Chapter 4. Electrostatics of Macroscopic Meia 4.1 Multipole Expansion Approximate potentials at large istances 3 x' x' (x') x x' x x Fig 4.1 We consier the potential in the far-fiel region (see Fig. 4.1 where x = r a) ue to a localize charge istribution (ρ(x ) = 0 for r > a). If the total charge is q, it is a goo approximation to treat the charge istribution as a point charge, so Φ 1 q 4πε 0 r. Even if q is zero, the potential oes not vanish, but it ecays much faster than 1/r. We will iscuss more etails about how the potential behaves in the far-fiel region. Electric ipole a r q x r - Fig 4.. An electric ipole consists of two equal an opposite charges q an q separate by a isplacement. -q We begin with a simple, yet exceeingly important case of charge istribution. Two equal an opposite charges separate by a small istance form an electric ipole. Suppose that q an q are separate by a isplacement vector as shown in Figure 4., then the potential at x is 1 1 Φ(x) = 1 ( q q ) = q 1 [(1 cos θ 4πε 0 r r 4πε 0 r r 4r ) (1 cos θ r 4r ) ] 1 (4.1)
In the far-fiel region for r = x, Φ(x) q 1 4πε 0 r [(1 r cos θ) (1 1 q cos θ cos θ)] = r 4πε 0 r This reuces to the coorinate inepenent expression Φ(x) = 1 p x 4πε 0 r 3 (4.) (4.3) where p = q is the electric ipole moment. For the ipole p along the z-axis, the electric fiels take the form E r = 1 p cos θ 4πε 0 r 3 E θ = 1 p sin θ 4πε 0 r 3 { E φ = 0 p E(r, θ) = 4πε 0 r 3 ( cos θ e r sin θ e θ ) From this, we can obtain the coorinate inepenent expression where n = x/r is a unit vector. E(x) = 1 3n(p n) p 4πε 0 r 3 (4.4) (4.5) Fig 4.3. Fiel of an electric ipole
Multipole expansion We can expan the potential ue to the charge istribution ρ(x ) using Eq. 3.68 1 x x = 4π l 1 Φ(x) = 1 4πε 0 ρ(x ) x x 3 x l=0 In the far-fiel region, r > r. Then we fin r < l l l1 r > m= l Y lm (θ, φ )Y lm (θ, φ) Φ(x) = 1 4π 4πε 0 l 1 [ Y lm (θ, φ )r l ρ(x ) 3 x ] Y lm(θ, φ) r l1 We can rewrite the equation where the coefficients l,m Φ(x) = 1 4π 4πε 0 l 1 q lm l,m Y lm (θ, φ) r l1 q lm = Y lm (θ, φ )r l ρ(x ) 3 x (1.1) (3.68) (4.6) (4.7) (4.8) are calle multipole moments. This is the multipole expansion of Φ in powers of 1/r. The first term (l = 0) is the monopole contribution ( 1/r); the secon (l = 1) is the ipole ( 1/r ); the thir is quarupole; an so on. Monopole moment or total charge q (l = 0, q = 4πq 00 ): q = ρ(x ) 3 x (4.9) Electric ipole moment p (l = 1, linear combinations of q 1m ): p = x ρ(x ) 3 x (4.10) Quarupole moment tensor Q ij (l =, linear combinations of q m ): Q ij = (3x i x j r δ ij )ρ(x ) 3 x (4.11) 3
The expansion of Φ(x) in rectangular coorinates Φ(x) = 1 [ q p x 4πε 0 r r 3 1 Q ij ij x i x j r 5 ] (4.1) Energy of a charge istribution in an external fiel If a localize charge istribution ρ(x) is place in an external potential Φ ext (x), the electrostatic energy of the system is W = ρ(x)φ ext 3 x (4.13) If Φ(x) is slowly varying over the region of ρ(x), we can expan it in a Taylor series Φ(x) = Φ ext (0) x Φ ext (0) 1 i,j x ix j Φ ext x i x j (0) = Φ ext (0) x E ext (0) 1 x ix j E ext,j x i (0) i,j ( 1 6 r E(0) = 0) = Φ ext (0) x E ext (0) 1 (3x 6 ix j r δ ij ) E ext,j i,j (0) x i (4.14) Then, the energy takes the form W = qφ ext (0) p E ext (0) 1 6 Q ij i,j E ext,j x i (0) (4.15) 4. Polarization an Electric Displacement in Macroscopic Meia Dielectrics Properties of an ieal ielectric material It has no free charges. Instea, all charges are attache to specific atoms or molecules. Electric fiels can inuce only small isplacements from their equilibrium positions. In a macroscopic scale, the effects of the electric fiels can be visualize as a isplacement of the entire positive charge in the ielectric relative to the negative charge. The ielectric is sai to be polarize. Electric Polarization If an electric fiel is applie to a meium compose of many atoms an molecules, each atom or molecule forms a ipole pi ue to the fiel inuce isplacements of the boun charges (see Fig. 4.4). Typically, this inuce ipole moment is approximately proportional to the fiel: p = γ a E (4.16) 4
where γ a is calle atomic polarizability. These little ipoles are aligne along the irection of the fiel, an the material becomes polarize. An electric polarization P is efine as ipole moment per unit volume: P = 1 V p i i = Np a V is a volume element which contains many atoms, yet it is infinitesimally small in the macroscopic scale. N is the number of atoms per unit volume an p a is the average ipole moment of the atoms. (4.17) E p i V P 1 V i p i Fig 4.4. An external electric fiel inuces electric polarization in a ielectric meium. Boun charges The ipole moment of V is PV = P(x ) 3 x, so the total electric potential (see Eq. 4.3) is We can rewrite this equation as Integrating by parts gives Φ(x) = 1 4πε 0 P(x ) (x x ) x x 3 3 x Φ(x) = 1 4πε 0 P(x ) 1 x x 3 x Φ(x) = 1 4πε 0 { [ P(x ) x x ] 3 x P(x ) x x 3 x } (4.18) (4.19) (4.0) Using the ivergence theorem Φ(x) = 1 4πε 0 { P(x ) n x x a P(x ) x x 3 x } (4.1) 5
where a is a surface element an n is the normal unit vector. Here we efine surface an volume charge ensities: an σ b = P n ρ b = P Then, the potential ue to the boun charges becomes Φ(x) = 1 4πε 0 σ b(x ) x x a ρ b(x ) x x 3 x (4.) (4.3) (4.4) b Fig 4.5. Origin of boun charge ensity. b Electric isplacement When a material system inclues free charges ρ as well as boun charges ρ b, the total charge ensity ρ total can be written: An Gauss s law reas ρ total = ρ ρ b ε 0 E = ρ P With the efinition of the electric isplacement D, Equation 4.6 becomes D = ε 0 E P D = ρ When an averaging is mae of the homogeneous equation, E micro = 0, the same equation E = 0 hols for the macroscopic, electric fiel E. This means that the electric fiel is still erivable from a potential in electrostatics. Equations 4.8 an 4.9 are the two electrostatic equations in the macroscopic scale. 6 (4.5) (4.6) (4.7) (4.8) (4.9)
Electric susceptibility, permittivity, an ielectric constant For many substances (we suppose that the meia are isotropic), the polarization is proportional to the fiel, provie E is not too strong: P = ε 0 χ e E (4.30) The constant χ e is calle the electric susceptibility of the meium. The isplacement D is therefore proportional to E, D = ε 0 (1 χ e )E = εe (4.31) where ε is electric permittivity an ε r = ε/ε 0 = 1 χ e is calle the ielectric constant or relative electric permittivity. Bounary conitions on the fiel vectors S S n 1 D 1 D 1 l E 1 E L Fig 4.6. Bounary conitions on the fiel vectors at the interface between two meia may be obtaine by applying Gauss s law to surface S an integrating E l aroun the path L. Consier two meia, 1 an, in contact as shown in Fig. 4.6. We shall assume that there is a surface charge ensity σ. Applying the Gauss s law to the small pill box S, we obtain D n 1 ΔS D 1 n 1 ΔS = σδs (4.3) This leas to i.e., (D D 1 ) n 1 = σ D n D 1n = σ Thus the iscontinuity in the normal component of D is given by the surface ensity of free charge on the interface. (4.33) (4.34) The line integral of E l aroun the path L must be zero: This gives i.e, E l E 1 ( l) = 0 (E E 1 ) n 1 = 0 E t = E 1t Thus the tangential component of the electric fiel is continuous across an interface. (4.35) (4.36) (4.37) 7
4.3 Bounary-Value Problems with Dielectrics If the ielectrics of interest are linear, isotropic, an homogeneous, D = εe (Eq. 4.31), where ε is a constant characteristic of the material, an we may write E = 1 ε ρ (4.38) Since E = 0 still hols, the electric fiel is erivable from a scalar potential Φ, i.e., E = Φ, so that Φ = 1 ε ρ (4.39) Thus the potential in the ielectric satisfies the Poisson s equation; the only ifference between this equation an the corresponing equation for the potential in vacuum is that ε replaces ε 0 (vacuum permittivity). In most cases of interest ielectrics contains no charge, i.e., ρ = 0. In those circumstances, the potential satisfies Laplaces equation throughout the boy of ielectric: Φ = 0 (4.40) An electrostatic problem involving linear, isotropic, an homogeneous ielectrics reuces, therefore, to fining solutions of Laplace s equation in each meium an joining the solutions in the various meia by means of the bounary conitions. We treat a few examples of the various techniques applie to ielectric meia. Point charge near a plane interface of ielectric meia x 1 q z Fig 4.7. We consier a point charge q embee in a semi-infinite ielectric ε 1 a istance away from a plane interface (z = 0) that separates the first meium from another semi-infinite ielectric ε as shown in Fig. 4.7. From Eqs. 3.34 an 3.37, we obtain the bounary conitions: E x z=0 = E x z=0 { E y z=0 = E y z=0 ε 1 E z z=0 = ε E z z=0 8 (4.41)
We apply the metho of images to fin the potential satisfying these bounary conitions (see Fig. 4.8). For the potential in the region z > 0, we locate an image charge q at z =. Then the potential at a point P escribe by cylinrical coorinates (ρ, φ, z) is where Φ = 1 4πε 1 ( q R 1 q R ), z > 0 (4.4) R 1 = ρ ( z) an R = ρ ( z) For the potential in the region z < 0, we locate an image charge q at z =. Then the potential at a point P is Φ = 1 4πε q" R 1, z < 0 (4.43) (4.44) (a) In the region z>0 (b) In the region z<0 q 1 1 P R 1 q z R R 1 P q z ' ", 1 q Fig 4.8. (a) The potential for z > 0 is ue to q an an image charge q at z =. (b) The potential for z < 0 is ue to an image charge q at z =. The first two bounary conitions in Eq. 4.41 are for the tangential components of the electric fiel:, 1 q 1 4πε 1 ρ ( q R 1 q R ) z=0 = 1 4πε q" ρ R 1 z=0 1 qρ [ q ρ ε 1 (ρ ) 3/ (ρ ) 3/] = 1 q"ρ ε (ρ ) 3/ 1 ε 1 (q q ) = 1 ε q The thir bounary conition in Eq. 4.41 is for the normal component of the isplacement: (4.45) 9
1 4π z ( q R 1 q R ) z=0 = 1 4π q" z R 1 z=0 q q = q" (ρ ) 3/ (ρ ) 3/ (ρ ) 3/ q q = q (4.46) From Eqs. 4.45 an 4.46, we obtain the image charges q an q : q = ( ε ε 1 ε ε 1 ) q (4.47) { q" = ( ε ε ε 1 ) q Figure 4.8 shows the lines of D for two cases ε > ε 1 (q < 0) an ε < ε 1 (q > 0) for q > 0. 1 1 Fig 4.8. Lines of electric isplacement The surface charge ensity is given by σ b = P n (Eq. 4.). Therefore, the polarization-surfacecharge ensity on the interface is σ b = P 1 ( e z ) P e z = P 1z P z (4.48) Since P z = (ε ε 0 )E z, σ b = (ε 1 ε 0 ) 1 ( q q (ε 4πε 1 z R 1 R ε 0 ) ) z=0 1 q" 4πε z R 1 z=0 = (ε 1 ε 0 ) 1 4πε 1 (q q ) (ρ ) 3/ (ε ε 0 ) 1 4πε q" (ρ ) 3/ σ b = q ε 0 (ε ε 1 ) π ε 1 (ε ε 1 ) (ρ ) 3/ (4.49) In the limit ε ε 1 (ε behaves like a conuctor) an ε 1 = ε 0, Eq. 4.49 becomes equivalent to Eq.. for a point charge in front of a conucting surface. 10
Dielectric sphere in a uniform electric fiel A ielectric sphere of raius a an permittivity ε is place in a region of space containing an initially uniform electric fiel E 0 = E 0 e z as shown in Fig. 4.9. The origin of our coorinate system is taken at the center of the sphere, an the electric fiel is aligne along the z-axis. We shoul like to etermine how the electric fiels are moifie by the ielectric sphere. r P a E0 E0 z Fig 4.9. Insie an outsie potential From the azimuthal symmetry of the geometry we can take the solution to be of the form: (i) Outsie: Φ out = [B l r l l=0 C l r l1 ] P l(cos θ) = E 0 r cos θ C l r l1 P l(cos θ) At large istances from the sphere, i.e., for the region r a, the potential is given by Φ out (r, θ) E 0 z = E 0 r cos θ Accoringly, we can immeiately set all B l except for B 1 (= E 0 ) equal to zero. (ii) Insie: Φ in = A l r l P l (cos θ) l=0 Since Φ in is finite at r = 0, r (l1) terms must vanish. Bounary conitions at r = a l=0 (4.50) (4.51) (4.5) (i) Tangential E: 1 a Φ in θ r=a = 1 a Φ out θ r=a (4.53) or Φ in (r = a) = Φ out (r = a) (4.54) (ii) Normal D: ε Φ in r r=a = ε 0 Φ out r r=a (4.55) 11
Applying bounary conition (i) (Eq. 4.54) tells us that A l a l P l (cos θ) = E 0 a cos θ C l a l1 P l(cos θ) l=0 We euce from this that l=0 (4.56) { A 1 = E 0 C 1, l = 1 a 3 A l = C l, l 1 l1 We apply bounary conition (ii) results in a ε A l la l 1 P l (cos θ) = ε 0 E 0 cos θ ε 0 (l 1)C l P l (cos θ) l=0 We euce from this that { ε l=0 ε 0 A 1 = E 0 C 1 a 3, l = 1 ε ε 0 la l = (l 1) The equations 4.57 an 4.60 can be satisfie only if a C l l1, l 1 A 1 = ( 3 ) E ε { r 0 C 1 = ( ε r 1 ) ε r a3 E 0 (4.57) (4.58) a l (4.59) (4.60) (4.61) (4.6) (4.63) where ε r = ε/ε 0 is the ielectric constant (or relative electric permittivity). From Eqs. 4.58 an 4.61, we can euce that A l = C l = 0 for all l 1. The potential is therefore Electric fiel an polarization Φ in = ( 3 ε r ) E 0r cos θ Φ out = E 0 r cos θ ( ε r 1 ε r ) E 0 a 3 cos θ r Equation 4.64 tells us that the fiel insie the sphere is a constant in the z irection: 3 E in = ( ε r ) E 0 (4.64) (4.65) (4.66) 1
For ε = 1 (no ielectric), this reuces as expecte to E 0 e z. The fiel outsie the ielectric is clearly compose of the original constant fiel E 0 e z an a fiel which has a characteristic ipole istribution with ipole moment of p = 4πε 0 ( ε r 1 ε r ) a3 E 0 (4.67) We compare this with that from integrating the polarization P over the sphere. Insier the ielectric we have P = (ε ε 0 )E in = 3ε 0 ( ε r 1 ε r ) E 0 Since P is constant, we obtain the total ipole moment Surface charge ensity p = ( 4π 3 a3 ) P = 4πε 0 ( ε r 1 ε r ) a3 E 0 which is equal to Eq. 4.67. (4.68) (a) polarization (b) Electric fiel ue to surface charge E 0 P E 0 Fig. 4.10 The uniform external electric fiel inuces the constant polarization insie a ielectric sphere (Eq. 4.68), an the inuce polarization gives rise to surface charge which prouces opposing electric fiel if ε > ε 0, as illustrate in Fig. 4.10. The surface charge ensity (Eq. 4.) is Spherical cavity in a ielectric meium σ b = P n = P ( r r ) = 3ε 0 ( ε r 1 ε r ) E 0 cos θ (4.69) a 0 E 0 z Fig. 4.11 Figure 4.11 sketches the problem of a spherical cavity of raius a in a ielectric meium (ε) with an external fiel E 0 = E 0 e z. We can obtain the solution of this problem by switching ε an ε 0 in 13
the solution of the previous problem (i.e., ε r = ε/ε 0 1/ε r = ε 0 /ε). For example, the fiel insie the cavity is constant in the z irection: E in = ( 3ε r ε r 1 ) E 0 (4.70) The fiel outsie the ielectric is compose of the original constant fiel E 0 e z an a fiel of the ipole moment p = 4πε 0 ( ε r 1 ε r 1 ) a3 E 0 (4.71) which is oriente oppositely to the applie fiel if ε > ε 0. 4.4 Microscopic Theory of Dielectrics We now examine the molecular nature of the ielectric, an see how the electric fiel responsible for polarizing the molecule is relate to the macroscopic electric fiel. Our iscussion is in terms of simple classical moels of the molecular properties, although a proper treatment necessarily woul involve quantum mechanical consieration. On the basis of a simple molecular moel it is possible to unerstan the linear behavior that is characteristic of a large class of ielectric materials. Molecular polarizability an electric susceptibility Molecular fiel an macroscopic fiel The electric susceptibility χ e is efine through the relation P = ε 0 χ e E (Eq. 4.30), where E is the macroscopic electric fiel. The electric fiel responsible for polarizing a molecule of the ielectric is calle the molecular fiel E m. E m is ifferent from E because the polarization of other molecules gives rise to an internal fiel E i, so that we can write E m = E E i. b p mol E E m Fig 4.1. The ielectric outsie the cavity is replace by a system of polarization charges σ b. Internal fiel In orer to fin out E i, we consier an imaginary sphere which contains neighboring molecules. It is much larger than the molecules, yet infinitesimally small in the macroscopic scale. The geometry is shown in Fig. 4.1. Then we can ecompose E i into two terms: E i = E near E S, where E near is the fiel ue to the neighboring molecules close to the given molecule an E S is 14
the contribution from all the other molecules. E S arises from surface charge ensity σ b = P n on the cavity surface. Using spherical coorinates, we obtain E S = 1 r 4πε 0 r 3 σ bs = 1 r (P cos θ) 4πε 0 r3 The x an y components vanish because they inclue the integrals of π 0 sin φ φ = 0, respectively. Therefore, E S = 1 4πε 0 0 π 0 π 0 π Pe z cos θ 0 π r sin θ θ φ π 0 sin θ θ φ = 1 3ε 0 P cos φ φ = 0 an Now we consier the term, E near. If the many molecules are ranomly istribute in position, then E near = 0. This is the case if the ielectric is a gas or a liqui. If the ipoles in the cavity are locate at the regular atomic positions of a cubic crystal, then again E near = 0 (you may refer to the proof in the textbook, pp. 160-161). We restrict further iscussion to the rather large classs of materials in which E near = 0. Then, Polarization an molecular polarizability The polarization vector is efine as E m = E E i = E E S = E 1 3ε 0 P P = Np mol where N is the number of molecules per unit volume an p mol is the ipole moment of the molecules. We efine the molecular polarizability as Combining Eqs. 4.73, 4.74, an 4.75, we obtain p mol = ε 0 γ mol E m (4.7) (4.73) (4.74) (4.75) P = Nε 0 γ mol E m = Nγ mol (ε 0 E 1 3 P) (4.76) Using P = ε 0 χ e E (Eq. 4.30), we fin Nγ mol χ e = 1 1 3 Nγ mol (4.77) as the relation between susceptibility (the macroscopic parameter) an molecular polarizability (the macroscopic parameter). 15
Using ε r = ε/ε 0, we fin γ mol = 3 N (ε r 1 ε r ) (4.78) This is calle the Clausius-Mossotti equation. Moels for the molecular polarizability The molecules of a ielectric may be classifie as polar or nonpolar. A polar molecule such as HO an CO has a permanent ipole moment, even in the absence of a polarizing fiel Em. In nonpolar molecules, the centers of gravity of the positive an negative charge istributions normally coincie. Symmetrical molecules such as O, monoatomic molecules such as He, an monoatomic solis such as Si fall into this category. We will iscuss simple moels for these polar an nonpolar molecules. Inuce ipoles: simple harmonic oscillator moel The application of an electric fiel causes a relative isplacement of the positive an negative charges in nonpolar molecules, an the molecular ipoles so create are calle inuce ipoles. To estimate the inuce ipole moments we consier a simple harmonic oscillator moel of boun charges (electrons an ions). Each charge e is boun uner the action of a restoring force by an applie electric fiel F = ee = mω 0 x (4.79) where m is the mass of the charge, an ω 0 is the frequency of oscillation about equilibrium. Consequently the inuce ipole moment is p mol = ex = e mω 0 E (4.80) Therefore the polarizability is e γ = ε 0 mω 0 (4.81) For a boun electron, a typical oscillation frequency is in the optical range, i.e., ω 0 ~10 16 Hz. Then the electronic contribution is γ el ~10 9 m 3. For gases at NTP, N =.7 10 5 m -3, so that their susceptibilities, χ e ~Nγ el (see Eq. 4.77), are of the orer of 10 3 at best. For example, the experimental value of ielectric constant for air is ε r = 1 χ air = 1.00054. For solis or liqui ielectrics, N~10 8 10 9 m -3, therefore the susceptibility can be of the orer of unity. 16
Polar molecules: Langevin-Debye formula In the absence of an electric fiel a macroscopic piece of polar ielectric is not polarize, since thermal agitation keeps the molecules ranomly oriente. If the polar ielectric is subjecte to an electric fiel, the iniviual ipoles experience torques which ten to align them with the fiel. The average effective ipole moment per molecule may be calculate by means of a principle from statistical mechanics. At temperature T the probability of fining a particular molecular energy or Hamiltonian H is proportional to e H kt (4.8) For a polar molecule in the presence of an electric fiel E = Ee z, the Hamiltonian inclues the potential energy (see Eq. 4.16), U = p 0 E = p 0 E cos θ (4.83) Where p 0 is a permanent ipole moment. Then the Hamiltonian is given by H = H 0 p 0 E cos θ where H 0 is a function of only the internal coorinates of the molecule (e.g., kinetic energy) so that it is inepenent of the applie fiel. Using the Boltzmann factor Eq. 4.8 we can write the average ipole moment as: p mol = p 0 cos θ e p0e cos θ kt e p 0E cos θ kt Ω Ω = p 0 [coth ( p 0E kt ) kt p 0 E ] Here the components of p 0 not parallel to E vanish. In general, the ipole potential energy p 0 E is much smaller than the thermal energy kt except at very low temperature. Then p 0 p mol 1 3 kt E Therefore the polarizability of the polar molecule is γ mol = 1 p 0 3ε 0 kt In general, inuce ipole effects are also present in polar molecules, yet they are inepenent of temperature. Then, the total molecular polarizability is γ mol = γ i 1 p 0 3ε 0 kt (4.84) (4.85) (4.86) (4.87) (4.88) 17
4.5 Electrostatic Energy in Dielectric Meia an Forces on Dielectrics Energy in ielectric systems We iscuss the electrostatic energy of an arbitrary istribution of charge in ielectric meia characterize by the macroscopic charge ensity ρ(x). The work one to make a small change δρ in ρ is δw = δρ(x)φ(x) 3 x (4.89) Where Φ(x) is the potential ue to the charge ensity ρ(x) alreay present. Since D = ρ, δρ = δd, where δd is the resulting change in D, so δw = ( δd)φ 3 x (4.90) Now ( δd)φ = (δdφ) δd Φ = (δdφ) δd E an hence (integrating by parts) δw = (δdφ) 3 x δd E 3 x (4.91) The ivergence theorem turns the first term into a surface integral, which vanishes if ρ(x) is localize an we integrate over all of space. Therefore, the work one is equal to δw = δd E 3 x (4.9) So far, this applies to any material. Now, if the meium is a linear ielectric, then D = εe so Thus δd E = εδe E = 1 εδ(e E) = 1 δ(d E) δw = δ ( 1 D E 3 x) (4.93) (4.94) The total work one, then, as we buil the free charge up from zero to the final configuration, is W = 1 D E 3 x Parallel-plate capacitor fille with a ielectric meium (4.95) A Q V Fig 4.13. -Q 18
We shall fin the electrostatic energy store in a parallel-plate capacitor. Its geometry is shown in Fig. 4.13: two conucting plates of area A (charge with Q an -Q) is separate by (we assume that is very small compare with the imensions of the plates), an the gap is fille with ielectric (ε). (i) Capacitance The electric fiel between the plates is E = σ ε = Q εa (4.96) The potential ifference V = E. Therefore, C = Q V = εa (ii) Electrostatic energy Using Eq. 1.40, we obtain the electrostatic energy store in the capacitor. This is consistent with Eq. 4.95: W = 1 CV = 1 εa (E) = 1 εe A (4.97) (4.98) W = 1 εe 3 x = 1 εe A Forces on ielectrics We have just evelope a proceure for calculating the electrostatic energy of a charge system incluing ielectric meia. We now iscuss how the force on one of the objects in the charge system may be calculate from this electrostatic energy. We assume all the charge resies on the surfaces on the conuctors. Constant total charge Let us suppose we are ealing with an isolate system compose of a number of parts (conuctors, point charges, ielectrics) an allow one of these parts to make a small isplacement x uner the influence of the electrical forces F acting upon it. The work performe by the electrical force on the system is F r = F x x F y y F z z Because the system is isolate, this work is one at the expense of the electrostatic energy W; in other wors, the change in the electrostatic energy is W = F r. Therefore, F i = ( W, i = x, y, z x i )Q where the subscript Q has been ae to enote that the system is isolate, an hence its total charge remains constant uring the isplacement r. 19 (4.99) (4.100)
Fixe potential We assume that all the conuctors of the system are maintaine at fixe potentials, V i, by means of external sources of energy (e.g., by means of batteries). Then, the work performe F r = W W b (4.101) where W b is the work supplie by the batteries. The electrostatic energy W of the system (see Eq. 1.36) is given as Since V i s are constant, W = 1 V iq i W = 1 V iq i Furthermore, the work supplie by the batteries is the work require to move each of the charge increments Q i from zero potential to the potential of the appropriate conuctor, therefore, Consequently, W = F r, an hence W b = V i Q i i F i = ( W x i )V i i = W, i = x, y, z Here the subscript V is use to enote that all potentials are maintaine constant. Dielectric slab within a parallel-plate capacitor As an example of the energy metho, we consier a parallel-plate capacitor in which a ielectric slab (ε) is partially inserte. The imensions of each plate are length l an with w. The separation between them is. The geometry is illustrate in Fig. 4.14. We shall calculate the force tening to pull the ielectric slab back into place. We consier two cases of (i) a constant potential ifference V an (ii) a constant total charge Q. x Q (4.10) (4.103) (4.104) (4.105) -Q V Fig 4.14. Dielectric slab partially withrawn from the gap between two charge plates. l 0
(i) Constant potential ifference V Since the electric fiel E = V/ is the same everywhere between the plates, we fin W = 1 ε(x)e 3 x = 1 ε (V ) wx 1 ε 0 ( V ) w(l x) (4.106) The force may be calculate from Eq. 4.106: F x = 1 (ε ε 0)w V = 1 (ε r 1)ε 0 E w (ii) Constant total charge Q The energy store in the capacitor (see Eq. 1.4) is an the capacitance in this case is C = εwx We apply Eq. 4.101 to obtain the force: Since C x = w (ε ε 0), we fin W = 1 CV = 1 ε 0w(l x) F x = W x = 1 F x = 1 (ε ε 0) w Q C = w [(ε ε 0)x ε 0 l] Q C C x Q C = 1 (ε ε 0)w V (4.107) (4.108) (4.109) (4.110) (4.111) Eq. 4.111 has the same expression with Eq. 4.107, but the force of constant charge (Eq. 4.111) is a function of x (C varies with x) while the force of constant potential (Eq. 4.107) is inepenent of x. 1