5-4 Electrostatic Boundary Value Problems

Transcription

1 11/8/4 Section 54 Electrostatic Bounary Value Problems blank 1/ 5-4 Electrostatic Bounary Value Problems Reaing Assignment: pp Q: A: We must solve ifferential equations, an apply bounary conitions to fin a unique solution. In EE an CoE, we typically use a voltage source to apply bounary conitions on electric potential function V ( r ). This process is best emonstrate with a series of examples: Example: Dielectric Fille Parallel Plates

2 11/8/4 Section 54 Electrostatic Bounary Value Problems blank / Example: Charge Fille Parallel Plates Example: The Electrostatic Fiels of a Coaxial Line

3 11/8/4 Example Dielectric Fille Parallel Plates 1/8 Example: Dielectric Fille Parallel Plates Consier two infinite, parallel conucting plates, space a istance apart. The region between the plates is fille with a ielectric ε. Say a voltage V is place across these plates. z=- + V ε z= z Q: What electric potential fiel V ( r ), electric fiel E ( r ) an charge ensity ( r ) is prouce by this situation? s A: We must solve a bounary value problem! We must fin solutions that: a) Satisfy the ifferential equations of electrostatics (e.g., Poisson s, Gauss s). b) Satisfy the electrostatic bounary conitions.

4 11/8/4 Example Dielectric Fille Parallel Plates /8 Q: Yikes! Where o we even start? A: We might start with the electric potential fiel V ( r ), since it is a scalar fiel. a) The electric potential function must satisfy Poisson s equation: ( r ) v r V = b) It must also satisfy the bounary conitions: V z = = V V z = = Consier first the ielectric region ( < z < ). Since the region is a ielectric, there is no free charge, an: ε r = Therefore, Poisson s equation reuces to Laplace s equation: V r = v This problem is greatly simplifie, as it is evient that the solution V ( r ) is inepenent of coorinates x an y. In other wors, the electric potential fiel will be a function of coorinate z only: V r = V z

5 11/8/4 Example Dielectric Fille Parallel Plates 3/8 This make the problem much easier! Laplace s equation becomes: ( z) ( z) V = r V = V z Integrating both sies of Laplace s equation, we get: = V ( z) z = z z V ( z) = C1 z An integrating again we fin: V ( z) z = C z 1 z V ( z) = C z + C 1 We fin that the equation V ( z) = C z + C will satisfy Laplace s 1 equation (try it!). We must now apply the bounary conitions to etermine the value of constants C 1 an C. We know that the value of the electrostatic potential at every point on the top (z =-) plate is V (-)=V, while the electric potential on the bottom plate (z =) is zero (V () = ). Therefore:

6 11/8/4 Example Dielectric Fille Parallel Plates 4/8 V z = = C + C = V 1 V z = = C + C = 1 Two equations an two unknowns (C 1 an C )! Solving for C 1 an C we get: V C = an C = 1 an therefore, the electric potential fiel within the ielectric is foun to be: Vz V z ( r ) = ( ) Before we procee, let s o a sanity check! In other wors, let s evaluate our answer at z = an z = -, to make sure our result is correct: an ( ) V = = = V z V V ( z ) V = = =

7 11/8/4 Example Dielectric Fille Parallel Plates 5/8 Now, we can fin the electric fiel within the ielectric by taking the graient of our result: E V ( r) = V ( r ) = ˆa ( z ) z An thus we can easily etermine the electric flux ensity by multiplying by the ielectric of the material: εv D E ˆ a z z ( r ) = ε ( r ) = ( ) Finally, we nee to etermine the charge ensity that actually create these fiels! Q: Charge ensity!?! I thought that we alreay r is equal to zero? etermine that the charge ensity v A: We know that the free charge ensity within the ielectric is zero but there must be charge somewhere, otherwise there woul be no fiels!

8 11/8/4 Example Dielectric Fille Parallel Plates 6/8 Recall that we foun that at a conuctor/ielectric interface, the surface charge ensity on the conuctor is relate to the electric flux ensity in the ielectric as: ( r) ( r) D = ˆa D = n n s First, we fin that the electric flux ensity on the bottom surface of the top conuctor (i.e., at z = ) is: V V D ε ε ˆ ˆ = a = r z a z= z z= For every point on bottom surface of the top conuctor, we fin that the unit vector normal to the conuctor is: ˆa n = ˆa z Therefore, we fin that the surface charge ensity on the bottom surface of the top conuctor is: ( r) = ˆa D ( r) s + n z = εv = ˆaz ˆaz εv ( z ) = =

9 11/8/4 Example Dielectric Fille Parallel Plates 7/8 Likewise, we fin the unit vector normal to the top surface of the bottom conuctor is (o you see why): ˆa n = ˆa z Therefore, evaluating the electric flux ensity on the top surface of the bottom conuctor (i.e., z = ), we fin: ( r) = ˆa D ( r) s n z = εv = ˆaz ˆaz εv ( z ) = = We shoul note several things about these solutions: 1) E x r = ) D V r = an r = 3) ( r ) an ( r) D E are normal to the surface of the conuctor (i.e., their tangential components are equal to zero). 4) The electric fiel is precisely the same as that given by using superposition an eq. 4. in section 4-5!

10 11/8/4 Example Dielectric Fille Parallel Plates 8/8 I.E.: E ε ε s+ s ( r ) = ˆa ˆa = ˆa ( < z < ) V z z z In other wors, the fiels ( r ), ( r ), an V ( r) attributable to charge ensities ( r ) an ( r ) E D are s+ s. s + ( r ) z=- V + E ( r ) ε s ( r ) z= z

11 11/8/4 Example Charge Fille Parallel Plates 1/4 Example: Charge Fille Parallel Plates Consier now a problem similar to the previous example (i.e., ielectric fille parallel plates), with the exception that the space between the infinite, conucting parallel plates is fille with free charge, with a ensity: v r = z ε (- < z < ) z=- + V v ( r ) z= z Q: How o we etermine the fiels within the parallel plates for this problem? A: Same as before! However, since the charge ensity between the plates is not equal to zero, we recognize that the electric potential fiel must satisfy Poisson s equation: v r V ( r ) ε =

12 11/8/4 Example Charge Fille Parallel Plates /4 For the specific charge ensity ( r ) v = z v r V ( r ) ε = = z ε : Since both the charge ensity an the plate geometry are inepenent of coorinates x an y, we know the electric potential fiel will be a function of coorinate z only (i.e., V r = V z ). Therefore, Poisson s equation becomes: V ( z) V ( z) = = We can solve this ifferential equation by first integrating both sies: V ( z) z = z z z V ( z) z = + C1 z z An then integrating a secon time: z ( r ) V z z = + C z 1 z 3 z V ( r ) = + C z + C 1 6

13 11/8/4 Example Charge Fille Parallel Plates 3/4 Note that this expression for V ( r ) satisfies Poisson s equation for this case. The question remains, however: what are the values of constants C an C? 1 We fin them in the same manner as before bounary conitions! Note the bounary conitions for this problem are: V z = = V V ( z ) = = Therefore, we can construct two equations with two unknowns: ( ) 3 3 V z = = V = + C + C 1 6 V z = = = + C + C 1 6 It is evient that C =, therefore constant C 1 is: C 1 V = + 6 The electric potential fiel between the two plates is therefore:

14 11/8/4 Example Charge Fille Parallel Plates 4/4 3 z V V z z 6 6 ( r ) = + ( < < ) Performing our sanity check, we fin: an ( ) 3 V 6 6 V z = - = = + V + = V 3 V V ( z = ) = = + + = From this result, we can etermine the electric fiel E ( r ), the electric flux ensity D ( r ), an the surface charge ensity s ( r ), as before. Note, however, that the permittivity of the material between the plates is ε, as the ielectric between the plates is freespace.

15 11/8/4 Example The Electorostatic Fiels of a Coaxial Line 1/1 Example: The Electrostatic Fiels of a Coaxial Line A common form of a transmission line is the coaxial cable. Outer Conuctor a b ε + V - Inner Conuctor Coax Cross-Section The coax has an outer iameter b, an an inner iameter a. The space between the conuctors is fille with ielectric material of permittivity ε. Say a voltage V is place across the conuctors, such that the electric potential of the outer conuctor is zero, an the electric potential of the inner conuctor is V.

16 11/8/4 Example The Electorostatic Fiels of a Coaxial Line /1 The potential ifference between the inner an outer conuctor is therefore V = V volts. Q: What electric potential fiel V ( r ), electric fiel E ( r ) an charge ensity ( r ) is prouce by this situation? s A: We must solve a bounary-value problem! We must fin solutions that: a) Satisfy the ifferential equations of electrostatics (e.g., Poisson s, Gauss s). b) Satisfy the electrostatic bounary conitions. Yikes! Where o we start? We might start with the electric potential fiel V ( r ), since it is a scalar fiel. a) The electric potential function must satisfy Poisson s equation: ( r ) v r V = b) It must also satisfy the bounary conitions: ( ) ( ) V = a = V V = b = ε

17 11/8/4 Example The Electorostatic Fiels of a Coaxial Line 3/1 Consier first the ielectric region (a < < b ). Since the region is a ielectric, there is no free charge, an: r = v Therefore, Poisson s equation reuces to Laplace s equation: V = r This particular problem (i.e., coaxial line) is irectly solvable because the structure is cylinrically symmetric. Rotating the coax aroun the z-axis (i.e., in the â φ irection) oes not change the geometry at all. As a result, we know that the electric potential fiel is a function of only! I.E.,: ( r ) = V V This make the problem much easier. Laplace s equation becomes: Be very careful uring this step! Make sure you implement the gul urn Laplacian operator correctly. ( ) ( ) V = r V = 1 V + + = V ( ) =

18 11/8/4 Example The Electorostatic Fiels of a Coaxial Line 4/1 Integrating both sies of the resulting equation, we fin: ( ) V = V ( ) = C1 where C 1 is some constant. Rearranging the above equation, we fin: V ( ) C1 = Integrating both sies again, we get: ( ) V C1 = p V = C + C ln[ ] 1 We fin that this final equation ( ( ) ln [ ] V = C + C ) will 1 satisfy Laplace s equation (try it!). We must now apply the bounary conitions to etermine the value of constants C 1 an C. * We know that on the outer surface of the inner conuctor (i.e., = a ), the electric potential is equal to V (i.e., V ( = a) = V ).

19 11/8/4 Example The Electorostatic Fiels of a Coaxial Line 5/1 * An, we know that on the inner surface of the outer conuctor (i.e., = b ) the electric potential is equal to zero (i.e., V ( b ) = = ). Therefore, we can write: V ( = a) = C ln [ a] + C = V ( ) [ ] 1 V = b = C ln b + C = 1 Two equations an two unknowns (C 1 an C )! Solving for C 1 an C we get: C C 1 V V = = ln [ b] ln [ a] ln b/a [ ] V ln b = ln b/a an therefore, the electric potential fiel within the ielectric is foun to be: [ ] V [ ] V V ( r ) = + b > > a ln b/a ln b/a ln ln b ( )

20 11/8/4 Example The Electorostatic Fiels of a Coaxial Line 6/1 Before we move on, we shoul o a sanity check to make sure we have one everything correctly. Evaluating our result at = a, we get: V ( a ) V = = + V = [ ] V ln [ b] ln a ln b/a ln b/a ( ln [ b] ln [ a] ) ln b/a ( ln b/a ) V = ln b/a = V Likewise, we evaluate our result at = b: V ( b ) Our result is correct! [ ] V [ ] V ln b ln b = = + ln b/a ln b/a ( ln [ b] ln [ b] ) V = ln b/a = Now, we can etermine the electric fiel within the ielectric by taking the graient of the electric potential fiel: V 1 E ˆ ln b/a ( r) = V ( r ) = a ( b > > a)

21 11/8/4 Example The Electorostatic Fiels of a Coaxial Line 7/1 Note that electric flux ensity is therefore: D εv ( r) = ε E( r ) = ˆa ( b > > a) 1 ln b/a Finally, we nee to etermine the charge ensity that actually create these fiels! Q1: Just where is this charge? After all, the ielectric (if it is perfect) will contain no free charge. A1: The free charge, as we might expect, is in the conuctors. Specifically, the charge is locate at the surface of the conuctor. Q: Just how o we etermine this surface r? charge s A: Apply the bounary conitions! Recall that we foun that at a conuctor/ielectric interface, the surface charge ensity on the conuctor is relate to the electric flux ensity in the ielectric as: ( r) ( r) D = ˆa D = n n s

22 11/8/4 Example The Electorostatic Fiels of a Coaxial Line 8/1 First, we fin that the electric flux ensity on the surface of the inner conuctor (i.e., at = a ) is: D ( r ) = ˆa = a εv 1 ln b/a εv 1 = ˆa ln b/a a For every point on outer surface of the inner conuctor, we fin that the unit vector normal to the conuctor is: = a ˆa n = ˆ a Therefore, we fin that the surface charge ensity on the outer surface of the inner conuctor is: sa ( r) = ˆa D ( r) n = a εv 1 = ˆa ˆa ln b/a a εv 1 ln b/a a ( a) = =

23 11/8/4 Example The Electorostatic Fiels of a Coaxial Line 9/1 Likewise, we fin the unit vector normal to the inner surface of the outer conuctor is (o you see why?): ˆa n = ˆ a Therefore, evaluating the electric flux ensity on the inner surface of the outer conuctor (i.e., = b ), we fin: sb ( r) = ˆa D ( r) n = b εv 1 = ˆa ˆa ln b/a b εv 1 ln b/a b ( b) = = Note the charge on the outer conuctor is negative, while that of the inner conuctor is positive. Hence, the electric fiel points from the inner conuctor to the outer E ( r )

24 11/8/4 Example The Electorostatic Fiels of a Coaxial Line 1/1 We shoul note several things about these solutions: 1) E x r = ) D V r = an r = 3) ( r ) an ( r) D E are normal to the surface of the conuctor (i.e., their tangential components are equal to zero). 4) The electric fiel is precisely the same as that given by eq in section 4-5! E a ε V ( r ) = sa ˆa = ˆa ( b > > a) 1 ln b/a In other wors, the fiels E( r ), D ( r ), an V ( r) are attributable to free charge ensities ( r ) an ( r). sa sb