Electric Fields in Dielectrics Any kind of matter is full of positive and negative electric charges. In a dielectric, these charges cannot move separately from each other through any macroscopic distance, so when an electric field is applied there is no net electric current. However, the field does push the positive charges just a tiny bit in the direction of E while the negative charges are pushed in the opposite directions. Consequently, the atoms and the molecules comprising the dielectric acquire tiny electric dipole moments in the direction of E. 1
To see the the net effect of all these dipole moments on the macroscopic scale, imagine smearing all the positive charges into a large uniform charge density +ρ and likewise all the negative charges into uniform charge density ρ. Without the electric field, these densities overlap each other over the whole dielectric, so the net charge density cancels out. But when we turn on the field, the positive density moves a tiny bit in the direction of E while the negative density moves in the opposite direction: positive charges move left positive charges move right ρ only overlap +ρ only 2
As the result of this move, the bulk of the dielectric where the positive and negative charges continue to overlap each other remains electrically neutral. But in a thin surface layer on the left side of the dielectric there are only negative charges while in a similar layer on the right side there are only positive charges. Altogether, the dielectric develops surface charges as shown on the picture below: neutral 3
Note the signs of the surface charges induced by the electric field: On the side of the positive external electrode, the induced surface charge is negative, while on the side of the negative external electrode, the induced surface charge is positive. Consequently, the induced surface charges reduce the electric field in the dielectric from what it would have been without them: The black lines here are the electric field lines. There are fewer of them inside the dielectric then outside it, which indicates a weaker electric field inside the dielectric. 4
The surface charges on the dielectric are proportional to the electric field, which is in turn proportional to the external charges on the capacitor plates. In the Electrodynamics class (PHY 352 K) you should take in your junior year, you will learn how this works in some detail. For the moment, let me simply state that for a dielectric slab between parallel capacitor plates Q surface = κ 1 Q plate κ for some constant κ > 1, and therefore the overall charge on one side of the dielectric is Q plate + Q surface = + 1 κ Q plate (2) same sign as the charge on the plate, but smaller. Consequently, the electric field in the dielectric is κ times weaker than the electric field in the air gaps between the plates and the dielectric. Specifically, for the plates of area A, E[in the gaps] = Q plate ǫ 0 A, (3) E[in the dielectric] = Q plate +Q surface ǫ 0 A (1) = 1 κ Q plate ǫ 0 A, (4) = 1 E[in the gaps]. (5) κ Theconstantκ whichdependsonadielectricinquestionbutnotonthecharges iscalled the relative permittivity or the dielectric constant. Table 26.1 on textbook page 841 gives the values of κ for some common dielectrics; a similar table is available on a nyperphysics web page http://hyperphysics.phy-astr.gsu.edu/hbase/tables/diel.html. Eq. (4) has immediate application to capacitors. Suppose almost all the space between the plates is filled with a dielectric with relative permittivity κ. If this space has width D, then the potential difference between the capacitor plates is V = D E[in the dielectric] = D 1 κ Q plate ǫ 0 A = Q plate D κǫ 0 A. (6) As far as the capacitor is concerned, the charge Q it stores is Q = +Q left plate = Qright plate (7) since its the plates which are connected by wires to the outside world, while the induced 5
charges on the dielectric s surface are not connected to anything. Consequently, the capacitance is κ times more than one would get without the dielectric. C = Q V = κǫ 0A D, (8) For example, consider a ceramic capacitor with plates of area A = 1.03 cm 2 separated by 25.0 µm gap filled with barium titanate, BaTiO 3. This particular ceramics has very high permittivity κ ranging from 1,000 to 10,000; for our example let κ = 2500. Then the capacitor in question has C = κǫ 0A D = 2500 ( 8.85 10 12 F/m) 1.03 10 4 m 2 25.0 10 6 m 91 nf. (9) But without the high-permittivity dielectric, this capacitor would have a much smaller capacitance of only a few picofarads. Now let s go beyond the capacitors and consider a point charge +Q stuck in the middle of a dielectric medium. The electric field of the point charge induces microscopic dipole moments throughout the dielectric, although the dipole moment density is no longer uniform in space but decreases with the distance from the charge. Nevertheless, the collective effect of all the microscopic dipole moments is to move the negative charges towards the point charge while the positive charges move outward; these charges cancel in the bulk of the dielectric, but there is a net negative screening charge Q screening = κ 1 κ Q (10) immediately surrounding the point charge Q, and the opposite positive charge on the outer surface of the dielectric: Q surface = + κ 1 κ Q (11) 6
Let s assume the dielectric piece has large size and focus on the neighborhood of the point charge Q, where we are much closer to Q than to the dielectric s surface. In this case, the electric field is dominated by the Q and by the screening charge immediately surrounding it, thus E = Q+Q screening 4πǫ 0 ı r r 2 = Q 4πκǫ 0 ı r r 2, (12) 1/κ of the Coulomb field charge Q would create in the vacuum. A point of terminology: In dielectric context, a free charge is a charge outside the dielectric (as on a capacitor plate) or put inside the dielectric (like an ion into water), but not not a charged induced in the dielectric itself by the electric field. Thus, the Coulomb 7
field in a dielectric is and the Coulomb force between two free charges is In particular, an ion dissolved in water has free charge E = Qfree 4πκǫ 0 ı r r 2, (13) F = Qfree 1 Q free 2 4πκǫ 0 r 2. (14) Q free = e valence, (15) for example Q free = +2e for the calcium Ca ++ ion or Q free = e for the chlorine Cl ion. Thus, the Coulomb force between two such ions is for κ(water) 80. F = e2 4πκǫ 0 valence 1 valence 2 r 2, (16) In terms of free charges, the Gauss Law for a uniform dielectric becomes Φ E [S] = S E d 2 Area = 1 κǫ 0 Q free [enclosed within S]. (17) This formula helps to find electric fields in a symmetric setups. For example, consider a coaxial cable where the inner conductor has charge +Q, the outer conductor has charge Q, and the space between the conductors is filled with a dielectric with relative permittivity κ, then the electric field in the dielectric is E = 1 Q/L 2πκǫ 0 r (18) where L is the cable s length. Consequently, the voltage between the conductors is V = Q/L 2πκǫ 0 R 2 R 1 dr r = Q/L 2πκǫ 0 ln R 2 R 1 (19) where R 1 is the outer radius of the inner conductor while R 2 is the inner radius of the outer 8
conductor. Consequently, the cable has capacitance C = Q V = 2πκǫ 0 L. (20) ln(r 2 /R 1 ) For example, for R 2 = 4.0 R 1 and polystyrene dielectric with κ 2.5, we get C L 1.0 pf/cm. (21) Note: in MKSA units, the relative permittivity κ of a dielectric always appears in combination κ ǫ 0. This product is called the absolute permittivity of the dielectric. In this context, the vacuum permittivity ǫ 0 is simply the absolute permittivity of vacuum viewed as just another dielectric! A point of notation: In many books and articles, the relative permittivity of a dielectric is denoted ǫ rather than κ. But sometimes ǫ is used for the absolute permittivity κ ǫ 0, although this is quite uncommon in modern literature. 9
Let me conclude these notes by defining the electric displacement field D = κǫ 0 E, (22) or more generally where P is the density of the induced dipole moment, D = ǫ 0 E + P (23) P = dipole moment vector volume. (24) In terms of the displacement field D, the Gauss Law in a dielectric becomes D d 2 Area = Q free [enclosed within S]. (25) S This formulation of the Gauss Law is quite general and works for all closed surfaces S, even if part of it is inside a dielectric an another part is outside. Or if several dielectrics with different κ s are involved, we may have different parts of S be inside different dielectrics. In particular, letting S be a squat cylinder spanning a border between two dielectrics D 1 S D 2 and applying the Gauss Law (25) tells us that for the electric field to the border, D 1 = D 2 = κ 1 E 1 = κ 2 E 2. (26) Finally, for non-uniform dielectrics where κ varies from place to place, it is convenient to use the differential form of the Gauss Law for D(r) = κ(r)ǫ 0 E(r): D = ρ free. (27) 10