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?

2.14 COMBINATION OF CAPACITORS We can combine several capacitors of capacitance C1, C2,…, Cn to obtain a system with some effective capacitance C. The effective capacitance depends on the way the individual capacitors are combined. Two simple possibilities are discussed below. 2.14.1 Capacitors in series Figure 2.26 shows capacitors C1 and C2 FIGURE 2.26 Combination of two combined in series. capacitors in series. The left plate of C1 and the right plate of C2 are connected to two terminals of a battery and have charges Q and –Q , respectively. It then follows that the right plate of C1 has charge –Q and the left plate of C2 has charge Q. If this was not so, the net charge on each capacitor would not be zero. This would result in an electric field in the conductor connecting C1and C2. Charge would flow until the net charge on both C1 and C2 is zero and there is no electric field in the conductor connecting C1 and C2. Thus, in the series combination, charges on the two plates FIGURE 2.27 Combination of n (±Q) are the same on each capacitor. The total capacitors in series. 71 Reprint 2025-26 Physics potential drop V across the combination is the sum of the potential drops V1 and V2 across C1 and C2, respectively. Q Q + (2.55) V = V1 + V2 = C1 C 2 V 1 1 i.e., = + , (2.56) Q C1 C 2 Now we can regard the combination as an effective capacitor with charge Q and potential difference V. The effective capacitance of the combination is Q C = (2.57) V We compare Eq. (2.57) with Eq. (2.56), and obtain 1 1 1 = + (2.58) C C1 C 2 The proof clearly goes through for any number of capacitors arranged in a similar way. Equation (2.55), for n capacitors arranged in series, generalises to Q Q Q V = V1 + V 2 + ... + V n = + + ... + (2.59) C1 C 2 C n Following the same steps as for the case of two capacitors, we get the general formula for effective capacitance of a series combination of n capacitors: 1 1 1 1 1 = + + + ... + (2.60) C C1 C 2 C 3 C n 2.14.2 Capacitors in parallel Figure 2.28 (a) shows two capacitors arranged in parallel. In this case, the same potential difference is applied across both the capacitors. But the plate charges (±Q1) on capacitor 1 and the plate charges (±Q2) on the capacitor 2 are not necessarily the same: Q1 = C1V, Q2 = C2V (2.61) The equivalent capacitor is one with charge Q = Q1 + Q2 (2.62) and potential difference V. Q = CV = C1V + C2V (2.63) The effective capacitance C is, from Eq. (2.63), C = C1 + C2 (2.64) The general formula for effective capacitance C for parallel combination of n capacitors [Fig. 2.28 (b)] follows similarly, Q = Q1 + Q2 + ... + Qn (2.65) FIGURE 2.28 Parallel combination of i.e., CV = C1V + C2V + ... CnV(2.66) (a) two capacitors, (b) n capacitors. which gives C = C1 + C2 + ... Cn (2.67) 72 Reprint 2025-26 Electrostatic Potential and Capacitance Example 2.9 A network of four 10 mF capacitors is connected to a 500 V supply, as shown in Fig. 2.29. Determine (a) the equivalent capacitance of the network and (b) the charge on each capacitor. (Note, the charge on a capacitor is the charge on the plate with higher potential, equal and opposite to the charge on the plate with lower potential.) FIGURE 2.29 Solution (a) In the given network, C1, C2 and C3 are connected in series. The effective capacitance C¢ of these three capacitors is given by 1 1 1 1 = + + C ′ C1 C 2 C 3 For C1 = C2 = C3 = 10 mF, C¢ = (10/3) mF. The network has C¢ and C4 connected in parallel. Thus, the equivalent capacitance C of the network is  10  C = C¢ + C4 =  3 + 10 mF =13.3mF (b) Clearly, from the figure, the charge on each of the capacitors, C1, C2 and C3 is the same, say Q. Let the charge on C4 be Q¢. Now, since the potential difference across AB is Q/C1, across BC is Q/C2, across CD is Q/C3 , we have Q Q Q + + = 500 V . C1 C 2 C 3 Also, Q¢/C4 = 500 V. This gives for the given value of the capacitances, 10 −3 Q = 500 V × µ F = 1.7 × 10 C and EXAMPLE 3 Q ′ = 500 V × 10 µ F = 5.0 × 10 −3 C 2.9