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 = 2 rˆ i ∆S (1.28) 4 πε0 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 DS and ˆr have the same direction. Therefore, FIGURE 1.22 Flux q ∆φ = 2 ∆ S (1.29) through a sphere 4 πε0 r 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 Reprint 2025-26 Physics q φ = Σ 2 ∆ S all ∆S 4 π ε0 r Since each area element of the sphere is at the same distance r from the charge, FIGURE 1.23 Calculation of the q q S φ = Σ ∆ S = flux of uniform electric field all ∆S 4 πε0 r 2 4 πεo r 2 through the surface of a cylinder. Now S, the total area of the sphere, equals 4pr 2. Thus, q 2 q φ = 2 × 4 πr = (1.30) 4 πε0 r 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. Reprint 2025-26 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 fL= EL.DS = ∆ S E L ⋅ nˆ L =EL DS cosq = –EL DS, since q = 180° = –ELa2 EXAMPLE fR= ER.DS = ER DS cosq = ER DS, since q = 0° = ERa2 1.10 Net flux through the cube 31 Reprint 2025-26 Physics = fR + fL = ERa2 – ELa2 = a2 (ER – EL) = aa2 [(2a)1/2 – a1/2] = aa5/2 2 – 1 ( ) = 800 (0.1)5/2 2 – 1 ( ) 1.10 = 1.05 N m2 C–1 (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, EXAMPLE 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 ˆi N/C for x > 0 and E = –200 ˆi 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 ˆi i ∆S fL= E.DS = – 200 = + 200 DS, since ˆi i ∆ 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 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 EXAMPLE = 2.78 × 10–11 C Reprint 2025-26 Electric Charges and Fields

2.4 A spherical conductor of radius 12 cm has a charge of 1.6 × 10–7C distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the centre of the sphere?

2.9 ELECTROSTATICS OF CONDUCTORS Conductors and insulators were described briefly in Chapter 1. Conductors contain mobile charge carriers. In metallic conductors, these charge carriers are electrons. In a metal, the outer (valence) electrons part away from their atoms and are free to move. These electrons are free within the metal but not free to leave the metal. The free electrons form a kind of ‘gas’; they collide with each other and with the ions, and move randomly in different directions. In an external electric field, they drift against the direction of the field. The positive ions made up of the nuclei and the bound electrons remain held in their fixed positions. In electrolytic 61conductors, the charge carriers are both positive and negative ions; but Reprint 2025-26 Physics the situation in this case is more involved – the movement of the charge carriers is affected both by the external electric field as also by the so-called chemical forces (see Chapter 3). We shall restrict our discussion to metallic solid conductors. Let us note important results regarding electrostatics of conductors. 1. Inside a conductor, electrostatic field is zero Consider a conductor, neutral or charged. There may also be an external electrostatic field. In the static situation, when there is no current inside or on the surface of the conductor, the electric field is zero everywhere inside the conductor. This fact can be taken as the defining property of a conductor. A conductor has free electrons. As long as electric field is not zero, the free charge carriers would experience force and drift. In the static situation, the free charges have so distributed themselves that the electric field is zero everywhere inside. Electrostatic field is zero inside a conductor. 2. At the surface of a charged conductor, electrostatic field must be normal to the surface at every point If E were not normal to the surface, it would have some non-zero component along the surface. Free charges on the surface of the conductor would then experience force and move. In the static situation, therefore, E should have no tangential component. Thus electrostatic field at the surface of a charged conductor must be normal to the surface at every point. (For a conductor without any surface charge density, field is zero even at the surface.) See result 5. 3. The interior of a conductor can have no excess charge in the static situation A neutral conductor has equal amounts of positive and negative charges in every small volume or surface element. When the conductor is charged, the excess charge can reside only on the surface in the static situation. This follows from the Gauss’s law. Consider any arbitrary volume element v inside a conductor. On the closed surface S bounding the volume element v, electrostatic field is zero. Thus the total electric flux through S is zero. Hence, by Gauss’s law, there is no net charge enclosed by S. But the surface S can be made as small as you like, i.e., the volume v can be made vanishingly small. This means there is no net charge at any point inside the conductor, and any excess charge must reside at the surface. 4. Electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface This follows from results 1 and 2 above. Since E = 0 inside the conductor and has no tangential component on the surface, no work is done in moving a small test charge within the conductor and on its surface. That is, there is no potential difference between any two points inside or on 62 the surface of the conductor. Hence, the result. If the conductor is charged, Reprint 2025-26 Electrostatic Potential and Capacitance electric field normal to the surface exists; this means potential will be different for the surface and a point just outside the surface. In a system of conductors of arbitrary size, shape and charge configuration, each conductor is characterised by a constant value of potential, but this constant may differ from one conductor to the other. 5. Electric field at the surface of a charged conductor σ E = nˆ (2.35) ε0 where s is the surface charge density and ˆn is a unit vector normal to the surface in the outward direction. To derive the result, choose a pill box (a short cylinder) as the Gaussian surface about any point P on the surface, as shown in Fig. 2.17. The pill box is partly inside and partly outside the surface of the conductor. It has a small area of cross section d S and negligible height. Just inside the surface, the electrostatic field is zero; just outside, the field is normal to the surface with magnitude E. Thus, the contribution to the total flux through the pill box comes only from the outside (circular) cross-section of the pill box. This equals ± EdS (positive for s > 0, negative for s < 0), since over the small area dS, E may be considered constant and E and dS are parallel or antiparallel. The charge enclosed by the pill box is sdS. By Gauss’s law σδS EdS = ε0 σ E = (2.36) ε0 Including the fact that electric field is normal to the FIGURE 2.17 The Gaussian surface surface, we get the vector relation, Eq. (2.35), which (a pill box) chosen to derive Eq. (2.35) is true for both signs of s. For s > 0, electric field is for electric field at the surface of a normal to the surface outward; for s < 0, electric field charged conductor. is normal to the surface inward. 6. Electrostatic shielding Consider a conductor with a cavity, with no charges inside the cavity. A remarkable result is that the electric field inside the cavity is zero, whatever be the size and shape of the cavity and whatever be the charge on the conductor and the external fields in which it might be placed. We have proved a simple case of this result already: the electric field inside a charged spherical shell is zero. The proof of the result for the shell makes use of the spherical symmetry of the shell (see Chapter 1). But the vanishing of electric field in the (charge-free) cavity of a conductor is, as mentioned above, a very general result. A related result is that even if the conductor 63 Reprint 2025-26 Physics is charged or charges are induced on a neutral conductor by an external field, all charges reside only on the outer surface of a conductor with cavity. The proofs of the results noted in Fig. 2.18 are omitted here, but we note their important implication. Whatever be the charge and field configuration outside, any cavity in a conductor remains shielded from outside electric influence: the field inside the cavity is always zero. This is known as electrostatic shielding. The effect can be made use of in protecting sensitive instruments from FIGURE 2.18 The electric field inside a outside electrical influence. Figure 2.19 gives a cavity of any conductor is zero. All summary of the important electrostatic properties charges reside only on the outer surface of a conductor.of a conductor with cavity. (There are no charges placed in the cavity.) FIGURE 2.19 Some important electrostatic properties of a conductor. Example 2.7 (a) A comb run through one’s dry hair attracts small bits of paper. Why? What happens if the hair is wet or if it is a rainy day? (Remember, a paper does not conduct electricity.) (b) Ordinary rubber is an insulator. But special rubber tyres of aircraft are made slightly conducting. Why is this necessary? (c) Vehicles carrying inflammable materials usually have metallic ropes touching the ground during motion. Why? (d) A bird perches on a bare high power line, and nothing happens to the bird. A man standing on the ground touches the same line and gets a fatal shock. Why? Solution (a) This is because the comb gets charged by friction. The molecules 2.7 in the paper gets polarised by the charged comb, resulting in a net force of attraction. If the hair is wet, or if it is rainy day, friction between hair and the comb reduces. The comb does not get EXAMPLE charged and thus it will not attract small bits of paper. 64 Reprint 2025-26 Electrostatic Potential and Capacitance (b) To enable them to conduct charge (produced by friction) to the ground; as too much of static electricity accumulated may result in spark and result in fire. EXAMPLE (c) Reason similar to (b). (d) Current passes only when there is difference in potential. 2.7