2.4 POTENTIAL DUE TO AN ELECTRIC DIPOLE As we learnt in the last chapter, an electric dipole consists of two charges q and –q separated by a (small) distance 2a. Its total charge is zero. It is characterised by a dipole moment vector p whose magnitude is q × 2a and which points in the direction from –q to q (Fig. 2.5). We also saw that the electric field of a dipole at a point with position vector r depends not just on the magnitude r, but also on the angle between r and p. Further, 49 Reprint 2025-26 Physics the field falls off, at large distance, not as 1/r 2 (typical of field due to a single charge) but as 1/r3. We, now, determine the electric potential due to a dipole and contrast it with the potential due to a single charge. As before, we take the origin at the centre of the dipole. Now we know that the electric field obeys the superposition principle. Since potential is related to the work done by the field, electrostatic potential also follows the superposition principle. Thus, the potential due to the dipole is the sum of potentials due to the charges q and –q 1  q q  V = − (2.9)FIGURE 2.5 Quantities involved in the calculation 4 πε0  r1 r2  of potential due to a dipole. where r1 and r2 are the distances of the point P from q and –q, respectively. Now, by geometry, r12 = r 2 + a 2 − 2ar cosq r22 = r 2 + a 2 + 2ar cosq (2.10) We take r much greater than a ( r  a ) and retain terms only upto the first order in a/r 2 2  2a cosθ a 2  r1 = r 1 − + 2  r r  2  2a cosθ (2.11) ≅ r  1 − r  Similarly, 2 2  2a cosθ (2.12) r2 ≅ r 1 + r  Using the Binomial theorem and retaining terms upto the first order in a/r ; we obtain, 1 a 1  2a cos θ − 1 / 2 1   cos θ ≅ 1 − ≅ 1 + [2.13(a)] r1 r  r  r  r  1 a 1  2a cos θ − 1 / 2 1   cos θ ≅ 1 + ≅ 1 − [2.13(b)] r2 r  r  r  r  Using Eqs. (2.9) and (2.13) and p = 2qa, we get q 2 acosθ p cos θ V = = 4 πε0 r 2 4 πε0r 2 (2.14) 50 Now, p cos q = p.rˆ Reprint 2025-26 Electrostatic Potential and Capacitance where ˆr is the unit vector along the position vector OP. The electric potential of a dipole is then given by 1 p.rˆ V = 2 ; (r >> a) (2.15) 4 πε0 r Equation (2.15) is, as indicated, approximately true only for distances large compared to the size of the dipole, so that higher order terms in a/r are negligible. For a point dipole p at the origin, Eq. (2.15) is, however, exact. From Eq. (2.15), potential on the dipole axis (q = 0, p ) is given by 1 p V = ± 2 (2.16) 4 πε0 r (Positive sign for q = 0, negative sign for q = p.) The potential in the equatorial plane (q = p/2) is zero. The important contrasting features of electric potential of a dipole from that due to a single charge are clear from Eqs. (2.8) and (2.15): (i) The potential due to a dipole depends not just on r but also on the angle between the position vector r and the dipole moment vector p. (It is, however, axially symmetric about p. That is, if you rotate the position vector r about p, keeping q fixed, the points corresponding to P on the cone so generated will have the same potential as at P.) (ii) The electric dipole potential falls off, at large distance, as 1/r 2, not as 1/r, characteristic of the potential due to a single charge. (You can refer to the Fig. 2.5 for graphs of 1/r 2 versus r and 1/r versus r, drawn there in another context.)

4.3 Bond Parameters c = 99 pm 1984.3.1 Bond Length r pmBond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule. Bond lengths are measured by spectroscopic, X-ray diffraction and electron-diffraction techniques about which you will learn in higher classes. Each atom of the bonded pair contributes to the rvdw= bond length (Fig. 4.1). In the case of a covalent 180 bond, the contribution from each atom is pm called the covalent radius of that atom. pm 360 The covalent radius is measured approximately as the radius of an atom’s Fig. 4.2 Covalent and van der Waals radii incore which is in contact with the core of an a chlorine molecule. The inner circles adjacent atom in a bonded situation. The correspond to the size of the chlorine covalent radius is half of the distance between atom (rvdw and rc are van der Waals and two similar atoms joined by a covalent bond covalent radii respectively). Reprint 2025-26 108 chemistry Some typical average bond lengths for Table 4.2 Average Bond Lengths for Some single, double and triple bonds are shown in Single, Double and Triple Bonds Table 4.2. Bond lengths for some common Covalent Bondmolecules are given in Table 4.3. Bond Type Length (pm) The covalent radii of some common O–H 96elements are listed in Table 4.4. C–H 107 4.3.2 Bond Angle N–O 136 C–O 143It is defined as the angle between the orbitals C–N 143 containing bonding electron pairs around the C–C 154 central atom in a molecule/complex ion. Bond C=O 121 angle is expressed in degree which can be N=O 122 experimentally determined by spectroscopic C=C 133 methods. It gives some idea regarding the C=N 138 distribution of orbitals around the central C≡N 116 atom in a molecule/complex ion and hence it C≡C 120 helps us in determining its shape. For Table 4.3 Bond Lengths in Some Commonexample H–O–H bond angle in water can be Moleculesrepresented as under : Molecule Bond Length (pm) H2 (H – H) 74 F2 (F – F) 144 4.3.3 Bond Enthalpy Cl2 (Cl – Cl) 199 It is defined as the amount of energy required Br2 (Br – Br) 228 to break one mole of bonds of a particular I2 (I – I) 267 type between two atoms in a gaseous state. N2 (N ≡ N) 109 The unit of bond enthalpy is kJ mol–1. For O2 (O = O) 121 example, the H – H bond enthalpy in hydrogen HF (H – F) 92 molecule is 435.8 kJ mol–1. HCl (H – Cl) 127 HBr (H – Br) 141H2(g) → H(g) + H(g); ∆aH = 435.8 kJ mol–1 HI (H – I) 160 Similarly the bond enthalpy for molecules Table 4.4 Covalent Radii, *rcov/(pm)containing multiple bonds, for example O2 and N2 will be as under : O2 (O = O) (g) → O(g) + O(g); ∆aH = 498 kJ mol–1 N2 (N ≡ N) (g) → N(g) + N(g); ∆aH = 946.0 kJ mol–1 It is important that larger the bond dissociation enthalpy, stronger will be the bond in the molecule. For a heteronuclear diatomic molecules like HCl, we have HCl (g) → H(g) + Cl (g); ∆aH = 431.0 kJ mol–1 In case of polyatomic molecules, the measurement of bond strength is more complicated. For example in case of H2O * The values cited are for single bonds, except where molecule, the enthalpy needed to break the otherwise indicated in parenthesis. (See also Unit 3 for two O – H bonds is not the same. periodic trends). Reprint 2025-26 Chemical Bonding And Molecular Structure 109 H2O(g) → H(g) + OH(g); ∆aH1 = 502 kJ mol–1 OH(g) → H(g) + O(g); ∆aH2 = 427 kJ mol–1 The difference in the ∆aH value shows that the second O – H bond undergoes some change because of changed chemical environment. This is the reason for some difference in energy of the same O – H bond in different molecules like C2H5OH (ethanol) and water. Therefore in polyatomic molecules the term mean or average bond enthalpy is used. It is obtained by dividing total bond dissociation enthalpy by the number of bonds broken as explained below in case of water molecule, 502 + 427 Fig. 4.3 Resonance in the O3 molecule Average bond enthalpy = 2 (structures I and II represent the two canonical = 464.5 kJ mol–1 forms while the structure III is the resonance hybrid) 4.3.4 Bond Order In the Lewis description of covalent bond, In both structures we have a O–O single the Bond Order is given by the number bond and a O=O double bond. The normal of bonds between the two atoms in a O–O and O=O bond lengths are 148 pm molecule. The bond order, for example in and 121 pm respectively. Experimentally H2 (with a single shared electron pair), in O2 determined oxygen-oxygen bond lengths in (with two shared electron pairs) and in N2 the O3 molecule are same (128 pm). Thus the (with three shared electron pairs) is 1,2,3 oxygen-oxygen bonds in the O3 molecule are respectively. Similarly in CO (three shared intermediate between a double and a single electron pairs between C and O) the bond bond. Obviously, this cannot be represented order is 3. For N2, bond order is 3 and its by either of the two Lewis structures shown is 946 kJ mol–1; being one of the highest above. for a diatomic molecule. The concept of resonance was introduced Isoelectronic molecules and ions have to deal with the type of difficulty experienced identical bond orders; for example, F2 and in the depiction of accurate structures of O22– have bond order 1. N2, CO and NO+ have molecules like O3. According to the concept bond order 3. of resonance, whenever a single Lewis structure cannot describe a molecule A general correlation useful for accurately, a number of structures withunderstanding the stablities of molecules is that: with increase in bond order, similar energy, positions of nuclei, bonding bond enthalpy increases and bond length and non-bonding pairs of electrons are decreases. taken as the canonical structures of the hybrid which describes the molecule 4.3.5 Resonance Structures accurately. Thus for O3, the two structures It is often observed that a single Lewis structure shown above constitute the canonical is inadequate for the representation of a structures or resonance structures and molecule in conformity with its experimentally their hybrid i.e., the III structure represents determined parameters. For example, the the structure of O3 more accurately. This is ozone, O3 molecule can be equally represented also called resonance hybrid. Resonance is by the structures I and II shown below: represented by a double headed arrow. Reprint 2025-26 110 chemistry Some of the other examples of resonance structures are provided by the carbonate ion and the carbon dioxide molecule. Fig. 4.5 Resonance in CO2 molecule, I, II Problem 4.3 and III represent the three canonical Explain the structure of CO32– ion in forms. terms of resonance. Solution In general, it may be stated that The single Lewis structure based on • Resonance stabilizes the molecule as the the presence of two single bonds and energy of the resonance hybrid is less one double bond between carbon than the energy of any single cannonical and oxygen atoms is inadequate to structure; and, represent the molecule accurately as it • R e s o n a n c e a v e r a g e s t h e b o n d represents unequal bonds. According characteristics as a whole. to the experimental findings, all carbon to oxygen bonds in CO32– are equivalent. Thus the energy of the O3 resonance Therefore the carbonate ion is best hybrid is lower than either of the two described as a resonance hybrid of the cannonical froms I and II (Fig. 4.3). canonical forms I, II, and III shown below. Many misconceptions are associated with resonance and the same need to be dispelled. You should remember that : • The cannonical forms have no real existence. • The molecule does not exist for a certain fraction of time in one cannonical form and for other fractions of time in other Fig. 4.4 Resonance in CO32–, I, II and III cannonical forms. represent the three canonical • There is no such equilibrium between forms. the cannonical forms as we have Problem 4.4 between tautomeric forms (keto and enol) in tautomerism. Explain the structure of CO2 molecule. • The molecule as such has a single Solution structure which is the resonance The experimentally determined carbon hybrid of the cannonical forms and to oxygen bond length in CO 2 is which cannot as such be depicted by 115 pm. The lengths of a normal a single Lewis structure. carbon to oxygen double bond (C=O) and carbon to oxygen triple bond (C≡O) 4.3.6 Polarity of Bonds are 121 pm and 110 pm respectively. The existence of a hundred percent ionic or The carbon-oxygen bond lengths in covalent bond represents an ideal situation. CO2 (115 pm) lie between the values for C=O and C≡O. Obviously, a single In reality no bond or a compound is either Lewis structure cannot depict this completely covalent or ionic. Even in case of position and it becomes necessary to covalent bond between two hydrogen atoms, write more than one Lewis structures there is some ionic character. and to consider that the structure of When covalent bond is formed between CO2 is best described as a hybrid of two similar atoms, for example in H2, O2, the canonical or resonance forms I, II and III. Cl2, N2 or F2, the shared pair of electrons is equally attracted by the two atoms. As a result Reprint 2025-26 Chemical Bonding And Molecular Structure 111 electron pair is situated exactly between the In case of polyatomic molecules the dipole two identical nuclei. The bond so formed is moment not only depend upon the individual called nonpolar covalent bond. Contrary to dipole moments of bonds known as bond this in case of a heteronuclear molecule like dipoles but also on the spatial arrangement HF, the shared electron pair between the two of various bonds in the molecule. In such atoms gets displaced more towards fluorine case, the dipole moment of a molecule is the since the electronegativity of fluorine (Unit 3) vector sum of the dipole moments of various is far greater than that of hydrogen. The bonds. For example in H2O molecule, which resultant covalent bond is a polar covalent has a bent structure, the two O–H bonds are bond. oriented at an angle of 104.50. Net dipole As a result of polarisation, the molecule moment of 6.17 × 10–30 C m (1D = 3.33564 possesses the dipole moment (depicted × 10–30 C m) is the resultant of the dipole below) which can be defined as the product of moments of two O–H bonds. the magnitude of the charge and the distance between the centres of positive and negative charge. It is usually designated by a Greek letter ‘µ’. Mathematically, it is expressed as follows : Dipole moment (µ) = charge (Q) × distance of separation (r) Dipole moment is usually expressed in Net Dipole moment, µ = 1.85 D Debye units (D). The conversion factor is = 1.85 × 3.33564 × 10–30 C m = 6.17 ×10–30 C m 1 D = 3.33564 × 10–30 C m The dipole moment in case of BeF2 is zero.where C is coulomb and m is meter. This is because the two equal bond dipoles Further dipole moment is a vector quantity point in opposite directions and cancel the and by convention it is depicted by a small effect of each other. arrow with tail on the negative centre and head pointing towards the positive centre. But in chemistry presence of dipole moment is represented by the crossed arrow ( ) put on Lewis structure of the molecule. The In tetra-atomic molecule, for example in cross is on positive end and arrow head is on BF3, the dipole moment is zero although thenegative end. For example the dipole moment B – F bonds are oriented at an angle of 120o of HF may be represented as : to one another, the three bond moments give a net sum of zero as the resultant of any two H F is equal and opposite to the third. This arrow symbolises the direction of the shift of electron density in the molecule. Note that the direction of crossed arrow is opposite to the conventional direction of dipole moment vector. Peter Debye, the Dutch chemist received Nobel prize in 1936 for Let us study an interesting case of NH3 his work on X-ray diffraction and and NF3 molecule. Both the molecules have dipole moments. The magnitude pyramidal shape with a lone pair of electrons of the dipole moment is given in Debye units in order to honour him. on nitrogen atom. Although fluorine is more electronegative than nitrogen, the resultant Reprint 2025-26 112 chemistry dipole moment of NH3 (4.90 × 10–30 C m) is • The smaller the size of the cation and the greater than that of NF3 (0.8 × 10–30 C m). This larger the size of the anion, the greater is because, in case of NH3 the orbital dipole the covalent character of an ionic bond. due to lone pair is in the same direction as • The greater the charge on the cation, the the resultant dipole moment of the N – H greater the covalent character of the ionic bonds, whereas in NF3 the orbital dipole is in bond. the direction opposite to the resultant dipole • For cations of the same size and charge,moment of the three N–F bonds. The orbital the one, with electronic configurationdipole because of lone pair decreases the effect (n-1)dnnso, typical of transition metals, isof the resultant N – F bond moments, which more polarising than the one with a noble results in the low dipole moment of NF3 as gas configuration, ns2 np6, typical of alkalirepresented below : and alkaline earth metal cations. The cation polarises the anion, pulling the electronic charge toward itself and thereby increasing the electronic charge between the two. This is precisely what happens in a covalent bond, i.e., buildup of electron charge density between the nuclei. The polarising power of the cation, the polarisability of the anion and the extent of distortion (polarisation) of anion are the factors, which determine the per Dipole moments of some molecules are cent covalent character of the ionic bond. shown in Table 4.5. Just as all the covalent bonds have 4.4 The Valence Shell Electron some partial ionic character, the ionic Pair Repulsion (VSEPR) Theory bonds also have partial covalent character. As already explained, Lewis concept is unable The partial covalent character of ionic to explain the shapes of molecules. This bonds was discussed by Fajans in terms of theory provides a simple procedure to predict the following rules: the shapes of covalent molecules. Sidgwick Table 4.5 Dipole Moments of Selected Molecules Dipole Type of Molecule Example Geometry Moment, µ(D) Molecule (AB) HF 1.78 linear HCl 1.07 linear HBr 0.79 linear Hl 0.38 linear H2 0 linear Molecule (AB2) H2O 1.85 bent H2S 0.95 bent CO2 0 linear Molecule (AB3) NH3 1.47 trigonal-pyramidal NF3 0.23 trigonal-pyramidal BF3 0 trigonal-planar Molecule (AB4) CH4 0 tetrahedral CHCl3 1.04 tetrahedral CCl4 0 tetrahedral Reprint 2025-26 Chemical Bonding And Molecular Structure 113 and Powell in 1940, proposed a simple theory result in deviations from idealised shapes and based on the repulsive interactions of the alterations in bond angles in molecules. electron pairs in the valence shell of the For the prediction of geometrical shapes atoms. It was further developed and redefined of molecules with the help of VSEPR theory, by Nyholm and Gillespie (1957). it is convenient to divide molecules into The main postulates of VSEPR theory are two categories as (i) molecules in which as follows: the central atom has no lone pair and • The shape of a molecule depends upon (ii) molecules in which the central atom the number of valence shell electron pairs has one or more lone pairs. (bonded or nonbonded) around the central Table 4.6 (page114) shows the atom. arrangement of electron pairs about a • Pairs of electrons in the valence shell repel central atom A (without any lone pairs) and one another since their electron clouds are geometries of some molecules/ions of the type negatively charged. AB. Table 4.7 (page 115) shows shapes of some simple molecules and ions in which the central• These pairs of electrons tend to occupy atom has one or more lone pairs. Table 4.8 such positions in space that minimise (page 116) explains the reasons for the repulsion and thus maximise distance distortions in the geometry of the molecule. between them. As depicted in Table 4.6, in the• The valence shell is taken as a sphere compounds of AB2, AB3, AB4, AB5 and AB6, with the electron pairs localising on the the arrangement of electron pairs and the spherical surface at maximum distance B atoms around the central atom A are : from one another. linear, trigonal planar, tetrahedral, • A multiple bond is treated as if it is a single trigonal-bipyramidal and octahedral, electron pair and the two or three electron respectively. Such arrangement can be seen pairs of a multiple bond are treated as a in the molecules like BF3 (AB3), CH4 (AB4) and single super pair. PCl5 (AB5) as depicted below by their ball and • Where two or more resonance structures stick models. can represent a molecule, the VSEPR model is applicable to any such structure. The repulsive interaction of electron pairs decrease in the order: Lone pair (lp) – Lone pair (lp) > Lone pair (lp) – Bond pair (bp) > Bond pair (bp) – Bond pair (bp) Fig. 4.6 The shapes of molecules in which central atom has no lone pair Nyholm and Gillespie (1957) refined the VSEPR model by explaining the important The VSEPR Theory is able to predict difference between the lone pairs and bonding geometry of a large number of molecules, pairs of electrons. While the lone pairs are especially the compounds of p-block elements localised on the central atom, each bonded accurately. It is also quite successful in pair is shared between two atoms. As a result, determining the geometry quite-accurately the lone pair electrons in a molecule occupy even when the energy difference between more space as compared to the bonding pairs possible structures is very small. The of electrons. This results in greater repulsion theoretical basis of the VSEPR theory regarding between lone pairs of electrons as compared the effects of electron pair repulsions on to the lone pair - bond pair and bond pair - molecular shapes is not clear and continues bond pair repulsions. These repulsion effects to be a subject of doubt and discussion. Reprint 2025-26 114 chemistry Table 4.6 Geometry of Molecules in which the Central Atom has No Lone Pair of Electrons Reprint 2025-26 Chemical Bonding And Molecular Structure 115 Table 4.7 Shape (geometry) of Some Simple Molecules/Ions with Central Ions having One or More Lone Pairs of Electrons(E). Reprint 2025-26 116 chemistry Table 4.8 Shapes of Molecules containing Bond Pair and Lone Pair Molecule No. of No. of Arrangement Shape Reason for the type bonding lone of electrons shape acquired pairs pairs AB2E 4 1 Bent Theoretically the shape should have been triangular planar but actually it is found to be bent or v-shaped. The reason being the lone pair- bond pair repulsion is much more as compared to the bond pair-bond pair repulsion. So the angle is reduced to 119.5° from 120°. AB3E 3 1 Trigonal Had there been a bp in place of lp the shape would have pyramidal been tetrahedral but one lone pair is present and due to the repulsion between lp-bp (which is more than bp-bp repulsion) the angle between bond pairs is reduced to 107° from 109.5°. Bent The shape should have been AB2E2 2 2 tetrahedral if there were all bp but two lp are present so the shape is distorted tetrahedral or angular. The reason is lp-lp repulsion is more than lp-bp repulsion which is more than bp-bp repulsion. Thus, the angle is reduced to 104.5° from 109.5°. AB4E 4 1 See- In (a) the lp is present at axial saw position so there are three lp—bp repulsions at 90°. In(b) the lp is in an equatorial position, and there are two lp—bp repulsions. Hence, arrangement (b) is more stable. The shape shown in (b) is described as a distorted tetrahedron, a folded square (More stable) or a see-saw. Reprint 2025-26 Chemical Bonding And Molecular Structure 117 Molecule No. of No. of Arrangement Shape Reason for the type bonding lone of electrons shape acquired pairs pairs AB3E2 3 2 T-shape In (a) the lp are at equatorial position so there are less lp- bp repulsions as compared to others in which the lp are at axial positions. So structure (a) is most stable. (T-shaped).

1.10 ELECTRIC DIPOLE An electric dipole is a pair of equal and opposite point charges q and –q, separated by a distance 2a. The line connecting the two charges defines a direction in space. By convention, the direction from –q to q is said to be the direction of the dipole. The mid-point of locations of –q and q is called the centre of the dipole. The total charge of the electric dipole is obviously zero. This does not mean that the field of the electric dipole is zero. Since the charge q and –q are separated by some distance, the electric fields due to them, when added, do not exactly cancel out. However, at distances much larger than the separation of the two charges forming a dipole (r >> 2a), the fields due to q and –q nearly cancel out. The electric field due to a dipole therefore falls off, at large distance, faster than like 1/r 2 (the dependence on r of the field due to a single charge q). These qualitative ideas are borne out by the explicit calculation as follows: 1.10.1 The field of an electric dipole The electric field of the pair of charges (–q and q) at any point in space can be found out from Coulomb’s law and the superposition principle. The results are simple for the following two cases: (i) when the point is on the dipole axis, and (ii) when it is in the equatorial plane of the dipole, i.e., on a plane perpendicular to the dipole axis through its centre. The electric field at any general point P is obtained by adding the electric fields E–q due to the charge –q and E+q due to the charge q, by the parallelogram law of vectors. (i) For points on the axis Let the point P be at distance r from the centre of the dipole on the side of the charge q, as shown in Fig. 1.17(a). Then q E − q = − 2 pˆ [1.13(a)] 4 πε0 (r + a ) where ˆp is the unit vector along the dipole axis (from –q to q). Also q E + q = pˆ [1.13(b)] 23 4 π ε0 (r − a )2 Reprint 2025-26 Physics The total field at P is q  1 1  pˆ − E = E + q + E − q =   (r + a )2 4 π ε0  (r − a )2  q 4 a r = ˆp (1.14) 4 π εo ( r 2 − a 2 )2 For r >> a 4 q a E = 3 pˆ (r >> a) (1.15) 4 πε0 r (ii) For points on the equatorial plane The magnitudes of the electric fields due to the two charges +q and –q are given by q 1 E + q = 2 2 [1.16(a)] 4 πε0 r + a q 1 E – q = 2 2 [1.16(b)] 4 πε0 r + a FIGURE 1.17 Electric field of a dipole and are equal. at (a) a point on the axis, (b) a point The directions of E+q and E–q are as shown in on the equatorial plane of the dipole. Fig. 1.17(b). Clearly, the components normal to the dipole p is the dipole moment vector of axis cancel away. The components along the dipole axis magnitude p = q × 2a and add up. The total electric field is opposite to ˆp. We have directed from –q to q. E = – (E +q + E –q ) cosq ˆp 2 q a = − pˆ (1.17) 4 π εo (r 2 + a 2 )3 / 2 At large distances (r >> a), this reduces to 2 q a E = − pˆ (r >> a ) (1.18) 4 π εo r 3 From Eqs. (1.15) and (1.18), it is clear that the dipole field at large distances does not involve q and a separately; it depends on the product qa. This suggests the definition of dipole moment. The dipole moment vector p of an electric dipole is defined by p = q × 2a ˆp (1.19) that is, it is a vector whose magnitude is charge q times the separation 2a (between the pair of charges q, –q) and the direction is along the line from –q to q. In terms of p, the electric field of a dipole at large distances takes simple forms: At a point on the dipole axis 2 p E = 3 (r >> a) (1.20) 4 πεor At a point on the equatorial plane p 3 (r >> a) (1.21) 24 E = −4 πεor Reprint 2025-26 Electric Charges and Fields Notice the important point that the dipole field at large distances falls off not as 1/r 2 but as1/r 3. Further, the magnitude and the direction of the dipole field depends not only on the distance r but also on the angle between the position vector r and the dipole moment p. We can think of the limit when the dipole size 2a approaches zero, the charge q approaches infinity in such a way that the product p = q × 2a is finite. Such a dipole is referred to as a point dipole. For a point dipole, Eqs. (1.20) and (1.21) are exact, true for any r. 1.10.2 Physical significance of dipoles In most molecules, the centres of positive charges and of negative charges* lie at the same place. Therefore, their dipole moment is zero. CO2 and CH4 are of this type of molecules. However, they develop a dipole moment when an electric field is applied. But in some molecules, the centres of negative charges and of positive charges do not coincide. Therefore they have a permanent electric dipole moment, even in the absence of an electric field. Such molecules are called polar molecules. Water molecules, H2O, is an example of this type. Various materials give rise to interesting properties and important applications in the presence or absence of electric field. Example 1.9 Two charges ±10 mC are placed 5.0 mm apart. Determine the electric field at (a) a point P on the axis of the dipole 15 cm away from its centre O on the side of the positive charge, as shown in Fig. 1.18(a), and (b) a point Q, 15 cm away from O on a line passing through O and normal to the axis of the dipole, as shown in Fig. 1.18(b). EXAMPLE FIGURE 1.18 1.9 * Centre of a collection of positive point charges is defined much the same way ∑ q i ri as the centre of mass: rcm = i . ∑ q i 25 i Reprint 2025-26 Physics Solution (a) Field at P due to charge +10 mC 10 −5 C 1 = − 12 2 −1 −2 × 2 −4 2 4 π (8.854 × 10 C N m ) (15 − 0.25) × 10 m = 4.13 × 106 N C–1 along BP Field at P due to charge –10 mC 10 –5 C 1 = −12 2 −1 −2 × 2 − 4 2 4 π (8.854 × 10 C N m ) (15 + 0.25) × 10 m = 3.86 × 106 N C–1 along PA The resultant electric field at P due to the two charges at A and B is = 2.7 × 105 N C–1 along BP. In this example, the ratio OP/OB is quite large (= 60). Thus, we can expect to get approximately the same result as above by directly using the formula for electric field at a far-away point on the axis of a dipole. For a dipole consisting of charges ± q, 2a distance apart, the electric field at a distance r from the centre on the axis of the dipole has a magnitude 2 p E = 3 (r/a >> 1) 4 πε0 r where p = 2a q is the magnitude of the dipole moment. The direction of electric field on the dipole axis is always along the direction of the dipole moment vector (i.e., from –q to q). Here, p =10–5 C × 5 × 10–3 m = 5 × 10–8 C m Therefore, 2 × 5 × 10 − 8 C m 1 E = −12 2 −1 −2 × 3 −6 3 = 2.6 × 105 N C–1 4 π (8.854 × 10 C N m ) (15) × 10 m along the dipole moment direction AB, which is close to the result obtained earlier. (b) Field at Q due to charge + 10 mC at B 10 −5 C 1 = −12 2 −1 − 2 × 2 2 −4 2 4 π (8.854 × 10 C N m ) [15 + (0.25) ] × 10 m = 3.99 × 106 N C–1 along BQ Field at Q due to charge –10 mC at A 10 −5 C 1 = − 12 2 −1 − 2 × 2 2 −4 2 4 π (8.854 × 10 C N m ) [15 + (0.25) ] × 10 m = 3.99 × 106 N C–1 along QA. Clearly, the components of these two forces with equal magnitudes cancel along the direction OQ but add up along the direction parallel to BA. Therefore, the resultant electric field at Q due to the two charges at A and B is 0.25 6 –1 = 2 × 2 2 × 3.99 × 10 N C along BA 1.9 15 + (0.25) = 1.33 × 105 N C–1 along BA. As in (a), we can expect to get approximately the same result by directly using the formula for dipole field at a point on the normal to EXAMPLE 26 the axis of the dipole: Reprint 2025-26 Electric Charges and Fields p E = 3 (r/a >> 1) 4 π 0ε r 5 × 10 −8 Cm 1 = −12 2 –1 –2 × 3 −6 3 4 π (8.854 × 10 C N m ) (15) × 10 m = 1.33 × 105 N C–1. The direction of electric field in this case is opposite to the direction EXAMPLE of the dipole moment vector. Again, the result agrees with that obtained before. 1.9