* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download electrostatic 2014-2015(chap1)
Survey
Document related concepts
Metric tensor wikipedia , lookup
Casimir effect wikipedia , lookup
Introduction to gauge theory wikipedia , lookup
Four-vector wikipedia , lookup
Speed of gravity wikipedia , lookup
Fundamental interaction wikipedia , lookup
Electromagnetism wikipedia , lookup
Magnetic monopole wikipedia , lookup
Potential energy wikipedia , lookup
Work (physics) wikipedia , lookup
Centripetal force wikipedia , lookup
Maxwell's equations wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Field (physics) wikipedia , lookup
Lorentz force wikipedia , lookup
Transcript
Institute of Technology of Cambodia Department of Foundation Year Electricity Part I Electrostatics Chapter I Field, Potential, and Energy of Electrostatics I- Coulomb’s law Coulomb’s law explain the interaction of charges like points Supposing, there are two point charges at P and M P q1 F21 F12 M q2 The charge q1 exerces the force F12 on q2, and the charge q2 exerces the force F21 on q1. F12 and F21 are defined by: II- Charge Distribution 1) Discontinuous Distribution of Charge A system consists of many point charges. The Total charge of the system is defined by the sum of each point charge: n Q qi i 1 2) Continuous Distribution of Charge There are three kinds of distributions: a) Volume distribution Let ρ is the density of charge in a volume, (C/m3) So that the charge contained within the volume is: b)Surface Charge distribution Let σ is a charge density on a surface So that the charge contained on the surface is: c) Line charge distribution Let λ is density of charge (C/m) So that the charge contained the line is: III Electrostatics Field 1) Electrostatics field of a point charge At P we place a charge of q, at M we place charges of q1, q2, ……., qn . They are called test charges. If we place q1at M, it experiences a force F1 by q P q F1 M q1 If we place q2at M, it experiences a force F2 by q P q F2 M q2 If we place qn at M, it experiences a force Fn by q Fn M qn P q The Force Fn is defined by: Fn 1 qq2 4 0 r PM 3 F 1 q PM n qn 4 0 r PM 3 PM So that the last expression is called the “Electrostatics Field, E” of a point charge q P q M EM In general a charge q at point P, creates an Electrostatics field at M. The vector field E is defined by: EM 1 q 40 r PM 3 PM 2) Superposition of many point charges There are n point charges q1, q2, ….., qn are placed at point P1, P2, ….., Pn which radius vector r1, r2, ….., rn respectively. M is an other point with radius vector r. The resultant of electrostatics field at point M is defined by the superposition of the fields created by each charge. The expression of electric field at M is defined by: 3) Electric Field of a Continuous charge distribution For volume charge The element charge dq at point P creates An element field of dE at point M. The resultant field at point M is defined by intergral of dE The same for surface and Line charge distribution 4) Tube line and streamline If (C) is a closed contour, the vector field which across this contour form a tube. We call Tube Line (បំពង់ដែន) All point, the vector field is always tangent to the dr By solving this system equation we obtain the equation of streamline(សមីការដសែដែន) IV Gauss’ theorem 1- Orientation of a surface - n is called normal vector of the surface (S) - n is define by the rule of Maxwell or screw rule (For an open surface) - For a clossed surface direction of n is from - inside to out side 2) Flux of a vector Field The flux of the arbitrary vector field A through an arbitrary surface is the average normal component of ector A over the surface multiplied by the area of the surface. Flux across an element surface of dS is dφ 3) Solid Angle By definition; the Solid Angle is defined by normal area divided by square of radius Solid Angle of a closed surface In case of observation point O is out side the surface We obtain the total solid angle by integral : Which means that if the observation point is out side a closed surface the total solid angle is zero. In case of the observation point O is inside the closed surface Firstly, we would calculate the solid angle of a sphere of radius R center O and the observation point in coincide to the center O of the sphere. By supposing that we have a closed surface(Σ), the pink one. The observation point is O (inside Σ) then we chose a sphere center O, Radius R For all sharp of any closed surface if the observation point inside the surface, the solid angle equal 4π 4) Apply Gauss’s Theorem By supposing that we have a charge which distributes over a volume. At point P the element charge dq creates an element field dE at M on an arbitrary enclosed surface The element of flux across the surface dS is 1 d d dE.n ds PM . n ds 3 4 0 PM 2 d 2 1 4 0 dd Flux of E throught a surface element ds is: d 1 40 d d v Q 40 d Total Flux of E throught a closed surface S is: Qin 4 0 Qin Qin s d 4 0 4 0 Where Qin is total charge continue in a closed surface. By putting ε0E = D; is called Electric Flux Density Vector, we obtain: D.nds Qin s V - Electrostatics work and Electrostatics Potential 1) Expression of elementary work of an electrostatics force . charge dq at point P created at point M an electrostatics filed dEM dEM dEM d 1 40 r P M i 3 PM ; d 3 (r r ' ) 40 r r r ' 1 q is placed at M, the force dF applies on q is dF q d 3 (r r ' ) 40 r r r ' 1 The elementary work effected by the elementary force dF is: 1 d W dF .dr q 3 (r r ' ).dr 40 r r ' 2 Or 2W q 1 d . d r r' 2 4 0 r r ' Or 1 W q d .d 4 0 r r' 2 1 The Elementary work effected by the force F=qE is: 1 W q .d d 40 V r r' d 1 W qd qdV 40 r r ' V 1 2) Electrostatics Potential a) For a point charge There is a point charge at point P. The electrostatics potential at M is defined by: Note: Potential is a scalar quantity, it can be positive or negative b) Superposition of Potential There are n point charges q1, q2, ….., qn are placed at point P1, P2, ….., Pn which radius vector r1, r2, ….., rn respectively. M is an other point with radius vector is r. The potential at M is defined by the sum of Potential of each point charge. V 1 40 n qi i 1 Pi M c) For a Continuous Charge VI Relationship between E and V 1) Gradient Field Let M is a point in free space: OM r xux yu y zuz rur Which means that E is a conservation field and rotE = 0 From the above relation we obtain: Is called electric potential 2) Gradient of V in the Three Coordinate Systems 3)Stock’s Theorem The circulation of an Electrostatics field along a closed contour is zero. VII- Local Equation of electromagnetism Divergence of a vector field By definition the flux of a vector A is defined by: and when the vector field is the electric flux density: = div D Maxwell’s first equation By Green Ostrogradsky By Gauss’Theorem: Qin 1 d s E.nds 0 0 v Divergence Expressions in the Three Coordinate Systems VIII Potential Energy of interaction 1) Electric work of a point charge in an External Field The is a point charge is placed in an external electric field E. The charge experiences an electric force F=qE. Work done by the electric force to move the charge q from A to B is: By putting UP(A)=qVA; UP(B)=qVB are called potential energies at A and B If a man (external force) want to move the charge q from A to B, he need a force Fex= - F = -qE, and the work done by himself is Wex: Wex Fex .dr qE.dr B B Wex q E.dr q dV qVB VA U P ( B ) U P A) A A This means that UP(a) and UP(B) is the work done by the electric force to take the charge q from point A and B to infinite, or the work done by the man (external force) to take the charge q from infinite to point A and B . The meaning of potential energy at a point is the work done by the man (external force) to move the charge q from infinite to that point disused. 2) Potential Energy of Interaction of a Point Charge Distribution Supposing there is a system consists of three point charges q1; q2; q3 are located with radius vectors r1; r2; r3 respectively. The system produces potential interaction energy. It equal the work done by the man to take them from infinite to each point Firstly, to bring q1 from infinite to r1 we don’t need work, W(ex1) = 0 Secondly, to bring q2 from infinite to r1 we need work W2 1 q1q2 Wex 2 F12 .dr 4 0 r r1 Wex 2 1 4 0 q1q2 . d r r1 2 r r1 r r1 .dr 3 Wex 2 W( ex 2 ) Wex 2 q1 q 2 d 4 0 r r1 1 q1 q 2 d 4 0 r r1 q1 q 2 1 1 q1 q 2 4 0 r r1 4 0 r12 1 r2 Thirdly, to bring q3 from infinite to r3 we need work W(ex3) q2 q3 1 q1q3 4 0 r3 r1 r3 r2 1 q1q3 q2 q3 Wex 3 4 0 r13 r23 Wex 3 The total work to bring each charge to each place we need W(ex)= W(ex1)+ W(ex2)+ W(ex3) 1 q1q2 q1q3 q2 q3 Wex 4 0 r12 r13 r23 Wex qi q j 40 i 1 j 1 rij 1 3 3 j i In case of many point charges can reduce this relation to: As we know the potential energy interaction equal to the work done by the external (man) to bring all charge from infinite to each point. We obtain; IX Electrostatics Energy of a continuous charge 1) Electrostatics Energy Interm of Potential We have; For a continuous charge we just change Σ to ʃ; to dq; and V(ri) to V(r) We obtain; 2) Electrostatics Energy Interm of Electric Field From 1 U e Vd 2 v And divE 0 1 U e 0 V .divEd 2 v div (VE ) VdivE gradV .E VdivE E 2 0 0 U e div (VE )d E 2d 2 v 2 v By Osrograsky Green; div (V .E ).d V .E.ds v s As; r ,V 0 V .E.ds 0 s We obtain; 1 U e 0 E 2 d 2 all space X) Boundary Condition Supposing, there are two deference medium separated by a surface, and their permittivity are ε1 and ε2 respectively. An electric field traverses from medium (1) to medium (2). We want to study about their Normal component and Tangent components. 1) Tangent Components of Field We have; c d a E1n E1t drab En Et drbc E2 n E2t drcd En Et drda 0 b a b c c d a E1t drab En drbc E2t drcd En drda 0 b a b c d d a En drbc En drda 0 c b d d E2t drab E1t drcd 0 b a c E1t l E2t l 0 2) Normal Components of a Field The total flux of field D across a closed cylindrical is zero. D.nds ds s s D1n .n1ds1 D2 n .n2 ds2 ds s s D2 n D1n n21ds ds s s s D2 n D1n We obtain; 2 E2n 1E1n