* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Physics 169
Survey
Document related concepts
Fundamental interaction wikipedia , lookup
Elementary particle wikipedia , lookup
Introduction to gauge theory wikipedia , lookup
Circular dichroism wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Work (physics) wikipedia , lookup
Vector space wikipedia , lookup
Maxwell's equations wikipedia , lookup
Field (physics) wikipedia , lookup
Metric tensor wikipedia , lookup
Magnetic monopole wikipedia , lookup
Euclidean vector wikipedia , lookup
Centripetal force wikipedia , lookup
Four-vector wikipedia , lookup
Lorentz force wikipedia , lookup
Transcript
Physics 169 Kitt Peak National Observatory Monday, January 30, 17 Luis anchordoqui 1 we see that like charges repel. Conversely, a rubbed balloon and your hair, which do not have the same kind of charge, are attracted to one another. Thus, unlike charges attract. 1.1 Electric Charge (a) n across the balloon charged and Two he other . 6 (b) (a) If you rub a balloon across your hair on a dry day the balloon and your hair become charged and attract each other (b) Two charged balloons, on the other hand, repel each other . The two balloons must have the same kind of charge because each became charged in the same way Because two charged balloons repel one another we see that like charges repel Conversely ☛ a rubbed balloon and your hair which do not have the same kind of charge are attracted to one another ☛ unlike charges attract Monday, January 30, 17 2 are applied. Another important aspect of electricity that arises from experimental obs Electric charge is conserved tions is that electric charge is always conserved in an isolated system. Th when one object is rubbed against another, charge is not created in the process. Charge is electrified conserved quantized stateand is due to a transfer of charge from one object to the other. object gains some amount of negative charge while the other gains an equal am of a positive example, when When glass charge. rod is For rubbed with silka glass rod is rubbed with silk, as in Fi 23.2, the silk obtains a negative charge that is equal in magnitude to the – electrons are transferred from the glass to the silk + – + tive charge on the glass rod. We now know from our understanding of atomic s – +++ Because of conservation of charge – ture that electrons are transferred from the glass to the silk in the rubbing pro + – eachSimilarly, electron adds negative charge to electrons the silk are transferred from the f when rubber is rubbed with fur, – rubber,positive giving the rubberisa left net negative and the an equal charge behind charge on theand rodthe fur a net po This process is consistent with the fact that neutral, uncharged m Also charge. ☛ because charges are transferred in discrete bundles contains as many positive charges (protons within atomic nuclei) as neg charges on (electrons). the two objects are charges In 1909, Robert Millikan (1868–1953) discovered that electric charge a Figure 23.2 When a glass rod is occurs as some integral multiple of a fundamental amount of charge e (see Se rubbed with silk, electrons are 25.7). In modern terms, the electric charge q is said to be quantized, where q i transferred from the glass to the standard symbol used for charge as a variable. That is, electric charge exis silk. Because of conservation of left charge, each electron adds negadiscrete “packets,” and we can write q ! Ne, where N is some integer. Other ex tive charge to the silk, and an equal A negatively rubber rod " e and the pr ments in the same period showed that charged the electron has a charge positive charge is left behind on has a charge of equalsuspended magnitude but # e. Some particles, suc by aopposite threadsign is attracted the rod. Also, because the charges the neutron,Rubber have no charge. to a positively charged glass rod are transferred in discrete bundles, Rubber From our discussion thus far, we conclude that electric charge has the followin the charges on the two objects are Fportant properties: $ e, or $ 2e, or $ 3e, and so on. ±e, ±2e, ±3e, · · · Fields –– – F – – F + + + + + + + –– – Glass – (a) – – – –– – – – Rubber F (b) Figure 23.1 (a) A negatively charged rubber rod suspended by a thread is attracted to a positively charged glass rod. (b) A negatively charged rubber rod is repelled by another negatively charged rubber rod. Monday, January 30, 17 right A negatively charged rubber rod is repelled by another negatively charged rubber rod 3 2.1 Electric Force 1.2 Electric Force q1 and Electric force force between two two charges The electric between charges q1 and byq2☛ Coulomb’s can be described by described Law Coulomb’s Law. q2 F~12 = Force on q1 exerted by q2 1 q1 q2 · 2 · r̂12 r12q2 0 F~12 = F orce on q14⇡✏ exerted by ~r12 ☛ unit vector which locates 1 relative to particle 2 = qparticle q2 1 1 ~ |~r12 | F12 = 4º≤0 · r2 · r̂12 r12 = ~r1 12~r2 i.e. ~ F~12 = r̂12 ~r12 where r̂12 = is the unit vector which locates particle 1 relative to particle 2. • q1 , q2 are|~relectrical charges in units of Coulomb (C) | 12 19 • Charge is quantized ☛i.e. electron carries 1.602 ⇥ 10 C ~r12 = ~r1 ° ~r2 • Permittivity of free space ✏0 = 8.85 ⇥ 10 12 C2 /Nm2 • q1 , q2 are electrical charges in units of Coulomb(C) • Charge is quantized Monday, January 30, 17 Recall 1 electron carries 1.602 £ 10°19 C 4 °12 2 2 COULOMB’S LAW: (1) q1 , q2 can be either positive or negative (2) If q1 , q2 are of same sign force experienced by q2 is in direction away from q1 i.e. ☛ repulsive (3) Force on q2 exerted by q1 : F~21 = BUT 1 q2 q1 · 2 · r̂21 4⇡✏0 r21 r12 = r21 = distance between q1 , q2 r̂21 ~r21 ~r2 ~r1 ~r12 = = = = r21 r21 r12 F~21 = Monday, January 30, 17 r̂12 F~12 Newton’s 3rd Law 5 SYSTEM WITH MANY CHARGES: 2.2 SYSTEM WITH MANY CHARGES: The total force experienced by charge q1 is the vector sum of the forces on q1 exerted by other charges. Total force experienced by charge q1 vector sum of forces on q1 exerted by other charges ~ F1 = Force experienced by q1 ~ F1 = Force experienced by q1 = F~1,2 + F~1,3 + F~1,4 + · · · + F~1,N = F~1,2 + F~1,3 + F~1,4 + · · · + F~1,N PRINCIPLE OF SUPERPOSITION: N X PRINCIPLE OF SUPERPOSITION F~1 = F~1,j j=2 Monday, January 30, 17 The Electric Field F~1 = N X F~1,j j=2 6 1.3 Electric Field While we need two charges to quantify electric force we define electric field for any single charge distribution 2.2. THE ELECTRIC FIELD to describe its effect on other charges Total force F~ = F~1 + F~2 + · · · + F~N Total force F~ = F~1 + F~2 + · · · + F~N The electric field is defined as Electric field is defined as F ~ ~ F ~ = lim E q0 !0 q0 lim q0 !0 q0 ~ =E (a) E-field due to a single charge qi : Monday, January 30, 17 7 (a) E-field due to a single charge qi : (i) E-field due to a single charge qi From definitions of Coulomb’s Law ~ = Recall E where r̂0,i experienced at location of qthe 0 (point P ) Fromforce the definitions of Coulomb’s Law, q0 qiP) force experienced at location 1 of q0 (point ~ F0,i = · 2 · r̂0,i 4⇡✏0 r0,i r̂0,i ☛ unit vector along direction from charge qi to q0 ~0,i = 1 · q0 qi · r̂0,i F ~ F 2 4º≤ r 0 ~ 0,i ) -field due to qi at point P lim E q0 !0 q0 along the direction from charge qi to q0 , is the unit vector r̂0,i = = 1 qi ~ Ei = Unit vector from charge qi to ·point P· r̂i 2 4⇡✏ r r̂i (radical unit vector from0qi ) i ~ri ☛ vector pointing from qi to point P r̂i ☛ unit vector pointing from qi to pointP F~ ~ Recall E = lim q0 !0 q0 ) E-field due to qi at point P: Note: ~ -field is a vector ~ i = 1 · qi · r̂i E (1) E 4º≤0 ri2 ~ -field depends on both position of (2) Direction of E where ~ri = Vector pointing from qi to point P, thus r̂i = Unit vector pointing from qi to point P Monday, January 30, 17 Note: (1) E-field is a vector. P and sign of qi 8 RIC FIELD ~ -field due to system of charges: (ii) E ole 11 Principle ofX Superposition 1 X qi ~ ~ Ei = r̂ ~ 2 i E N charges ☛ total -field due to all charges 4º≤0 i ri ~ -field due to individual charges vector sum of E X X qi 1 ~i = ~ = E r̂i i.e. ☛ E 2 4⇡✏0 ri i.e. E= In a system with i (iii) Electric Dipole i i d opposite charges tance d. System of equal2.1: and An opposite by of a distance d Figure electriccharges dipole.separated (Direction d~ from negative to positive charge) Electric Dipole Moment ☛ p ~ = q d~ = qddˆ p = qd Electric Dipole Moment Monday, January 30, 17 p~ = q d~ = qddˆ 9 p = qd ~ due mple: Example: E dipole along x-axis ~ to E due to dipole along x -axis P at distance x along perpendicular axis of dipole p~ nsider point P at distance x along the perpendicular axis of the dipole p~ : ice: Consider point ~ E ~+ E ~E ~ =E ~+ + + E ~° = E " " (E-field (E-field due to +q) °q)due to q -field due to +q dueEto E-field ~ + and E ~ ° cancel out. Horizontal E-field components of E Monday, January 30, 17 10 due to °q) due to +q) Notice: ~ + and E ~ ° cancel out. Horizontal E-field components of E ~ cancel out ~ + and E ~-field components of E Notice: Horizontal E ) Net E-field points along the axis opp site to the dipole moment vector. 2.3. CONTINUOUS CHARGE DISTRIBUTION 12 ~ -field points along axis opposite to dipole moment vector ) Net E Magnitude of E-field = 2E+ cos µ ~ -field = 2E+ cos ✓ Magnitude of E ) E But = 2 ! 1 q · 2 cos ✓ 4⇡✏0 r | E{z magnitude! } or EE+ or E °magnitude + r⇣ ⌘µ z 1 }| q{ ∂ 2 ) E =d2 cos µ 2· 2 r = + x 2 4º≤0 r d/2 cos ✓ = But r r= s ≥ d ¥2 + x2 2 1 p d/2 2 3 )E = · µ 2=+ d ] 2 4⇡✏0 cos[x r2 (p = qd) Monday, January 30, 17 )E = 1 p · 4º≤0 [x2 + ( d2 )2 ] 32 11 Special case ☛ When x ⇥ d x + 2 2 • Binomial Approximation d 2 ⇤ 32 = x 3 ⇥ d 1+ 2x (1 + y)n ⇡ 1 + ny ~ E • Compare with if y⌧1 1 1 p field of dipole ' · 3 / 3 4⇡✏0 x x 1 ~ -field for single charge E 2 r • Result also valid for point Monday, January 30, 17 2 ⇤ 32 P along any axis with respect to dipole 12 2.3 Continuous Charge Distribution 1.4 Continuous Charge Distribution E -field at point P due to dq E-field ~at point1P duedq to dq: dE = · 2 · r̂ 4⇡✏0 r 1 dq · 2 · r̂ ) E -field due to charge 4º≤distribution r 0 ~ = dE ~ = E Z ~ = dE Z 1 dq · 2 · r̂ 4⇡✏0 r (1) Take advantage of symmetry of system to simplify integral (2) To write down small charge element dq ☛ 1 D dq = ds = linear charge density 2 D dq = dA = surface charge density 3 D dq = ⇢ dV Monday, January 30, 17 ⇢ = volume charge density ds = small length element dA = small area element dV = small volume element 13 1-D 2-D 3-D dq = ∏ ds dq = æ dA dq = Ω dV ∏ = linear charge density æ = surface charge density Ω = volume charge density ds = small length element dA = small area element dV = small volume element Example 1: Uniform line of charge Uniform line of charge Example 1 charge per unit length = ∏ per unit charge (1) Symmetry considered ☛E -field from +z and length = z directions cancel along (1) Symmetry considered: The E-field from +z and °z directions cancel along z-direction, ) Only horizontal components needcomponents to be considered. ) Only E-field z -direction, horizontal need E-field to be considered (2) For each element of length dz, charge dq = ∏dz (2) For each dzdz, charge ) Horizontal E-field element at point P of due length to element = dq =∏dz dz 1 ~ cos µ = E| · cos µ to element ) Horizontal E-field at point P |ddue 4º≤0 r2dz is | {z } 1 dz dEdz ~ dz = |dE| cos ✓ = to entire · 2 cos ✓ ) E-field due charge at point P 4⇡✏0 rline | {z } Monday, January 30, 17 dEdz ˆL/2 E = °L/2 1 ∏dz · 2 cos µ 4º≤0 r 14 ⌥ cos ⇥ = |dE| 4⌅ ⇤ 0 ⇥ · dEdz r2 ) E -field due to entire line charge at point P E-field due to entire line charge a Z L/2 1 dz E = · L/2 cos ✓ 2 rˆ L/2 4⇡✏0 1 ⇤dz E = · 2 cos Z L/2 4⌅ 0 r dz = 2 · L/22 cos ✓ 4⇡✏0 rL/2 0 ˆ ⇤ dz = 2 · 2 cos 4⌅ 0 r To calculate this integral ☛ 0 • First, notice that x is fixed, but z, r, ✓ all varies • Change of variable (from z to ✓ ) Monday, January 30, 17 15 (1) z = x tan ✓ x = r cos ✓ z = 0 ) dz = x sec2 ✓ d✓ ) r2 = x2 sec2 ✓ ✓ = 0 L/2 ✓ = ✓0 where tan ✓0 = (2) When z = L/2 Z ✓0 x 2 x sec ✓ d✓ E = 2· · cos ✓ 2 2 4⇡✏0 0 x sec ✓ Z ✓0 1 = 2· · cos ✓ d✓ 4⇡✏0 0 x ✓0 1 = 2· · · (sin ✓) 4⇡✏0 x 0 1 = 2· · · sin ✓0 4⇡✏0 x 1 L/2 r = 2· · · ⇣ ⌘2 4⇡✏0 x x2 + L2 Monday, January 30, 17 16 1 E = · 4⇡✏0 x r L x2 + Important limiting cases (1) But (2) x L: ⇣ ⌘2 along x-direction L 2 1 L E + · 2 4⇡✏0 x L = Total charge on rod ) System behave like a point charge L x: 1 L E + · 4⇡✏0 x · L2 Ex = 2⇡✏0 x ELECTRIC FIELD DUE TO INFINITELY LONG LINE OF CHARGE Monday, January 30, 17 17 2.3. CONTINUOUS CHARGE DISTRIBUTION Example 2: 15 Ring of Charge Example 2 Ring of Charge z above at a height at a height z abovea aring ring of of charge of radius R E-field E-field charge of radius R (1) Symmetry considered: For every charge element dq considered, there exists (1) Symmetry considered ☛ For every charge element dq considered, ⌃ field components dq⇤ where the horizontal E cancel. 0 ~ field components cancel ⇥there Overall E-fielddq lies along z-direction. exists where horizontal E (2) For each element of length dz, charge Monday, January 30, 17 18 (1) Symmetry considered: For every charge element dq considered, there exists ⌃ field components cancel. dq⇤ where the horizontal E ⇥ Overall E-field lies along z-direction. (2) For each element of length dz , charge (2) For each element of length dz, charge dq = dq = ⇤ · ds · ds Linear Circular Linear Circular charge density element charge density length length element dqdq = ⇤=· R d⇧, · R d where ⇧ is the angle measured on the ring plane where is angle measured on ring plane Net along z-axis duezto dq: due to dq ) E-field Net E -field along -axis 1 dq dE = · 2 · cos ⇥ 4⌅ 0 r 1 dq dE = · 2 · cos ✓ 4⇡✏0 r Monday, January 30, 17 19 Total E -field = = Z Z 0 dE 2⇡ 1 · 4⇡✏0 Rd r2 · cos ✓ Note: Here in this case, ✓, R and r are fixed as BUT we want to convert r, ✓ to z (cos ✓ = ) r varies! R, z 1 Rz E = · 3 4⇡✏0 r Z 2⇡ d 0 1 (2⇡R)z E = · 2 4⇡✏0 (z + R2 )3/2 BUT ☛ along z -axis (2⇡R) = total charge on ring Monday, January 30, 17 20 ample 3: 4⌅ BUT: 0 (z + R ) ⇤(2⌅R) = total charge on the ring Example 3 a disk E-field fromdensity a disk of surface charge density E-field from of surface charge ⇧ We find E-field of a disk by integrating concentric rings of charges 2.3. CONTINUOUS CHARGE DISTRIBUTION Total We find the E-field of a disk by charge of ring integrating concentric rings of charges. dq = Total charge of ring 17 view from top · (2⇡r | {zdr}) Area of ring dq = ⇤ · ( 2⇥r⌥ dr⌦ ) Area of the ring Recall from Example 2: E-field from ring: dE = Monday, January 30, 17 1 E = 4⇥ 0 ˆ 0 ˆ R 1 dq z · 2 4⇥ 0 (z + r2 )3/2 2⇥⇤r dr · z (z 2 + r2 )3/2 21 Recall from Example 2 1 dq z · 2 4⇡✏0 (z + r2 )3/2 Z R 1 2⇡ r dr · z ) E = 4⇡✏0 0 (z 2 + r2 )3/2 E-field from ring ☛ dE = 1 = 4⇡✏0 Z R 0 r dr 2⇡ z 2 (z + r2 )3/2 • Change of variable: u = z 2 + r2 ) ) du = 2r dr ) Monday, January 30, 17 (z 2 + r2 )3/2 = u3/2 1 r dr = du 2 22 • Change of integration limit: ⇢ BUT Z u = z2 u = z 2 + R2 Z z2 +R2 1 1 ) E= · 2⇡ z u 4⇡✏0 2 z2 u Monday, January 30, 17 3/2 r = 0 r = R du = u 1/2 = 2u 3/2 du 1/2 1/2 z 2 +R2 1 ) E = z ( u 1/2 ) 2✏0 z2 ! 1 1 1 p = z + 2 2 2✏0 z z +R " # z p E = 1 2✏0 z 2 + R2 23 VERY IMPORTANT LIMITING CASE: VERY IMPORTANT LIMITING CASE If R ⇤ z, that is if we have an infinite sheet of charge with charge density ⇥: z , that is if we have an infinite sheet of charge with charge If R ⇤ ⌅ "⇥ z # density E = 1 ⇧z 2 2 2 z + R 0 p ⇥ E = 1 z2 + R2 2✏0 ⇥ z ⌅ 1 "2 0 #R ' 2✏0 1 z R E ⇡ 2✏0 ⇥ E⇥ 20 E-field is normal to the charged surface E -field is normal to charged surface Figure 2.2: E-field due to an infinite sheet of charge, charge density = ⇥ Monday, January 30, 17 24 1.5 Electric Field Lines To visualize electric field we can use a graphical tool called electric field lines Conventions 1. Start on positive charges and end on negative charges 2. Direction of E-field at any point is given by tangent of E-field line 3. Magnitude of E-field at any point proportional to number of E-field lines per unit area perpendicular to lines Monday, January 30, 17 25 2.4. ELECTRIC FIELD LINES Uniform E-field Monday, January 30, 17 19 Non-uniform E-field 26 ~P > E ~P E 1 2 Monday, January 30, 17 ~ = E +q r̂ 2 4⇡✏0 r 27 Infinite sheet of charge E = Monday, January 30, 17 2✏0 28 2.4. ELECTRIC FIELD LINES Monday, January 30, 17 20 29 ~ at point O = 0 E Monday, January 30, 17 30 2.5 Point Charge in E-field When we place a chargein q in E-field an E-field E, the force experienced by the charge is 1.6 Point Charge ~ma E = qE = When we place a charge q in an EF-field ,force experienced by charge is Applications: ~ ~ F printer, = q ETV=cathoderay m~a tube. Ink-jet Applications ☛ Ink-jet printer, TV cathode ray tube Example: Example Ink particle has mass m, charge q (q < 0 here) Ink particle has mass m & charge q (q < 0 here) that mass inkdrop is is small, what’s thes deflection y of the Assume Assume that mass of of inkdrop small, what’ deflection of charge? charge? Solution: Monday, January 30, 17 First, the charge carried by the inkdrop is negtive, i.e. q < 0. 31 st, the charge carried by the inkdrop is negtive, i.e. q < 0. Solution ☛ Charge carried by inkdrop is negative ☛ q<0 ~ ~ points in qopposite Note: Note: direction of E qE E points in opposite Horizontal motion ☛ Net force = 0 ) L = vt rizontal motion: ~ Vertical motion ☛ |q E| Net force = 0 ) Net force = |q|E )a= m |m~g |, q is negative |q|E = ma ☛ Newton’s 2nd Law L = vt 1 2 Vertical distance travelled ☛ y = at 2 Monday, January 30, 17 32 di Review everything for next class BUT don’t forget Monday, January 30, 17 33 Monday, January 30, 17 34 HOMEWORK Vector Review these slides B4Algebra watching superbowl 1.1 Definitions A vector consists of two components magnitude and direction (e.g. force, velocity,) pressure) A scalar consists of magnitude only (e.g. mass, charge, density) Euclidean vector, a geometric entity endowed with magnitude and direction as well as a positive-definite inner product; an element of a Euclidean vector space! In physics, Euclidean vectors are used to represent physical quantities that have both magnitude and direction, such as force, in contrast to scalar quantities, which have no direction Monday, January 30, 17 35 (e.g. force, velocity, pressure) A scalar consists of magnitude only. (e.g. mass, charge, density) 1.2 Vector Algebra 1.2 Vector Algebra ~b +algebra ~aFigure + ~b1.1:=Vector ~a ~a + (~b + ~c) = (~a + ~b) + ~c ~a + ~b = ~b + ~a ~ = (~a + ~c) + d~ ~a + (~c + d) Monday, January 30, 17 36 1.3. COMPONENTS OF VECTORS 2 1.3 1.3 Components of of Vectors Components Vectors 1.3 Components of Vectors Usually vectors are expressed according to coordinate can Usually vectors are expressed according to coordinatesystem. system. Each Each vector vector can be expressed in terms of components. be expressed in terms of components. Usually vectors are expressed according to coordinate system The most common system: The most common coordinate system: Cartesian Each vector cancoordinate be expressed in Cartesian terms of components! The most common coordinate system Cartesian ~a = ax + ay + az ~a = ~a + ~a + ~a ~a = ~ax x+ ~ayy+ ~azz Magnitude of Magnitude of ~a = |~aq | = a, ~a = |~a| = a Magnitude of ~a = |~a| = a, 2 q= a a + a =q a2x + a2y +xa2z a = a2x + a2y + a2z a2y + a2z ~a = ax + ay q a = a2x + a2y ~a = ~ax + ~ay axq = a cos 2 a2x~a+ ~a a== ~ax + y ay aqya cos¡; = a sin a = x a = a2x + a2yay = a sin¡ ay tan = aayy /a tan¡ = ax = a cos¡; = x a sin¡ ax Figure 1.2: ¡ measured anti-clockwise tan¡ = Monday, January 30, 17 from position x-axis Figure 1.2: ¡ measured anti-clockwise from position x-axis ay ax 37 Unit vectors have magnitude of 1 ~a â = = unit vector along ~a direction |~a| ı̂ |ˆ k̂ x y z are unit vectors along directions ~a = ax ı̂ + ay |ˆ + az k̂ Monday, January 30, 17 38 1.3. COMPONENTS OF VECTORS 1. Polar Coordinates 1. Polar Coordinate: ! ~a = ar r̂ + a✓ ✓ˆ ~a = ar r̂ + aµ µ̂ Figure 1.3: Polar Coordinates 2. Cylindrical Monday, January 30, 17Coordinates: 39 Figure 1.3: Polar Coordinates 2. Cylindrical Cylindrical Coordinates:Coordinates ! ~a = ar r̂ + aµ µ̂ + az ẑ ~a = ar r̂ + a✓ ✓ˆ + az ẑ r̂ originated from r̂ originated from nearest point on nearest point on z-axis (Point z-axis (Point O’)O’)! igure 1.4: Cylindrical Coordinates pherical Coordinates: Monday, January 30, 17 40 Figure 1.4: Cylindrical Coordinates 3. Spherical Coordinates 3. Spherical Coordinates: ! ~a = ar r̂ +~aa=✓ a✓ˆr r̂++aaµ µ̂ˆ+ a¡ ¡ˆ r̂ originated from Originfrom O r̂ originated Origin O Figure 1.5: Spherical Coordinates Monday, January 30, 17 41 1.4 Multiplication of Vectors 1. Scalar multiplication If ~b = m ~a ~b, ~a are vectors; m is a scalar then b = m a (Relation between magnitude) bx = m a x by = m a y i.e. } ~a = ax Components also follow relation ı̂ + ay m~a = max ı̂ + may Monday, January 30, 17 |ˆ + az k̂ |ˆ + maz k̂ 42 2. Dot Product (Scalar Product) Cont’d ı̂ · ı̂ = |ı̂| |ı̂| cos 0 = 1 · 1 · 1 = 1 ı̂ · |ˆ = |ı̂| |ˆ || cos 90 = 1 · 1 · 0 = 0 ı̂ · ı̂ = |ˆ · |ˆ = k̂ · k̂ = 1 ı̂ · |ˆ = |ˆ · k̂ = k̂ · ı̂ = 0 If then ~a = ax ı̂ + ay |ˆ + az k̂ ~b = bx ı̂ + by |ˆ + bz k̂ ~a · ~b = ax bx + ay by + az bz ~a · ~a = |~a| · |~a| cos 0 = a · a = a2 Monday, January 30, 17 43 1.4. MULTIPLICATION OF VECTORS 5 3. Cross Product (Vector Product): 2. Cross Product (Vector Product) ~c = ~a ⇥ ~b If If ~c = ~a £ ~b, then c = |~c| = a b sin¡ c| = a b sin then c = |~ ~a £ ~b 6= ~b £ ~a !!! ~a ⇥ ~b 6= ~b ⇥ ~a ~a £ ~b = °~b £ ~a ~a ⇥ ~b = !!! ~b ⇥ ~a Figure 1.7: Note: How angle ¡ is measured • Direction of cross product determined from right hand rule • Direction of cross product determined from right hand rule. • Also, ~ a ⇥ ~b ~is ? to ~a ~and ~b i.e. • Also, ~a £ b is ? to ~a and b, i.e. ~a · (~a ⇥ ~b) = 0 ~b · (~a ⇥ ~b) = 0 ~a · (~a £ ~b) = 0 ~b · (~a £ ~b) = 0 • IMPORTANT: Monday, January 30, 17 ~a £ ~a = a · a sin0± = 0 44 • IMPORTANT ~a ⇥ ~a = a · a sin 0 = 0 ± £ î| = |ı̂|î|⇥|î| sin0 ı̂| = |ı̂| |ı̂| sin=0 1=· 1 1 ·· 0 1 ·= 0 0 = ± £ ĵ| = |î| |ĵ| sin90 = 1 · 1 · 1 = 1 0 |ı̂ ⇥ |ˆ| = |ı̂| |ˆ || sin 90 = 1 · 1 · 1 = 1 ⇥ îı̂ == ĵ|ˆ£⇥ĵ|ˆ==k̂k̂£⇥k̂k̂==0 0 îı̂ £ î £ ĵ = k̂; ĵ £ k̂ = î; k̂ £ î = ĵ ı̂ ⇥ |ˆ = k̂; |ˆ ⇥ k̂ = ı̂; k̂ ⇥ ı̂ = |ˆ ~a ⇥ ~b = Ø î Ø Ø = Ø ax Ø bx ı̂ ax bx |ˆ k̂ ay az Ø b y Ø bz = (ay bz az by ) ı̂ + (az bx ax bz ) |ˆ + (ax by ĵ k̂ Ø (ay bz ° az by ) î ay az ØØ = +(az bx ° ax bz ) ĵ by bz ˆ Monday, January 30, 17 ay bx ) k̂ 45 4. Vector identities ~a ⇥ (~b + ~c) = ~a ⇥ ~b + ~a ⇥ ~c ~a · (~b ⇥ ~c) = ~b · (~c ⇥ ~a) = ~c · (~a ⇥ ~b) ~a ⇥ (~b ⇥ ~c) = (~a · ~c) ~b (~a · ~b)~c 1.5 Vector Field (Physics Point of View) ~ (x, y, z) is a mathematical function A vector field F which has a vector output for a position input (Scalar field U(x, y, z) ) Monday, January 30, 17 46 1.6 Other Topics 1.6 Analytic Geometry Tangential Vector Tangential Vector igure 1.8: d~l is a vector that is always tangential to the curve C with infinitesimal ength dl urface Vector Surface Vector Figure 1.8: d~l is a vector that is always tangential to the curve C with infinitesimal length dl Surface Vector igure 1.9: d~a uncertainty! is a vector that to the surface S with Some ( d~ais always versus perpendicular d~a ) nfinitesimal area da Monday, January 30, 17 Figure 1.9: d~a is a vector that is always perpendicular to the surface S with 47 Two conventions: Some uncertainty! (d~a versus ° d~a) • Area formed from a closed curve Two Two conventions: conventions: • •Area formed from afrom closedacurve Area formed closed curve Figure 1.10: Direction of d~a determined from right-hand rule • Closed surface enclosing a volume Figure 1.10: Direction of d~a determined from right-hand rule • Closed surface enclosing a volume • Closed surface enclosing a volume Monday, January 30, 1.11: 17 Figure Direction of d~a going from inside to outside 48