1.6 FORCES BETWEEN MULTIPLE CHARGES The mutual electric force between two charges is given by Coulomb’s law. How to calculate the force on a charge where there are not one but several charges around? Consider a system of n stationary charges q1, q2, q3, ..., qn in vacuum. What is the force on q1 due to q2, q3, ..., qn? Coulomb’s law is not enough to answer this question. Recall that forces of mechanical origin add according to the parallelogram law of addition. Is the same true for forces of electrostatic origin? Experimentally, it is verified that force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges. This is termed as the principle of superposition. To better understand the concept, consider a system of three charges q1, q2 and q3, as shown in Fig. 1.5(a). The force on one charge, say q1, due to two other charges q2, q3 can therefore be obtained by performing a vector addition of the forces due to each one of these charges. Thus, if the force on q1 due to q2 is denoted by F12, F12 is given by Eq. (1.3) even though other charges are present. FIGURE 1.5 A system of 1 q1q 2 (a) three charges Thus, F12 = 2 ˆr12 (b) multiple charges. 11 4 πε0 r12 Reprint 2025-26 Physics In the same way, the force on q1 due to q3, denoted by F13, is given by 1 q1q 3 F13 = 2 rˆ13 4 πε0 r13 which again is the Coulomb force on q1 due to q3, even though other charge q2 is present. Thus the total force F1 on q1 due to the two charges q2 and q3 is given as 1 q1q 2 1 q1q 3 F1 = F12 + F13 = rˆ12 + rˆ13 (1.4) 4 π ε0 r122 4 πε0 r132 The above calculation of force can be generalised to a system of charges more than three, as shown in Fig. 1.5(b). The principle of superposition says that in a system of charges q1, q2, ..., qn, the force on q1 due to q2 is the same as given by Coulomb’s law, 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  q1q 2 q1q 3 q1q n  F1 = F12 + F13 + ...+ F1n =  2 rˆ12 + 2 rˆ13 + ... + 2 1rˆ n  4 πε0  r12 r13 r1n  q1 n q i = ˆr1i 2 (1.5) 4πε0 =∑i 2 r1 i 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 1.5 Consider three charges q1, q2, q3 each equal to q at the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in Fig. 1.6? FIGURE 1.6 1.5 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 EXAMPLE 12 from A is (2/3) AD = (1/ 3 ) l. By symmatry AO = BO = CO. Reprint 2025-26 Electric Charges and Fields Thus, 3 Qq Force F1 on Q due to charge q at A = 2 along AO 4 πε0 l 3 Qq 2 along BOForce F2 on Q due to charge q at B = 4 πε0 l 3 Qq 2 along COForce F3 on Q due to charge q at C = 4 πε0 l 3 Qq 2 along OA, by theThe resultant of forces F2 and F3 is 4 πε0 l 3 Qq parallelogram law. Therefore, the total force on Q = 2 ( rˆ − rˆ ) 4 πε0 l = 0, where ˆr is the unit vector along OA. It is clear also by symmetry that the three forces will sum to zero. Suppose that the resultant force was non-zero but in some direction. EXAMPLE Consider what would happen if the system was rotated through 60° about O. 1.5 Example 1.6 Consider the charges q, q, and –q placed at the vertices of an equilateral triangle, as shown in Fig. 1.7. What is the force on each charge? FIGURE 1.7 Solution The forces acting on charge q at A due to charges q at B and –q at C are F12 along BA and F13 along AC respectively, as shown in Fig. 1.7. By the parallelogram law, the total force F1 on the charge q at A is given by F1 = F 1ˆr where 1ˆr is a unit vector along BC. The force of attraction or repulsion for each pair of charges has the q 2 same magnitude F = 4 π ε0 l 2 EXAMPLE The total force F2 on charge q at B is thus F2 = F ˆr 2, where ˆr 2 is a unit vector along AC. 1.6 13 Reprint 2025-26 Physics Similarly the total force on charge –q at C is F3 = 3 F ˆn , where ˆn is the unit vector along the direction bisecting the ÐBCA. It is interesting to see that the sum of the forces on the three charges 1.6 is zero, i.e., F1 + F2 + F3 = 0 The result is not at all surprising. It follows straight from the fact that Coulomb’s law is consistent with Newton’s third law. The proof EXAMPLE is left to you as an exercise. 1.7 ELECTRIC FIELD Let us consider a point charge Q placed in vacuum, at the origin O. If we place another point charge q at a point P, where OP = r, then the charge Q will exert a force on q as per Coulomb’s law. We may ask the question: If charge q is removed, then what is left in the surrounding? Is there nothing? If there is nothing at the point P, then how does a force act when we place the charge q at P. In order to answer such questions, the early scientists introduced the concept of field. According to this, we say that the charge Q produces an electric field everywhere in the surrounding. When another charge q is brought at some point P, the field there acts on it and produces a force. The electric field produced by the charge Q at a point r is given as 1 Q 1 Q rˆ rˆ = E ( r ) = (1.6) 4 πε0 r 2 4 πε0 r 2 where rˆ = r/r, is a unit vector from the origin to the point r. Thus, Eq.(1.6) specifies the value of the electric field for each value of the position vector r. The word “field” signifies how some distributed quantity (which could be a scalar or a vector) varies with position. The effect of the charge has been incorporated in the existence of the electric field. We obtain the force F exerted by a charge Q on a charge q, as 1 Qq F = 2 rˆ (1.7) 4 πε0 r Note that the charge q also exerts an equal and opposite force on the charge Q. The electrostatic force between the charges Q and q can be looked upon as an interaction between charge q and the electric field of Q and vice versa. If we denote the position of charge q by the vector r, it experiences a force F equal to the charge q multiplied by the electric field E at the location of q. Thus, F(r) = q E(r) (1.8) Equation (1.8) defines the SI unit of electric field as N/C*. Some important remarks may be made here: (i) From Eq. (1.8), we can infer that if q is unity, the electric field due to FIGURE 1.8 Electric a charge Q is numerically equal to the force exerted by it. Thus, the field (a) due to a electric field due to a charge Q at a point in space may be defined charge Q, (b) due to a as the force that a unit positive charge would experience if placed charge –Q. 14 * An alternate unit V/m will be introduced in the next chapter. Reprint 2025-26 Electric Charges and Fields at that point. The charge Q, which is producing the electric field, is called a source charge and the charge q, which tests the effect of a source charge, is called a test charge. Note that the source charge Q must remain at its original location. However, if a charge q is brought at any point around Q, Q itself is bound to experience an electrical force due to q and will tend to move. A way out of this difficulty is to make q negligibly small. The force F is then negligibly small but the ratio F/q is finite and defines the electric field:  F  E = lim (1.9) q → 0  q  A practical way to get around the problem (of keeping Q undisturbed in the presence of q) is to hold Q to its location by unspecified forces! This may look strange but actually this is what happens in practice. When we are considering the electric force on a test charge q due to a charged planar sheet (Section 1.14), the charges on the sheet are held to their locations by the forces due to the unspecified charged constituents inside the sheet. (ii) Note that the electric field E due to Q, though defined operationally in terms of some test charge q, is independent of q. This is because F is proportional to q, so the ratio F/q does not depend on q. The force F on the charge q due to the charge Q depends on the particular location of charge q which may take any value in the space around the charge Q. Thus, the electric field E due to Q is also dependent on the space coordinate r. For different positions of the charge q all over the space, we get different values of electric field E. The field exists at every point in three-dimensional space. (iii) For a positive charge, the electric field will be directed radially outwards from the charge. On the other hand, if the source charge is negative, the electric field vector, at each point, points radially inwards. (iv) Since the magnitude of the force F on charge q due to charge Q depends only on the distance r of the charge q from charge Q, the magnitude of the electric field E will also depend only on the distance r. Thus at equal distances from the charge Q, the magnitude of its electric field E is same. The magnitude of electric field E due to a point charge is thus same on a sphere with the point charge at its centre; in other words, it has a spherical symmetry. 1.7.1 Electric field due to a system of charges Consider a system of charges q1, q2, ..., qn with position vectors r1, r2, ..., rn relative to some origin O. Like the electric field at a point in space due to a single charge, electric field at a point in space due to the system of charges is defined to be the force experienced by a unit test charge placed at that point, without disturbing the original positions of charges q1, q2, ..., qn. We can use Coulomb’s law and the superposition principle to determine this field at a point P denoted by 15position vector r. Reprint 2025-26 Physics Electric field E1 at r due to q1 at r1 is given by 1 q1 E1 = 2 ˆr1P 4 πε0 r1 P where ˆr1P is a unit vector in the direction from q1 to P, and r1P is the distance between q1 and P. In the same manner, electric field E2 at r due to q2 at r2 is 1 q 2 E2 = 2 ˆr2P 4 πε0 r2 P where ˆr2P is a unit vector in the direction from q2 to P FIGURE 1.9 Electric field at a point and r2P is the distance between q2 and P. Similar due to a system of charges is the expressions hold good for fields E3, E4, ..., En due to vector sum of the electric fields at charges q3, q4, ..., qn. the point due to individual charges. By the superposition principle, the electric field E at r due to the system of charges is (as shown in Fig. 1.9) E(r) = E1 (r) + E2 (r) + … + En(r) 1 q1 1 q 2 1 q n = 2 rˆ1P + 2 rˆ2 P + ... + 2 rˆn P 4 πε0 r1P 4 πε0 r2 P 4 πε0 rn P 1 n q i E(r) = 2 ˆri P (1.10) 4 πε0 =∑i 1 ri P E is a vector quantity that varies from one point to another point in space and is determined from the positions of the source charges. 1.7.2 Physical significance of electric field You may wonder why the notion of electric field has been introduced here at all. After all, for any system of charges, the measurable quantity is the force on a charge which can be directly determined using Coulomb’s law and the superposition principle [Eq. (1.5)]. Why then introduce this intermediate quantity called the electric field? For electrostatics, the concept of electric field is convenient, but not really necessary. Electric field is an elegant way of characterising the electrical environment of a system of charges. Electric field at a point in the space around a system of charges tells you the force a unit positive test charge would experience if placed at that point (without disturbing the system). Electric field is a characteristic of the system of charges and is independent of the test charge that you place at a point to determine the field. The term field in physics generally refers to a quantity that is defined at every point in space and may vary from point to point. Electric field is a vector field, since force is a vector quantity. The true physical significance of the concept of electric field, however, emerges only when we go beyond electrostatics and deal with time- dependent electromagnetic phenomena. Suppose we consider the force between two distant charges q1, q2 in accelerated motion. Now the greatest speed with which a signal or information can go from one point to another 16 is c, the speed of light. Thus, the effect of any motion of q1 on q2 cannot Reprint 2025-26 Electric Charges and Fields arise instantaneously. There will be some time delay between the effect (force on q2) and the cause (motion of q1). It is precisely here that the notion of electric field (strictly, electromagnetic field) is natural and very useful. The field picture is this: the accelerated motion of charge q1 produces electromagnetic waves, which then propagate with the speed c, reach q2 and cause a force on q2. The notion of field elegantly accounts for the time delay. Thus, even though electric and magnetic fields can be detected only by their effects (forces) on charges, they are regarded as physical entities, not merely mathematical constructs. They have an independent dynamics of their own, i.e., they evolve according to laws of their own. They can also transport energy. Thus, a source of time- dependent electromagnetic fields, turned on for a short interval of time and then switched off, leaves behind propagating electromagnetic fields transporting energy. The concept of field was first introduced by Faraday and is now among the central concepts in physics. Example 1.7 An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 104 N C–1 [Fig. 1.10(a)]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Fig. 1.10(b)]. Compute the time of fall in each case. Contrast the situation with that of ‘free fall under gravity’. FIGURE 1.10 Solution In Fig. 1.10(a) the field is upward, so the negatively charged electron experiences a downward force of magnitude eE where E is the magnitude of the electric field. The acceleration of the electron is ae = eE/me where me is the mass of the electron. Starting from rest, the time required by the electron to fall through a 2 h 2 h m e distance h is given by t e = = a e e E For e = 1.6 × 10–19C, me = 9.11 × 10–31 kg, E = 2.0 × 104 N C–1, h = 1.5 × 10–2 m, te = 2.9 × 10–9s In Fig. 1.10 (b), the field is downward, and the positively charged proton experiences a downward force of magnitude eE. The acceleration of the proton is ap = eE/mp EXAMPLE where mp is the mass of the proton; mp = 1.67 × 10–27 kg. The time of fall for the proton is 1.7 17 Reprint 2025-26 Physics 2 h 2 h m p –7 t p = = = 1. 3 × 10 s a p e E Thus, the heavier particle (proton) takes a greater time to fall through the same distance. This is in basic contrast to the situation of ‘free fall under gravity’ where the time of fall is independent of the mass of the body. Note that in this example we have ignored the acceleration due to gravity in calculating the time of fall. To see if this is justified, let us calculate the acceleration of the proton in the given electric field: e E a p = m p (1 . 6 × 10 − 19 C) × (2. 0 × 10 4 N C −1 ) = −27 1 .67 × 10 kg 12 –2 1.7 = 1 .9 × 10 m s which is enormous compared to the value of g (9.8 m s–2), the acceleration due to gravity. The acceleration of the electron is even greater. Thus, the effect of acceleration due to gravity can be ignored EXAMPLE in this example. Example 1.8 Two point charges q1 and q2, of magnitude +10–8 C and –10–8 C, respectively, are placed 0.1 m apart. Calculate the electric fields at points A, B and C shown in Fig. 1.11. FIGURE 1.11 Solution The electric field vector E1A at A due to the positive charge q1 points towards the right and has a magnitude (9 × 10 9 Nm 2 C -2 ) × (10 −8 C) E1A = 2 = 3.6 × 104 N C–1 (0.05m) 1.8 The electric field vector E2A at A due to the negative charge q2 points towards the right and has the same magnitude. Hence the magnitude of the total electric field EA at A is EA = E1A + E2A = 7.2 × 104 N C–1 EXAMPLE EA is directed toward the right.18 Reprint 2025-26 Electric Charges and Fields The electric field vector E1B at B due to the positive charge q1 points towards the left and has a magnitude (9 × 10 9 Nm 2 C –2 ) × (10 −8 C) E1B = 2 = 3.6 × 104 N C–1 (0.05 m) The electric field vector E2B at B due to the negative charge q2 points towards the right and has a magnitude (9 × 10 9 Nm 2 C –2 ) × (10 −8 C) E 2B = 2 = 4 × 103 N C–1 (0.15 m) The magnitude of the total electric field at B is EB = E1B – E2B = 3.2 × 104 N C–1 EB is directed towards the left. The magnitude of each electric field vector at point C, due to charge q1 and q2 is (9 × 10 9 Nm 2 C –2 ) × (10 − 8 C) E1C = E 2C = 2 = 9 × 103 N C–1 (0.10 m) The directions in which these two vectors point are indicated in Fig. 1.11. The resultant of these two vectors is π π EXAMPLE E C = E1c cos + E 2 c cos = 9 × 103 N C–1 3 3 1.8 EC points towards the right.

1.20 A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C and points radially inward, what is the net charge on the sphere? 43 Reprint 2025-26 Physics 1.21 A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 mC/m2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere? 1.22 An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density. 1.23 Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10–22 C/m2. What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates? Reprint 2025-26 Chapter Two ELECTROSTATIC POTENTIAL AND CAPACITANCE 2.12.12.12.12.1 IIINTRODUCTIONIINTRODUCTIONNTRODUCTIONNTRODUCTIONNTRODUCTION In Chapters 5 and 7 (Class XI), the notion of potential energy was introduced. When an external force does work in taking a body from a point to another against a force like spring force or gravitational force, that work gets stored as potential energy of the body. When the external force is removed, the body moves, gaining kinetic energy and losing an equal amount of potential energy. The sum of kinetic and potential energies is thus conserved. Forces of this kind are called conservative forces. Spring force and gravitational force are examples of conservative forces. Coulomb force between two (stationary) charges is also a conservative force. This is not surprising, since both have inverse-square dependence on distance and differ mainly in the proportionality constants – the masses in the gravitational law are replaced by charges in Coulomb’s law. Thus, like the potential energy of a mass in a gravitational field, we can define electrostatic potential energy of a charge in an electrostatic field. Consider an electrostatic field EEEEE due to some charge configuration. First, for simplicity, consider the field E due to a charge Q placed at the origin. Now, imagine that we bring a test charge q from a point R to a point P against the repulsive force on it due to the charge Q. With reference Reprint 2025-26 Physics to Fig. 2.1, this will happen if Q and q are both positive or both negative. For definiteness, let us take Q, q > 0. Two remarks may be made here. First, we assume that the test charge q is so small that it does not disturb the original configuration, namely the charge Q at the origin (or else, we keep Q fixed at the origin by some unspecified force). Second, in bringing the charge q fromFIGURE 2.1 A test charge q (> 0) is moved from the point R to the R to P, we apply an external force Fext just enough to point P against the repulsive counter the repulsive electric force FE (i.e, Fext= –FE). force on it by the charge Q (> 0) This means there is no net force on or acceleration of placed at the origin. the charge q when it is brought from R to P, i.e., it is brought with infinitesimally slow constant speed. In this situation, work done by the external force is the negative of the work done by the electric force, and gets fully stored in the form of potential energy of the charge q. If the external force is removed on reaching P, the electric force will take the charge away from Q – the stored energy (potential energy) at P is used to provide kinetic energy to the charge q in such a way that the sum of the kinetic and potential energies is conserved. Thus, work done by external forces in moving a charge q from R to P is WRP = – = (2.1) This work done is against electrostatic repulsive force and gets stored as potential energy. At every point in electric field, a particle with charge q possesses a certain electrostatic potential energy, this work done increases its potential energy by an amount equal to potential energy difference between points R and P. Thus, potential energy difference ∆U = U P − U R = W RP (2.2) (Note here that this displacement is in an opposite sense to the electric force and hence work done by electric field is negative, i.e., –WRP .) Therefore, we can define electric potential energy difference between two points as the work required to be done by an external force in moving (without accelerating) charge q from one point to another for electric field of any arbitrary charge configuration. Two important comments may be made at this stage: (i) The right side of Eq. (2.2) depends only on the initial and final positions of the charge. It means that the work done by an electrostatic field in moving a charge from one point to another depends only on the initial and the final points and is independent of the path taken to go from one point to the other. This is the fundamental characteristic of a conservative force. The concept of the potential energy would not be meaningful if the work depended on the path. The path-independence of work done by an electrostatic field can be proved using the 46 Coulomb’s law. We omit this proof here. Reprint 2025-26 Electrostatic Potential and Capacitance (ii) Equation (2.2) defines potential energy difference in terms of the physically meaningful quantity work. Clearly, potential energy so defined is undetermined to within an additive constant.What this means is that the actual value of potential energy is not physically significant; it is only the difference of potential energy that is significant. We can always add an arbitrary constant a to potential energy at every point, since this will not change the potential energy difference: (U P + α) − (U R + α) = U P − U R Put it differently, there is a freedom in choosing the point where potential energy is zero. A convenient choice is to have electrostatic potential energy zero at infinity. With this choice, if we take the point R at infinity, we get from Eq. (2.2) Count Alessandro Volta (1745 – 1827) Italian W ∞ P = U P − U ∞ = U P (2.3) physicist, professor at Since the point P is arbitrary, Eq. (2.3) provides us with a Pavia. Volta established that the animal electri- COUNTdefinition of potential energy of a charge q at any point. city observed by LuigiPotential energy of charge q at a point (in the presence of field Galvani, 1737–1798, indue to any charge configuration) is the work done by the experiments with frog external force (equal and opposite to the electric force) in muscle tissue placed in bringing the charge q from infinity to that point. contact with dissimilar metals, was not due to 2.2 ELECTROSTATIC POTENTIAL any exceptional property of animal tissues but ALESSANDROConsider any general static charge configuration. We define was also generated potential energy of a test charge q in terms of the work done whenever any wet body on the charge q. This work is obviously proportional to q, since was sandwiched between the force at any point is qE, where E is the electric field at that dissimilar metals. This VOLTA point due to the given charge configuration. It is, therefore, led him to develop the convenient to divide the work by the amount of charge q, so first voltaic pile, orthat the resulting quantity is independent of q. In other words, battery, consisting of a (1745 work done per unit test charge is characteristic of the electric large stack of moist disks of cardboard (electro-field associated with the charge configuration. This leads to lyte) sandwiched the idea of electrostatic potential V due to a given charge between disks of metal –1827) configuration. From Eq. (2.1), we get: (electrodes). Work done by external force in bringing a unit positive charge from point R to P  U P − U R  = VP – VR = (2.4)  q  where VP and VR are the electrostatic potentials at P and R, respectively. Note, as before, that it is not the actual value of potential but the potential difference that is physically significant. If, as before, we choose the potential to be zero at infinity, Eq. (2.4) implies: Work done by an external force in bringing a unit positive charge from infinity to a point = electrostatic potential (V ) at that point. 47 Reprint 2025-26 Physics In other words, the electrostatic potential (V ) at any point in a region with electrostatic field is the work done in bringing a unit positive charge (without acceleration) from infinity to that point. The qualifying remarks made earlier regarding potential energy also apply to the definition of potential. To obtain the work done per unit test charge, we should take an infinitesimal test charge FIGURE 2.2 Work done on a test charge q dq, obtain the work done dW in bringing it from by the electrostatic field due to any given infinity to the point and determine the ratio charge configuration is independent dW/dq. Also, the external force at every point of the of the path, and depends only on path is to be equal and opposite to the electrostatic its initial and final positions. force on the test charge at that point. 2.3 POTENTIAL DUE TO A POINT CHARGE Consider a point charge Q at the origin (Fig. 2.3). For definiteness, take Q to be positive. We wish to determine the potential at any point P with position vector r from the origin. For that we must calculate the work done in bringing a unit positive test charge from infinity to the point P. For Q > 0, the work done against the repulsive force on the test charge is positive. Since work done is independent of the path, we choose a convenient path – along the radial direction from infinity to the point P. At some intermediate point P¢ on the path, the electrostatic force on a unit positive charge is FIGURE 2.3 Work done in bringing a unit positive test charge from infinity to the Q × 1 rˆ ′ (2.5) point P, against the repulsive force of 2 4 πε0r ' charge Q (Q > 0), is the potential at P due to the charge Q. where ˆ′r is the unit vector along OP¢. Work done against this force from r¢ to r¢ + Dr¢ is Q ∆W = − 2 ∆′r (2.6) 4 πε0r ' The negative sign appears because for Dr¢ < 0, DW is positive. Total work done (W) by the external force is obtained by integrating Eq. (2.6) from r¢ = ¥ to r¢ = r, r Q Q r Q = dr ′ = ε 0r ′ 2 4 πε0r ′ ∞ 4 πε0r (2.7) W = − ∫4∞ π This, by definition is the potential at P due to the charge Q Q V (r ) = (2.8) 48 4 πε0r Reprint 2025-26 Electrostatic Potential and Capacitance Equation (2.8) is true for any sign of the charge Q, though we considered Q > 0 in its derivation. For Q < 0, V < 0, i.e., work done (by the external force) per unit positive test charge in bringing it from infinity to the point is negative. This is equivalent to saying that work done by the electrostatic force in bringing the unit positive charge form infinity to the point P is positive. [This is as it should be, since for Q < 0, the force on a unit positive test charge is attractive, so that the electrostatic force and the displacement (from infinity to P) are FIGURE 2.4 Variation of potential V with r [in units of in the same direction.] Finally, we (Q/4pe0) m-1] (blue curve) and field with r [in units of (Q/4pe0) m-2] (black curve) for a point charge Q.note that Eq. (2.8) is consistent with the choice that potential at infinity be zero. Figure (2.4) shows how the electrostatic potential ( 1/r) and the electrostatic field (1/r 2 ) varies with r. Example 2.1 (a) Calculate the potential at a point P due to a charge of 4 × 10–7C located 9 cm away. (b) Hence obtain the work done in bringing a charge of 2 × 10–9 C from infinity to the point P. Does the answer depend on the path along which the charge is brought? Solution (a) = 4 × 104 V (b) W = qV = 2 × 10–9C × 4 × 104V = 8 × 10–5 J No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements: One along EXAMPLE r and another perpendicular to r. The work done corresponding to the later will be zero. 2.1

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?