5.8 THE CONSERVATION OF MECHANICAL a ball of mass m being dropped from a cliff of ENERGY height H. For simplicity we demonstrate this important principle for one-dimensional motion. Suppose that a body undergoes displacement ∆x under the action of a conservative force F. Then from the WE theorem we have, ∆K = F(x) ∆x If the force is conservative, the potential energy function V(x) can be defined such that − ∆V = F(x) ∆x The above equations imply that Fig. 5.5 The conversion of potential energy to kinetic ∆K + ∆V = 0 energy for a ball of mass m dropped from a ∆(K + V ) = 0 (5.10) height H. Reprint 2025-26 WORK, ENERGY AND POWER 79 The total mechanical energies E0, Eh, and EH Answer (i) There are two external forces on of the ball at the indicated heights zero (ground the bob : gravity and the tension (T ) in the level), h and H, are string. The latter does no work since the displacement of the bob is always normal to the EH = mgH (5.11 a) string. The potential energy of the bob is thus 1 2 E h = mgh + mv h (5.11 b) associated with the gravitational force only. The 2 2 total mechanical energy E of the system is E0 = (1/2) mvf (5.11 c) conserved. We take the potential energy of the The constant force is a special case of a spatially system to be zero at the lowest point A. Thus, dependent force F(x). Hence, the mechanical at A : energy is conserved. Thus EH = E0 1 2 1 2 E = mv0 (5.12) or, mgH = mv f 2 2 v f = 2 gH [Newton’s Second Law] a result that was obtained in section 5.7 for a where TA is the tension in the string at A. At thefreely falling body. highest point C, the string slackens, as the Further, tension in the string (TC) becomes zero. EH = Eh Thus, at Cwhich implies, 2 1 2 v h = 2 g(H − h) (5.11 d) E = mv c + 2mgL (5.13) 2 and is a familiar result from kinematics. At the height H, the energy is purely potential. mvc2It is partially converted to kinetic at height h and mg = [Newton’s Second Law] (5.14) L is fully kinetic at ground level. This illustrates the conservation of mechanical energy. where vC is the speed at C. From Eqs. (5.13) ⊳ and (5.14) Example 5.7 A bob of mass m is suspended 5 by a light string of length L . It is imparted a E = mgL horizontal velocity vo at the lowest point A 2 such that it completes a semi-circular Equating this to the energy at A trajectory in the vertical plane with the string 5 m 2 becoming slack only on reaching the topmost mgL = v 0 point, C. This is shown in Fig. 5.6. Obtain an 2 2 expression for (i) vo; (ii) the speeds at points or, v 0 = 5 gL B and C; (iii) the ratio of the kinetic energies (ii) It is clear from Eq. (5.14) (KB/KC) at B and C. Comment on the nature of the trajectory of the bob after it reaches vC = gL the point C. At B, the energy is 1 2 E = mv B + mgL 2 Equating this to the energy at A and employing the result from (i), namely v 02 = 5 gL , 1 2 1 2 mv B + mgL = mv 0 2 2 5 = m g L Fig. 5.6 2 Reprint 2025-26 80 PHYSICS ∴ vB = 3 gL k x m2 W = + (5.16) 2 (iii) The ratio of the kinetic energies at B and C is : 1 2 mv B K B 2 3 = = 2 1 K C 1 mvC 2 At point C, the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise the bob will continue on its circular path and complete the revolution. ⊳

10.2 HUYGENS PRINCIPLE We would first define a wavefront: when we drop a small stone on a calm pool of water, waves spread out from the point of impact. Every point on the surface starts oscillating with time. At any instant, a photograph of the surface would show circular rings on which the disturbance is maximum. Clearly, all points on such a circle are oscillating in phase because they are at the same distance from the source. Such a locus of points, which oscillate in phase is called a wavefront; thus a wavefront is defined as a surface of constant FIGURE 10.1 (a) Aphase. The speed with which the wavefront moves outwards from the diverging spherical source is called the speed of the wave. The energy of the wave travels wave emanating from in a direction perpendicular to the wavefront. a point source. The If we have a point source emitting waves uniformly in all directions, wavefronts are then the locus of points which have the same amplitude and vibrate in spherical. the same phase are spheres and we have what is known as a spherical wave as shown in Fig. 10.1(a). At a large distance from the source, a small portion of the sphere can be considered as a plane and we have what is known as a plane wave [Fig. 10.1(b)]. Now, if we know the shape of the wavefront at t = 0, then Huygens principle allows us to determine the shape of the wavefront at a later time t. Thus, Huygens principle is essentially a geometrical construction, which given the shape of the wafefront at any time allows us to determine the shape of the wavefront at a later time. Let us consider a diverging FIGURE 10.1 (b) At a wave and let F1F2 represent a portion of the spherical wavefront at t = 0 large distance from (Fig. 10.2). Now, according to Huygens principle, each point of the the source, a small wavefront is the source of a secondary disturbance and the wavelets portion of the emanating from these points spread out in all directions with the speed spherical wave can of the wave. These wavelets emanating from the wavefront are usually be approximated by a plane wave.referred to as secondary wavelets and if we draw a common tangent to all these spheres, we obtain the new position of the wavefront at a later time. FIGURE 10.2 F1F2 represents the spherical wavefront (with O as centre) at t = 0. The envelope of the secondary wavelets emanating from F1F2 produces the forward moving wavefront G1G2. The backwave D1D2 does not exist. 257 Reprint 2025-26 Physics Thus, if we wish to determine the shape of the wavefront at t = t, we draw spheres of radius vt from each point on the spherical wavefront where v represents the speed of the waves in the medium. If we now draw a common tangent to all these spheres, we obtain the new position of the wavefront at t = t. The new wavefront shown as G1G2 in Fig. 10.2 is again spherical with point O as the centre. The above model has one shortcoming: we also have a backwave which is shown as D1D2 in Fig. 10.2. Huygens argued that the amplitude of the secondary wavelets is maximum in the forward direction and zero in the backward direction; by making this adhoc assumption, Huygens could explain the absence of the backwave. However, this adhoc assumption is not satisfactory and the absence of the backwave is really justified from more rigorous wave theory. In a similar manner, we can use Huygens principle to determine the shape of the wavefront for a plane wave propagating through a medium (Fig. 10.3). FIGURE 10.3 Huygens geometrical construction for a 10.3 REFRACTION AND REFLECTION OF plane wave PLANE WAVES USING HUYGENS PRINCIPLE propagating to the right. F1 F2 is the 10.3.1 Refraction of a plane wave plane wavefront at t = 0 and G1G2 is the We will now use Huygens principle to derive the laws of refraction. Let PP¢ wavefront at a later represent the surface separating medium 1 and medium 2, as shown in time t. The lines A1A2, Fig. 10.4. Let v1 and v2 represent the speed of light in medium 1 and B1B2 … etc., are medium 2, respectively. We assume a plane wavefront AB propagating in normal to both F1F2 the direction A¢A incident on the interface at an angle i as shown in the and G1G2 and figure. Let t be the time taken by the wavefront to travel the distance BC. represent rays. Thus, BC = v1 t FIGURE 10.4 A plane wave AB is incident at an angle i on the surface PP¢ separating medium 1 and medium 2. The plane wave undergoes refraction and CE represents the refracted wavefront. The figure corresponds to v2 < v1 so that the refracted waves bends towards the 258 normal. Reprint 2025-26 Wave Optics In order to determine the shape of the refracted wavefront, we draw a sphere of radius v2t from the point A in the second medium (the speed of the wave in the second medium is v2). Let CE represent a tangent plane drawn from the point C on to the sphere. Then, AE = v2 t and CE would represent the refracted wavefront. If we now consider the triangles ABC and AEC, we readily obtain BC v1τ sin i = = (10.1) AC AC and AE v 2τ sin r = = (10.2) AC AC ChristiaanChristiaanChristiaanChristiaanChristiaan HuygensHuygensHuygensHuygensHuygens where i and r are the angles of incidence and refraction, (1629(1629(1629(1629(1629 ––––– 1695)1695)1695)1695)1695) Dutch respectively. Thus we obtain physicist, astronomer, mathematician and the sin i v1 founder of the wave = sin r v 2 (10.3) theory of light. His book, Treatise on light, makes From the above equation, we get the important result CHRISTIAAN fascinating reading even that if r < i (i.e., if the ray bends toward the normal), the today. He brilliantly speed of the light wave in the second medium (v2) will be explained the double less then the speed of the light wave in the first medium refraction shown by the (v1). This prediction is opposite to the prediction from mineral calcite in this the corpuscular model of light and as later experiments work in addition to HUYGENSshowed, the prediction of the wave theory is correct. Now, reflection and refraction. if c represents the speed of light in vacuum, then, He was the first to analyse circular and c simple harmonic motion (1629 n1 = – v1 (10.4) and designed and built improved clocks and and telescopes. He discovered c the true geometry of 1695) n2 = v 2 (10.5) Saturn’s rings. are known as the refractive indices of medium 1 and medium 2, respectively. In terms of the refractive indices, Eq. (10.3) can be written as n1 sin i = n2 sin r (10.6) This is the Snell’s law of refraction. Further, if l1 and l 2 denote the wavelengths of light in medium 1 and medium 2, respectively and if the distance BC is equal to l1 then the distance AE will be equal to l2 (because if the crest from B has reached C in time t, then the crest from A should have also reached E in time t ); thus, λ1 BC v1 = = λ2 AE v 2 or v1 v 2 = (10.7) 259 λ1 λ2 Reprint 2025-26 Physics The above equation implies that when a wave gets refracted into a denser medium (v1 > v2) the wavelength and the speed of propagation decrease but the frequency n (= v/l) remains the same. 10.3.2 Refraction at a rarer medium We now consider refraction of a plane wave at a rarer medium, i.e., v2 > v1. Proceeding in an exactly similar manner we can construct a refracted wavefront as shown in Fig. 10.5. The angle of refraction will now be greater than angle of incidence; however, we will still have effect n1 sin i = n2 sin r . We define an angle ic by the following equation n 2 sin i c = (10.8) Doppler n1 and Thus, if i = ic then sin r = 1 and r = 90°. Obviously, for i > ic, there can not be any refracted wave. The angle ic is known as the critical angle and for all angles of incidence greater than the critical angle, we will not have any refracted wave and the wave will undergo what is known as total internal reflection. The phenomenon of total internal reflection and its resonance applications was discussed in Section 9.4. refraction, diffraction, interference, of FIGURE 10.5 Refraction of a plane wave incident on a rarer medium for which v2 > v1. The plane wave bends away from the normal. Demonstration http://www.falstad.com/ripple/ 10.3.3 Reflection of a plane wave by a plane surface We next consider a plane wave AB incident at an angle i on a reflecting surface MN. If v represents the speed of the wave in the medium and if t represents the time taken by the wavefront to advance from the point B to C then the distance BC = vt In order to construct the reflected wavefront we draw a sphere of radius vt from the point A as shown in Fig. 10.6. Let CE represent the tangent plane drawn from the point C to this sphere. Obviously 260 AE = BC = vt Reprint 2025-26 Wave Optics FIGURE 10.6 Reflection of a plane wave AB by the reflecting surface MN. AB and CE represent incident and reflected wavefronts. If we now consider the triangles EAC and BAC we will find that they are congruent and therefore, the angles i and r (as shown in Fig. 10.6) would be equal. This is the law of reflection. Once we have the laws of reflection and refraction, the behaviour of prisms, lenses, and mirrors can be understood. These phenomena were discussed in detail in Chapter 9 on the basis of rectilinear propagation of light. Here we just describe the behaviour of the wavefronts as they undergo reflection or refraction. In Fig. 10.7(a) we consider a plane wave passing through a thin prism. Clearly, since the speed of light waves is less in glass, the lower portion of the incoming wavefront (which travels through the greatest thickness of glass) will get delayed resulting in a tilt in the emerging wavefront as shown in the figure. In Fig. 10.7(b) we consider a plane wave incident on a thin convex lens; the central part of the incident plane wave traverses the thickest portion of the lens and is delayed the most. The emerging wavefront has a depression at the centre and therefore the wavefront becomes spherical and converges to the point F which is known as the focus. In Fig. 10.7(c) a plane wave is incident on a concave mirror and on reflection we have a spherical wave converging to the focal point F. In a similar manner, we can understand refraction and reflection by concave lenses and convex mirrors. From the above discussion it follows that the total time taken from a point on the object to the corresponding point on the image is the same measured along any ray. For example, when a convex lens focusses light to form a real image, although the ray going through the centre traverses a shorter path, but because of the slower speed in glass, the time taken is the same as for rays travelling near the edge of the lens. FIGURE 10.7 Refraction of a plane wave by (a) a thin prism, (b) a convex lens. 261 (c) Reflection of a plane wave by a concave mirror. Reprint 2025-26 Physics Example 10.1 (a) When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency. Explain why? (b) When light travels from a rarer to a denser medium, the speed decreases. Does the reduction in speed imply a reduction in the energy carried by the light wave? (c) In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity of light in the photon picture of light. Solution (a) Reflection and refraction arise through interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators, which take up the frequency of the external agency (light) causing forced oscillations. The frequency of light emitted by a charged oscillator equals its frequency of oscillation. Thus, the frequency of scattered light equals the frequency of incident light. (b) No. Energy carried by a wave depends on the amplitude of the 10.1 wave, not on the speed of wave propagation. (c) For a given frequency, intensity of light in the photon picture is determined by the number of photons crossing an unit area per EXAMPLE unit time. 10.4 COHERENT AND INCOHERENT ADDITION OF WAVES In this section we will discuss the interference pattern produced by the superposition of two waves. You may recall that we had discussed the superposition principle in Chapter 14 of your Class XI textbook. Indeed the entire field of interference is based on the superposition (a) (b) principle according to which at a particular point in the medium, the resultant FIGURE 10.8 (a) Two needles oscillating in displacement produced by a number ofphase in water represent two coherent sources. (b) The pattern of displacement of water waves is the vector sum of the displace- molecules at an instant on the surface of water ments produced by each of the waves. showing nodal N (no displacement) and Consider two needles S1 and S2 moving antinodal A (maximum displacement) lines. periodically up and down in an identical fashion in a trough of water [Fig. 10.8(a)]. They produce two water waves, and at a particular point, the phase difference between the displacements produced by each of the waves does not change with time; when this happens the two sources are said to be coherent. Figure 10.8(b) shows the position of crests (solid circles) and troughs (dashed circles) at a given instant of time. Consider a point P for which 262 S1 P = S2 P Reprint 2025-26 Wave Optics Since the distances S1 P and S2 P are equal, waves from S1 and S2 will take the same time to travel to the point P and waves that emanate from S1 and S2 in phase will also arrive, at the point P, in phase. Thus, if the displacement produced by the source S1 at the point P is given by y1 = a cos wt then, the displacement produced by the source S2 (at the point P) will also be given by y2 = a cos wt Thus, the resultant of displacement at P would be given by y = y1 + y2 = 2 a cos wt Since the intensity is proportional to the square of the amplitude, the resultant intensity will be given by I = 4 I0 where I0 represents the intensity produced by each one of the individual sources; I0 is proportional to a2. In fact at any point on the perpendicular bisector of S1S2, the intensity will be 4I0. The two sources are said to FIGURE 10.9interfere constructively and we have what is referred to as constructive (a) Constructive interference. We next consider a point Q [Fig. 10.9(a)] interference at a for which point Q for which the S2Q –S1Q = 2l path difference is 2l. (b) Destructive The waves emanating from S1 will arrive exactly two cycles earlier interference at a than the waves from S2 and will again be in phase [Fig. 10.9(a)]. Thus, if point R for which the the displacement produced by S1 is given by path difference is 2.5 l. y1 = a cos wt then the displacement produced by S2 will be given by y2 = a cos (wt – 4p) = a cos wt where we have used the fact that a path difference of 2l corresponds to a phase difference of 4p. The two displacements are once again in phase and the intensity will again be 4 I0 giving rise to constructive interference. In the above analysis we have assumed that the distances S1Q and S2Q are much greater than d (which represents the distance between S1 and S2) so that although S1Q and S2Q are not equal, the amplitudes of the displacement produced by each wave are very nearly the same. We next consider a point R [Fig. 10.9(b)] for which S2R – S1R = –2.5l The waves emanating from S1 will arrive exactly two and a half cycles later than the waves from S2 [Fig. 10.10(b)]. Thus if the displacement FIGURE 10.10 Locus produced by S1 is given by of points for which y1 = a cos wt S1P – S2P is equal to zero, ±l, ± 2l, ± 3l. then the displacement produced by S2 will be given by y2 = a cos (wt + 5p) = – a cos wt 263 Reprint 2025-26 Physics where we have used the fact that a path difference of 2.5l corresponds to a phase difference of 5p. The two displacements are now out of phase and the two displacements will cancel out to give zero intensity. This is referred to as destructive interference. To summarise: If we have two coherent sources S1 and S2 vibrating in phase, then for an arbitrary point P whenever the path difference, S1P ~ S2P = nl (n = 0, 1, 2, 3,...) (10.9) we will have constructive interference and the resultant intensity will be 4I0; the sign ~ between S1P and S2 P represents the difference between S1P and S2 P. On the other hand, if the point P is such that the path difference, 1 S1P ~ S2P = (n+ ) l (n = 0, 1, 2, 3, ...) (10.10) 2 we will have destructive interference and the resultant intensity will be zero. Now, for any other arbitrary point G (Fig. 10.10) let the phase difference between the two displacements be f. Thus, if the displacement produced by S1 is given by y1 = a cos wt then, the displacement produced by S2 would be interference y2 = a cos (wt + f) wave and the resultant displacement will be given by on y = y1 + y2 = a [cos wt + cos (wt +f)] = 2 a cos (f/2) cos (wt + f/2) experiments Tank The amplitude of the resultant displacement is 2a cos (f/2) and therefore the intensity at that point will be I = 4 I0 cos2 (f/2) (10.11) Ripple http://phet.colorado.edu/en/simulation/legacy/wave-interference If f = 0, ± 2 p, ± 4 p,… which corresponds to the condition given by Eq. (10.9) we will have constructive interference leading to maximum intensity. On the other hand, if f = ± p, ± 3p, ± 5p … [which corresponds to the condition given by Eq. (10.10)] we will have destructive interference leading to zero intensity. Now if the two sources are coherent (i.e., if the two needles are going up and down regularly) then the phase difference f at any point will not change with time and we will have a stable interference pattern; i.e., the positions of maxima and minima will not change with time. However, if the two needles do not maintain a constant phase difference, then the interference pattern will also change with time and, if the phase difference changes very rapidly with time, the positions of maxima and minima will also vary rapidly with time and we will see a “time-averaged” intensity distribution. When this happens, we will observe an average intensity that will be given by I = 2 I0 (10.12) 264 at all points. Reprint 2025-26 Wave Optics When the phase difference between the two vibrating sources changes rapidly with time, we say that the two sources are incoherent and when this happens the intensities just add up. This is indeed what happens when two separate light sources illuminate a wall.

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