8.2 A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to 213 a 230 V ac supply with a (angular) frequency of 300 rad s–1. Reprint 2025-26 Physics (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates. FIGURE 8.6 8.3 What physical quantity is the same for X-rays of wavelength 10–10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500m? 8.4 A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength? 8.5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band? 8.6 A charged particle oscillates about its mean equilibrium position with a frequency of 10 9 Hz. What is the frequency of the electromagnetic waves produced by the oscillator? 8.7 The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave? 8.8 Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is n = 50.0 MHz. (a) Determine, B0,w, k, and l. (b) Find expressions for E and B. 8.9 The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hn (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation? 8.10 In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m–1. (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals the average energy density of the B field. [c = 3 × 108 m s–1.] Reprint 2025-26

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?

2.13 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