7.5 ACCELERATION DUE TO GRAVITY OF THE EARTH 4π 3 density and hence its mass is M E = R E ρ 3 The earth can be imagined to be a sphere made where ME is the mass of the earth RE is its radiusof a large number of concentric spherical shells and ρ is the density. On the other hand thewith the smallest one at the centre and the largest one at its surface. A point outside the 4π 3 ρr and mass of the sphere Mr of radius r isearth is obviously outside all the shells. Thus, 3 Reprint 2025-26 GRAVITATION 133 hence its distance from the centre of the earth is (RE + h ). If F (h) denoted the magnitude of the force on the point mass m , we get from G m M E Eq. (7.5) : = 3 r (7.10) R E If the mass m is situated on the surface of GM E m F (h ) = earth, then r = RE and the gravitational force on ( R E + h )2 (7.13) it is, from Eq. (7.10) The acceleration experienced by the point M E m F = G 2 (7.11) mass is F (h )/ m ≡ g (h ) and we get R E The acceleration experienced by the mass F (h ) GM E . g (h ) = = (7.14)m, which is usually denoted by the symbol g is m ( R E + h )2 related to F by Newton’s 2nd law by relation This is clearly less than the value of g on the F = mg. Thus GM E . g = surface of earth : GM F For h << R E , we can E R E2 g = = 2 (7.12) m R E expand the RHS of Eq. (7.14) : E Acceleration g is readily measurable. RE is a g (h ) = 2 GM 2 = g (1 + h / R E )−2known quantity. The measurement of G by R E (1 + h / R E ) Cavendish’s experiment (or otherwise), combined h << 1 , using binomial expression,with knowledge of g and RE enables one to For R E estimate ME from Eq. (7.12). This is the reason  2h why there is a popular statement regarding g (h ) ≅ g 1 − . (7.15)Cavendish : “Cavendish weighed the earth”.  RE  7.6 ACCELERATION DUE TO GRAVITY BELOW Equation (7.15) thus tells us that for small AND ABOVE THE SURFACE OF EARTH heights h above the value of g decreases by a factor (1 − 2h / R E ).Consider a point mass m at a height h above the Now, consider a point mass m at a depthsurface of the earth as shown in Fig. 7.8(a). The d below the surface of the earth (Fig. 7.8(b)),radius of the earth is denoted by RE . Since this so that its distance from the centre of thepoint is outside the earth, earth is ( R E − d ) as shown in the figure. The earth can be thought of as being composed of a smaller sphere of radius (RE – d ) and a spherical shell of thickness d. The force on m due to the outer shell of thickness d is zero because the result quoted in the previous section. As far as the smaller sphere of radius ( RE – d ) is concerned, the point mass is outside it and hence according to the result quoted earlier, the force due to this smaller sphere is just as if the entire mass of the smaller sphere is concentrated at the centre. If Ms is the mass of the smaller sphere, then, Ms/ME = ( RE – d)3 / RE3 ( 7.16) Since mass of a sphere is proportional to be Fig. 7.8 (a) g at a height h above the surface of the cube of its radius. earth. Reprint 2025-26 134 PHYSICS close to the surface of earth, at distances from the surface much smaller than the radius of the earth. In such cases, the force of gravity is practically a constant equal to mg, directed towards the centre of the earth. If we consider a Ms ME point at a height h1 from the surface of the earth and another point vertically above it at a height h2 from the surface, the work done in lifting the particle of mass m from the first to the second position is denoted by W12 Fig. 7.8 (b) g at a depth d. In this case only the smaller W12 = Force × displacement = mg (h2 – h1) (7.20) sphere of radius (RE–d) contributes to g. Thus the force on the point mass is If we associate a potential energy W(h) at a point at a height h above the surface such that F (d) = G Ms m / (RE – d ) 2 (7.17) W(h) = mgh + Wo (7.21) Substituting for Ms from above , we get (where Wo = constant) ; F (d) = G ME m ( RE – d ) / RE 3 (7.18) then it is clear that and hence the acceleration due to gravity at W12 = W(h2) – W(h1) (7.22) a depth d, The work done in moving the particle is just the difference of potential energy between its F ( d ) final and initial positions.Observe that the g(d) = is m constant Wo cancels out in Eq. (7.22). Setting h = 0 in the last equation, we get W ( h = 0 ) = Wo. F (d ) GM E g ( d ) = = 3 ( R E − d ) . h = 0 means points on the surface of the earth. m R E Thus, Wo is the potential energy on the surface of the earth. R E − d = g = g (1 − d / R E ) (7.19) If we consider points at arbitrary distance R E from the surface of the earth, the result just Thus, as we go down below earth’s surface, derived is not valid since the assumption that the acceleration due gravity decreases by a factor the gravitational force mg is a constant is no (1 − d / R E ). The remarkable thing about longer valid. However, from our discussion we know that a point outside the earth, the force of acceleration due to earth’s gravity is that it is gravitation on a particle directed towards the maximum on its surface decreasing whether you centre of the earth is go up or down. G ME m F = 2 (7.23)7.7 GRAVITATIONAL POTENTIAL ENERGY r where ME = mass of earth, m = mass of theWe had discussed earlier the notion of potential particle and r its distance from the centre of theenergy as being the energy stored in the body at earth. If we now calculate the work done inits given position. If the position of the particle lifting a particle from r = r1 to r = r2 (r2 > r1) alongchanges on account of forces acting on it, then a vertical path, we get instead of Eq. (7.20) the change in its potential energy is just the amount of work done on the body by the force. r2 G M m W12 2 d rAs we had discussed earlier, forces for which the =∫r1 r work done is independent of the path are the conservative forces.  1 1  = − G M E m − (7.24) The force of gravity is a conservative force  r2 r1  and we can calculate the potential energy of a In place of Eq. (7.21), we can thus associate body arising out of this force, called the a potential energy W(r) at a distance r, such that gravitational potential energy. Consider points Reprint 2025-26 GRAVITATION 135 G M E m W (r ) =− + W1 , (7.25) r valid for r > R , so that once again W12 = W(r2) – W(r1). Setting r = infinity in the last equation, we get W ( r = infinity ) = W1 . Thus, W1 is the potential energy at infinity. One should note that only the difference of potential energy between two points has a definite meaning from Eqs. (7.22) and (7.24). One conventionally sets W1 equal to zero, so that the potential energy at a point is just the amount of work done in displacing the particle from infinity to that point. We have calculated the potential energy at a point of a particle due to gravitational forces Fig. 7.9 on it due to the earth and it is proportional to the mass of the particle. The gravitational The gravitational potential at the centre of potential due to the gravitational force of the the square r = 2 l/2 is ( )earth is defined as the potential energy of a particle of unit mass at that point. From the G m U (r ) = − 4 2 . ⊳earlier discussion, we learn that the gravitational l potential energy associated with two particles of masses m1 and m2 separated by distance by a 7.8 ESCAPE SPEED distance r is given by If a stone is thrown by hand, we see it falls back Gm 1m 2 to the earth. Of course using machines we can V = – (if we choose V = 0 as r →∞) r shoot an object with much greater speeds and It should be noted that an isolated system of with greater and greater initial speed, the object particles will have the total potential energy that scales higher and higher heights. A natural equals the sum of energies (given by the above query that arises in our mind is the following: equation) for all possible pairs of its constituent ‘can we throw an object with such high initial particles. This is an example of the application speeds that it does not fall back to the earth?’ of the superposition principle. The principle of conservation of energy helps us to answer this question. Suppose the object ⊳ Example 7.3 Find the potential energy of did reach infinity and that its speed there was a system of four particles placed at the Vf. The energy of an object is the sum of potential vertices of a square of side l. Also obtain and kinetic energy. As before W1 denotes that the potential at the centre of the square. gravitational potential energy of the object at infinity. The total energy of the projectile at infinity then is Answer Consider four masses each of mass m at the corners of a square of side l; See Fig. 7.9. 2 mV fWe have four mass pairs at distance l and two E ( ∞=) W1 + (7.26) 2 diagonal pairs at distance 2 l If the object was thrown initially with a speed Hence, Vi from a point at a distance (h+RE) from the G m 2 G m 2 centre of the earth (RE = radius of the earth), its W (r ) = − 4 − 2 l 2 l energy initially was 2 2 1 2 GmM E 2 G m  1  G m E (h + R E ) = mVi – + W1 (7.27) 5.41 = − 2 (h + R E ) 2 +  = − l  l  2  Reprint 2025-26 136 PHYSICS By the principle of energy conservation ⊳ Eqs. (7.26) and (7.27) must be equal. Hence Example 7.4 Two uniform solid spheres 2 2 of equal radii R, but mass M and 4 M have mV f mVi GmM E a centre to centre separation 6 R, as shown – = (7.28) 2 (h + R E ) 2 in Fig. 7.10. The two spheres are held fixed. The R.H.S. is a positive quantity with a A projectile of mass m is projected from the minimum value zero hence so must be the L.H.S. surface of the sphere of mass M directly Thus, an object can reach infinity as long as Vi towards the centre of the second sphere. is such that Obtain an expression for the minimum speed v of the projectile so that it reaches mVi 2 GmM E the surface of the second sphere. – ≥ 0 (7.29) 2 (h + R E ) The minimum value of Vi corresponds to the case when the L.H.S. of Eq. (7.29) equals zero. Thus, the minimum speed required for an object to reach infinity (i.e. escape from the earth) Fig. 7.10corresponds to 1 2 GmM E Answer The projectile is acted upon by two m = (7.30) mutually opposing gravitational forces of the two ( Vi )min 2 h + R E spheres. The neutral point N (see Fig. 7.10) is defined as the position where the two forces If the object is thrown from the surface of cancel each other exactly. If ON = r, we have the earth, h = 0, and we get G M m 4 G M m = 2GM E r 2 (6 R −r )2 (Vi )min = (7.31) (6R – r)2 = 4r2 R E 6R – r = ±2r r = 2R or – 6R. 2 The neutral point r = – 6R does not concern Using the relation g = GM E / R E , we get us in this example. Thus ON = r = 2R. It is sufficient to project the particle with a speed (Vi )min = 2 gR E (7.32) which would enable it to reach N. Thereafter, the greater gravitational pull of 4M would suffice. The mechanical energy at the surface Using the value of g and RE, numerically of M is (Vi)min≈11.2 km/s. This is called the escape 1 2 G M m 4 G M mspeed, sometimes loosely called the escape E i = m v − − . velocity. 2 R 5 R Equation (7.32) applies equally well to an At the neutral point N, the speed approaches object thrown from the surface of the moon with zero. The mechanical energy at N is purely g replaced by the acceleration due to Moon’s potential. gravity on its surface and rE replaced by the G M m 4 G M m − EN = − .radius of the moon. Both are smaller than their 2 R 4 R values on earth and the escape speed for the From the principle of conservation of moon turns out to be 2.3 km/s, about five times mechanical energy smaller. This is the reason that moon has no atmosphere. Gas molecules if formed on the 1 2 GM 4GM GM GMsurface of the moon having velocities larger than v − − = − − this will escape the gravitational pull of the 2 R 5 R 2R R moon. or Reprint 2025-26 GRAVITATION 137 + h) with speed V. Its 2 2 G M  4 1  traverses a distance 2π(RE v = − time period T therefore is R  5 2  2π( R E + h ) 2π( R E + h )3 / 2 T = = (7.37)  3 G M 1/2 V ⊳ G M E v =  5 R  on substitution of value of V from Eq. (7.35). A point to note is that the speed of the projectile Squaring both sides of Eq. (7.37), we get is zero at N, but is nonzero when it strikes the 2 T = k ( RE + h)3 (where k = 4 π2 / GME) (7.38)heavier sphere 4 M. The calculation of this speed is left as an exercise to the students. which is Kepler’s law of periods, as applied to motion of satellites around the earth. For a 7.9 EARTH SATELLITES satellite very close to the surface of earth h can be neglected in comparison to RE in Eq. (7.38). Earth satellites are objects which revolve around Hence, for such satellites, T is To, where the earth. Their motion is very similar to the motion of planets around the Sun and hence T 0 = 2π R E / g (7.39) Kepler’s laws of planetary motion are equally If we substitute the numerical values applicable to them. In particular, their orbits g ≃ 9.8 m s-2 and RE = 6400 km., we get around the earth are circular or elliptic. Moon is the only natural satellite of the earth with a near 6.4 × 10 6 T 0 = 2π s circular orbit with a time period of approximately 9.8

7.3 Universal law of above fall towards the earth and there are many other such gravitation phenomena. Historically it was the Italian Physicist Galileo

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 Reprint 2025-26 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 = 2 πε0r Vectorially, E at any point is given by λ E = nˆ (1.32) 2 πε0r where ˆn 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 ˆa , 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 ˆa if A > 0 and opposite to ˆa 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, Reprint 2025-26 Electric Charges and Fields 2 EA = sA/e0 or, E = s/2e0 Vectorically, σ E = nˆ (1.33) 2ε0 where ˆn 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 2 E × 4 p r 2 = 4 π R ε0 σ R 2 q Or, E = 2 = 2 ε0 r 4 π ε0 r where q = 4 p R2 s is the total charge on the spherical shell. Vectorially, q FIGURE 1.28 Gaussian E = 2 rˆ (1.34) 4 πε0 r 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 Reprint 2025-26 Physics The flux through the Gaussian surface, calculated as before, is E × 4 p r2. 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 π R 3 ρ= 0– Ze 3 3 Ze or ρ= − 3 4 π R 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 spherical surface centred at the nucleus. We consider two situations, 1.12 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 EXAMPLE * Compare this with a uniform mass shell discussed in Section 7.5 of Class XI 36 Textbook of Physics. Reprint 2025-26 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 π r 3 i.e., q = Z e + ρ 3 Substituting for the charge density r obtained earlier, we have r 3 q = Z e − Z e 3 R Gauss’s law then gives, Z e 1 r E (r ) = 2 − 3 ; r < R 4 π ε0 r R The electric field is directed radially outward. (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, EXAMPLE E (r) × 4 p r 2 = 0 or E (r) = 0; r > R At r = R, both cases give the same result: E = 0. 1.12 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 Reprint 2025-26 Physics 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 (q1q 2 ) F21 = force on q2 due to q1 = 2 rˆ21 r21 1 where ˆr21 is a unit vector in the direction from q1 to q2 and k = 4 πε0 is the constant of proportionality. In SI units, the unit of charge is coulomb. The experimental value of the constant e0 is e0 = 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 k e 2 39 ≅ 2 . 4 × 10 G m e m p 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 the vector sum of the force on q1 due to q2, the force on q1 due to q3, and so on. For each pair, the force is given by the Coulomb’s law for two charges stated earlier. 8. The electric field E at a point due to a charge configuration is the force on a small positive test charge q placed at the point divided by the magnitude of the charge. Electric field due to a point charge q has a magnitude |q|/4pe0r2; it is radially outwards from q, if q is positive, and radially inwards if q is negative. Like Coulomb force, electric field also satisfies superposition principle. 9. An electric field line is a curve drawn in such a way that the tangent at each point on the curve gives the direction of electric field at that point. The relative closeness of field lines indicates the relative strength of electric field at different points; they crowd near each other in regions of strong electric field and are far apart where the electric field is weak. In regions of constant electric field, the field lines are uniformly spaced parallel straight lines. 10. Some of the important properties of field lines are: (i) Field lines are continuous curves without any breaks. (ii) Two field lines cannot cross each other. (iii) Electrostatic field lines start at positive charges and end at negative charges —they cannot form closed loops. 11. An electric dipole is a pair of equal and opposite charges q and –q separated by some distance 2a. Its dipole moment vector p has magnitude 2qa and is in the direction of the dipole axis from –q to q. Reprint 2025-26 Electric Charges and Fields 12. Field of an electric dipole in its equatorial plane (i.e., the plane perpendicular to its axis and passing through its centre) at a distance r from the centre: − p 1 E = 2 2 3/2 4 πεo (a + r ) − p ≅ 3 , for r >> a 4 πεo r Dipole electric field on the axis at a distance r from the centre: 2 p r E = 2 2 2 4 πε0 (r − a ) 2 p ≅ 3 for r >> a 4 πε0 r The 1/r3 dependence of dipole electric fields should be noted in contrast to the 1/r 2 dependence of electric field due to a point charge. 13. In a uniform electric field E, a dipole experiences a torque τ given by τ = p × E but experiences no net force. 14. The flux Df of electric field E through a small area element DS is given by Df = E.DS The vector area element DS is DS = DS ˆn where DS is the magnitude of the area element and ˆn is normal to the area element, which can be considered planar for sufficiently small DS. For an area element of a closed surface, ˆn is taken to be the direction of outward normal, by convention. 15. Gauss’s law: The flux of electric field through any closed surface S is 1/e0 times the total charge enclosed by S. The law is especially useful in determining electric field E, when the source distribution has simple symmetry: (i) Thin infinitely long straight wire of uniform linear charge density l λ E = nˆ 2 πε0 r where r is the perpendicular distance of the point from the wire and ˆn is the radial unit vector in the plane normal to the wire passing through the point. (ii) Infinite thin plane sheet of uniform surface charge density s σ E = nˆ 2 ε0 where ˆn is a unit vector normal to the plane, outward on either side. 39 Reprint 2025-26 Physics (iii) Thin spherical shell of uniform surface charge density s q E = 2 rˆ (r ≥ R ) 4 πε0 r E = 0 (r < R) where r is the distance of the point from the centre of the shell and R the radius of the shell. q is the total charge of the shell: q = 4pR2s. The electric field outside the shell is as though the total charge is concentrated at the centre. The same result is true for a solid sphere of uniform volume charge density. The field is zero at all points inside the shell. Physical quantity Symbol Dimensions Unit Remarks Vector area element D S [L2] m2 DS = DS ˆn Electric field E [MLT–3A–1] V m–1 Electric flux f [ML3 T–3A–1] V m Df = E.DS Dipole moment p [LTA] C m Vector directed from negative to positive charge Charge density: linear l [L–1 TA] C m–1 Charge/length surface s [L–2 TA] C m–2 Charge/area volume r [L–3 TA] C m–3 Charge/volume POINTS TO PONDER 1. You might wonder why the protons, all carrying positive charges, are compactly residing inside the nucleus. Why do they not fly away? You will learn that there is a third kind of a fundamental force, called the strong force which holds them together. The range of distance where this force is effective is, however, very small ~10-14 m. This is precisely the size of the nucleus. Also the electrons are not allowed to sit on top of the protons, i.e. inside the nucleus, due to the laws of quantum mechanics. This gives the atoms their structure as they exist in nature. 2. Coulomb force and gravitational force follow the same inverse-square law. But gravitational force has only one sign (always attractive), while Reprint 2025-26 Electric Charges and Fields Coulomb force can be of both signs (attractive and repulsive), allowing possibility of cancellation of electric forces. This is how gravity, despite being a much weaker force, can be a dominating and more pervasive force in nature. 3. The constant of proportionality k in Coulomb’s law is a matter of choice if the unit of charge is to be defined using Coulomb’s law. In SI units, however, what is defined is the unit of current (A) via its magnetic effect (Ampere’s law) and the unit of charge (coulomb) is simply defined by (1C = 1 A s). In this case, the value of k is no longer arbitrary; it is approximately 9 × 109 N m2 C–2. 4. The rather large value of k, i.e., the large size of the unit of charge (1C) from the point of view of electric effects arises because (as mentioned in point 3 already) the unit of charge is defined in terms of magnetic forces (forces on current–carrying wires) which are generally much weaker than the electric forces. Thus while 1 ampere is a unit of reasonable size for magnetic effects, 1 C = 1 A s, is too big a unit for electric effects. 5. The additive property of charge is not an ‘obvious’ property. It is related to the fact that electric charge has no direction associated with it; charge is a scalar. 6. Charge is not only a scalar (or invariant) under rotation; it is also invariant for frames of reference in relative motion. This is not always true for every scalar. For example, kinetic energy is a scalar under rotation, but is not invariant for frames of reference in relative motion. 7. Conservation of total charge of an isolated system is a property independent of the scalar nature of charge noted in point 6. Conservation refers to invariance in time in a given frame of reference. A quantity may be scalar but not conserved (like kinetic energy in an inelastic collision). On the other hand, one can have conserved vector quantity (e.g., angular momentum of an isolated system). 8. Quantisation of electric charge is a basic (unexplained) law of nature; interestingly, there is no analogous law on quantisation of mass. 9. Superposition principle should not be regarded as ‘obvious’, or equated with the law of addition of vectors. It says two things: force on one charge due to another charge is unaffected by the presence of other charges, and there are no additional three-body, four-body, etc., forces which arise only when there are more than two charges. 10. The electric field due to a discrete charge configuration is not defined at the locations of the discrete charges. For continuous volume charge distribution, it is defined at any point in the distribution. For a surface charge distribution, electric field is discontinuous across the surface. 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. 41 Reprint 2025-26 Physics 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 mC due to another small sphere of charge –0.8 mC 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 mC, qB = –5 mC, qC = 2 mC, and qD = –5 mC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 mC 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 mC and qB = –3 mC 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 qA = 2.5 × 10–7 C and qB = –2.5 × 10–7 C located 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 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 Figure 1.30 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle 42 has the highest charge to mass ratio? Reprint 2025-26 Electric Charges and Fields FIGURE 1.30