SplitPDFFile_27_to_37.pdf

Electric Charges and Fields p E = (r/a >> 1) 4 πε 0 r 3 5 × 10−8 C m...

Electric Charges and Fields p E = (r/a >> 1) 4 πε 0 r 3 5 × 10−8 C m 1 = × 4 π (8.854 ×10−12 C2 N –1 m –2 ) (15)3 × 10 −6 m 3 EXAMPLE 1.9 = 1.33 × 105 N C–1. The direction of electric field in this case is opposite to the direction of the dipole moment vector. Again, the result agrees with that obtained before. 1.11 DIPOLE IN A UNIFORM EXTERNAL FIELD Consider a permanent dipole of dipole moment p in a uniform external field E, as shown in Fig. 1.19. (By permanent dipole, we mean that p exists irrespective of E; it has not been induced by E.) There is a force qE on q and a force –qE on –q. The net force on the dipole is zero, since E is uniform. However, the charges are separated, so the forces act at different points, resulting in a torque on the dipole. When the net force is zero, the torque (couple) is independent of the origin. Its magnitude equals the magnitude of FIGURE 1.19 Dipole in a each force multiplied by the arm of the couple (perpendicular uniform electric field. distance between the two antiparallel forces). Magnitude of torque = q E × 2 a sinq = 2 q a E sinq Its direction is normal to the plane of the paper, coming out of it. The magnitude of p × E is also p E sinq and its direction is normal to the paper, coming out of it. Thus, t =p×E (1.22) This torque will tend to align the dipole with the field E. When p is aligned with E, the torque is zero. What happens if the field is not uniform? In that case, the net force will evidently be non-zero. In addition there will, in general, be a torque on the system as before. The general case is involved, so let us consider the simpler situations when p is parallel to E or antiparallel to E. In either case, the net torque is zero, but there is a net force on the dipole if E is not uniform. Figure 1.20 is self-explanatory. It is easily seen that when p is parallel to E, the dipole has a net force in the direction of increasing field. When p is antiparallel to E, the net force on the dipole is in the direction of decreasing field. In general, the force depends on the orientation of p with respect to E. This brings us to a common observation in frictional electricity. A comb run through dry hair attracts pieces of FIGURE 1.20 Electric force on a paper. The comb, as we know, acquires charge through dipole: (a) E parallel to p, (b) E friction. But the paper is not charged. What then explains antiparallel to p. the attractive force? Taking the clue from the preceding 27 2024-25  Physics discussion, the charged comb ‘polarises’ the piece of paper, i.e., induces a net dipole moment in the direction of field. Further, the electric field due to the comb is not uniform. This non-uniformity of the field makes a dipole to experience a net force on it. In this situation, it is easily seen that the paper should move in the direction of the comb! 1.12 CONTINUOUS CHARGE DISTRIBUTION We have so far dealt with charge configurations involving discrete charges q1, q2,..., qn. One reason why we restricted to discrete charges is that the mathematical treatment is simpler and does not involve calculus. For many purposes, however, it is impractical to work in terms of discrete charges and we need to work with continuous charge distributions. For example, on the surface of a charged conductor, it is impractical to specify the charge distribution in terms of the locations of the microscopic charged constituents. It is more feasible to consider an area element DS (Fig. 1.21) on the surface of the conductor (which is very small on the macroscopic scale but big enough to include a very large number of electrons) and specify the charge DQ on that element. We then define a surface charge density s at the area element by ∆Q σ= (1.23) ∆S We can do this at different points on the conductor and thus arrive at a continuous function s, called the surface charge density. The surface charge density s so defined ignores the quantisation of charge and the discontinuity in charge distribution at the microscopic level*. s represents macroscopic surface charge density, which in a sense, is a smoothed out average of the microscopic charge density over an area element DS which, as said before, is large microscopically but small macroscopically. The units for s are C/m2. FIGURE 1.21 Similar considerations apply for a line charge distribution and a volume Definition of linear, charge distribution. The linear charge density l of a wire is defined by surface and volume ∆Q charge densities. λ = (1.24) In each case, the ∆l element (Dl, DS, DV ) where Dl is a small line element of wire on the macroscopic scale that, chosen is small on however, includes a large number of microscopic charged constituents, the macroscopic and DQ is the charge contained in that line element. The units for l are scale but contains C/m. The volume charge density (sometimes simply called charge density) a very large number is defined in a similar manner: of microscopic constituents. ∆Q ρ= (1.25) ∆V where DQ is the charge included in the macroscopically small volume element DV that includes a large number of microscopic charged constituents. The units for r are C/m3. The notion of continuous charge distribution is similar to that we adopt for continuous mass distribution in mechanics. When we refer to 28 * At the microscopic level, charge distribution is discontinuous, because they are discrete charges separated by intervening space where there is no charge. 2024-25  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 r. Choose any convenient origin O and let the position vector of any point in the charge distribution be r. The charge density r may vary from point to point, i.e., it is a function of r. Divide the charge distribution into small volume elements of size DV. The charge in a volume element DV is rDV. Now, consider any general point P (inside or outside the distribution) with position vector R (Fig. 1.21). Electric field due to the charge rDV is given by Coulomb’s law: 1 ρ ∆V ∆E = rˆ' (1.26) 4πε 0 r' 2 where r¢ is the distance between the charge element and P, and r̂ ¢ 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 ρ ∆V E≅ Σ rˆ' (1.27) 4 πε 0 all ∆V r' 2 Note that r, r¢, rˆ ′ all can vary from point to point. In a strict mathematical method, we should let DV®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. 1.13 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. 1.22. The flux through an area element DS is q ∆φ = E i ∆ S = rˆ i ∆S (1.28) 4 πε 0 r 2 where we have used Coulomb’s law for the electric field due to a single charge q. The unit vector r̂ 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 DS and r̂ have the same direction. Therefore, q FIGURE 1.22 Flux ∆φ = ∆S (1.29) through a sphere 4 πε 0 r 2 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: 29 2024-25  Physics q φ= Σ ∆S all ∆S 4 π ε0 r 2 Since each area element of the sphere is at the same distance r from the charge, FIGURE 1.23 Calculation of the q q flux of uniform electric field φ= Σ ∆S = S 4 πεo r 2 all ∆S 4 πε 0 r 2 through the surface of a cylinder. Now S, the total area of the sphere, equals 4pr 2. Thus, q q φ= × 4 πr 2 = (1.30) 4 πε 0 r 2 ε0 Equation (1.30) is a simple illustration of a general result of electrostatics called Gauss’s law. We state Gauss’s law without proof: Electric flux through a closed surface S = q/e0 (1.31) q = total charge enclosed by S. The law implies that the total electric flux through a closed surface is zero if no charge is enclosed by the surface. We can see that explicitly in the simple situation of Fig. 1.23. Here the electric field is uniform and we are considering a closed cylindrical surface, with its axis parallel to the uniform field E. The total flux f through the surface is f = f1 + f2 + f3, where f1 and f2 represent the flux through the surfaces 1 and 2 (of circular cross-section) of the cylinder and f3 is the flux through the curved cylindrical part of the closed surface. Now the normal to the surface 3 at every point is perpendicular to E, so by definition of flux, f3 = 0. Further, the outward normal to 2 is along E while the outward normal to 1 is opposite to E. Therefore, f1 = –E S1, f2 = +E S2 S1 = S2 = S where S is the area of circular cross-section. Thus, the total flux is zero, as expected by Gauss’s law. Thus, whenever you find that the net electric flux through a closed surface is zero, we conclude that the total charge contained in the closed surface is zero. The great significance of Gauss’s law Eq. (1.31), is that it is true in general, and not only for the simple cases we have considered above. Let us note some important points regarding this law: (i) Gauss’s law is true for any closed surface, no matter what its shape or size. (ii) The term q on the right side of Gauss’s law, Eq. (1.31), includes the sum of all charges enclosed by the surface. The charges may be located anywhere inside the surface. (iii) In the situation when the surface is so chosen that there are some charges inside and some outside, the electric field [whose flux appears on the left side of Eq. (1.31)] is due to all the charges, both inside and outside S. The term q on the right side of Gauss’s law, however, 30 represents only the total charge inside S. 2024-25  Electric Charges and Fields (iv) The surface that we choose for the application of Gauss’s law is called the Gaussian surface. You may choose any Gaussian surface and apply Gauss’s law. However, take care not to let the Gaussian surface pass through any discrete charge. This is because electric field due to a system of discrete charges is not well defined at the location of any charge. (As you go close to the charge, the field grows without any bound.) However, the Gaussian surface can pass through a continuous charge distribution. (v) Gauss’s law is often useful towards a much easier calculation of the electrostatic field when the system has some symmetry. This is facilitated by the choice of a suitable Gaussian surface. (vi) Finally, Gauss’s law is based on the inverse square dependence on distance contained in the Coulomb’s law. Any violation of Gauss’s law will indicate departure from the inverse square law. Example 1.10 The electric field components in Fig. 1.24 are Ex = ax1/2, Ey = Ez = 0, in which a = 800 N/C m1/2. Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that a = 0.1 m. FIGURE 1.24 Solution (a) Since the electric field has only an x component, for faces perpendicular to x direction, the angle between E and DS is ± p/2. Therefore, the flux f = E.DS is separately zero for each face of the cube except the two shaded ones. Now the magnitude of the electric field at the left face is EL = ax1/2 = aa1/2 (x = a at the left face). The magnitude of electric field at the right face is ER = a x1/2 = a (2a)1/2 (x = 2a at the right face). The corresponding fluxes are EXAMPLE 1.10 f = E.DS = ∆S E L ⋅ n L L ˆ L =E DS cosq = –E DS, since q = 180° L L = –ELa2 fR= ER.DS = ER DS cosq = ER DS, since q = 0° = ERa2 Net flux through the cube 31 2024-25  Physics = fR + fL = ERa2 – ELa2 = a2 (ER – EL) = aa2 [(2a)1/2 – a1/2] = aa5/2 ( 2 –1) EXAMPLE 1.10 = 800 (0.1)5/2 ( 2 –1) 2 –1 = 1.05 N m C (b) We can use Gauss’s law to find the total charge q inside the cube. We have f = q/e0 or q = fe0. Therefore, q = 1.05 × 8.854 × 10–12 C = 9.27 × 10–12 C. Example 1.11 An electric field is uniform, and in the positive x direction for positive x, and uniform with the same magnitude but in the negative x direction for negative x. It is given that E = 200 î N/C for x > 0 and E = –200 î N/C for x < 0. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at x = +10 cm and the other is at x = –10 cm (Fig. 1.25). (a) What is the net outward flux through each flat face? (b) What is the flux through the side of the cylinder? (c) What is the net outward flux through the cylinder? (d) What is the net charge inside the cylinder? Solution (a) We can see from the figure that on the left face E and DS are parallel. Therefore, the outward flux is f = E.DS = – 200 ˆii ∆S L = + 200 DS, since ˆii ∆S = – DS = + 200 × p (0.05)2 = + 1.57 N m2 C–1 On the right face, E and DS are parallel and therefore fR = E.DS = + 1.57 N m2 C–1. (b) For any point on the side of the cylinder E is perpendicular to DS and hence E.DS = 0. Therefore, the flux out of the side of the cylinder is zero. (c) Net outward flux through the cylinder f = 1.57 + 1.57 + 0 = 3.14 N m2 C–1 FIGURE 1.25 EXAMPLE 1.11 (d) The net charge within the cylinder can be found by using Gauss’s law which gives q = e0f = 3.14 × 8.854 × 10–12 C = 2.78 × 10–11 C 32 2024-25  Electric Charges and Fields 1.14 APPLICATIONS OF 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. 1.14.1 Field due to an infinitely long straight uniformly charged wire Consider an infinitely long thin straight wire with uniform linear charge density l. 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 l > 0, inward if l < 0). This is clear from Fig. 1.26. 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. 1.26(b). Since the field is everywhere radial, flux through the two ends of the cylindrical Gaussian surface is zero. At the cylindrical FIGURE 1.26 (a) Electric field due to an part 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 2prl, where l is the length of the cylinder. 33 2024-25  Physics Flux through the Gaussian surface = flux through the curved cylindrical part of the surface = E × 2prl The surface includes charge equal to l l. Gauss’s law then gives E × 2prl = ll/e0 λ i.e.,E = 2πε 0r Vectorially, E at any point is given by λ E= ˆ n (1.32) 2πε0r where n̂ is the radial unit vector in the plane normal to the wire passing through the point. E is directed outward if l is positive and inward if l is negative. Note that when we write a vector A as a scalar multiplied by a unit vector, i.e., as A = A â , the scalar A is an algebraic number. It can be negative or positive. The direction of A will be the same as that of the unit vector â if A > 0 and opposite to â 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 (ll ) 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. 1.14.2 Field due to a uniformly charged infinite plane sheet Let s be the uniform surface charge density of an infinite plane sheet (Fig. 1.27). 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.DS through both the surfaces are equal and add up. Therefore FIGURE 1.27 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 sA. 34 Therefore by Gauss’s law, 2024-25  Electric Charges and Fields 2 EA = sA/e0 or, E = s/2e0 Vectorically, σ E= ˆ n (1.33) 2ε 0 where n̂ is a unit vector normal to the plane and going away from it. E is directed away from the plate if s is positive and toward the plate if s is negative. Note that the above application of the Gauss’ law has brought out an additional fact: E is independent of x also. For a finite large planar sheet, Eq. (1.33) is approximately true in the middle regions of the planar sheet, away from the ends. 1.14.3 Field due to a uniformly charged thin spherical shell Let s be the uniform surface charge density of a thin spherical shell of radius R (Fig. 1.28). The situation has obvious spherical symmetry. The field at any point P, outside or inside, can depend only on r (the radial distance from the centre of the shell to the point) and must be radial (i.e., along the radius vector). (i) Field outside the shell: Consider a point P outside the shell with radius vector r. To calculate E at P, we take the Gaussian surface to be a sphere of radius r and with centre O, passing through P. All points on this sphere are equivalent relative to the given charged configuration. (That is what we mean by spherical symmetry.) The electric field at each point of the Gaussian surface, therefore, has the same magnitude E and is along the radius vector at each point. Thus, E and DS at every point are parallel and the flux through each element is E DS. Summing over all DS, the flux through the Gaussian surface is E × 4 p r 2. The charge enclosed is s × 4 p R 2. By Gauss’s law σ E × 4 p r2 = 4 π R2 ε0 σ R2 q Or, E = = ε0 r 2 4 π ε0 r 2 where q = 4 p R2 s is the total charge on the spherical shell. Vectorially, q FIGURE 1.28 Gaussian E= rˆ (1.34) 4 πε 0 r 2 surfaces for a point with (a) r > R, (b) r < R. The electric field is directed outward if q > 0 and inward if q < 0. This, however, is exactly the field produced by a charge q placed at the centre O. Thus for points outside the shell, the field due to a uniformly charged shell is as if the entire charge of the shell is concentrated at its centre. (ii) Field inside the shell: In Fig. 1.28(b), the point P is inside the shell. The Gaussian surface is again a sphere through P centred at O. 35 2024-25  Physics The flux through the Gaussian surface, calculated as before, is E × 4 p r 2. However, in this case, the Gaussian surface encloses no charge. Gauss’s law then gives E × 4 p r2 = 0 i.e., E = 0 (r < R ) (1.35) that is, the field due to a uniformly charged thin shell is zero at all points inside the shell*. This important result is a direct consequence of Gauss’s law which follows from Coulomb’s law. The experimental verification of this result confirms the 1/r2 dependence in Coulomb’s law. Example 1.12 An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negative charge up to a radius R. The atom as a whole is neutral. For this model, what is the electric field at a distance r from the nucleus? FIGURE 1.29 Solution The charge distribution for this model of the atom is as shown in Fig. 1.29. The total negative charge in the uniform spherical charge distribution of radius R must be –Z e, since the atom (nucleus of charge Z e + negative charge) is neutral. This immediately gives us the negative charge density r, since we must have 4 πR3 ρ = 0 – Ze 3 3 Ze or ρ= − 4 π R3 To find the electric field E(r) at a point P which is a distance r away from the nucleus, we use Gauss’s law. Because of the spherical symmetry of the charge distribution, the magnitude of the electric field E(r) depends only on the radial distance, no matter what the direction of r. Its direction is along (or opposite to) the radius vector r from the origin to the point P. The obvious Gaussian surface is a EXAMPLE 1.12 spherical surface centred at the nucleus. We consider two situations, namely, r < R and r > R. (i) r < R : The electric flux f enclosed by the spherical surface is f = E (r ) × 4 p r 2 * Compare this with a uniform mass shell discussed in Section 7.5 of Class XI 36 Textbook of Physics. 2024-25  Electric Charges and Fields where E (r ) is the magnitude of the electric field at r. This is because the field at any point on the spherical Gaussian surface has the same direction as the normal to the surface there, and has the same magnitude at all points on the surface. The charge q enclosed by the Gaussian surface is the positive nuclear charge and the negative charge within the sphere of radius r, 4 πr3 i.e., q = Z e + ρ 3 Substituting for the charge density r obtained earlier, we have r3 q = Ze−Ze R3 Gauss’s law then gives, Z e 1 r E (r ) = − 3 ; r < R 4 π ε0 r 2 R The electric field is directed radially outward. EXAMPLE 1.12 (ii) r > R: In this case, the total charge enclosed by the Gaussian spherical surface is zero since the atom is neutral. Thus, from Gauss’s law, E (r ) × 4 p r 2 = 0 or E (r ) = 0; r > R At r = R, both cases give the same result: E = 0. 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 37 2024-25