGURE 1.33 1.15 Consider a uniform electric field E = 3 × 103 î N/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis? 1.16 What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes? 1.17 Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 103 Nm2/C. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not? 1.18 A point charge +10 µC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.) FIGURE 1.34 Physics 1.32 (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable. (b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart. 1.33 A particle of mass m and charge (–q) enters the region between the two charged plates initially moving along x-axis with speed vx (like particle 1 in Fig. 1.33). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL2/(2m v x 2). Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics. 1.34 Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vx = 2.0 × 106 m s–1. If E between the plates separated by 0.5 cm is 9.1 × 102 N/C, where will the electron strike the upper plate? (|e|=1.6 × 10–19 C, me = 9.1 × 10–31 kg.)

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