11/4/2004 Section 5_3 Dielectrics blank.doc 1/ Dielectrics

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1 11/4/4 Section 5_3 ielectrics blank.doc 1/3 5-3 ielectrics Reading Assignment: pp Recall that if a dielectric material is immersed in an electric field, each atom/molecule in the material will form an electric dipole! HO: The Polarization Vector A. Polarization Charge istribution Q: A: HO: Polarization Charge istributions

2 11/4/4 Section 5_3 ielectrics blank.doc /3 B. lectric Flux ensit HO: lectric Flux ensit C. Field quations in ielectrics Q: A: HO: lectrostatic Field quations in ielectrics. lectric Boundar Conditions Q:, 1 1??,?? ε 1 ε

3 11/4/4 Section 5_3 ielectrics blank.doc 3/3 A: HO: ielectric Boundar Conditions HO: Boundar Conditions on Perfect Conductors xample: ielectric Boundar Conditions

4 11/4/4 The Polarization Vector.doc 1/3 The Polarization Vector Recall that in dielectric materials (i.e., insulators), the charges are bound. p - + As a result, atoms/molecules form electric dipoles when an electric field is present! Note that even for some small volume v, there are man atoms/molecules present; therefore there will be man electric dipoles.

5 11/4/4 The Polarization Vector.doc /3 We will therefore define an average dipole moment, per unit P r. volume, called the Polarization Vector ( ) pn dipole moment C P v unit volume m where p n is one of N dipole moments in volume v, centered at position r. Note the polarization vector is a vector field. As a result, the direction and magnitude of the Polarization vector can change as function of position (i.e., a function of r ). Q: How are vector fields P and related?? A: Recall that the direction of each dipole moment is the same r r have the as the polarizing electric field. Thus P ( ) and ( ) same direction. There magnitudes are related b a unitless χ r, called electric susceptibilit: scalar value ( ) e ε χ P e lectric susceptibilit is a material parameter indicating the stretchabilit of the dipoles.

6 11/4/4 The Polarization Vector.doc 3/3 Q: Can we determine the fields created b a polarized material? A: Recall the electric potential field created b one dipole is: V p ( r-r ) 4πε r-r 3 Therefore, using d p P dv, the electric potential field created b a distribution of dipoles (i.e., P ) across some volume V is (see fig. 5.9): V P V ( r ) ( r-r ) 4πε r-r 3 dv Q: But I thought scalar charge distributions ρ v and ρ s created the electric potential field V. Now ou are saing that electric fields are created b the vector field P!?! A: As we will soon see, the polarization vector P creates equivalent charge distributions we will get the V r from either source! correct answer for ( )

7 11/4/4 Polarization Charge istributions.doc 1/6 Polarization Charge istributions Consider a hunk of dielectric material with volume V. Sa this dielectric material is immersed in an electric field r P r. ( ), therefore creating atomic dipoles with densit ( ) Q: What electric potential field V is created b these diploes? A: We know that: V P V ( r ) ( r-r ) 4πε r-r 3 dv But, it can be shown that (p. 135): V V P ( r ) ( r-r ) 4πε r-r 3 dv ( r ) P( r ) ( r) 1 P 1 n dv ds 4πε + r-r 4πε r-r V S where S is the closed surface that surrounds volume V, and a r is the unit vector normal to surface S (pointing outward). ˆ n ( )

8 11/4/4 Polarization Charge istributions.doc /6 This complicated result is onl important when we compare it to ρ r the electric potential created b volume charge densit v ( ) and surface charge densit ρ : s V V ( r ) 1 ρv 4πε r-r V ( r ) 1 ρs 4πε r-r S dv ds If both volume and surface charge are present, the total electric potential field is: ( r ) ρ ( r ) 1 ρv 1 s V dv ds 4πε + r-r 4πε r-r V S Compare this expression to the previous integral involving the P r. It is evident that the two expressions polarization vector ( ) are equal if the following relations are true: vp ( r) ( r) ρ P and ( ) P ( ) ρ r r sp n

9 11/4/4 Polarization Charge istributions.doc 3/6 The subscript p (e.g., ρ, ρ ) indicates that these functions vp sp represent equivalent charge densities due to the due to the dipoles created in the dielectric. In other words, the electric potential field V (and thus electric field ) created b the dipoles in the dielectric (i.e., P ) is indistinguishable from the electric potential field created b the equivalent charge densities ρ and ρ ( r)! vp sp For example, consider a dielectric material immersed in an P r is: electric field, such that its polarization vector ( ) P C m ( r) 3 z z

10 11/4/4 Polarization Charge istributions.doc 4/6 Note since the polarization vector is a constant, the equivalent volume charge densit is zero: ρ vp ( r) P ( r) 3ˆ a z On the top surface of the dielectric ( surface charge is: n ), the equivalent z ( r) P ( r) ρ sp 3 3 On the bottom of the dielectric ( surface charge is: z z n n C m ), the equivalent z ( r) P ( r) ρ sp 3 3 z n z C m On the sides of the dielectric material, the surface charge is zero, since. z n

11 11/4/4 Polarization Charge istributions.doc 5/6 This result actuall makes phsical sense! Note at the top of dielectric, there is a laer of positive charge, and at the bottom, there is a laer of negative charge. ρ sp 3 C/m z ρ vp ρ 3 C/m sp In the middle of the dielectric, there are positive charge laers on top of negative charge laers. The two add together and cancel each other, so that equivalent volume charge densit is zero. Finall, recall that there is no perfect dielectric, all materials σ r. will have some non-zero conductivit ( ) As a result, we find that the total charge densit within some material is the sum of the polarization charge densit and the free charge (i.e., conducting charge) densit:

12 11/4/4 Polarization Charge istributions.doc 6/6 ( ) ( ) + ( ) ρ ρ ρ vt r v r vp r Where: ρ vt ρ ρ v vp total charge densit free charge densit polarization charge densit This is likewise (as well as more frequentl!) true for surface charge densit: ( ) ( ) + ( ) ρ ρ ρ st r s r sp r

13 11/4/4 lectric Flux ensit.doc 1/5 lectric Flux ensit Yikes! Things have gotten complicated! In free space, we found that charge v field. Prett simple! ρ v ρ creates an electric But, if dielectric material is present, we find that charge ρ v creates an initial electric field i. This electric field in turn polarizes the material, forming bound charge ρ. This bound charge, however, then creates its own electric field s (sometimes called a secondar field), which modifies the initial electric field! Not so simple! ρ ρ v i vp vp s The total electric field created b free charge when dielectric r r + r. material is present is thus ( ) ( ) ( ) i s Q: Isn t there some easier wa to account for the effect of dielectric material?? A: Yes there is! We use the concept of dielectric permittivit, and a new vector field called the r. electric flux densit ( )

14 11/4/4 lectric Flux ensit.doc /5 To see how this works, first consider the point form of Gauss s Law: ρ ( r vt ) ε where ( r vt ) free charge densit ρ and bound charge densit ρ is the total charge densit, consisting of both the v vt ( r) ( r) + ( r) ρ ρ ρ v vp ρ : vp Therefore, we can write Gauss s Law as: ( ) ( ) ( ) ε r ρ r r v + ρvp Recall the bound charge densit is equal to: vp ( r) ( r) ρ P Inserting into the above equation: And rearranging: ( ) ( ) ( ) ε r ρ r r v P ( r) P( r) ρv ( r) ( r) P( r) ρ ( r) ε + ε + v

15 11/4/4 lectric Flux ensit.doc 3/5 Note this final result sas that the divergence of vector field ε ( r ) + P ( r) is equal to the free charge densit ρ v. Let s r : define this vector field the electric flux densit ( ) electric flux densit ε + P C m Therefore, we can write a new form of Gauss s Law: ( r) ( r) ρ v This equation sas that the electric flux densit diverges from free charge ρ v flux densit is free charge ρ --and free charge onl!. In other words, the source of electric v * The electric field is created b both free charge and bound charge within the dielectric material. * However, the electric flux densit is created b free charge onl the bound charge within the dielectric material makes no difference with regard to r! ( )

16 11/4/4 lectric Flux ensit.doc 4/5 But wait! We can simplif this further. Recall that the polarization vector is related to electric field b susceptibilit χ r : e ( ) ( r) ( r) ( r) P ε χ e Therefore the electric flux densit is: ( r) ε ( r) + ε χ e ( r) ( r) ε 1+ χ r ( ) ( ) e We can further simplif this b defining the permittivit of the medium (the dielectric material): ( ) ( ) + ( ) permittivit ε r ε 1 χ e r And can further define relative permittivit: ε relative permittivit εr 1 + χe r ε ( ) Note therefore that ε ( r) εr ( r) ε.

17 11/4/4 lectric Flux ensit.doc 5/5 We can thus write a simple relationship between electric flux densit and electric field: ( r) ε ( r) ( r) ε ε ( r) ( r) r Like conductivit σ, permittivit ε is a fundamental material parameter. Also like conductivit, it relates the electric field to another vector field. Thus, we have an alternative wa to view electrostatics: 1. Free charge ρ creates electric flux densit v.. The electric field can be then determined b simpl r ε r (i.e., dividing ( ) b the material permittivit ( ) ( r) ( r) ε ( r) ). ρ v

18 11/4/4 lectrostatic Field quations in ielectrics.doc 1/ lectrostatic Field quations in ielectrics The electrostatic equations for fields in dielectric materials are: ( ) x r ( r) ( r) ρ v ( r) ε ( r) ( r) In integral form, these equations are: S C ( ) r d ds Q enc ( r) ε ( r) ( r)

19 11/4/4 lectrostatic Field quations in ielectrics.doc / Likewise, for free charge located in some homogeneous (i.e., constant) material with permittivit ε, we get the following relations: Q r-r r (for point charge Q) 4πε r-r ( ) V ( r ) 1 ρv 4πε r-r V dv ( ) ρv r V In other words, for homogenous materials, replace ε (the permittivit of free-space) with the more general permittivit value ε. ε Prett simple! For example: If the media is free-space, use the permittivit of freespace. If the media is, for example, plastic, then use the permittivit of plastic.

20 11/4/4 ielectric Boundar Conditions.doc 1/4 ielectric Boundar Conditions Consider the interface between two dissimilar dielectric regions: Ε, ( r) 1 1 ε 1 Ε, ( r) ε Sa that an electric field is present in both regions, thus r ε r ). producing also an electric flux densit ( ( ) ( ) Q: How are the fields in dielectric region 1 (i.e., Ε r, r ) related to the fields in region (i.e., ( ) ( ), ( r) 1 1 Ε )? A: The must satisf the dielectric boundar conditions!

21 11/4/4 ielectric Boundar Conditions.doc /4 First, let s write the fields at the dielectric interface in terms r r ) vector of their normal ( ( )) and tangential ( ( ) components: n t Ε1 n ( r) ( r) + ( r) Ε Ε Ε 1 1t 1n ε 1 n Ε n Ε1 t Ε t ( r) ( r) + ( r) Ε Ε Ε t n ε Our first boundar condition states that the tangential component of the electric field is continuous across a boundar. In other words: 1t b t b where r b denotes an point on the boundar (e.g., dielectric interface).

22 11/4/4 ielectric Boundar Conditions.doc 3/4 The tangential component of the electric field at one side of the dielectric boundar is equal to the tangential component at the other side! We can likewise consider the electric flux densities on the dielectric interface in terms of their normal and tangential components: 1 n ( r) ε Ε ( r) ε 1 n n 1 t t ( r) ε Ε ( r) ε The second dielectric boundar condition states that the normal vector component of the electric flux densit is continuous across the dielectric boundar. In other words: 1n b n b where r b denotes an point on the dielectric boundar (i.e., dielectric interface).

23 11/4/4 ielectric Boundar Conditions.doc 4/4 Since ( r) ε Ε ( r) expressed as:, these boundar conditions can likewise be 1t b t b 1t b t b ε ε 1 and as: ε 1n b n b ε 1 1n b n b MAK SUR YOU UNRSTAN THIS: These boundar conditions describe the relationships of the vector fields at the dielectric interface onl (i.e., at points r r b )!!!! The sa nothing about the value of the fields at points above or below the interface.

24 11/4/4 Boundar Conditions on Perfect Conductors.doc 1/5 Boundar Conditions on Perfect Conductors Consider the case where region is a perfect conductor: Ε 1 ε 1 n ( ) Ε r σ (i.e., perfect conductor) Recall ( r) that both the tangential and normal component of ( ) equal to zero: in a perfect conductor. This of course means ( r) ( r) t n are also And, since the tangential component of the electric field is continuous across the boundar, we find that at the interface: 1t b t b r

25 11/4/4 Boundar Conditions on Perfect Conductors.doc /5 Think about what this means! The tangential vector component in the dielectric (at the dielectric/conductor boundar) is zero. Therefore, the electric field at the boundar onl has a normal component: 1 b 1n b Therefore, we can sa: The electric field on the surface of a perfect conductor is orthogonal (i.e., normal) to the conductor. Q1: What about the electric flux densit 1? A1: The relation ( r) ( r) ε is still of course valid, so that the electric flux densit at the surface of the conductor must also be orthogonal to the conductor. Q: But, we learned that the normal component of the electric flux densit is continuous across an interface. If n, wh isn t 1n ( r b)? A: Great question! The answer comes from a more general form of the boundar condition.

26 11/4/4 Boundar Conditions on Perfect Conductors.doc 3/5 Consider again the interface of two dissimilar dielectrics. This time, however, there is some surface charge distribution ρ r (i.e., free charge!) at the dielectric interface: s ( ) b Ε, ( r) 1 1 ε 1 n ρ s b Ε, ( r) ε The boundar condition for this situation turns out to be: ( ) ( ) ρ ( ) ( ) ( ) ρ ( ) ˆ n 1 n b n b s b a r r r r r r 1n b n b s b where ˆ a is the scalar component of n b n n b the units of each side are C/m!). (note Note that if ρ s ( rb), this boundar condition returns (both phsicall and mathematicall) to the case studied earlier the normal component of the electric flux densit is continuous across the interface. n b

27 11/4/4 Boundar Conditions on Perfect Conductors.doc 4/5 This more general boundar condition is useful for the dielectric/conductor interface. Since in the conductor, we find that: ( ) ( ) ρ ( ) ˆ an 1n ( rb) ρs ( rb) ρ ˆ an 1 n rb n rb s rb 1n b s b In other words, the normal component of the electric flux densit at the conductor surface is equal to the charge densit on the conductor surface. Note in a perfect conductor, there is plent of free charge available to form this charge densit! Therefore, we find in general that 1 at the surface of a conductor. n 1 b ε 1 n ρs ( rb) ( ) r σ (i.e., perfect conductor)

28 11/4/4 Boundar Conditions on Perfect Conductors.doc 5/5 Summarizing, the boundar conditions for the tangential components field components at a dielectric/conductor interface are: 1t 1t b b but for the normal field components: ( ) ρ ( ) r r 1n b s b 1n b ρ s ε 1 b Again, these boundar conditions describe the fields at the conductor/dielectric interface. The sa nothing about the value of the fields at locations above this interface.

29 11/4/4 xample Boundar Conditions.doc 1/1 xample: Boundar Conditions Two slabs of dissimilar dielectric material share a common boundar, as shown below. It is known that the electric field in the lower dielectric region is: x + 6 V m and it is known that the electric field in the top region is likewise some constant field: ( ) r + 1 x1 x 1 V m ε 6ε 1 ( ) r + 1 x1 x 1 x ε 3ε ( ) r x + 6

30 11/4/4 xample Boundar Conditions.doc /1 In each dielectric region, let s determine (in terms of ε ): 1) the electric field ) the electric flux densit 3) the bound volume charge densit (i.e., the equivalent polarization charge densit) within the dielectric. 4) the bound surface charge densit (i.e., the equivalent polarization charge densit) at the dielectric interface Since we alread know the electric field in the region, let s evaluate region first. We can easil determine the electric flux densit within the region: ε ( ˆ x a ) 3ε + 6 6ε + 18ε x C m Likewise, the polarization vector within the region is: P ε χ e ( ε 1) ( ˆ r a 6ˆ x a ) ( 3 1) ( x 6 ) ε + ε + 4ε + 1ε x C m

31 11/4/4 xample Boundar Conditions.doc 3/1 Q: Wh did we determine the polarization vector? It is not one of the quantities this problem asked for! A: True! But the problem did ask for the equivalent bound charge densities (both volume and surface) within the P r to find dielectric. We need to know polarization vector ( ) this bound charge! Recall the bound volume charge densit is: vp ( r) ( r) ρ P and the bound surface charge densit is: ( ) P ( ) ρ r r sp Since the polarization vector P is a constant (i.e., it has precisel the same magnitude and direction at ever point with region ), we find that the divergence of P is zero, and thus the volume bound charge densit is zero within the region: ρ ( r) P ( r) vp ( 4ε 1 x ε ˆ a ) + C m 3 However, we find that the surface bound charge densit is not zero! n

32 11/4/4 xample Boundar Conditions.doc 4/1 Note that the unit vector normal to the surface of the bottom dielectric slab is ˆ a n aˆ : ˆ a n a ˆ x Since the polarization vector is constant, we know that its value at the dielectric interface is likewise equal to 4ε 1 x + ε ˆ a. Thus, the equivalent polarization (i.e., bound) surface charge densit on the top of region (at the dielectric interface) is ρ ( ) P ( ) r r ˆ a sp b b n ( 4ε 1ε ) ˆ a + ˆ a ˆ a 1ε x C m Now, let s determine these same quantities for region 1 (i.e., the top dielectric slab). Q1: How the heck can we do this? We don t know anthing about the fields in region 1! A1: True! We don t know 1 or 1 or even 1 However, we know the next best thing we know ( r ) ( ) and even P ( )! r r P. and

33 11/4/4 xample Boundar Conditions.doc 5/1 Q: Huh!?! A: We can use boundar conditions to transfer our solutions from region into region 1! First, we note that at the dielectric interface, the vector components of the electric fields tangential to the interface r a r a : are ( ) and ( ) ˆ 1t b 1x x ˆ t b x ( ) ˆ 1t rb 1x ax ˆ t b ax x Thus, appling the boundar condition 1t b t b ˆ a ˆ a 1x x x ˆ a ˆ a ˆ a ˆ a 1x x x x x 1x, we find: 1t b t b Likewise, we note that at the dielectric interface, the vector components of the electric fields normal to the interface are r a r 6 a : ( ) and ( ) ˆ 1n b 1 ˆ n b

34 11/4/4 xample Boundar Conditions.doc 6/1 ( ) ˆ 1n rb 1 a ˆ n b 6 a x Here, we can appl a second boundar condition, ε r ε r : ( ) ( ) 1 1n b n b ε ε 1 1n b n b 6ε ˆ a 3ε 6 ˆ a 1 ˆ a 3 ˆ a 1 ˆ a ˆ a 3 ˆ a ˆ a Thus, we have concluded using boundar conditions that x1 and 1 3, or the electric field in the top region is: ( ) 1 r x + 3 V m Likewise, we can find the electric flux densit b multipling b the permittivit of region 1 ( ε 1 6ε ): ε ε + 18ε x C m

35 11/4/4 xample Boundar Conditions.doc 7/1 Note we could have solved this problem another wa! Instead of appling boundar conditions to applied them to electric flux densit : ε ε C m x, we could have We know that the electric flux densit within region 1 must be constant, i.e.: 1 ˆ C x1a ˆ x + 1a m 1 r r at the interface are related b the boundar conditions: and that the vector fields ( ) and ( ) and 1t b t b ε ε 1 1n b n b It is evident that for this problem: ( ) r 1t b x1 x ( ) r 1n b 1 and for region :

36 11/4/4 xample Boundar Conditions.doc 8/1 1ε t b 18ε n b x Combining the results, we find the two boundar conditions are: and: 1t b t b ε 1 ˆ ˆ 1 a 6εa 6ε 3ε x x x ˆ a 1ε ˆ a 1x x x ˆ a ˆ a 1ε ˆ a ˆ a 1x x x x x ε 1ε 1x 1n b n b ˆ a 18ε ˆ a 1 ˆ a ˆ a 18ε ˆ a ˆ a 1 18ε 1 Therefore, we find that the electric flux densit is: ε ε C m x Precisel the same result as before!

37 11/4/4 xample Boundar Conditions.doc 9/1 Likewise, we can find the electric field in region 1 b dividing b the dielectric permittivit: 1 1 ε 1 1ε + 18ε x 6ε + 3 x V m Again, the same result as before! Now, finishing this problem, we need to find the polarization vector P 1 : P r ε ε 1 r ( ) ( ) ( ) 1 r 1 1 ( 6 1)( ˆ a 3 ˆ x a ) ε + 1ε ˆ a + 15ε x ˆ a C m Thus, the volume charge densit of bound charge is again zero: ρ ( r) P ( r) vp1 1 ( 1 ε 15 x ε ˆ a ) + However, we again find that the surface bound charge densit is not zero!

38 11/4/4 xample Boundar Conditions.doc 1/1 Note that the unit vector normal to the bottom surface of the top dielectric slab points downward, i.e., ˆ a n 1 aˆ : ˆ a n 1 a ˆ x Since the polarization vector is constant, we know that its value at the dielectric interface is likewise equal to 1ε 15 x + ε ˆ a. Thus, the equivalent polarization (i.e., bound) surface charge densit on the bottom of region 1 (at the dielectric interface) is: ρ r P r ˆ a ( ) ( ) sp1 b 1 b n 1 ( 1ε ˆ ˆ ) ( ˆ ax 15εa a ) + 15ε C m Now, we can determine the net surface charge densit of bound charge that is ling on the dielectric interface: + ρ ρ ρ sp b sp1 b sp b 15ε + 1ε 3ε C m