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. ⊳

5.5 WORK DONE BY A VARIABLE FORCE A constant force is rare. It is the variable force, which is more commonly encountered. Fig. 5.3 is a plot of a varying force in one dimension. If the displacement ∆x is small, we can take the force F (x) as approximately constant and the work done is then ∆W =F (x) ∆x Fig. 5.3(a) Reprint 2025-26 76 PHYSICS The work done by the frictional force is Wf → area of the rectangle AGHI Wf = (−50) × 20 = − 1000 J The area on the negative side of the force axis has a negative sign. ⊳ 5.6 THE WORK-ENERGY THEOREM FOR A VARIABLE FORCE We are now familiar with the concepts of workFig. 5.3 (a) The shaded rectangle represents the work done by the varying force F(x), over and kinetic energy to prove the work-energy the small displacement ∆x, ∆W = F(x) ∆x. theorem for a variable force. We confine (b) adding the areas of all the rectangles we ourselves to one dimension. The time rate of find that for ∆x →0, the area under the curve change of kinetic energy is is exactly equal to the work done by F(x). d K d  1 2  =⊳ d t  2 m v  Example 5.5 A woman pushes a trunk on d t a railway platform which has a rough d v surface. She applies a force of 100 N over a = m v d t distance of 10 m. Thereafter, she gets progressively tired and her applied force = F v (from Newton’s Second Law) reduces linearly with distance to 50 N. The d x total distance through which the trunk has = F d t been moved is 20 m. Plot the force applied Thus by the woman and the frictional force, which dK = Fdx is 50 N versus displacement. Calculate the Integrating from the initial position (x i ) to final work done by the two forces over 20 m. position ( x f ), we have Answer K f x f F dx ∫ d K = ∫ K i x i where, Ki and K f are the initial and final kinetic energies corresponding to x i and x f. x f F d x or (5.8a) K f − K i = ∫ Fig. 5.4 Plot of the force F applied by the woman and x i the opposing frictional force f versus From Eq. (5.7), it follows that displacement. Kf − Ki = W (5.8b) The plot of the applied force is shown in Fig. 5.4. At x = 20 m, F = 50 N (≠ 0). We are given Thus, the WE theorem is proved for a variable that the frictional force f is |f|= 50 N. It opposes force. motion and acts in a direction opposite to F. It While the WE theorem is useful in a variety of is therefore, shown on the negative side of the problems, it does not, in general, incorporate the force axis. complete dynamical information of Newton’s The work done by the woman is second law. It is an integral form of Newton’s WF → area of the rectangle ABCD + area of second law. Newton’s second law is a relation the trapezium CEID between acceleration and force at any instant of 1 time. Work-energy theorem involves an integral WF = 100 × 10 + (100 + 50) × 10 over an interval of time. In this sense, the temporal 2 = 1000 + 750 (time) information contained in the statement of = 1750 J Newton’s second law is ‘integrated over’ and is Reprint 2025-26 WORK, ENERGY AND POWER 77 not available explicitly. Another observation is that are like ‘compressed springs’. They possess a Newton’s second law for two or three dimensions large amount of potential energy. An earthquake is in vector form whereas the work-energy results when these fault lines readjust. Thus, theorem is in scalar form. In the scalar form, potential energy is the ‘stored energy’ by virtue information with respect to directions contained of the position or configuration of a body. The in Newton’s second law is not present. body left to itself releases this stored energy in ⊳ the form of kinetic energy. Let us make our notion Example 5.6 A block of mass m = 1 kg, of potential energy more concrete. moving on a horizontal surface with speed The gravitational force on a ball of mass m is vi = 2 m s–1 enters a rough patch ranging mg . g may be treated as a constant near the earth from x = 0.10 m to x = 2.01 m. The retarding surface. By ‘near’ we imply that the height h of force Fr on the block in this range is inversely the ball above the earth’s surface is very small proportional to x over this range, compared to the earth’s radius RE (h <<RE) so that −k we can ignore the variation of g near the earth’s Fr = for 0.1 < x < 2.01 m surface*. In what follows we have taken the x upward direction to be positive. Let us raise the = 0 for x < 0.1m and x > 2.01 m ball up to a height h. The work done by the external where k = 0.5 J. What is the final kinetic agency against the gravitational force is mgh. This energy and speed vf of the block as it work gets stored as potential energy. crosses this patch ? Gravitational potential energy of an object, as a function of the height h, is denoted by V(h) and it Answer From Eq. (5.8a) is the negative of work done by the gravitational 2.01 ( −k ) force in raising the object to that height. d x V (h) = mgh K f = K i + ∫ x 0.1 If h is taken as a variable, it is easily seen that the gravitational force F equals the negative of 1 2 2.01 = mv i − k ln ( x ) 0.1 the derivative of V(h) with respect to h. Thus, 2 d F = − V(h) = −m g 1 2 d h = mv i − k ln (2.01/0.1) 2 The negative sign indicates that the = 2 − 0.5 ln (20.1) gravitational force is downward. When released, the ball comes down with an increasing speed. = 2 − 1.5 = 0.5 J Just before it hits the ground, its speed is given v f = 2K f / m = 1 m s−1 by the kinematic relation, v2 = 2gh This equation can be written as Here, note that ln is a symbol for the natural 1logarithm to the base e and not the logarithm to the base 10 [ln X = loge X = 2.303 log10 X]. ⊳ 2 m v2 = m g h which shows that the gravitational potential5.7 THE CONCEPT OF POTENTIAL ENERGY energy of the object at height h, when the object The word potential suggests possibility or is released, manifests itself as kinetic energy of capacity for action. The term potential energy the object on reaching the ground. brings to one’s mind ‘stored’ energy. A stretched Physically, the notion of potential energy is bow-string possesses potential energy. When it applicable only to the class of forces where work is released, the arrow flies off at a great speed. done against the force gets ‘stored up’ as energy. The earth’s crust is not uniform, but has When external constraints are removed, it discontinuities and dislocations that are called manifests itself as kinetic energy. Mathematically, fault lines. These fault lines in the earth’s crust (for simplicity, in one dimension) the potential * The variation of g with height is discussed in Chapter 7 on Gravitation. Reprint 2025-26 78 PHYSICS energy V(x) is defined if the force F(x) can be which means that K + V, the sum of the kinetic written as and potential energies of the body is a constant. Over the whole path, xi to xf, this means that d V F ( x ) = − d x Ki + V(xi ) = Kf + V(xf) (5.11) The quantity K +V(x), is called the totalThis implies that mechanical energy of the system. Individually xf Vf the kinetic energy K and the potential energy ∫ F(x) d x = − ∫ d V = Vi − V f V(x) may vary from point to point, but the sum x i Vi is a constant. The aptness of the term The work done by a conservative force such as ‘conservative force’ is now clear. gravity depends on the initial and final positions Let us consider some of the definitions of a only. In the previous chapter we have worked conservative force. on examples dealing with inclined planes. If an l A force F(x) is conservative if it can be derived object of mass m is released from rest, from the from a scalar quantity V(x) by the relation top of a smooth (frictionless) inclined plane of given by Eq. (5.9). The three-dimensional height h, its speed at the bottom generalisation requires the use of a vector is 2 gh irrespective of the angle of inclination. derivative, which is outside the scope of this book.Thus, at the bottom of the inclined plane it l The work done by the conservative forceacquires a kinetic energy, mgh. If the work done depends only on the end points. This can be or the kinetic energy did depend on other factors seen from the relation, such as the velocity or the particular path taken W = Kf – Ki = V (xi) – V(xf)by the object, the force would be called non- which depends on the end points. conservative. l A third definition states that the work done The dimensions of potential energy are by this force in a closed path is zero. This is [ML2T –2] and the unit is joule (J), the same as once again apparent from Eq. (5.11) since kinetic energy or work. To reiterate, the change xi = xf .in potential energy, for a conservative force, ∆V is equal to the negative of the work done by Thus, the principle of conservation of total mechanical energy can be stated asthe force ∆V = − F(x) ∆x (5.9) The total mechanical energy of a system is In the example of the falling ball considered in conserved if the forces, doing work on it, are this section we saw how potential energy was conservative. The above discussion can be made moreconverted to kinetic energy. This hints at an concrete by considering the example of theimportant principle of conservation in mechanics, gravitational force once again and that of thewhich we now proceed to examine. spring force in the next section. Fig. 5.5 depicts

2.7 POTENTIAL ENERGY OF A SYSTEM OF CHARGES Consider first the simple case of two charges q1and q2 with position vector r1 and r2 relative to some origin. Let us calculate the work done (externally) in building up this configuration. This means that we consider the charges q1 and q2 initially at infinity and determine the work done by an external agency to bring the charges to the given locations. Suppose, first the charge q1 is brought from infinity to the point r1. There is no external field against which work needs to be done, so work done in bringing q1 from infinity to r1 is zero. This charge produces a potential in space given by 1 q1 V1 = 4 πε0 r1P where r1P is the distance of a point P in space from the location of q1. From the definition of potential, work done in bringing charge q2 from infinity to the point r2 is q2 times the potential at r2 due to q1: 1 q1q 2 work done on q2 = 4 πε0 r12 55 Reprint 2025-26 Physics where r12 is the distance between points 1 and 2. Since electrostatic force is conservative, this work gets stored in the form of potential energy of the system. Thus, the potential energy of a system of two charges q1 and q2 is FIGURE 2.13 Potential energy of a 1 q1q 2 U = system of charges q1 and q2 is 4 πε0 r12 (2.22) directly proportional to the product of charges and inversely to the Obviously, if q2 was brought first to its present location and distance between them. q1 brought later, the potential energy U would be the same. More generally, the potential energy expression, Eq. (2.22), is unaltered whatever way the charges are brought to the specified locations, because of path-independence of work for electrostatic force. Equation (2.22) is true for any sign of q1and q2. If q1q2 > 0, potential energy is positive. This is as expected, since for like charges (q1q2 > 0), electrostatic force is repulsive and a positive amount of work is needed to be done against this force to bring the charges from infinity to a finite distance apart. For unlike charges (q1 q2 < 0), the electrostatic force is attractive. In that case, a positive amount of work is needed against this force to take the charges from the given location to infinity. In other words, a negative amount of work is needed for the reverse path (from infinity to the present locations), so the potential energy is negative. Equation (2.22) is easily generalised for a system of any number of point charges. Let us calculate the potential energy of a system of three charges q1, q2 and q3 located at r1, r2, r3, respectively. To bring q1 first from infinity to r1, no work is required. Next we bring q2 from infinity to r2. As before, work done in this step is 1 q1q 2 q 2 V1 ( r2 ) = (2.23) 4 πε0 r12 The charges q1 and q2 produce a potential, which at any point P is given by 1  q1 q 2  V1, 2 = + (2.24) 4 πε0  r1P r2 P  Work done next in bringing q3 from infinity to the point r3 is q3 times V1, 2 at r3 1  q1q 3 q 2 q 3  q 3 V1, 2 ( r3 ) = + (2.25) 4 πε0  r13 r23  The total work done in assembling the charges at the given locations is obtained by adding the work done in different steps [Eq. (2.23) and Eq. (2.25)], 1  q1q 2 q1q 3 q 2 q 3  U = + + (2.26) FIGURE 2.14 Potential energy of a 4 πε0  r12 r13 r23  system of three charges is given by Again, because of the conservative nature of the Eq. (2.26), with the notation given electrostatic force (or equivalently, the path in the figure. independence of work done), the final expression for U, Eq. (2.26), is independent of the manner in which 56 the configuration is assembled. The potential energy Reprint 2025-26 Electrostatic Potential and Capacitance is characteristic of the present state of configuration, and not the way the state is achieved. Example 2.4 Four charges are arranged at the corners of a square ABCD of side d, as shown in Fig. 2.15.(a) Find the work required to put together this arrangement. (b) A charge q0 is brought to the centre E of the square, the four charges being held fixed at its corners. How much extra work is needed to do this? FIGURE 2.15 Solution (a) Since the work done depends on the final arrangement of the charges, and not on how they are put together, we calculate work needed for one way of putting the charges at A, B, C and D. Suppose, first the charge +q is brought to A, and then the charges –q, +q, and –q are brought to B, C and D, respectively. The total work needed can be calculated in steps: (i) Work needed to bring charge +q to A when no charge is present elsewhere: this is zero. (ii) Work needed to bring –q to B when +q is at A. This is given by (charge at B) × (electrostatic potential at B due to charge +q at A)  q  q 2 = −q × − ε  4 πε0 d = 4 π 0 d (iii) Work needed to bring charge +q to C when +q is at A and –q is at B. This is given by (charge at C) × (potential at C due to charges at A and B)  + q −q  = + q + ε 0 d   4 πε0 d 2 4 π −q 2  1  = 4 πε0 d 1 − 2  (iv) Work needed to bring –q to D when +q at A,–q at B, and +q at C. This is given by (charge at D) × (potential at D due to charges at A, B and C)  + q −q q  = −q + + ε  4 π 0 d 4 πε0 d 2 4 πε0 d  EXAMPLE −q 2  1  = 4 πε0 d  2 − 2  2.4 57 Reprint 2025-26 Physics Add the work done in steps (i), (ii), (iii) and (iv). The total work required is −q 2   1   1   =  4 πε0 d ( 0 ) + (1) + 1 − 2 +  2 − 2    −q 2 = 4 − 2 ( ) 4 πε0 d The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges. (Students may try calculating same work/energy by taking charges in any other order they desire and convince themselves that the energy will remain the same.) 2.4 the(b) Thefourextrachargesworkarenecessaryat A, B, CtoandbringD ais chargeq0 × (electrostaticq0 to the pointpotentialE whenat E due to the charges at A, B, C and D). The electrostatic potential at E is clearly zero since potential due to A and C is cancelled by that due to B and D. Hence, no work is required to bring any charge to EXAMPLE point E. 2.8 POTENTIAL ENERGY IN AN EXTERNAL FIELD 2.8.1 Potential energy of a single charge In Section 2.7, the source of the electric field was specified – the charges and their locations - and the potential energy of the system of those charges was determined. In this section, we ask a related but a distinct question. What is the potential energy of a charge q in a given field? This question was, in fact, the starting point that led us to the notion of the electrostatic potential (Sections 2.1 and 2.2). But here we address this question again to clarify in what way it is different from the discussion in Section 2.7. The main difference is that we are now concerned with the potential energy of a charge (or charges) in an external field. The external field E is not produced by the given charge(s) whose potential energy we wish to calculate. E is produced by sources external to the given charge(s).The external sources may be known, but often they are unknown or unspecified; what is specified is the electric field E or the electrostatic potential V due to the external sources. We assume that the charge q does not significantly affect the sources producing the external field. This is true if q is very small, or the external sources are held fixed by other unspecified forces. Even if q is finite, its influence on the external sources may still be ignored in the situation when very strong sources far away at infinity produce a finite field E in the region of interest. Note again that we are interested in determining the potential energy of a given charge q (and later, a system of charges) in the external field; we are not interested in the potential energy of the sources producing the external electric field. The external electric field E and the corresponding external potential V may vary from point to point. By definition, V at a point P is the work 58 done in bringing a unit positive charge from infinity to the point P. Reprint 2025-26 Electrostatic Potential and Capacitance (We continue to take potential at infinity to be zero.) Thus, work done in bringing a charge q from infinity to the point P in the external field is qV. This work is stored in the form of potential energy of q. If the point P has position vector r relative to some origin, we can write: Potential energy of q at r in an external field = qV(r) (2.27) where V(r) is the external potential at the point r. Thus, if an electron with charge q = e = 1.6×10–19 C is accelerated by a potential difference of DV = 1 volt, it would gain energy of qDV = 1.6 × 10–19J. This unit of energy is defined as 1 electron volt or 1eV, i.e., 1 eV=1.6 × 10–19J. The units based on eV are most commonly used in atomic, nuclear and particle physics, (1 keV = 103eV = 1.6 × 10–16J, 1 MeV = 106eV = 1.6 × 10–13J, 1 GeV = 109eV = 1.6 × 10–10J and 1 TeV = 1012eV = 1.6 × 10–7J). [This has already been defined on Page 117, XI 2.8.2 Potential energy of a system of two charges in an external field Next, we ask: what is the potential energy of a system of two charges q1 and q2 located at r1and r2, respectively, in an external field? First, we calculate the work done in bringing the charge q1 from infinity to r1. Work done in this step is q1 V(r1), using Eq. (2.27). Next, we consider the work done in bringing q2 to r2. In this step, work is done not only against the external field E but also against the field due to q1. Work done on q2 against the external field = q2 V (r2) Work done on q2 against the field due to q1 q1q 2 = 4 πεo r12 where r12 is the distance between q1 and q2. We have made use of Eqs. (2.27) and (2.22). By the superposition principle for fields, we add up the work done on q2 against the two fields (E and that due to q1): Work done in bringing q2 to r2 q1q 2 = q 2 V ( r2 ) + (2.28) 4 πεo r12 Thus, Potential energy of the system = the total work done in assembling the configuration q1q 2 = q1V ( r1 ) + q 2 V ( r2 ) + (2.29) 4 πε0r12 Example 2.5 (a) Determine the electrostatic potential energy of a system consisting of two charges 7 mC and –2 mC (and with no external field) placed at (–9 cm, 0, 0) and (9 cm, 0, 0) respectively. EXAMPLE (b) How much work is required to separate the two charges infinitely 2.5 away from each other? 59 Reprint 2025-26 Physics (c) Suppose that the same system of charges is now placed in an external electric field E = A (1/r 2); A = 9 × 105 NC–1 m2. What would the electrostatic energy of the configuration be? Solution 1 q1q 2 9 7 × ( −2) × 10 −12 (a) U = = 9 × 10 × = –0.7 J. 4 πε0 r 0.18 (b) W = U2 – U1 = 0 – U = 0 – (–0.7) = 0.7 J. (c) The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find, 7 µ C −µ2 C q1V ( r1 ) + q 2 V ( r2 ) = A + A 0.09m 0.09m 2.5 and the net electrostatic energy is q1q 2 7 µC −µ2 C q1V ( r1 ) + q 2 V ( r2 ) + = A + A − 0.7 J 4 πε0r12 0.09 m 0.09 m EXAMPLE = 70 − 20 − 0.7 = 49.3 J 2.8.3 Potential energy of a dipole in an external field Consider a dipole with charges q1 = +q and q2 = –q placed in a uniform electric field E, as shown in Fig. 2.16. As seen in the last chapter, in a uniform electric field, the dipole experiences no net force; but experiences a torque ttttt given by ttttt ===== p × E (2.30) which will tend to rotate it (unless p is parallel or antiparallel to E). Suppose an external torque text is applied in such a manner that it just neutralises this torque and rotates it in the plane of paper from angle q0 to angle q1 at an infinitesimal angular speed and without angular acceleration. The amount of work done by the external torque will be given by FIGURE 2.16 Potential energy of a dipole in a uniform external field. = pE ( cosθ0 − cosθ1 ) (2.31) This work is stored as the potential energy of the system. We can then associate potential energy U(q) with an inclination q of the dipole. Similar to other potential energies, there is a freedom in choosing the angle where the potential energy U is taken to be zero. A natural choice is to take q0 = p / 2. (An explanation for it is provided towards the end of discussion.) We can then write, (2.32) 60 Reprint 2025-26 Electrostatic Potential and Capacitance This expression can alternately be understood also from Eq. (2.29). We apply Eq. (2.29) to the present system of two charges +q and –q. The potential energy expression then reads q 2 U ′ (θ) = q [V ( r1 ) − V ( r2 )] − (2.33) 4 πε0 × 2a Here, r1 and r2 denote the position vectors of +q and –q. Now, the potential difference between positions r1 and r2 equals the work done in bringing a unit positive charge against field from r2 to r1. The displacement parallel to the force is 2a cosq. Thus, [V(r1)–V (r2)] = –E × 2a cosq . We thus obtain, q 2 q 2 U ′ (θ) = − pE cosθ− = − p.E − (2.34) 4 πε0 × 2a 4 πε0 × 2a We note that U¢(q) differs from U(q ) by a quantity which is just a constant for a given dipole. Since a constant is insignificant for potential energy, we can drop the second term in Eq. (2.34) and it then reduces to Eq. (2.32). We can now understand why we took q0=p/2. In this case, the work done against the external field E in bringing +q and – q are equal and opposite and cancel out, i.e., q [V (r1) – V (r2)]=0. Example 2.6 A molecule of a substance has a permanent electric dipole moment of magnitude 10–29 C m. A mole of this substance is polarised (at low temperature) by applying a strong electrostatic field of magnitude 106 V m–1. The direction of the field is suddenly changed by an angle of 60º. Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. For simplicity, assume 100% polarisation of the sample. Solution Here, dipole moment of each molecules = 10–29 C m As 1 mole of the substance contains 6 × 1023 molecules, total dipole moment of all the molecules, p = 6 × 1023 × 10–29 C m = 6 × 10–6 C m Initial potential energy, Ui = –pE cos q = –6×10–6×106 cos 0° = –6 J Final potential energy (when q = 60°), Uf = –6 × 10–6 × 106 cos 60° = –3 J Change in potential energy = –3 J – (–6J) = 3 J EXAMPLE So, there is loss in potential energy. This must be the energy released by the substance in the form of heat in aligning its dipoles. 2.6