2.13 EFFECT OF DIELECTRIC ON CAPACITANCE With the understanding of the behaviour of dielectrics in an external field developed in Section 2.10, let us see how the capacitance of a parallel plate capacitor is modified when a dielectric is present. As before, we have two large plates, each of area A, separated by a distance d. The charge on the plates is ±Q, corresponding to the charge density ±s (with s = Q/A). When there is vacuum between the plates, σ E 0 = ε0 69 Reprint 2025-26 Physics and the potential difference V0 is V0 = E0d The capacitance C0 in this case is Q A C 0 = = ε0 (2.46) V 0 d Consider next a dielectric inserted between the plates fully occupying the intervening region. The dielectric is polarised by the field and, as explained in Section 2.10, the effect is equivalent to two charged sheets (at the surfaces of the dielectric normal to the field) with surface charge densities sp and –sp. The electric field in the dielectric then corresponds to the case when the net surface charge density on the plates is ±(s – sp). That is, σ − σP E = (2.47) ε0 so that the potential difference across the plates is σ − σP V = E d = d (2.48) ε0 For linear dielectrics, we expect sp to be proportional to E0, i.e., to s. Thus, (s – sp) is proportional to s and we can write σ σ − σP = (2.49) K where K is a constant characteristic of the dielectric. Clearly, K > 1. We then have σd Qd V = = (2.50) ε0 K Aε0 K The capacitance C, with dielectric between the plates, is then Q ε0KA C = = (2.51) V d The product e0K is called the permittivity of the medium and is denoted by e e = e0 K (2.52) For vacuum K = 1 and e = e0; e0 is called the permittivity of the vacuum. The dimensionless ratio ε K = (2.53) ε0 is called the dielectric constant of the substance. As remarked before, from Eq. (2.49), it is clear that K is greater than 1. From Eqs. (2.46) and (2. 51) C K = (2.54) C 0 Thus, the dielectric constant of a substance is the factor (>1) by which the capacitance increases from its vacuum value, when the dielectric is 70 inserted fully between the plates of a capacitor. Though we arrived at Reprint 2025-26 Electrostatic Potential and Capacitance Eq. (2.54) for the case of a parallel plate capacitor, it holds good for any type of capacitor and can, in fact, be viewed in general as a definition of the dielectric constant of a substance. Example 2.8 A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates? Solution Let E0 = V0/d be the electric field between the plates when there is no dielectric and the potential difference is V0. If the dielectric is now inserted, the electric field in the dielectric will be E = E0/K. The potential difference will then be 1 E 0 3 V = E 0 ( d ) + ( d ) 4 K 4 1 3 K + 3 = E 0 d ( + ) = V 0 4 4 K 4 K The potential difference decreases by the factor (K + 3)/4K while the free charge Q0 on the plates remains unchanged. The capacitance thus increases EXAMPLE Q 0 4 K Q 0 4 K C = = = C 0 V K + 3 V 0 K + 3 2.8

2.5 A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10–12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

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