Fundamentals of Electromagnetics Fields Assoc. Prof., Dr. Sc. Trần Hoài Linh School of Electrical and Electronics Engineering linh.tranhoai@hust.edu.vn, thlinh2000@yahoo.com Subject contents • Slides: Materials for tests. • Text book(s): Engineering Electromagnetics (recommended 6th ed.), William H. Hayt, John A. Buck. • Midterm test and Final test: 3 questions (9 points) + 1 point for presentation 1 Subject contents 1. Selected mathematics tools required 2. Static electrical field 3. Steady electrical field 4. Static magnetic field 5. Time-varying fields 6. Waves propagation and transmission lines 1. Selected mathematics tools required • Single variable, multi-variable functions • Derivative, partial derivatives • Integral (indefinite, definite) • Multi-level integrals • Vectors and vector functions • 3 basic systems of coordinations • Line integral, surface integral, volume integral • Stokes’, Green’s theorems • Integral of gradient 2 1.1. Single variable, multi-variable functions • Single variable function: y f x example : y x x 2 2 x sin x y f a ,b,c x example : ya ,b,c x ax 2 bx c • Multi-variable function : E E x, y , z example : E x, y , z x 2 xy sin z E f a ,b,c x, y , z example : Ea ,b ,c x, y , z ax 2 bxy c sin z 1.2. Derivative, partial derivatives • Derivative (of single variable functions): dy f x dx dy y f a ,b,c x y( x) f x dx y f x y( x) example : y x 2 2 x sin x y 2 x 2 cos x example : y ax 2 bx c y x 2ax b • Partial derivatives (of multi-variable functions): E x ... E E E x, y , z ... y E ... z E x 2 x y E 2 example : E x xy sin z x y E cos z z In partial derivatives, other variables are treated as constants. 3 1.3. Indefinite integral, definite integral • Indefinite integral: finding antiderivative function F ( x) f x dx F ( x) f x F ( x, y ) f x, y dx F x, y f x, y x Example: 1 ax b dx 2 ax 2 x y dx x 2 2 bx C yx C1 1 2 x y dy 2 xy 2 y 2 C2 1.3. Indefinite integral, definite integral • Definite integral: b b f x dx F x a F b F a a with F ( x ) f x Example: 2 ax b dx 1 2 1 2 ax bx C 2a 2b C 0.5a b C 1.5a b 2 1 4 1.3. Indefinite integral, definite integral • Definite integral: b b f x, y dx F x, y a F b, y F a, y a with F x, y x f x, y Example: 2 2 x y dx x 2 yx C1 1 2 2 x y dy 2 xy 1 2 1 4 2 y C1 1 y C1 3 y 2 1 2 y C2 4 x 2 C2 2 x 0.5 C2 2 x 1.5 2 1 1.4. Multi-level integrals • Usually deal with definite integrals. • In “Fundamentals of Electromagnetic Fields”: usually up to 3 levels (triple integral). • Example: Double integral: 2 I1 3 2 x y dx dy x 1 y 2 I2 2 x y dx dy S OAB with A 3,0 , B (0, 2). 5 1.4. Multi-level integrals 2 I1 3 2 x y dx dy x 1 y 2 I2 2 x y dx dy with A 3,0 , B (0, 2). S OAB 1.4. Multi-level integrals • Method of calculation: • Select an order of integration • Find the integral limits in the reverse order of the integration orders, where the limit of inner integral depends on the variables of outer integrals. Exercise: Rework the multilevel integral with different orders of integrations. 6 1.4. Multi-level integrals • Triple integral: 2 I1 3 1 2 x yz dx dy dz x 1 y 2 z 1 I2 2 x y dx dy dz with A 1,0, 0 , B 0,3, 0 , C 0,0, 2 . V OABC 1.5. Vector, vector function • Vector: defined by 4 parameters: • The origin (the tail): A • The direction: from A to B (from tail to head) • The magnitude: length of AB • Unit vector: the vector with length = 1 7 1.5. Vector, vector function • Vector: defined by 4 parameters: • The origin (the tail): A • The direction: from A to B (from tail to head) • The magnitude: length of AB • Unit vector: the vector with length = 1 AB AB AB a AB a AB 1.5. Vector, vector function • Vector addition and subtraction: AB AC AC CD AD AB AC CB since AC CB AB 8 1.5. Vector, vector function • The dot product (scalar product) of 2 vectors (with the same origin) AB AC AB AC cos Properties: • Commutative: AB AC AC AB • Distributive: A B C A B A C • • … 2 AB AC 0 1.5. Vector, vector function • The cross product (vector product) of two vectors (with the same origin) AB AC AD a AD AB AC sin with AD AB, AC and ( AB, AC, AD) form a right - hand triple (or right hand rule) 9 1.5. Vector, vector function • Vector function: the value of the function at a point is a vector example : E x, y, z x 2 a x xy a y sin z a z 1.6. Basic coordinate systems • 3 fundamental coordinate systems: • Cartesian (rectangular) coordinate system • Spherical coordinate system • Cylindrical coordinate system • For each coordinate system: • 3 coordinates/components, • The unit vectors for each coordinate, • Differential displacement (line) vector, • Differential surface vector • Differential volume (scalar) 10 1.6.1. Cartesian coordinate system • 3 coordinates: x, y, z • 3 “axis” x-y-z form a right-hand-triple 1.6.1. Cartesian coordinate system • 3 coordinates: x, y, z • 3 “axis” x-y-z form a right-hand-triple • Unit vectors (for each coordinate): ax, ay, az (other format (i, j, k) or (ix, iy, iz)) • Ranges of coordinates: x , y , z , 11 1.6.1. Cartesian coordinate system • Unit vectors at a given point (not only at the origin O(0,0,0)): The unit vector for 1 coordinate has the same direction as the differential displacement vector at the given point for the given coordinate! A Ax , Ay , Az A Ax , Ay dy , Az dl y AA A Ax , Ay , Az A Ax , Ay , Az dz dl z AA A Ax , Ay , Az A Ax dx, Ay , Az dl x AA 1.6.1. Cartesian coordinate system • Unit vectors at a given point (not only at the origin O(0,0,0)): The unit vector for 1 coordinate has the same direction as the differential displacement vector at the given point for the given coordinate! A Ax , Ay , Az A Ax dx, Ay , Az AA Ox a x Ox AA dx dl dx x dl x dx a x 12 1.6.1. Cartesian coordinate system • Unit vectors at a given point (not only at the origin O(0,0,0)): The unit vector for 1 coordinate has the same direction as the differential displacement vector at the given point for the given coordinate! A Ax , Ay , Az A Ax , Ay dy , Az AA Oy a y Oy AA dy dl y dy dl y dy a y 1.6.1. Cartesian coordinate system • Unit vectors at a given point (not only at the origin O(0,0,0)): The unit vector for 1 coordinate has the same direction as the differential displacement vector at the given point for the given coordinate! A Ax , Ay , Az A Ax , Ay , Az dz AA Oz a z Oz AA dz dl dz z dl z dz a z 13 1.6.1. Cartesian coordinate system • Unit vectors at a given point (not only at the origin O(0,0,0)): The unit vector for 1 coordinate has the same direction as the differential displacement vector at the given point for the given coordinate! ax a y ,a y az ,az ax a x a x 1,... a x a y 0,... a x a y a z ,... 1.6.1. Cartesian coordinate system With all 3 components determined, the general differential displacement vector is given by: dl dl x a x dl y a y dlz a z dx a x dy a y dz a z 14 1.6.1. Cartesian coordinate system • Differential surface vector: Limit consideration for surfaces with 1 coordinate is constant! At those cases, the differential surface vector at a point (A) is determined as follow: • The direction is normal (perpendicular) to the surface at A (i.e. parallel to the unit vector of the constant coordinate) • The direction is facing “outward” of the given surface • The magnitude equals the product of two differential displacements of the two varying coordinates 1.6.1. Cartesian coordinate system • For example: surface with x = const ds dl y dlz a x dy dz a x 15 1.6.1. Cartesian coordinate system • For example surface with y = const ds dl z dlx a y dz dx a y 1.6.1. Cartesian coordinate system • For example surface with z = const ds dl x dl y a z dx dy a z 16 1.6.1. Cartesian coordinate system • Differential volume at a point (A) = product of 3 differential displacements for all coordinates at A: dv dl x dl y dlz dx dy dz 1.6.2. Spherical coordinate system • 3 coordinates: r, , • Ranges of coordinates: • r – length of OA: r 0, • - angle from Oz to OA: 0, • - angle from Ox to OH: 0, 2 17 1.6.2. Spherical coordinate system • 3 coordinates: r, , • The unit vectors: ar, a, a • ar and a on zOH • a perpendicular to zOH • Follow right-hand-rule 1.6.2. Spherical coordinate system • The differential displacement vectors for each coordinates: A Ar , A , A A Ar , A d , A dl AA Ar d a r d a A Ar , A , A A Ar , A , A d A Ar , A , A A Ar dr , A , A dl r AA dr a r dl AA OH d a Ar sin A d a r sin d a The general differential displacement vector for cylindrical coordinate system: dl dlr a r dl a dl a dr a r r d a r sin d a 18 1.6.2. Spherical coordinate system • Differential surface vector: Limit consideration for surfaces with 1 coordinate is constant! At those cases, the differential surface vector at a point (A) is determined as follow: • The direction is normal (perpendicular) to the surface at A (i.e. parallel to the unit vector of the constant coordinate) • The direction is facing “outward” of the given surface • The magnitude equals the product of two differential displacements of the two varying coordinates 1.6.2. Spherical coordinate system • For example surface with r = const ds dl dl a r r d r sin d a r r 2 sin d d a r 19 1.6.2. Spherical coordinate system • For example surface with = const ds dlr dl a dr r sin d a r sin dr d a 1.6.2. Spherical coordinate system • For example surface with = const (please note that it’s only ½ plane) ds dlr dl a dr r d a r dr d a 20 1.6.2. Spherical coordinate system • Differential volume at a point (A) = product of 3 differential displacements for all coordinates at A: dv dlr dl dl dr r d r sin d r 2 sin dr d d 1.6.3. Cylindrical coordinate system • 3 coordinates: , , z • Ranges of coordinates: • – distance from A to Oz: 0, • - angle from Ox to OH: 0,2 • z – distance from A to xOy: z , 21 1.6.3. Cylindrical coordinate system • 3 coordinates: , , z • 3 unit vectors: a, a, az • a and az are on zOH • a is perpendicular to zOH • The 3 unit vectors follow a right-hand-rule 1.6.3. Cylindrical coordinate system • The differential displacement vectors for each coordinates: A A , A , Az A A , A d , Az dl AA d a A A , A , Az A A , A , Az dz dl z AA dz a z A A , A , Az A A d , A , Az dl AA d a The general differential displacement vector for cylindrical coordinate system: dl dl a dl a dl z a z d a d a dz a z 22 1.6.3. Cylindrical coordinate system • Differential surface vector: Limit consideration for surfaces with 1 coordinate is constant! At those cases, the differential surface vector at a point (A) is determined as follow: • The direction is normal (perpendicular) to the surface at A (i.e. parallel to the unit vector of the constant coordinate) • The direction is facing “outward” of the given surface • The magnitude equals the product of two differential displacements of the two varying coordinates 1.6.3. Cylindrical coordinate system • For example surface with = const ds dl dl z a d dz a d dz a 23 1.6.3. Cylindrical coordinate system • For example surface with = const ds dl dl z a d dz a 1.6.3. Cylindrical coordinate system • For example surface with z = const ds dl dl a z d d a z d d a z 24 1.6.3. Cylindrical coordinate system • Differential volume at a point (A) = product of 3 differential displacements for all coordinates at A: dv dl dl dlz d d dz d d dz 1.7. “Special” derivatives of vector functions a. Gradient: grad ( f ) f ... b. Divergence: div( v) ( v) ... c. Rotation (Curl): rot ( v ) ( v) ... d. Laplacian: lapl ( f ) f ... Remind: Vector function in a 3-component coordinate system , , : v , , v , , a v , , a v , , a including 3 scalar functions along 3 “axes”, each scalar function can depend on all 3 coordinates. Examples: v x, y , z y z a x x y a y sin z a z v r , , r sin a r cos a r 1 a v , , z z a z a z 1 a z 25 1.7. “Special” derivatives of vector functions a. Gradient: • In Cartesian coordinate system: grad ( f ) f f f f ax a y az x y z • In spherical coordinate system: grad ( f ) f f 1 f 1 f ar a a r r r sin • In cylindrical coordinate system: grad ( f ) f f 1 f f a a a z z 1.7. “Special” derivatives of vector functions a. Gradient: • Cartesian coordinate system: grad ( f ) f f x, y, z x 2 xy sin z f f f ax a y az x y z 26 1.7. “Special” derivatives of vector functions a. Gradient: • Spherical coordinate system: grad ( f ) f f r , , r sin f 1 f 1 f ar a a r r r sin 1.7. “Special” derivatives of vector functions a. Gradient: • Cylindrical coordinate system: grad ( f ) f f , , z sin z f 1 f f a a a z z 27 1.7. “Special” derivatives of vector functions b. Divergence : • Cartesian coordinate system: div( v) ( v) vx v y vz x y z • Spherical coordinate system: 2 1 r vr 1 sin v 1 v div( v) ( v ) 2 r r sin r sin r • Cylindrical coordinate system: div( v) ( v ) 1 v 1 v vz z 1.7. “Special” derivatives of vector functions b. Divergence : v x, y , z y z a x x y a y sin z a z v y vz v div( v) ( v) x x y z • Cartesian coordinate system: 28 1.7. “Special” derivatives of vector functions b. Divergence : • Spherical coordinate system: v r , , r sin a r cos a r 1 a 2 1 r vr 1 sin v 1 v div( v ) ( v ) 2 r r sin r sin r 1.7. “Special” derivatives of vector functions b. Divergence: • Cylindrical coordinate system: div( v ) ( v ) v , , z z a z a z 1 a z 1 v 1 v vz z 29 1.7. “Special” derivatives of vector functions c. Rotation (curl): • Cartesian coordinate system: v v y v y vx vx vz rot ( v) ( v) z a ax az y z x y z y x • Spherical coordinate system: rot ( v) ( v ) a v 1 sin v 1 1 vr rv ar r sin r sin r • Cylindrical coordinate system: 1 vz v z rot ( v ) ( v ) v vz a z 1 rv vr r r a 1 v v a a z 1.7. “Special” derivatives of vector functions c. Rotation (curl): • Cartesian coordinate system: v x, y , z y z a x x y a y sin z a z v v y v y vx vx vz rot ( v ) ( v) z a a x az y z x y z y x 30 1.7. “Special” derivatives of vector functions c. Rotation (curl): • Spherical coordinate system: v r , , r sin a r cos a r 1 a rot ( v) ( v ) a v 1 sin v 1 1 vr rv ar r sin r sin r 1 rv vr r r a 1.7. “Special” derivatives of vector functions c. Rotation (curl): • Cylindrical coordinate system: v , , z z a z a z 1 a z 1 vz v z rot ( v ) ( v) v v z a z 1 v v a a z 31 1.7. “Special” derivatives of vector functions d. Laplacian: • Cartesian coordinate system: 2 f 2 f 2 f lapl ( f ) f 2 2 2 x y z • Spherical coordinate system: lapl ( f ) f 1 2 f 1 f r 2 sin r r r r sin 1 2 f 2 2 r sin 2 • Cylindrical coordinate system: 1 f 1 2 f 2 f lapl ( f ) f 2 2 z 2 1.7. “Special” derivatives of vector functions d. Laplacian: • Cartesian coordinate system: f x, y, z x 2 xy sin z 2 f 2 f 2 f lapl ( f ) f 2 2 2 x y z 32 1.7. “Special” derivatives of vector functions d. Laplacian: • Spherical coordinate system: lapl ( f ) f f r , , r sin 1 2 f 1 f r 2 sin r r r r sin 1 2 f 2 2 r sin 2 1.7. “Special” derivatives of vector functions d. Laplacian: • Cylindrical coordinate system: f , , z sin z 1 f 1 2 f 2 f lapl ( f ) f 2 2 z 2 33 1.8. Line/Surface/Volume integrals a. Line integral: v dl P P – a Path (can be a closed path) v – a vector function (in any coordinate system) dl – differential line vector v dl – the dot (scalar) product between two vector functions General steps: - Use the dl for the same coordinate system that v to convert the dot product to a scalar function. - Split P into segments if necessary - Find the integrals of that scalar function for each segment and sum them up. 1.8. Line/Surface/Volume integrals a. Line integral: v dl P Example: P is a closed path OABO with A(3, 0) and B(0, 2). Function v is: v x, y , z y z a x x y a y sin z a z 34 1.8. Line/Surface/Volume integrals a. Line integral: v dl P Example: P is a closed path OABO with A(3, 0) và B(0, 3). Function v is: v r , , r sin ar cos a r 1 a 1.8. Line/Surface/Volume integrals a. Line integral: v dl P Example: P is a ABCD with OA=R, AOB = /4, BC=h. Function v is: v , , z z a z a z 1 a z 35 1.8. Line/Surface/Volume integrals v ds b. Surface integral: S S – a surface in the space (limit consideration the surface with 1 coordinate is constant, can be closed surface). v – vector function (in any coordinate system) ds – differential surface vector v ds – the dot (scalar) product between two vector functions General steps: - Use the ds for the same coordinate system that v to convert the dot product to a scalar function. - Split S into sub-surfaces if necessary - Find the integrals of that scalar function for each sub-surface (as double integrals) and sum them up. 1.8. Line/Surface/Volume integrals b. Surface integral : v ds S Example: S is the triangle OAB (on yOz) with A(0, 3, 0) and B(0, 0, 2). Function v is: v x, y , z y z a x x y a y sin z a z 36 1.8. Line/Surface/Volume integrals b. Surface integral : v ds S Example: S is ¼ of circle (surface) OAB with radius R on yOz. Function v is: v r , , r sin ar cos a r 1 a 1.8. Line/Surface/Volume integrals b. Surface integral : v ds S Example: S is the closed surface enclosing the cylinder of radius R, height h, bottom based on xOy. Function v is: v , , z z a z a z 1 a z 37 1.8. Line/Surface/Volume integrals f dv c. Volume integral: V V – a volume in space f – a scalar function (in any coordinate system) dv – differential volume General steps: Using the differential volume formula to convert a volume integral into a multi-level (triple) integral. 1.8. Line/Surface/Volume integrals c. Volume integral: f dv V Example: V is a tetrahedron OABC with A(1, 0, 0), B(0, 3, 0) and C(0, 0, 2). Function f is: f x, y , z 2 x y In Cartesian coordinate system: dv dx dy dz (similar to example in 1.4) 38 1.8. Line/Surface/Volume integrals c. Volume integral: f dv V Example: V is 1/8 sphere with radius R. Function f 1 1.8. Line/Surface/Volume integrals c. Volume integral: f dv V Example: V is a cylinder of radius R, height h, bottom based on xOy. Function f is: f , , z z 1 39 1.9. Stokes’, Green’s and gradient theorems a. Stokes’ theorem (line integral surface integral) b. Green’s (surface integral volume integral) c. Integral of gradient theorem 1.9. Stokes’, Green’s and gradient theorems a. Stokes theorems (line integral surface integral, Kelvin – Stokes’ theorem) v dl P rot v ds S P P – a closed path, S – surface “based” on the path P (the “outward” normal vector of S and the direction of the path P fulfill the right-hand-rule) ds – differential surface vector 40 1.9. Stokes’, Green’s and gradient theorems a. Stokes theorems (line integral surface integral, Kelvin – Stokes’ theorem) Example (compare with result from 1.8): P is a closed path OABO with A(3, 0) and B(0, 2). Function v is: v x, y , z y z a x x y a y sin z a z 1.9. Stokes’, Green’s and gradient theorems a. Stokes theorems (line integral surface integral, Kelvin – Stokes’ theorem) Example (compare with result from 1.8): P is a closed path OABO with A(3, 0) and B(0, 3). Function v is: v r , , r sin ar cos a r 1 a 41 1.9. Stokes’, Green’s and gradient theorems b. Green’s theorem (surface integral volume integral) v ds S div v dv V S S – a closed surface, V – the volume enclosed in S dv – differential volume 1.9. Stokes’, Green’s and gradient theorems b. Green’s theorem (surface integral volume integral) Example: Find the surface integral v ds where: S S – closed surface enclosing the cube with edge = a, v – vector function: v x, y , z y z a x x y a y sin z a z 42 1.9. Stokes’, Green’s and gradient theorems c. Integral of gradient theorem grad f dl f ( B ) f ( A) P:A B f – scalar function, P – any path connecting A and B dl – differential line vector 2. Static electrical field 1. Coulomb force 2. Vector of electrical field intensity 3. Field intensity for selected charge distributions 4. Streamlines and sketches of fields 5. Gauss’s law, vector of electric flux density, electric flux 6. Energy and potential, voltage drop 7. Dipole 8. Dielectric (insulator) and conductors 9. Border conditions between layers of materials 10. Capacitance(s) of a closed system of conductors 11. Poisson’s – Laplace’s equations 43 2.1. Coulomb force (Charles) Coulomb’s experiment on the interaction between two “point” charges: • Force acted on Q1 (F21) and force acted on Q2 (F12) have the same magnitude, opposite direction on the line connecting two “point” charges Q1 and Q2. • Two charges with same sign repelling each other, different signs attracting each other • Force is proportional to charges and inversely proportional to square of distance. • When the charges are put in a oil (dielectrics) the force is smaller. 2.1. Coulomb force Coulomb’s experiment on the interaction between two “point” charges: F12 F21 k Q1Q2 d2 Assume the unit vector for the direction from Q1Q2 is a12 F12 F12 a12 k where: k 1 Q1Q2 a 12 d2 4 0 44 2.1. Coulomb force Constant 0 – permittivity of free space 0 C2 1 109 8,854 1012 2 36 N m F 8,854 1012 m Permittivity of air is about the same! 2.2. Electric field and vector of electric field intensity The idea of a “field” (electric field): • Q1 generates an (electric) field around it, • When a new charge (Q2) appears in the field, it’s the field which creates a force acted on the new charge. Remark: - Q2 also has its own field, and Q2 field creates the force F21 acted on Q1. - At this moment, we don’t consider yet the propagation of the field! 45 2.2. Electric field and vector of electric field intensity Electric field intensity (of the field generated by Q1): depends only on Q1 (located at A) and the position of B E( B ) F B 1 Q1 a12 Q2 4 0 d 2 If we assume Q1 is located at the origin (the spherical coordinate system): E( r ) 1 Q1 ar 4 0 r 2 2.2. Electric field and vector of electric field intensity The force fulfills the superposition properties The field intensity also fulfills the superposition properties : F (C ) F1q F2 q E C E1 E2 1 Q1 4 0 AC 2 a AC 1 Q2 4 0 BC 2 a BC Remark: The directions AC and BC are different vector summation is not too easy! 46 2.2. Electric field and vector of electric field intensity To sum 2 (or more) vectors, we can use the projection property: Fq F1q F2 q Fq F1q F2 q Fq F1q F2q where the sum F1q + F2q is an algebraic sum of scalars! 2.3. Basic charge distributions The charge 1C is a huge value (the charge of an electron = ………………)? A charge consists of big number of fundamental charges (for example electrons), spreading “relatively” uniform in the space. There are 3 basic distributions: • Line distribution of charges: When 1 dimension (the length) of the system is significantly bigger than other dimensions charge density (C/m) • Surface distribution of charges: When 2 dimensions (the surface) of the system of charges are significantly bigger than other dimension charge density (C/m2). • Volume distribution of charges: When 3 dimensions of the system of charge are not significantly different charge density (C/m3). 47 2.3. Basic charge distributions • Line charge distribution Line charge density (C/m) l scalar function • Surface charge distribution Surface charge density (C/m2) s scalar function • Volume charge distribution Volume charge density (C/m3) v scalar function Remark: With the density function, the total charge in the system is: Q l dl P Q s ds S Q v dv V 2.3. Basic charge distributions Example: Find the field intensity of points on the bisecting line of a constant line charge distribution of length L: l const. We can assume that the line is placed symmetrically on Oz the bisecting lines are on xOy let’s consider the points on Ox. The coordinate used for calculation is the cylindrical system. Due to the symmetry of the system: • The intensity doesn’t depend on angle . • For A belonging to Ox, EA || Ox. 48 2.3. Basic charge distributions Consider a differential length dl around point B this piece contains a charge of: dq l dl This charge will create a differential field intensity dE at A with magnitude equal: dE 1 dq 4 0 BA2 The projection of the differential field intensity dE at A on Ox has magnitude equal: dE x 1 dq cos 4 0 BA2 2.3. Basic charge distributions Summing for all points along the line taking integral along the line The total intensity of the field at A has magnitude equal: Ex dE x whole line L 2 L 2 1 l dl BA2 1 dq L dEx L 4 0 BA2 cos zB L 4 0 zB L 2 2 z B 2 cos 2 Along the line, we have: dl dz 49 2.3. Basic charge distributions To calculate the single integral, let’s transform all variables to 1 variable. For example, let’s transform to functions of : BA EA OA OA ; z B OA tan dz d cos cos 2 max max max 1 4 0 4 0 1 max OA d cos 2 cos 2 OA cos l 1 l 4 0 OA cos d 1 l 4 0 OA sin max max l 2sin max OA 2.3. Basic charge distributions The full form of the vector of field intensity is (for points on the bisecting plane): EA 1 4 0 l 2sin max OA E , ,0 l sin max a 2 0 For “infinite” line: E , , z l a 2 0 Exercise: Find the vector of field intensity for points not belonging to the bisecting plane 50 2.3. Basic charge distributions Example: Find the vector of field intensity on the symmetric perpendicular axis of a circular surface of radius R, charged with constant surface charge density s const. Let the disc located at the origin, on the plane of xOy. Let’s use the cylindrical coordinate system. Assume A on Oz, due to the symmetry of the system, we can predict that EA || Oz and EA doesn’t depend on . 2.3. Basic charge distributions Consider a differential area ds around B This area contains a charge of: dq s ds s d d s d d This charge will create a differential intensity dE at A with magnitude equal: dE 1 dq 4 0 BA2 The projection of the intensity dE at A on Oz has magnitude equal: dE z 1 dq cos 4 0 BA2 51 2.3. Basic charge distributions Summing for all points along the surface surface integral The total intensity of the field at A has magnitude equal: Ez R dEz 2 0 0 1 4 0 2 R dE z 0 0 disc R 2 1 dq cos 2 4 BA 0 0 0 s d d BA2 cos Transform the function to the variables of cylindrical system: BA OA2 2 ;cos OA OA BA OA2 2 2.3. Basic charge distributions Ez A R 2 s d d OA 2 2 4 0 OA OA2 2 0 0 OA s 4 0 OA s 2 0 Ez A OA s 2 0 1 R 0 OA R 0 2 d 2 3 2 2 0 d OA2 2 3 2 d OA s 2 0 1 1 OA 2 OA R 2 1 OA2 2 R 0 s OA 1 2 0 OA2 R 2 52 2.3. Basic charge distributions Full form of the vector of field intensity for points on Oz: Ez A s OA 1 2 0 OA2 R 2 s z 1 2 0 z 2 R2 For infinite surface: E , , z l a z 2 0 E 0, , z a z Exercise: Find the intensity at the points not belonging to the perpendicular symmetric axis ( 0). Exercise: Find the intensity vector around a rectangular disc. 2.3. Basic charge distributions Example: Find the vector of field intensity around a sphere of radius R, charged with constant volume charge density v const. Let’s the sphere be at the origin and a point A on Oz. The coordinate system is the spherical system. From the symmetry we can predict that the intensity doesn’t depend on and . 53 2.3. Basic charge distributions Consider a differential volume dv around B this volume contains a charge of: dq v dv v dr r d r sin d v r 2 sin dr d d This charge will create a differential intensity dE at A with magnitude equal: dE 1 dq 4 0 BA2 The projection of the intensity dE at A on Oz has magnitude equal: dE z 1 dq cos 4 0 BA2 2.3. Basic charge distributions Summing for all points with the volume volume integral. Total intensity of the field at point A (rA > R) has magnitude equal: Ez R dE z R 2 r 0 0 0 sphere 2 r 0 0 0 1 4 0 R dE z 2 r 0 0 0 1 4 0 BA2 v r 2 sin dr d d BA2 dq cos cos Transform the function to the spherical coordinates: BA OA2 r 2 2 OA r cos ; cos OA2 BA2 r 2 2OA2 2 OA r cos OA r cos 2 OA BA 2 OA OA2 r 2 2 OA r cos OA2 r 2 2 OA r cos 54 2.3. Basic charge distributions R Ez 2 1 r 0 0 0 BA2 cos R 4 0 v r 2 sin dr d d v r 2 sin dr d 1 2 0 OA2 r 2 2 OA r cos r 0 0 OA r cos 2 OA r 2 2 OA r cos Do: r OA cos OA2 r 2 2 OA r cos nên: Ez R OA2 sin OA r cos OA2 r 2 2 OA r cos 2 1 v r dr r OA cos 2 2 2 OA 0 OA r 2 2 OA r cos r 0 3 0 v r dr r OA r OA v r 2 dr 2 OA2 OA2 r 2 2 OA r OA2 r 2 2 OA r OA2 0 r 0 0 r 0 R ... R 2 2.3. Basic charge distributions Ez A R 1 v r 2 dr r 0 0 E r , , OA2 R v R3 1 3 r 0OA2 3 r 0 3 0OA2 v v R3 ar 3 0 r 2 Exercise: 1. Find the intensity of the thin cylinder charged with constant surface density. 2. Find the intensity of the whole cylinder charged with constant volume density. 55 2.4. Streamlines and sketches of fields Streamlines are a type of graphic tool to represent an (electrical) field with following assumptions: • Streamlines are lines, that at each point, the tangent line have the same direction as the intensity vector at that point. • The density of the lines around a point is proportional to the magnitude of the intensity at that point. Remarks: • The streamlines flows away from the positive charge, flows into the negative charges. • The streamlines don’t cross each other. 2.4. Streamlines and sketches of fields Streamlines are a type of graphic tool to represent an (electrical) field with following assumptions: • Streamlines are lines, that at each point, the tangent line have the same direction as the intensity vector at that point. • The density of the lines around a point is proportional to the magnitude of the intensity at that point. The field around a point charge: E( r ) 1 Q ar 4 0 r 2 Positive charge Negative charge 56 2.4. Streamlines and sketches of fields Other examples: 2.5. Gauss’s law, electric flux density, electric flux Faraday experiment (~1837): • The inner sphere had total charge +Q • The two outer half-sphere were connected and earthed shortly (to let Qexternal = 0) • After earth connection was disconnect, the total charge of the outer surface was –Q Conclusion: There were some kind of “movement/displacement” from the inner sphere to the outer sphere. This displacement flux is called the electric flux and: Qinside Vector of displacement flux density D: D ds S At each point on the surface, D has the direction as same as the streamlines of the electric field (i.e. D is parallel to E at that point) and has the magnitude proportional to the density of the streamlines at that point (i.e. proportional to the magnitude of the intensity at that point). 57 2.5. Gauss’s law, electric flux density, electric flux Consider the field of a point charge (located at the origin), let S is a spherical surface of radius R. Find the integral: E(r ) ds S The intensity of the field is: E( r ) 1 Q ar 4 0 r 2 On the sphere (r=R=const), the differential surface vector is: ds dl dl ar ( R d ) R sin d a r R 2 sin d d a r 2.5. Gauss’s law, electric flux density, electric flux Then: 1 Q E(r ) ds 4 0 R2 R S 2 sin d d S Q 4 0 0 sin d 2 0 d Q 4 0 2 2 Q 0 Remark: • The result doesn’t depend on the radius R. • Can extend (using the superposition property): The equation is fulfilled for all closed surface và all system of charges: S : E(r ) ds S Qinner 0 58 2.5. Gauss’s law, electric flux density, electric flux In vacuum/air: D(r ) ds Qinside S E(r ) ds S Qinside 0 D(r ) 0E(r ) Remark: Gauss’s law can be used to find the intensity vector in some symmetrical configurations of charges distributions. The main idea is select a surface S, such as: • It’s easy to find Qinside • It’s easy to find the surface integral Dds 2.5. Gauss’s law, electric flux density, electric flux Example: The field around an infinite long straight line, charged with constant line charge density l: Due to the symmetry of the problem: E , , z E ( ) a Select the closed surface S enclosing a cylinder of radius , height h, with the symmetric axis along Oz. Total charge inside S: Qinside l h S consists of 2 circular base surfaces (S1 and S2), 1 curved surface S3: E(r ) ds 0; E(r ) ds 0 S1 because d s E S2 59 2.5. Gauss’s law, electric flux density, electric flux (because ds E) E( ) ds E ( ) ds S3 S3 = E( ) ds (because E( ) is const on S3 ) S3 E ( ) Area S3 E ( ) 2 h Summing up together: E ( ) 2 h E( ) l h l E ( ) 0 2 0 l a 2 0 2.5. Gauss’s law, electric flux density, electric flux Example: Find the electrical intensity around an infinite hyperplane charged with constant surface density s: Due to the symmetry: E x, y, z E ( z ) a z and E ( z ) E ( z ) The selected closed surface S is enclosing a cylinder of two bases S1=S2, height 2h (located such that the charged hyperplane is besecting the cylinder). Total charge within S: Qinside s S1 S consists of 2 bases S1 and S2, curved surface S3: E( z ) ds 0 because ds E S3 60 2.5. Gauss’s law, electric flux density, electric flux S1 S 2 E( z ) ds S1 S2 = E( z ) E ( z ) ds (because ds E) ds (because E( z ) = const on S1 and S 2 ) S1 S2 E ( z ) Area S1 S2 E ( z ) 2 S1 Summing up: s S1 E ( z) s 0 2 0 E( x, y, z ) s a z independent on (x, y, z) 2 0 E ( z ) 2S1 2.5. Gauss’s law, electric flux density, electric flux Example: Find the field intensity of spherical surface of radius R, charged with constant surface density s: Due to the symmetry: E r , , E ( r ) a r Let S is also a spherical surface of radius r. Total charge inside S: 0 Q inside Q 4 R 2 s rR rR Question: What for r = R? 61 2.5. Gauss’s law, electric flux density, electric flux E(r ) ds E (r ) ds S (because ds || E) S = E( r ) ds (because E( r ) = const on S ) S E (r ) Area S E (r ) 4 r 2 Summing up: When r < R, E(r) = 0. When r > R: E ( r ) 4 r 2 E (r ) s 4 R 2 Qinside 0 0 1 Qinside 4 0 E( r , , ) r2 s R2 0 r2 1 Qinside a r (similar to the field of a point charge) 4 0 r 2 Exercise: 1. Find the field of a sphere charged with volume density v! 2. Find the field of an infinite cylinder T charged with volume density v! 2.5. Gauss’s law, electric flux density, electric flux Deduction from Gauss’s law – 1st equation of Maxwell D ds Qtrong S Transform of both sides: S : D ds S Qinside In vacuum/air: div D dv V S div D v v dv V S div E v 0 62 2.6. Energy, potential, voltage drop Reconsidering the system of two charges Q1 and Q2. Let Q1 be fixed, under action of the field’s force, Q2 is pushed away The field acted a work! F12 1 Q1 Q2 a AB Q2 E( B ) ( N ) 4 0 AB 2 When the charge is moved by dl, work created by the electrical field is: dWelectric field F12 dl Q2 E( B) dl ( J ) Q2 E( B) dl (or we can say that the work that external sources need to generate is dW = –dW electric field). 2.6.1. Potential Generalizing, in the electric field E, to move a charge Q from A to B, the work needed is: B B A A W F dl Q E dl ( J ) Remark: In the electric field, the work doesn’t depend on the shape of the path, but only the two end points the field is a potential field! Example: Find the work generate by the field when moving the charge 1C through A C B and through A D B. 63 2.6.1. Potential Potential: The potential of a point in the space is defined as the work needed to move a charge of 1C from that point to “infinity”. V ( A) E dl (V ) A Example: In the field of a point charge placed at the origin 1 Q ar 4 0 r 2 dl dr a r r d a r sin d a E( r ) Q 1 Q 1 Q V ( A) dr 2 4 0 r 4 0 r r r 4 0 rA A A 1 2.6.1. Potential Assumption: - If nothing else is stated, the infinity point has the potential equal “0”. - We can select 1 point to be the “GROUND” – the point with potential equal 0. "0" V ( A) E dl (V ) A 64 2.6.1. Potential Potential has the superposition property! E C E1 E2 1 Q1 a AC 1 Q2 4 0 AC 4 0 BC 2 1 Q1 1 Q2 V C V1 V2 4 0 AC 4 0 BC 2 a BC 2.6.1. Potential Example: Find the potential at point A on Oz generated by a ring charged with constant line density l = const. Consider a differential length dl at B: the charge contained in that length: dq l Rd The potential created at A by this differential length: dVA 1 dq 1 4 0 BA 4 0 l Rd R 2 OA2 65 2.6.1. Potential Total potential at A: VA dVA Bcircle 0 1 4 0 l R 1 4 0 R 2 OA2 l R 1 2 0 l Rd R 2 OA2 2 0 R 2 OA2 In the formal form: VA 0, 0, z 2 l R 2 0 R 2 z 2 2.6.2. Equi-potential surface Equi-potential surface/line: - contain the points with the same potential. - usually drawn with the field streamlines 66 2.6.2. Equi-potential surface Examples the equi-potential surface/lines: 2.6.2. Equi-potential surface Examples the equi-potential surface/lines: 67 2.6.3. Potential difference (voltage drop) The difference of the potentials of two points (also called as the voltage drop): VAB VA VB "0" "0" B A B A E dl E dl E dl The electric field is a potential field: closed path P : E dl 0 P → Following Stokes’ theorem: closed path P, S P : 0 E dl rot E ds rot E 0 P S 2.6.3. Potential difference (voltage drop) Example: The potential difference between A, B around an infinite wire charged constantly. The field generated by the wire: E , , z l a 2 0 dl d a d a dz a z The potential difference between A-B: B B l d l ln B A 2 0 2 0 A VAB E dl A l ln B ln A l ln B 2 0 2 0 A 68 2.6.4. Gradient of potential Gradient of potential: B VAB VA VB E dl • From the definition of potential: A • From the theorem of integral of gradient: B scalar function f : grad ( f ) dl f B f A A all path from A to B : B B B A A A grad ( f ) dl f B f A grad (V ) dl VB VA E dl E grad V V 2.6.5. Energy in a charge distribution Energy (stored) in a charge distribution: • Assume an empty space at the beginning we can move q1 to its location without any work. • To move q2 to its location we need a work: W2 q2V21 q2 q1 4 r12 The work used to move to charge will be come the “potential” energy of the charges! • To move q3 to its location we need a work: q1 q2 W3 q3V31,2 q3 4 r13 4 r23 69 2.6.5. Energy in a charge distribution • Total energy stored in the system (example for N = 3): W W1 W2 W3 q1 q1 q2 q1 0 q2 q3 4 r12 4 r13 4 r23 • Present the energy in the symmetrical form: W q1 q1 q3 q3 q2 1 q2 q1 q2 q3 2 4 r21 4 r31 4 r12 4 r32 4 r13 4 r23 1 q1V1* q2V2* q3V3* 2 • Generalize for a “discrete” distribution of charges: W 1 qiVi 2 i 2.6.5. Energy in a charge distribution • Generalize for a “discrete” distribution of charges : W 1 qiVi 2 i • Generalize for a “continuous” distribution of charges : dq v dv or s ds or l dl W 1 V , , dq , , 2 Area with charges • It can be proved for limited distribution of charges: W 1 1 D E dv 0 E2 dv 2 Area with charges 2 Area with charges We can define the function of energy density: w 1 0 E2 2 J / m 3 70 2.7. Dipole • Good approximation for models of atoms. • Popular used to explain many physical phenomenas • Dipole consists of 2 charges of opposite signs, same absolute value, placed very close to each other (d 0) 2.7. Dipole • Good approximation for models of atoms. • Popular used to explain many physical phenomenas • Dipole consists of 2 charges of opposite signs, same absolute value, placed very close to each other (d 0) 71 2.7. Dipole • Good approximation for models of atoms. • Popular used to explain many physical phenomenas • Dipole consists of 2 charges of opposite signs, same absolute value, placed very close to each other (d 0) V V V Q 1 1 4 0 R R Q R R 4 0 R R Q d cos 4 0 r 2 2.7. Dipole • Potential created by a dipole: V r , , Q d cos 4 0 r 2 • Field intensity: V 1 V 1 V E grad V ar a a r r sin r Q d 2 cos Q d sin ar a 0 a 3 2 r r 4 0 r 4 0 Qd 2 cos ar sin a 4 0 r 3 72 2.7. Dipole • Potential created by a dipole: V r , , Q d cos 4 0 r 2 • Field intensity: V 1 V 1 V E grad V ar a a r r sin r Q d 2 cos Q d sin ar a 0 a 3 2 3 r 4 0 r r 4 0 Qd 4 0 r 3 2 cos ar sin a 2.7. Dipole • Moment of a dipole: p Q AB Q d Because: C m d a r d cos V Q d cos 4 0 r 2 Q d ar 4 0 r 2 p ar 4 0 r 2 73 2.8. Dielectric (insulator) and conductors • Materials with strong interaction between electrons – nucleus, number of “free” ions limited an insulator/dielectric • Materials with weak interaction between electrons – nucleus, number of “free” ions huge (almost infinite) conductors (Ideal conductors has unlimited “free” ions (such as electrons) No consideration yet of semi-conductors in this class! 2.8.1. Dielectric • When there is no external electric field the atom is neutral (total charge = 0) • When there is an external electric field the orbitals of electrons and nuclei changed the atom can be approximated by a dipole. 74 2.8.1. Dielectric • When there is no external electric field the atom is neutral (total charge = 0) • When there is an external electric field the orbitals of electrons and nuclei changed the atom can be approximated by a dipole. Note: When a dipole appears New electrical field (to be superimposed with existing fields) with the trend to: - Outer area of dipole:………………. - Inner area of dipole:………………. 2.8.1. Dielectric Analyzing the differential thin outer layer of the material: • Some dipoles will have 1 pole located inside the material, 1 pole outside of the material (so called the border dipoles). • What is the sign of the poles outside of the material?.............. • What is the thickness of the layer containing the border dipoles?......................... • Only the border dipoles can affect the total inside charge Qinside. • The dipoles will create Qsurface, the material may be charged with some Qinside before. Qtotal Qsurface Qinside 75 2.8.1. Dielectric • Assume P (C/m3) – vector of dipole density (number of dipoles formed in an unit of volume) within a differential volume dv there is: dv dp P dv • For a volume with surface S, thickness dcos : v S d cos p total P v P S d cos P S d On the other hand: ptotal Qsurface d Qsurface P ds → S P ds closed surface enclosing S 2.8.1. Dielectric • Reconsidering the definition of vector of electric flux density: Qinside Qtotal Qsurface 0 E ds P ds 0E P ds Ddielectric 0 E P S S S Limit the consideration for isotropic materials, which have P E Ddielectric 0 E P 0 r E 0 1 e E Where: r – relative permittivity (1) of the material. e – electric susceptibility of the material. Note : 0 r E E where – the (absolute) permittivity of the material. 76 2.8.1. Dielectric Relative permittivity of some common dielectrics: 2.8.1. Dielectric In dielectric, the magnitude of Coulomb’s force, the field intensity and the potential are reduced by r times: FQ1 ,Q2 E( r ) V (r ) Q1Q2 1 4 r 0 r 1 Q1 4 r 0 r 2 2 ar 1 Q1Q2 4 r 2 1 Q1 4 r 2 ar 1 Q 1 Q 4 r 0 r 4 r 77 2.8.2. Border condition between two layers of dielectric Based on two integrals: 1. P : E dl 0 P 2. S : D ds Qinside S E dl P h w AB AB BC CD DA E1 AB E2 CD 0 CD E1 E2 D D 1 2 1 2 2.8.2. Border condition between two layers of dielectric Based on two integrals: 1. P : E dl 0 P 2. S : D ds Qinside S D ds Qinside S D1 ds D2 ds s S S D1 S S D2 S s S D1 D2 s 1 E1 2 E2 s In most of the cases s = 0: D1 D2 0 1E1 2 E2 78 2.8.2. Border condition between two layers of dielectric Name the angles form by E1 and E2 with the normal vector is 1 and 2 E1 sin 1 E2 sin 2 1E1 cos 1 2 E2 cos 2 tan 1 1 tan 2 2 tan 1 1 tan 2 2 2.8.3. Conductor Conductor has “unlimited” amount of free ions! When conductors are charged additionally with free charges the free charges moves to the surface Inside the conductor is charge-free! 79 2.8.3. Conductor When conductors are put in an external electrical field: The free ions are moved due to the action of the external field The relocation of the charges will create new field. When the movement (relocation) stops? In this class, we don’t consider yet the details process of relocation of charges (the transient state), but only the steady states! 2.8.3. Conductor The field intensity inside the conductors = 0: Conductors are equi-potential! In the steady state, the charges are “static” the direction of the electric field is perpendicular to the surface (same as the normal vector of the surface) If the conductor is hollow the hollow space has no electric field! (this property helps to create the protection cage against the external electric fields) 80 2.8.4. Border condition between a dielectric layer and a conductor layer Based on two integrals: E dl 0 1. closed P : P 2. closed S : D ds Qinside S E dl P h w AB AB BC CD DA E1 AB E2 CD 0 CD E1 E2 0 D D 1 2 0 1 2 2.8.4. Border condition between a dielectric layer and a conductor layer Based on two integrals: 1. closed P : E dl 0 P 2. closed S : D ds Qinside S D ds Qtrong S D1 ds D2 ds s S S S S D2 S s S D1 s D1 D2 s s 1 E1 2 E2 s E1 1 D1 81 2.9. Capacitance(s) in a closed system • A closed system of conductors a system with total charge equal 0: • If limit the consideration to system of 2 conductors one conductor is charged +Q, the other charged with –Q • Since the conductors are equi-potential There exists a potential difference between the conductors The system is called a capacitor. The conductors are called the plates of the capacitor. • Experiments showed that the potential difference is proportional to the charge Q on each plate. The ration Q/U is called the capacitance of the capacitor: C Q (F ) U Note: 1F is a huge value, more common are capacitance of pF, nF, F. 2.9. Capacitance(s) in a closed system Example: A capacitor with two “infinite” flat plates charged with constant surface density s. • The absolute intensity of the field generated by each of the plate: E E s Q 2 2 S • The intensity of the field in the area between the plates: E 2E Q S 82 2.9. Capacitance(s) in a closed system Example: A capacitor with two “infinite” flat plates charged with constant surface density s. • Let the plates of the capacitor be parallel to Oz: E E ay Q S ay • The potential difference between the plates: negative plate V negative plate E dl positive plate positive plate negative plate E negative plate E a y dl dy E d positive plate E dy positive plate Qd S • The capacitance of the capacitor: C Q S U d Note: Range of the capacitance = ? 2.9. Capacitance(s) in a closed system • The capacitance of the “flat” capacitor: C S d • The energy stored in the capacitor: WE E const 1 1 E 2 dv E 2 dv 2 capacitor's 2 capacitor's space space 1 E 2 The volume of the area 2 2 1 1 Q 1 Q2 d E2 S d = Sd = 2 2 S 2 S Simplified formula: WE = 1 Q2d 1 1 1 Q2 C U 2 Q U 2 S 2 2 2 C 83 2.9. Capacitance(s) in a closed system Exercises: Find the potential difference between the capacitor’s plates for following configurations: • Two coaxial cylinders of radius R1 < R2 • Two cylinders of radius R1 < R2, placed parallel to each other with the distance between two axes d >> R2 • Two concentric spheres of radius R1 < R2 • Two spheres of radius R1 < R2 , distanced from each other by d >> R2 2.10. Poisson’s – Laplace’s equations What we known until now: div D v E grad (V ) div grad V v 0 D 0E div grad V lapl (V ) V Laplacian of a function: • In Cartesian coordinates: lapl ( f ) f 2 f 2 f 2 f x 2 y 2 z 2 • In spherical coordinates: lapl ( f ) f 1 2 f 1 f 1 2 f r 2 sin 2 2 r r r r sin r sin 2 • In cylindrical coordinates: lapl ( f ) f 1 f 1 2 f 2 f 2 2 z 2 84 2.10. Poisson’s – Laplace’s equations Poisson’s equation: div grad V v 0 If in the domain of the equation there is no free charge v = 0 Laplace’s equation: div grad V 0 Theorem of unique solution: For two scalar functions V1 and V2 satisfying: • div(grad(V1)) = div(grad(V2)) or lapl(V1) = lapl(V2) • Same boundary conditions: V1(x) = V2(x) for x at the boundaries. → two func ons V1 and V2 are identical! 2.10. Poisson’s – Laplace’s equations Methods for solving Poisson’s – Laplace’s equations: • Directly finding the antiderivative functions (analytic solution of the equations) • Method of separating variables (self reading) • Method of images (mirror imaging) • Method of finite difference • Method of finite elements (self reading) • … 85 2.10. Poisson’s – Laplace’s equations Method of mirror imaging: Transform a charge distribution to an equivalent distribution with simpler calculation. Example: Single charge placed near to an “infinite”, grounded plate. V2 ( P ) ? 1. v1 v 2 (for z 0) lapl V1 lapl V2 2. V1 z 0 V2 z 0 0 V1 V2 0 for zP 0 : V1 P V2 P = 1 Q Q 4 0 r r 2.10. Poisson’s – Laplace’s equations Example: Single charge placed near to a conducting, grounded sphere. • Charge q placed near a sphere of radius a, distance between centers d >a. • Remove the conducting sphere, add new qa charge q • We can prove that: d 1. v1 v 2 (for r a) lapl V1 lapl V2 2. V1 r a V2 r a 0 V1 V2 0 for r a : V1 P V2 P = 1 q q 4 0 r1 r2 86 2.10. Poisson’s – Laplace’s equations Example: A straight wire placed parallel to a conducting plate. • The wire is charged with a line density l, placed parallel to the conducting plate zOy • Remove the plate, add a new wire mirroring symmetrically (but with line density –l. • We can prove that the new configuration is equivalent to the original (for x<0) 2.10. Poisson’s – Laplace’s equations • First, consider the field generated by the positive charged line, assuming that the “GROUND” is at the origin O: l a 2 0 r r E VP rO E dl rP • Then, consider the field generated by the negative charged line, still assuming that the “GROUND” is at the origin O: VP rO E rP • l ln r ln(a) 2 0 dl l ln r ln(a ) 2 0 Superimpose two fields: VP VP VP r l ln r ln r l ln 2 0 2 0 r 87 2.10. Poisson’s – Laplace’s equations • Convert into Cartesian coordinates: y2 x a r l ln l ln 2 0 r 2 0 y 2 x a 2 2 VP • To find the equi-potential lines, we solve the equation: y2 x a 2 y2 x a 2 ln K1 l 1 ln K1 l VP 2 0 2 4 0 K1 2 2a K1 K1 1 4 K1a 2 2 x a y 2 K1 1 K1 1 K1 1 = equation for a circle of radius 2 K 1 , centered at a 1 , 0 K1 1 K1 1 2a K1 K1 = 1 the plane x = 0. 2.10. Poisson’s – Laplace’s equations • Using the Cartesian coordinates: y2 x a r l l VP ln ln 2 0 r 2 0 y 2 x a 2 2 Vector of field intensity: 2 2 2 l 2a y a x a x 4a x y a y E grad V 2 2 2 0 y2 x a y2 x a equation for the streamlines of the field : x 2 y a cot K 2 2 a2 Ey dy 2 xy dx E x y 2 a 2 x 2 sin 2 K 2 = Equation of a circle of radius a , centered at 0, a cot K 2 sin K 2 88 2.10. Poisson’s – Laplace’s equations • Equation for streamlines of the field: • Equation for equi-potential lines: x 2 y a cot K 2 2 2 2a K1 K1 1 4 K1a 2 2 xa y 2 K1 1 K1 1 K1 1 a2 sin 2 K 2 2 2.10. Poisson’s – Laplace’s equations Method of finite difference: (in the Cartesian coordinates, self-reading about the other two system of coordinates) Laplace’s equation: lapl (V ) 2V x 2 2V y 2 2V z 2 0 Assume the domain is on xOy (formulas for 3D can be developed in analogical way): 2V x 2 2V y 2 0 Finite difference formulas: f ( x ) f ( x h) f ( x) f ( x) h f ( x h) f ( x ) f ( x ) f ( x h) h h f ( x h) f ( x) f ( x ) f ( x h) f ( x h) 2 f ( x ) f ( x h) h h h h2 89 2.10. Poisson’s – Laplace’s equations Laplace’s equation for each of the nodes in the grid: V1 V3 2V0 x 2 V2 V4 2V0 y 2 0 1 1 V V V V 2V0 2 2 1 2 3 2 2 4 y x y x V1 V3 V2 V4 x y 2 V0 1 1 2 2 2 y x 2 Limit consideration to squared grid: x y V0 V1 V2 V3 V4 4 2.10. Poisson’s – Laplace’s equations General algorithm: For each node with unknown voltage, form 1 equation number of equations = number of unknown nodes! 90 2.10. Poisson’s – Laplace’s equations Solve the system of N equations, N unknowns: Gauss’s elimination,… 2.10. Poisson’s – Laplace’s equations Solve the system of N equations, N unknowns: Iterative method for x = F(x) Vd Vb 100 Va V V V 100 V e c a b Vb V f 100 Vc Vg Ve Va Vd 1 Ve V V V V d h f 4 b V f V Ve Vi c V V V g d h V Ve Vg Vi h V V Vi h f 91 2.10. Poisson’s – Laplace’s equations Solve the system of N equations, N unknowns: Iterative method for x = F(x) Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Step 8 Va = 0.25(Vd+Vb+100) Vb = 0.25(Va+Ve+Vc+100) Vc = 0.25(Vb+Vf+100) Vd = 0.25(Va+Ve+Vg) Ve = 0.25(Vb+Vd+Vf+Vh) Vf = 0.25(Vc+Ve+Vi) Vg = 0.25(Vd+Vh) Vh = 0.25(Ve+Vg+Vi) Vi = 0.25(Vf+Vh) 2.10. Poisson’s – Laplace’s equations • How to find the field intensity at the nodes? • How to find the potential and field intensity at any point? 92 3. Steady electric field 1. Flow of charges, vector of current density, current (intensity) 2. Steady electric field 3. Voltage difference between two points of a conductor 4. Conductor resistance Charges flowing inside a conductor! 3.1. Flow of charges and vector of current density Flow of charges: charges are moved along a conductor. The direction of the flow is the moving direction of the positive charges. Current I: is the total charge moved through a surface (the conductor crosssection) in 1 sec. I dQ ( A) dt A better description of the currents will be the vector of current density J: J ( A / m2 ) 93 3.1. Flow of charges and vector of current density Vector of current density J as a function of speed (average movement) and volume density: J v v From J to I: I J ds ( A) S Note: • It’s not easy or straight forward to find J from I • Movement direction of charges will follow the shapes of the conductors • Kirchhoff Current Law for currents! 3.1. Flow of charges and vector of current density Example: Conductor with constant cross-section with constant current density. I J ds J ds (because J ds ) S S J ds (because J const ) S J Area S = J S I S I J = az S J 94 3.1. Flow of charges and vector of current density Định luật bảo toàn điện tích: Tổng lượng điện tích chuyển ra khỏi một mặt kín bằng lượng điện tích thay đổi trong khối đó. dq J ds dt S d v dv dt V Sử dụng định lý Green: J ds div J dv S V div J v t 3.2. Steady electric field • The charges are moved due to forces generated from the power of the sources F v & Esteady v • E is proportional to J: Esteady F v J • For linear material: J Esteady where: or Esteady J S / m : conductivity of the material m : resistivity of the material 95 3.3. Potential difference (voltage drop) in steady field • The potential difference between 2 points in the (steady) electric field: B U AB Esteady dl A Note: • We can choose an arbitrary point to be the “GROUND” (with potential equal 0) • In static field, conductors are equi-potential, but in steady field conductors are not equipotential. Example: J= I 1 I a z Esteady J a S S z B B I l AB I dz S S A U AB Esteady dl A 3.3. Potential difference (voltage drop) in steady field Example: Find the potential difference between two layers of conductor using insulator with conductivity > 0. Let the length L of the wire be dominate, ignoring the influence of the terminals of the wire. Consider S – closed surface enclosing a coaxial cylinder of radius r (R1 < r < R2) S = 2 bases S1 and S2 + curved surface S3 Two bases S1 and S2 have normal vector perpendicular to J surface integral = 0 96 3.3. Potential difference (voltage drop) in steady field Example: Find the potential difference between two layers of conductor using insulator with conductivity > 0. I J ds J ds S3 S3 J ds S3 J Area S3 J 2 L Using the cylindrical coordinate system: J I 2 L J ... 3.3. Potential difference (voltage drop) in steady field Example: Find the potential difference between two layers of conductor using insulator with conductivity > 0. I I a 2 L 2 L I I d E dl d 2 L 2 L J U ab a E steady R2 I 2 L R1 d I 2 L R ln R2 1 I ln R2 ln R1 2 L R I ln 2 2 L R1 97 3.4. Conductor resistance (Ohm’s law for linear conductor): The potential difference between the two terminals of a conductor is proportional to the current flowing though that conductor. RAB U AB I AB Example: for cylindrical conductor U AB I l AB S RAB U AB l l AB AB I S S Note: The nonlinear resistors have a nonlinear relationship between u and i: U AB f I AB 3.4. Conductor resistance Example: The “leaking” resistance between two layers of conductors in coaxial cable: U AB RAB R ln 2 2 L R1 I R U AB 1 ln 2 I 2 L R1 In practice, it’s common to use the leaking resistivity (resistance per unit length): GAB 1 2 L RAB ln R2 R1 G0 G AB 2 L ln R2 R1 Exercise: Find the leaking resistivity between two cylindrical conductors (of radiuses R1 and R2) placed parallel to each other at distance d >> R1, R2 98 3.4. Conductor resistance Example: Find the “step” voltage around a leak point: Consider surface S enclosing half sphere of radius r > R (consisting of S1 is the half sphere surface, S2 is the base circle on the ground) I I ra J ds S J (r ) ds ignoring the leak to the air S1 J (r ) ds S1 J (r ) ds S1 J (r ) 2 r 2 J r I 2 r 2 3.4. Conductor resistance Example: Find the “step” voltage around a leak point: Let the sphere is located at the origin : J r I 2 r 2 Esteady r ar I 2 r 2 ar The potential difference between two points: b U ab Esteady dl a b a b I 2 r 2 dr a I dr 2 r 2 I 1 1 2 ra rb 99 3.4. Conductor resistance Exercise: Find the leak resistance of a flat capacitor. Find the leak resistance of a spherical capacitor. 4. Static magnetic field 1. Ampere’s and Lorentz’s forces 2. Biot – Savart’s law 3. Magnetic field intensity vector H, magnetic induction vector (magnetic flux density) B 4. The magnetic flux passing through a surface 5. Self inductance, mutual inductance 6. Law of electromagnetic induction 7. Magnetic circuits 100 4.1. Ampere’s and Lorentz’s forces • The Ampere’s force: Two wires with currents act on each other: if 2 currents are of same direction attract each other, if 2 currents are of opposite directions repel each other. • This is not the force in static electric field. • The force is similar to the force created by magnets. There exists a magnetic field “around” the wire (include the inner space of the wire) with currents flowing through. • The magnetic field is characterized by vector of magnetic intensity H (A/m) or vector of magnetic induction B (Tesla) • The Lorentz’s force: Fmagnetic Q v B 4.1. Ampere’s and Lorentz’s forces • Magnetic force acts on a moving charge: Fmagnetic Q v B • If the area also has an electric field: Ftotal QE Q v B Q E v B • The current I: 1A = 1C / 1s • Assume the line density of the moving charges is l I l v In some cases, we use vector of current: I I av 101 4.1. Ampere’s and Lorentz’s forces • Magnetic force acting on a current: Fmagnetic v B dq v B l dl I B dl Because: I dl Fmagnetic I dl B • For case with 1 conductor and 1 current I = const Fmagnetic I dl B dq l dl s ds v dv • Other type of current distributions: I dl K ds J dv 4.2. Biot – Savart’s law • Force between two currents: F12 where: 0 4 I1 dl1 I 2 dl 2 a12 r122 L1 L2 0 4 107 N / A2 • Vector of magnetic induction of the field of 1 wire: Br 0 4 L1 I ar r 2 dl 0 dl a r I 4 L r 2 1 T 1T 1 N Am 102 4.2. Biot – Savart’s law Example: Magnetic induction of a wire segment: P dl a r dl sin dl cos Because: l tan dl cos2 d d cos 2 0 dl cos 0 cos B P I I 4 L 4 L r2 2 1 1 cos 2 0 I cos d 4 L 1 0 I sin 2 sin 1 4 For an infinite wire: B 0 I B ... 2 4.2. Biot – Savart’s law Example: Find the magnetic induction on the symmetric perpendicular axis of a conducting ring: 103 4.2. Biot – Savart’s law Exercise: Find the vector of magnetic induction at point P. P (c) 4.3. Ampere’s law • Similar to Gauss’s law in electrostatic field • Let the closed path P be a circle of radius B dl ? P I I I B 0 a B dl 0 d 0 d 2 2 2 0 I B dl 2 d 0 I P P We can generalize to: • For all closed path P • For all combination of currents B dl 0 Iinside P 104 4.3. Ampere’s law Example: Using Ampere’s law to find the magnetic induction vector of the field around an infinite straight wire. 4.3. Ampere’s law Example: Using Ampere’s law to find the magnetic induction around a straight coil. 105 4.3. Ampere’s law Example: Using Ampere’s law to find the magnetic induction around a straight coil. 4.3. Ampere’s law Example: Using Ampere’s law to find the magnetic induction around a torus 106 4.4. Divergence and rotation of B • Divergence of B: B 0 I a div B 0 2 div B 0 Can be generalized for any static magnetic field: • Rotation of B: B dl 0 Iinside P B dl rot B ds P S P rot B 0 J I inside J ds S P 4.5. Vector of magnetic field intensity • Magnetic field intensity vector H: H 1 0 B A / m • Similarity between formulas for static electric field and static magnetic field: D 0 E B 0 H • Similarity between formulas for Gauss’s law in electric and magnetic fields: D ds Qinside B ds 0 S S 107 4.6. Vector of magnetic potential • (Remind) Potential (scalar) in a static magnetic field: V A "0" E dl; E grad V A • Divergence of B: div B 0 f : rot grad f 0 • Properties of rot() and grad() (for any functions): v : div rot v 0, There exists a vector function A: B rot A f : rot A grad ( f ) rot A B Select vector function A, such that: B rot A div A 0 4.6. Vector of magnetic potential • Rotation of B: B rot A rot B rot rot A grad div A lapl A lapl A 0 J General form of solution: A r 0 J r dv 4 r In Cartesian coordinates: A Ax a x Ay a y Az a z lapl A lapl Ax a x lapl Ay a y lapl Az a z lapl Ax 0 J x lapl Ay 0 J y lapl Az 0 J z 108 4.7. Magnetic material and magnetic induction • Magnetic induction in vacuum: B 0 H • When the magnetic material is placed in an external magnetic field there will be new inducing fields B 0 H M • Limit consideration to paramagnetic material (H and M are parallel, B is increased) B 0 H M r 0 H H r relative permeability 4.7. Magnetic material and magnetic inductance • Examples of B-H characteristics: 109 4.7. Magnetic material and magnetic inductance • Magnetic hysteresis effect: 4.8. Border conditions for magnetic fields Based on two integrals: 1. closed surface S : B ds 0 S 2. closed path P : H dl I inside P B ds 0 S B1 ds B2 ds 0 S B1 S S B2 S 0 B1 B2 1H1 2 H 2 110 4.8. Border conditions for magnetic fields Based on two integrals: 1. closed surface S : B ds 0 S 2. closed path P : H dl I inside P H dl P AB H1 BC AB h L CD DA H 2 CD AB Iinside K CD L H1 H 2 K B1 1 B2 2 K 4.9. Magnetic flux • In a magnetic field with B, for a surface S The magnetic flux passing through surface S is: B ds S Example: Flux through a rectangular area in the field of a straight current. 111 4.9. Magnetic flux Example: The total flux through a coil 4.9. Magnetic flux Example: The total flux through a torus 112 4.10. Self-inductance and mutual inductance • The flux passing through a surface depends on the current. If the system is linear ration Ψ/I is a constant • If the source B of Ψ is caused by the current flowing in the surface Self flux Ψ L self I self inductance (H) • If the source B of Ψ is caused by the current NOT flowing in the surface Mutual flux Ψ M mutual I mutual inductance (H) 4.11. Law of electromagnetic induction • Consider a closed path P and a surface S “based” on P. einduced d dt If there is a closed circuit, the induced current will create an induced field to counter the changes caused by the original field. Example: Find the direction of the induced current when the frame is moved. 113 4.11. Law of electromagnetic induction Example: Principle of rotating electric generators 4.11. Law of electromagnetic induction Example: The skin effect in conductors with AC currents 114 4.11. Law of electromagnetic induction Example: Induction cooker 4.12. Magnetic force acted on the currents • Lorentz’s force: F Q E v B • Differential magnetic force: dF dQ v B • When there is a current: dQ v dv • Represent J as function of volume (moving) charge density: J v v; J dv I dl Differential magnetic force: dF J B dv I dl B Magnetic force acted on a segment of wire: B B B A A A FAB dF I dl B I B dl 115 4.13. Magnetic circuit • Frames made of magnetic materials • Sources of magnetic fields are the coils with currents • Need to determine the fluxes in the frames For simplification, there are approximations needed: • Ignore the leak into air, the fringe field effects. • Ignore the variations at the “corners” of the frames, let assume that the frame can be divided into segments with constant fields. 4.13. Magnetic circuit Base formulas: • Flux through a cross-section: B ds B S • Flux through a closed surface: S B ds 0 S • Ampere’s law: For all closed paths Iinside H dl P • Characteristic curve of B-H (if nonlinear) or the parameters of B r 0 H 116 4.13. Magnetic circuit System of equations for a magnetic circuit: • Nr of equations: number of branches with unknown fluxes • Nr of equations for “nodes” = nr of high order nodes - 1 • Nr of equations for “loops” = total nr of equations – nr of equations for nodes. Example: • Nr of equations: 3 corresponding to Φ1, Φ2, Φ3 • Nr of equations for “nodes” = 2 – 1 = 1 • Nr of equations for “loops” = 3 – 2 = 1 4.13. Magnetic circuit System of equations for magnetic circuit: • Equation for node: 1 2 3 e • Equations for loops: f H ea lea H ax lax H xy l xy H yc l yc H ch lch H he lhe N I H af laf H fg l fg H gc l gc H cy lcy h g H yx l yx H xa lxa 0 General algorithm: Transform the equations for loops in to equations of fluxes using relationships B H,B H f S → System of equations for fluxes! 117 Example: A magnetic circuit with a single air gap is shown with cross-section area Ac = 1.8 x 10-3 m2, mean core length lc = 0.6 m, gap length: g = 2.3 x 10-3 m, N = 83 turns. The current in the coil is i = 1.5 A. Assume that the core is of infinite permeability ( m →∞) and neglect the effects of fringing fields at the air gap and leakage flux. (a) Calculate the reluctance of the core Re and the reluctance of the gap Rg. (b) The total flux ϕ (c) The flux linkages λ of the coil. (d) The coil inductance L. Example: A magnetic circuit with a single air gap is shown with cross-section area Ac = 1.8 x 10-3 m2, mean core length lc = 0.6 m, gap length: g = 2.3 x 10-3 m. Assume that the core is of infinite permeability ( m →∞) and neglect the effects of fringing fields at the air gap and leakage flux. (a) The number of turns required to achieve an inductance of 12 mH (b) The inductor current which will result in a core flux density of 1.0 T. 118 Example: In the given magnetic circuit, the relative permeability of the ferromagnetic material is 1200. Dimensions are given in cm, the cross-section is square. Determine: a) The air gap flux, b) The air gap flux density, c) The magnetic field intensity in the air gap. 119 Example: For the given magnetic circuit and the B-H curve, find the flux density and flux in each of the outer limbs and the central limbs. Assume the relative permeability of iron of the core to be: (a) ∞ (b) 4500. Example: The magnetic circuit is made of cast-steel core. The cross-sectional area of the central limb is 800 mm2 and that of each outer limb is 600 mm2. Calculate the exciting current needed to set up a flux of 0.8 mWb in the air gap. Neglect magnetic leakage and fringing. 120 Example: In the given magnetic circuit, the coil F1 is supplying 4000 Ampere-turn in the direction indicated. Find the Ampere-turn of coil F2 and current direction to produce air-gap flux of 4 mWb from top to bottom. The relative permeability of iron may be taken as 2500. Example: For the given magnetic circuit, the airgap flux is 0.24 mWb and the number of turns of the coil wound on the central limb is 1000. The magnetization characteristic of the core is given in the table. Calculate: (a) The flux in the central limb, (b) The current required. 121
