SI Units Coulomb s Law Gauss Law Voltage & Energy Poisson s Equation Capacitance Boundary Conditions MoI Summary Problems.

Transcription

1 S. R. Zinka School of Electronics Engineering Vellore Institute of Technology October 18, 2012

2 Outline 1 SI Units 2 Coulomb s Law 3 Gauss Law 4 Voltage & Energy 5 Poisson s Equation 6 Capacitance 7 Boundary Conditions 8 MoI 9 Summary 10 Problems

3 Outline 1 SI Units 2 Coulomb s Law 3 Gauss Law 4 Voltage & Energy 5 Poisson s Equation 6 Capacitance 7 Boundary Conditions 8 MoI 9 Summary 10 Problems

4 SI Base Units Unit Name Unit Symbol Quantity Name Quantity Symbol meter m length l kilogram kg mass m second s time t ampere A electric current I kelvin K thermodynamic temperature T candela cd luminous intensity I v mole mol amount of substance n

5 A Few Named Units Derived from SI Base Units Name Symbol Quantity Expression in terms of SI base units couloumb C electric charge s.a farad F electric capacitance kg 1.m 2.s 4.A 2 henry H inductance kg.m 2.s 2.A 2 hertz Hz frequency s 1 joule J energy, work, heat kg.m 2.s 2 newton N force, weight kg.m.s 2 ohm Ω electric resistance, impedance, reactance kg.m 2.s 3.A 2 radian rad angle dimensionless siemens S electrical conductance, susceptance kg 1.m 2.s 3.A 2 steradian sr solid angle dimensionless tesla T magnetic field, magnetic flux density kg.s 2.A 1 volt V voltage, electric potential, electromotive force kg.m 2.s 3.A 1 watt W power, radiant flux kg.m 2.s 3 weber Wb magnetic flux kg.m 2.s 2.A 1

6 Outline 1 SI Units 2 Coulomb s Law 3 Gauss Law 4 Voltage & Energy 5 Poisson s Equation 6 Capacitance 7 Boundary Conditions 8 MoI 9 Summary 10 Problems

7 First of all... One Important Vector Calculus Identity Let s find out the gradient for the following equation: ( ) ( ) 1 1 = r rˆr From the definition of gradient in spherical coordinate system, ( ) 1 w w = ˆq i h i i q i we get, ( ) 1 = r r ( ) 1 ˆr = 1 r r 2 ˆr = 1 r 2 r r = r r 3. (1)

8 Coulomb s Law - Statement The force F ir,stat (ir stands for irrotational) acting on the charge q located at r, due to the presence of the charge q 1 located at r 1 in an otherwise empty space, is given as F ir,stat ( r) = qq 1 r r 1 4πε 0 r r 1 3 = qq ( ) 1 1 4πε 0 r r 1 (2) where vacuum primittivity, ε 0 = 10 7 / ( 4πc 2) Fm 1 (SI unit).

9 Electric Field (Intensity) - Definition For many purposes it is useful to introduce the concept of electric field, where electric field E ir,stat is defined by the limiting process F E ir,stat ir,stat lim = q 1 r r 1 q 0 q 4πε 0 r r 1 3 (3) where F ir,stat is the electric force, as defined in equation (2), due to a net electric charge q 1 on the test particle with a small electric net electric charge q. Note that F ir,stat and thus E ir,stat are singular at r = r.

10 Electric Field Due to a Point Source - I...Positive Source... Negative Source

11 Electric Field Due to a Point Source - II...Positive Source... Negative Source

12 Electric Field Due to Several Discrete Charges In the presence of several discrete electric charges q i, located at the points r i, i = 1, 2, 3,..., respectively, in an otherwise empty space, the assumption of linearity of vacuum allows us to superimpose their individual electrostatic fields into a total electrostatic field E ir,stat total ir,stat ( Fi lim = 1 q 0 q 4πε i 0 i q i r r i r r i 3 ). (4)

13 Electric Field Due to a Pair of Charges...Electric Field... Magnitude of Electric Field

14 Electric Field Due to a Pair of Charges...Electric Field... Magnitude of Electric Field

15 Electric Field Due to a Pair of Charges...Opposite polarities... Same polarities

16 Electric Field Due to Continuous Charge Distribution de ir,stat = E ir,stat ir,stat = de V dq r r 4πε 0 r r 3 = ( ) ρ e,vol r dv r r 4πε 0 r r 1 V r r ( = 4πε 0 r r 3 ρ e,vol r ) dv (5) ( = 1 V ρ e,vol r ) 4πε 0 r r dv (6) 3

17 Why E produced due to electric charges is always irrotational?

18 Since [ α ( r)] 0 for any R 3 scalar field, we immediately find that in electrostatics E ir,stat = 1 4πε 0 = 1 4πε 0 ( ˆV ρ e,vol r ) r r dv ( ˆV ρ e,vol r ) dv r r = 0 (7) i.e., that E ir,stat is an irrotational (also known as curl-free / lamellar / conservative / gradient) field.

19 Irrotational & Rotational Fields - How do they look like?

20 y x...rotational Field... Corresponding Curl

21 y x...rotational Field... Corresponding Curl

22 Now... What About These Fields?...Point charge... A pair of charges

23 Dirac Delta Function - Heuristic Description The Dirac delta can be loosely thought of as a function on the real line which is zero everywhere except at the origin, where it is infinite, { +, x = 0 δ (x) = 0, x = 0 and which is also constrained to satisfy the identity ˆ + δ (x) dx = 1.

24 Dirac Delta Function - A Few Properties δ ( x) = δ (x) (Symmetry Property) + δ (αx) dx = + du δ (u) = 1 (Scaling Property) α α + f (x) δ (x x 0) dx = f (x 0 ) (Translation or Sifting Property) δ (x) 1

25 Volume Charge Densities of Point, Line, and Sheet Charges X, Y, Z Source coordinates X, Y, Z Test charge coordinates

26 Electric Field Due to Discrete Charges - Revisited Point Charge: Volume charge density corresponding to a point charge of strength q 1 at r 1 = x 1 ˆx + y 1 ŷ + z 1 ẑ can be represented using the Dirac delta function as, ( ) ( ) ρ e,vol r = q 1 δ r r 1 = q 1 δ (x x 1) δ (y y 1) δ (z z 1). Substituting the above volume charge density into (6) gives which is nothing but (3). E ir,stat = q 1 4πε 0 r r 1 r r 1 3, (8)

27 Electric Field Due to Discrete Charges - Revisited Multiple Point Charges: Volume charge density corresponding to a set of point charges can be represented as, ( ) ρ e,vol r = N i=1 ( ) q i δ r r i. Once again, substituting the above volume charge density into (6) gives which is nothing but (4). ( ) E ir,stat N q = i r r i 4πε i=1 0 r r i 3, (9)

28 Now... Some Special Cases

29 Electric Field Intensity Due to a Line Charge Electric field due to a small differential charge density at r = z ẑ is given as de ir,stat = = ρ L (z ) δ (x ) δ (y ) dx dy dz r r 4πε 0 r r 3 ρ L (z ) δ (x ) δ (y ) dx dy dz 4πε 0 xˆx + yŷ + (z z ) ẑ [ x 2 + y 2 + (z z ) 2] 3. 2 Converting the above equation into cylindrical coordinate system gives de ir,stat = ρ L (z ) δ (x ) δ (y ) dx dy dz 4πε 0 ρ ˆρ + (z z ) ẑ [ ρ 2 + (z z ) 2] 3. 2

30 Electric Field Intensity Due to a Line Charge To get the total electric filed, let s evaluate the volume integral given below: E ir,stat ir,stat = de V = = ρ L (z ) δ (x ) δ (y ) V 4πε 0 ˆ z =+ z = ρ L (z ) 4πε 0 ρ ˆρ + (z z ) ẑ [ ρ 2 + (z z ) 2] 3 dx dy dz 2 ρ ˆρ + (z z ) ẑ [ ρ 2 + (z z ) 2] 3 dz 2 For the simplest case, assuming ρ L (z ) = ρ L (where ρ L is a constant), E ir,stat = = ρ L ˆ z =+ 4πε 0 z = ρ L ˆ z =+ 4πε 0 z = ρ ˆρ [ ρ 2 + (z z ) 2] 3 2 ρ [ ρ 2 + (z z ) 2] 3 2 (z z ) ẑ [ ρ 2 + (z z ) 2] 3 dz 2 }{{} this term is zero ˆ z =+ dz + z = dz ˆρ (10)

31 Electric Field Intensity Due to a Line Charge The previous integral (10) can be calculated as shown below: ˆ z =+ ˆ ρ ξ= z [ = ρ 2 + (z z ) 2] 3 dz 1 = ρ 2 ξ=+ [ρ 2 + ξ 2 ] 3 2 ( dξ), where ξ = z z So, finally, from (10) and (11), E ir,stat total is given as ˆ ξ=+ 1 = ρ ξ= [ρ 2 + ξ 2 ] 3 dξ 2 ˆ τ=0 1 = ρ τ=π [ρ 2 + ρ 2 cot 2 τ] 3 2 = ˆ τ=0 ρ 2 τ=π 1 [ρ 2 csc 2 τ] 2 3 ( csc 2 τdτ ) = 1 ˆ τ=0 sin τdτ = 2 ρ τ=π ρ ( ρ csc 2 τdτ ), where ξ = ρcotτ (11) E ir,stat = ρl ˆρ (12) 2πε 0 ρ

32 Electric Field Intensity Due to a Sheet Charge - I *** Electric field due to a small differential charge density at r = x ˆx + z ẑ is given as de ir,stat = = ρ S (x, z ) δ (y ) dx dy dz r r 4πε 0 r r 3 ρ S (x, z ) δ (y ) dx dy dz 4πε 0 (x x ) ˆx + yŷ + (z z ) ẑ [ (x x ) 2 + y 2 + (z z ) 2] 3. 2

33 Electric Field Intensity Due to a Sheet Charge - I *** To get the total electric filed, let s evaluate the volume integral given below (also, let s assume the simplest case, ρ S (x, z ) = ρ S, where ρ S is a constant): E ir,stat ir,stat = de V = = = V ρ S (x, z ) δ (y ) dx dy dz ˆ z =+ ˆ x =+ z = x = 4πε 0 ρ S 4πε 0 (x x ) ˆx + yŷ + (z z ) ẑ [ (x x ) 2 + y 2 + (z z ) 2] 3 2 (x x ) ˆx + yŷ + (z z ) ẑ [ (x x ) 2 + y 2 + (z z ) 2] 3 dx dz 2 ˆ ˆ ρ S (x x ) ˆx 4πε 0 [ (x x ) 2 + y 2 + (z z ) 2] 3 dx dz 2 } {{ } this term is zero ˆ ˆ ρ + yŷ 4πε 0 [ (x x ) 2 + y 2 + (z z ) 2] 3 dx dz 2 ˆ ˆ ρ + (z z ) ẑ 4πε 0 [ (x x ) 2 + y 2 + (z z ) 2] 3 dx dz 2 } {{ } this term is zero

34 Electric Field Intensity Due to a Sheet Charge - I *** So, E ir,stat is given as E ir,stat = = = = = = ˆ ˆ ( 1 ρ [ (x x ) 2 + y 2 + (z z ) 2] 3 dx dz S ) y ŷ 2 4πε 0 ˆ ˆ x =+ ( 1 x [ ] dx ρ S ) y = (x x ) 2 3 dz ŷ, where Ω 2 = y 2 + (z z ) 2 + Ω 2 2 4πε 0 (ˆ z =+ ( ) ) ( 2 ρ Ω 2 dz S ) y ŷ, from (10) 4πε 0 z = (ˆ z =+ z = (ˆ ξ=+ ξ= [ 1 y tan 1 1 y 2 + (z z ) 2 dz ) ( ρ S y 1 y 2 + ξ 2 dξ ( ξ y 2πε 0 ) ] ξ=+ ( ρ S y ξ= ) ( ρ S y 2πε 0 ) ŷ 2πε 0 ) ŷ, where ξ = z z ) ( π ŷ = 2 + π ) ( ρ S ) ŷ = ρs ŷ. (13) 2 2πε 0 2ε 0

35 Electric Field Intensity Due to a Sheet Charge - II Electric field at point r = xˆx + yŷ due to the strip located at at r = x ˆx is given as de ir,stat strip = = ρ S dx r r r 2πε 0 r r r ( ) ( ) ρ S dx x ˆx + yŷ 2πε 0 x 2 + y 2.

36 Electric Field Intensity Due to a Sheet Charge - II To calculate the total electric filed, let s integrate along x as shown below: E ir,stat = = = = = ˆ x = ( ρ S x = 2πε 0 ( ) ρ S ˆ x = 2πε 0 x = ( ρ S ) (ˆ y x = ) ( ) x ˆx + yŷ x 2 + y 2 dx ˆ yŷ x = x 2 + y 2 dx ) 1 2πε 0 x = x 2 + y 2 dx ŷ ( ρ S ) [ ( y 1 x ) x ] =+ 2πε 0 y tan 1 ŷ y x = ( ρ S 2πε 0 x ˆx x = x 2 + y 2 dx }{{} this term is zero ) πŷ = ρs 2ε 0 ŷ. (14) We can see that both (14) and (13) are the same.

37 Outline 1 SI Units 2 Coulomb s Law 3 Gauss Law 4 Voltage & Energy 5 Poisson s Equation 6 Capacitance 7 Boundary Conditions 8 MoI 9 Summary 10 Problems

38 Faraday s Experiment & Electric Flux Density What Faraday did was, he enclosed the inner sphere of charge +Q with another bigger outer sphere.

39 Faraday s Experiment & Electric Flux Density Then, he discharged the outer sphere by connecting it to the ground momentarily.

40 Faraday s Experiment & Electric Flux Density When measured, outer sphere has a total charge of Q. Now, what does that mean??...

41 Faraday s Experiment & Electric Flux Density Flux: If we assume that there exists a concept called electric flux which represents the transfer of charge from one sphere to another, then the total flux ( ) is equal to Ψ = +Q. Flux Density: And electric flux density at any point r = rˆr is defined as D ir,stat = Q 4πr 2 ˆr.

42 Faraday s Experiment & Electric Flux Density Flux Density: A more general definition of electric flux density at any point r = rˆr due to a continuous source is given as D ir,stat = 1 ( 4π V ρ e,vol r ) r r r r 3 dv (15)

43 Gauss Law - Statement & Physical Interpretation Gauss Law: The electric flux passing through any closed surface is equal to the total charge enclosed by that surface. (ρ e,vol) dv = D ir,stat ds (16) V S Divergence Theorem: V ) ( F dv = F ds S Combining Gauss law and divergence theorem gives D ir,stat = ρ e,vol. (17)

44 Let s Go Back to Line Charge

45 Let s Go Back to Sheet Charge

46 One More Important Vector Calculus Identity *** ( ) 1 2 = 4πδ ( r) r ( ) 1 2 r ( ) 1 = r = ( 1r ) 2 ˆr = 1 [ (r 2 r 2 sin θ 1r )] sin θ r 2 = 0, when r = 0 Since the the term represented in red color has a singularity at r = 0, let us solve this problem from the definition of divergence itself... ( ) 1 2 = ( 1r ) r 2 ˆr = 4πr2 ( 1/r 2) dv [ ( )] [ ] 4πr 2 = lim r 0 r 2 lim dv 1 r 0 = 4πδ ( r) (18)

47 Gauss Law Derivation - From the Definition of Electric Flux Density *** According to Helmholtz s theorem, a well-behaved vector field is completely known if one knows its divergence and curl. Taking the divergence of (5) gives ( V ρ e,vol r ) r r dv ( = 1 V ρ r ) e,vol 4π 2 r r dv = 1 4π V 2 1 ( r r ρ e,vol r ( ) ( ) = δ r r ρ V e,vol r dv = δ V D ir,stat = 1 4π ) dv ( ( ) r r) ρ e,vol r dv = ρ e,vol ( r). (19)

48 Outline 1 SI Units 2 Coulomb s Law 3 Gauss Law 4 Voltage & Energy 5 Poisson s Equation 6 Capacitance 7 Boundary Conditions 8 MoI 9 Summary 10 Problems

49 Electrostatic Scalar Potential - Definition Since E ir,stat = 0, E ir,stat can be represented as E ir,stat = V stat (20) where V stat is usually defined as electrostatic scalar potential. Now, comparing (21) and (6) gives V stat = 1 4πε 0 ( ˆV ρ e,vol r ) r r dv (21)

50 The Gradient Theorem A line integral through a gradient or conservative vector field can be evaluated by evaluating the original scalar field at the endpoints of the curve: ˆ γ[ p, q] φ ( r) dr = φ ( q) φ ( p) (22) The gradient theorem implies that line integrals through irrotational vector fields are path independent.

51 Electrostatic Scalar Potential - Physical Interpretation Energy needed to bring a unit charge from to r in the presence of charge q at 0 is given as ˆ ˆ W E = E ir,stat dr = C 1 E ir,stat dr C 2 Evaluating the above expression gives ˆ E ir,stat dr = C 2 ˆ r ( ) q 4πε 0 r 2 dr = q 4πε 0 ( ) 1 r = q r 4πε 0 r = Vstat (23)

52 Electrostatic Scalar Potential - Physical Interpretation So, finally notations that we will use in this course: V stat A V stat AB VAB stat = = VA stat VB stat = V stat A = ˆ A E ir,stat dl B ˆ A E ir,stat dl

53 Analysis of a Dipole Voltage due to the charges +q and q at point P is, V stat = q [ 1 4πε 0 r 1 ] 1 r 2 [ r2 r 1 r 1 r 2 = q 4πε 0 For far-field approximation r 1 r 2 r, we get r 2 r 1 = d cos θ ]. (24) r 1 = r 2 = r. (25) Substituting the above equations in (24) gives V stat = qd cos θ 4πε 0 r 2. (26) If we define dipole momentum as P = q d, V stat = P ˆr 4πε 0 r 2. (27)

54 Analysis of a Dipole Now, we can obtain electric field at point P by using the formula E ir,stat = V stat as shown below: ( ) 3 E ir,stat 1 V = stat ˆq i h i=1 i q i ( V stat = ˆr + 1 r r ( = = 2qd cos θ 4πε 0 r 3 V stat θ ) ˆθ ) qd sin θ ˆr 4πε 0 r ˆθ 3 qd 4πε 0 r 3 ( 2 cos θˆr + sin θ ˆθ ).

55 Energy of an Electrostatic System Energy spent in bringing all these charges from is W stat E Also, it is easy to prove that q 2 V stat 21 = q 1 V stat 12, i.e., = 0 + q 2 V stat 21 + q 3 V stat 31 + q 3 V stat (28) q 2 V stat 21 = q 2 q 1 4πε 0 r 21 = q 1 q 2 4πε 0 r 12 = q 1 V stat 12.

56 Energy of an Electrostatic System So, (28) can be rewritten as W stat E Now, combining (28) and (29) gives = q 1 V stat 12 + q 1 V stat 13 + q 2 V stat (29) 2WE stat = ( q 2 V21 stat + q 1 V12 stat ) ( + q1 V13 stat + q 3 V31 stat ) ( + q2 V23 stat + q 3 V32 stat ) + = q 1 ( V stat 12 + V stat 13 + ) + q 2 ( V stat 21 + V stat 23 + ) + q 3 ( V stat 31 + V stat 32 + ) +. Each sum of potentials in parentheses is the combined potential due to all the charges except for the charge at the point where this combined potential is being found. In other words, 2W stat E = q 1 V stat 1 + q 2 V stat 2 + q 3 V stat 3 +. So, from the above equation, the total energy of the electrostatic system is given as WE stat = 1 N 2 q i Vi stat = 1 ρ e,vol V stat dv. (30) 2 i=0 V

57 Energy Density of an Electrostatic System From the previous equation, From the vector identity WE stat = 1 ρ e,vol V stat dv 2 V = 1 ( ) D ir,stat Vdv. 2 V (wa) = w A + A w, we get W E = 1 ( ) D ir,stat V stat dv 2 V = 1 [ ( V stat ) D ir,stat D ir,stat V stat] dv 2 V = 1 [ ( V stat )] D ir,stat dv 1 [ D ir,stat V stat] dv 2 V 2 V = 1 ( V stat ) D ir,stat dv 1 [ D ir,stat V stat] [ 1 dv = 2 S 2 V V 2 ε 0 ] E ir,stat 2 dv. (31) }{{} = 0, if S is sphere

58 Outline 1 SI Units 2 Coulomb s Law 3 Gauss Law 4 Voltage & Energy 5 Poisson s Equation 6 Capacitance 7 Boundary Conditions 8 MoI 9 Summary 10 Problems

59 Poisson s Equation - Derivation From (21) and (19), we get where (32) is known as Poisson s equation. E ir,stat = ρ e,vol ε 0 ( V stat) = ρ e,vol ε 0 2 V stat = ρ e,vol ε 0 (32)

60 Poisson s Equation - Uniqueness

61 Outline 1 SI Units 2 Coulomb s Law 3 Gauss Law 4 Voltage & Energy 5 Poisson s Equation 6 Capacitance 7 Boundary Conditions 8 MoI 9 Summary 10 Problems

62 Capacitor - Basics A Few Basic Equations: Q = t 0 idt = Cv (assuming that current doesn t exist for t < 0) i = dq dt = C dv dt Energy = t 0 (power)dt = t 0 vidt = t 0 vc dv dt dt = C v 0 vdv = 1 2 Cv2 I (jω) = jωcv (jω) = V(jω) I(jω) = 1 jωc = Z (jω) (using Fourier transformation)

63 Evaluation of Capacitance Starting with the Assumption of Charge

64 Parallel Plate Capacitor STEP1: Using the Gauss law and assuming that fringing fields are negligible, we can prove that D (x) = ρs ˆx. STEP2: From the definition of voltage, we have V 0d = V 0 V d = From the above equation Q = = = ρs d ε 0 ˆ x=0 E dl x=d ˆ x=0 ( ) ρ S dx x=d ε 0 = Qd Aε 0. ( ) Aε0 V 0d. (33) d

65 Cylindrical Capacitor STEP1: Using the Gauss law and assuming that fringing fields are negligible, we can prove that ρ S a D (ρ) = ρ ˆρ. STEP2: From the definition of voltage, we have ˆ ρ=a V ab = V a V b = E dl ρ=b ˆ ρ=a ρ = S a ρ=b ε 0 ρ dρ ( ) = ρs a b ln = Q ( ) ( ) a b ln ε 0 a 2πaL ε 0 a ( ) 1 2πLε 0 = Q ln ( ) (34) b a

66 Spherical Capacitor STEP1: Using the Gauss law, we can prove that ρ S a 2 D (r) = r 2 ˆr. STEP2: From the definition of voltage, we have V ab = V a V b = = ˆ r=a E dl r=b ˆ r=a ρ S a 2 r=b ε 0 r 2 dr 1 r=a ( ) ( Q a 2 1 ε 0 r = r=b 4πa 2 ε 0 a 1 ) b ( ) 1 4πε 0 1 a 1. (35) b = ρs a 2 = Q

67 Evaluation of Capacitance Starting with the Assumption of Voltage (i.e., using the Poisson s Equation)

68 Parallel Plate Capacitor STEP1: Using Poisson s equation and boundary conditions, one gets 2 V = 2 V x 2 = 0 V = Ax + B ( V = V 0 1 x ). d STEP2: From the the relation between electric field and voltage, we have E = V = V x ˆx = V 0 d ˆx. On the surface of a metal plate (PEC), charge density is given by D norm. So, Q = Aρ S = A V ( ) 0 A ε 0 d = V 0. (36) ε 0 d

69 Cylindrical Capacitor STEP1: Using Poisson s equation and boundary conditions, one gets 2 V = 1 ρ ( ρ V ) = 0 ρ ρ ρ V ρ = A V = A ln ρ + B = V 0 ln (b/ρ) ln (b/a). STEP2: From the the relation between E and V, V E = V = ρ ˆρ = V 0 1 ρ ln (b/a) ˆρ. On the surface of a metal plate (PEC), charge density is given by D norm. So, ( ) Q = Aρ S 2πLε0 = V 0. (37) ln (b/a)

70 Conical Capacitor (!) STEP1: Using Poisson s equation and boundary conditions, one gets 2 V = ( 1 r 2 sin θ V ) = 0 sin θ θ θ sin θ V θ = A ( V = A ln tan θ ) ln ( tan θ ) 2 + B = V 0 2 ln ( tan α ). 2 STEP2: From the the relation between E and V, 1 V E = V = r θ ˆθ 1 = V 0 r sin θ ln ( tan α ) ˆθ. 2 On the surface of a metal plate (PEC), charge density is given by D norm. So, ˆ 2π ˆ ( ) r Q = ρ S ε 0 V 0 ds = 0 0 r sin α ln ( tan α ) r sin αdφdr = 2πrε 0 2 ln ( cot α ) V 0 (38) 2

71 Outline 1 SI Units 2 Coulomb s Law 3 Gauss Law 4 Voltage & Energy 5 Poisson s Equation 6 Capacitance 7 Boundary Conditions 8 MoI 9 Summary 10 Problems

72 Boundary Conditions - Tangential Components Using the Stokes Theorem, (ˆ ˆ ˆ ˆ ) ( E ) dl (ˆ ˆ ) ( E ) + dl 1 3 (E 1 tangential E2 tangential ) l = 0 = ( E ) ds = 0, ( E = 0) E 1 tangential = E 2 tangential D 1 tangential ε r1 = D2 tangential ε r2.

73 Boundary Conditions - Normal Components Using the divergence theorem (and Gauss law), ( 5 ) ( D ) + + ds = Q e,cylinder 6 7 ( D 2 normal D1 normal) ds = ρe ds D 2 normal D1 normal = ρ e ε r2 E 2 normal ε r1e 1 normal = ρ e.

74 Outline 1 SI Units 2 Coulomb s Law 3 Gauss Law 4 Voltage & Energy 5 Poisson s Equation 6 Capacitance 7 Boundary Conditions 8 MoI 9 Summary 10 Problems

75 Method of Images

76 Outline 1 SI Units 2 Coulomb s Law 3 Gauss Law 4 Voltage & Energy 5 Poisson s Equation 6 Capacitance 7 Boundary Conditions 8 MoI 9 Summary 10 Problems

77 - Summary

78 Outline 1 SI Units 2 Coulomb s Law 3 Gauss Law 4 Voltage & Energy 5 Poisson s Equation 6 Capacitance 7 Boundary Conditions 8 MoI 9 Summary 10 Problems

79 Coulomb s Law 1 Find E at P (1, 1, 1) caused by four identical 3 nc charges located at P1 (1, 1, 0), P 2 ( 1, 1, 0), P 3 ( 1, 1, 0), and P 4 (1, 1, 0). Ans: 6.82ˆx ŷ ẑ V/m [H1, E2.2, P35] 2 An infinitely long uniform line charge is located at y = 3, z = 5. If ρ L = 30 nc/m, find electric field intensity at the origin. Ans: ŷ 79.3ẑ V/m 3 In a rectangular region 2 x 2, 3 y 3, z = 0, a surface charge density is given by ρ S = ( x 2 + y ) 3/2 C/m 2. If no other charge is present, find E at P(0, 0, 1). Ans: ẑ V/m

80 Gauss Law 1 Calculate the total charge within the universe for the given volume charge density, ρ v = e 2r r 2 C/m 3. Also, calculate the electric flux density D at r = rˆr. Ans: 6.28 C, 1 e 2r r 2 C/m 2 [H1, D2.4, P38] 2 Evaluate both sides of the Gauss Law for the field D = 2ρ (z + 1) cos φ ˆρ ρ (z + 1) sin φ ˆφ + ρ 2 cos φẑ µc/m 2, and the region 0 ρ 2, 0 φ 0.5φ, and 0 z 4. 3 A non-uniform volume charge density, ρ v = 120r C/m 3, lies within the spherical surface r = 1m, and ρ v = 0 elsewhere. (a) Find D r everywhere. (b) What surface charge density ρ s2 should be on the surface r = 2m so that D r r=2 = D r r=2 +?

81 Voltage - I 1 given the nonuniform field E = yˆx + xŷ + 2ẑ, determine the work expended in carrying 2C from A (1, 0, 1) to B (0.8, 0.6, 1) along the arc of the circle x 2 + y 2 = 1, z = 1. Ans: J [H1, E4.1, P87] 2 Again find the work required to carry 2 C from A to B in the same field, but this time use the straight-line path from B to A. Ans: J [H1, E4.2, P88] 3 Let E = yˆx at a certain instant of time, and calculate the work required to move a 3C charge from (1, 3, 5) to (2, 0, 3) along the straight line segments joining: (1, 3, 5) to (2, 3, 5) to (2, 0, 5) to (2, 0, 3) (1, 3, 5) to (1, 3, 3) to (1, 0, 3) to (2, 0, 3) Ans: -9 J; 0 J [H1, D4.3, P90]

82 Voltage - II 1 An electric field is expressed in Cartesian coordinates by E = 6x2 ˆx + 6yŷ + 4ẑ V/m. Find: V MN if points M and N are specified by M (2, 6, 1) and N ( 3, 3, 2) V M if V = 0 at Q (4, 2, 35) V N if V = 2 at P (1, 2, 4) Ans: -139 V; -120 V; 19 V [H1, D4.4, P93] 2 If V = x 2 y (z + 3) V, then find the electric field intensity E at point P Cart (3, 4, 6). Ans: 72ˆx + 27ŷ 36ẑ V/m 3 If E = 5ˆx + 2ŷ + 4ẑ kv/m in a medium with εr = 4, then find the corresponding volume energy density. Ans: µj/m 3 4 Given the potential V = 4xy + 3z 2 V, what is the volume charge density ρ v at point P Cart ( 4, 3, 6)? Ans: 6ε 0 C/m 3

83 Capacitance 1 Two parallel plates are separated by a dielectric of thickness 2mm with a dielectric constant of 6. If the area of each plate is 40 cm 2 and the potential difference between them is 1.5 kv, then find the magnitude of the electric field inside the dielectric medium. 2 A parallel-plate capacitor with area 0.30m 2 and separation 5.5 mm contains three dielectrics with interfaces normal to E and D as follows: εr1 = 3, ε r2 = 4, ε r3 = 6d 1 = 1 mm, d 2 = 2 mm, and d 3r1 = 2.5 mm. Find the capacitance. Ans: 2.12nF

84 Miscellaneous 1 Determine the value of E in a material for which the electric susceptibility is 2.5 and P = pc/m 2.