EE468G NOTES (3) Contents: Static Fields Vector operations Differential operatiors Integrals (volume, surface, line) Coulomb’s Law and Gauss’ Law Potential functions Boundary conditionSources for electromagnetic fields Charge (Q): Unit: Coulomb, [C] Polarization: positive, negative Elemental charge partical: electron and proton Electron charge = ][1060210.119C Point charges Charge distribution: volume, surface, line Volume charge density: V,  VVdVzyxQ ,, Surface charge density: S Line charge density: l Law of conservation of charge: The total charge in a closed system remain the same for all time. Current (I): Flow of charges (Rate of change in charge) Unit: Ampere, [A], or [C/s] Direction: positive charge moving direction. Current = integral of current density Volume current density: [2A/m], Symbol: J Surface current current density: [ A/m ], SJ Continuity relation: Integral format: tQSdJSenc, Differential format: tJVTwo point charges suspended in free-space will act a force to each other. Consider one charge in space. It has the ability to act a force to a charge if it is put into the space. We use a test charge to examine the force. If the testing charge has charge Q, the force acting on it is F, then, the electric field intensity is defined as: QFEQ0lim, [ N/C, or V/m] Lorentz force: The forces act on a test charge Q by the fields.  BuQEQBuEQF The first term is the electric force, it defines electric field intensity The second force is the magnetic force (acts on a moving charge), it defines magnetic flux density. 1Q 2Q 1F 2FCoulomb’s Law: Electric field created by a point charge Q at origin is given by  20ˆ4rQE r ar [V/m] Electric field unit: V/m Charge unit: Columb 1208.854 10 [F/m] is the free-space dielectric constant. Can use: 361090 [F/m] Features: (1) EQ (2) 21Er (3) E is in the direction of ˆra (4) E points outward for positive charge (5) E points inward for negative charge. Positive Charge Negative ChargePoint charge at location: zzyyxxrˆ'ˆ'ˆ'  generates electric field: R is called field-source distance. ˆRais the unit vector pointing from source point 'r to field (observation) point r. Features: (1) EQ (2) 21ER (3) E is in the direction of ˆRa (4) E points outward of 'rfor positive charge (5) E points inward of 'rfor negative charge. r 'r RRQaRQrER30204ˆ4)( , |'| rrR 'rrRExample: A point charge of 5 [nC] is located at P(2.0, 1.0, 4.0). Calculate the electric field it created at point P1(4.0,1.0,0.0). Solution: zyxrˆ4ˆˆ2' , yxrˆˆ4  zxrrRˆ4ˆ2'  20R     [V/m] ˆ95.7ˆ98.3ˆ4ˆ22036/10410540,1,439930zxzxRRQESuperposition theorem: The total electric field created by N charges together is the sum of the N electric fields created each of the charges individually.     '23202'13101222021210144ˆ4ˆ4rrRQrrRQaRQaRQrERR 1Q 2Q 1R 2RExample: Two charges Q1=1 [nC], located at P1(0.0, 1.5, 2.0), Q2=3 [nC], located at P2(2.0, 2.5, 3.0). Calculated the electric field at P(4,4,6) created by the two charges together. Solution: We first calculate the two individual electric fields. zyxzyxzyxaaaraaaraaarˆ3ˆ5.2ˆ2ˆ2ˆ5.1ˆ0ˆ6ˆ0.4ˆ4'2'1 For charge 1: 18.6,ˆ4ˆ5.2ˆ41'11 RzyxrrR     [V/m] ˆ153.0ˆ095.0ˆ153.0ˆ4ˆ5.2ˆ418.636/10410146,4,439913101zyxzyxRRQE For charge 2: 91.3,ˆ3ˆ5.1ˆ22'22 RzyxrrR     [V/m] ˆ355.1ˆ678.0ˆ903.0ˆ3ˆ5.1ˆ291.336/10410346,4,439923202zyxzyxRRQE The total field is      [V/m] ˆ508.1ˆ773.0ˆ056.16,4,46,4,46,4,421zyxEEEExample 2: Two charges Q1=1 [nC], located at P1(3,1,2), Q2 located at P2(2,1,1). Determine the value of Q2 so that the total field due to the two charges has no x component at point (0,0,0). Solution outline: Following the procedure of the previous example, express the total field at (0,0,0) with Q2 as a variable, set x component of the total field to 0 to solve for Q2. Solution: First determine the expressions of E1 and E2:  zyxRQEˆ2ˆˆ34)0,0,0(31011  zyxRQEˆˆˆ24)0,0,0(32022 04243)0,0,0()0,0,0()0,0,0(3202310121RQRQEEExxx Solve for Q2: nC][42.014142663231131322 QQRRQ Exercise: For the configuration of this problem, what is Q2 if the total field has no y-component or has no z-component at (0,0,0)? Answer: Q2= -0.28 [nC] if E to have no y-component at (0,0,0) Q2= -0.56 [nC] if E to have no z-component at (0,0,0). Exercise 2: Suspended point charges (p.112 of text book).Electric field created by distributed charge: Line charge density l, integrate over the line:       LLllrrrrdzrErrrrdzrEd''4'''4'3030 Consider the field at )0,,(: Field point: arˆ, Source point: zazrˆ''   2/3223'',ˆ'ˆ' zrrazarrz        aazdzzazdzarrdzazarrrrdzrEllzlllzLLlˆ224ˆ)'(''4ˆ)'('4ˆ'4'ˆ'ˆ''4'0202/3222/322003030 Feature: (1) Field magnitude /1, direction in aˆ x y 'rrR  dzIn the above configuration, if the length is 2L, then  aLLLLarEllˆ224ˆ2202220 Example: A line charge of infinitely long and with uniform