1 1. The discovery of electrification 2electric charge is an intrinsic characteristic of the fundamental particles making up matters. 1The phenomenon of electrification Glass rod is rubbed with silk—positive Amber rod is rubbed with fur--negative § 16.1 Coulomb’s force law for pointlike charges 2 3like charges(charges with the same sign) repel each other, and unlike charges(charges with opposite electrical signs) attract each other. § 16.1 Coulomb’s force law for pointlike charges § 16.1 Coulomb’s force law for pointlike charges 2. The properties of electric charge 1the charges are scalar; 2charges are quantization; e=1.602177*10 -19 C q=ne up quarks: 2e/3, down quarks: -e/3 3the total amount of electric charges is conserved in any process involving an isolated system. 3. The polarization and induction Why can comb rubbed with fur attract small pieces of paper? 3 § 16.1 Coulomb’s force law for pointlike charges § 16.1 Coulomb’s force law for pointlike charges + + + + + ? ? ? ? ? ? ?? ??? + + +++ + + + + + + ? ? ? ? ? ???? ?? induction + + + + + grounding conductor induction + + +++ + electrical polarization insulator +? +? +? +? +? +? +? Electric dipole 4 § 16.1 Coulomb’s force law for pointlike charges § 16.1 Coulomb’s force law for pointlike charges 5 § 16.1 Coulomb’s force law for pointlike charges 4. Coulomb’s force law 1Experiments: 2 1 r F ∝ 2 r Qq F ∝ 2 0 4 1 r Qq F πε = m)/(NC10854187817.8 212 0 ?×= ? ε Permitivity of free space q q q Q Q Q 21 2 210 ? 4 1 r r qQ F πε = r 2mathematical expression 3Compare Coulomb’s law and universal gravitational force § 16.1 Coulomb’s force law for pointlike charges 21 2 210 ? 4 1 r r qQ F πε = r 21 2 21 ? r r mM GF ?= r For two electrons: 2 2 0 elec 4 1 r e F πε = 2 2 grav r m GF e ?= 42 2 2 0grav elec 1017.4 4 1 ×== e Gm e F F πε 6 § 16.1 Coulomb’s force law for pointlike charges 5. The principle of superposition The total electrical force on given charge is the vector sum of the electrical forces caused by the other charges, calculated as if each acted alone. L rrr ++= 21 FFF total q 1 Q 3 Q 2 Q + + + ? 1 F r 3 F r 2 F r § 16.2 The electric field of static charge and electric field lines 1. The definition of electric field of static charge test q F E r r = electric field Test charge is small enough, does not alter the distribution of the charges creating the field. The electric field at a point in space is the force per Coulomb at that point(N/C), direct the direction of force exerted on a positive test charge . 7 Notice: 1The electric field depends on the shape and the distribution of the charges. 2the gravitational force is always parrallel to the gravitational field in the gravitational case, but the electrical force can be either parallel or antiparallel to the electric field. The force on a positive charge +q is parallel to the electric field, the force on a negative charge -q is antiparallel to the electric field. EqF rr = § 16.2 The electric field of static charge and electric field lines § 16.2 The electric field of static charge and electric field lines 2. The electric field of pointlike charge distribution Test charge r r F r q Q 21 2 210 ? 4 1 r r qQ F πε = r r r Q q F E ? 4 1 2 0 πε == r r r r Q q F E ? 4 1 2 0 πε ?== r r For +Q For -Q 8 § 16.2 The electric field of static charge and electric field lines 3. The principle of superposition The total electric field at the point when all the pointlike charges are present simultaneously at their fixed position is the vector sum of the individual fields. L rr L rrr r ++= ++== 21 21 EE q F q F q F E 4. Electric field lines A useful alternative geometric representation of electric field. i i i i r r Q E ? 4 2 0 ∑ = πε r or 1The electric field lines begin on positive charges and end on negative charges; The direction of the electric field at any point is tangent to the field line passing through that point and in the direction indicated by arrows on the field line. The properties of the electric field lines: § 16.2 The electric field of static charge and electric field lines 2The electric field is strong where field lines are close together and weak where they are far apart. The number of lines passing through a square meter oriented perpendicular to the lines is proportional to the magnitude of the electric field. 9 § 16.2 The electric field of static charge and electric field lines Examples: 3Field lines are never cross. The field at any point has a unique direction. § 16.2 The electric field of static charge and electric field lines 10 1. The electric field of the electric dipole § 16.3 The distribution of electric field and calculation +? d r dQp r r = Q? Q+ Electric dipole moment dQp r r = k d z d z Q EEE ? ] ) 2 ( 1 ) 2 ( 1 [ 4 22 0 + ? ? =+= ?+ πε rrr k d z zdQ ? ) 4 ( 2 4 2 2 2 0 ? = πε 3 0 4 2 z p E πε r r = dz >> 1along the axis of the dipole § 16.3 The distribution of electric field and calculation Q? Q+ z 2d + E r ? E r A o z y 11 3 0 3 0 3 0 3 0 3 0 4 4 )( 4 ) 4 ( 4 Q y p y dQ rr y Q r rQ r r EEE πε πεπε πεπε r r rr rr rrr ?= ?=?≈ ?+= += ?+ ? ? + + ?+ 2the perpendicular bisector of the dipole § 16.3 The distribution of electric field and calculation Q? Q+ B y + r r? r r o z y E r + E r ? E r d r 2. The electric dipole in the electric field EQ d F d r r r r r ×=×= 22 11 τ )( 22 22 EQ d F d r r r r r ?×?=×?=τ EpEdQ EQ d EQ d r r rr r r r r rrr ×=×= ×+×= += 22 21 τττ Q+ Q? § 16.3 The distribution of electric field and calculation 12 3. The electric field of continuous distributions of charge 1the charge distribution is imagined to be composed of a continuous sea of pointlike charges. 2the differential bit of charge dQ produces a differential bit of electric field E r d ? ? ? ? ? = V S l Q d d d d ρ σ λ r r Q E ? 4 d d 2 0 πε = r E r d Qd r r P § 16.3 The distribution of electric field and calculation 3the total electric field is the vector sum of all the contributions from all bits of charge. r r Q E ? d 4 1 dist. charge 2 0 ∫ = πε r Electric field in space depends on: ?The charge and its sign ?The geometric shape of the charge distribution ?The distance from the charge distribution ?The displacement of the point with respect to the charge distribution § 16.3 The distribution of electric field and calculation 13 Example 1: A ring of radius R has a total charge Q smeared out uniformly along its circumference. Calculate the electric field of the ring at a point P along the axis of the ring a distance z from the center of the ring. R o P z z Q § 16.3 The distribution of electric field and calculation s R Q sQ d 2 dd π λ == r r s r r Q E ? 4 d ? 4 d d 2 0 2 0 πε λ πε = = r 0d == ∫ ⊥⊥ EE Solution: r r Qd o E r d θ θ R z P z Q′d E′ r d r ′ r § 16.3 The distribution of electric field and calculation 14 k ? )(4 k ? d 2 1 4 k ? cos 2 d 4 1 2 3 22 0 2 0 2 0 2 0 //surviv Rz Qz s Rr z r Q R sQ r EE R + =?= ??== ∫ ∫ πε ππε θ ππε π rr max 2 then 0 d d from EE R z z E = ±== 0=E 0 →∞→ Ez 2 0 4 z Q ERz πε ≈>> At center of the ring 2 R ? 2 R O E z § 16.3 The distribution of electric field and calculation ∫ == == ?d ?d?d EE EQ rrQ d 2d σπ= 2 3 22 0 )(4 d d rz Qz E + ? = πε Example 2: Find the electric field at a distance z along the axis of a uniformly charged circular disk of radius R and charge σ per unit area. Solution: § 16.3 The distribution of electric field and calculation 15 k Rz z k Rz z k rz rrz E ? ] )( 1[2 4 1 ? ] )( 1[2 4 1 ? )(4 d2 2/122 0 2/122 0 0 2 3 22 0 + ?= + ?= + ?? = ∫ ∞ σ ε πσ πε πε πσ r kk Rz z E R R ? 2 ? ] )( 1[2 4 1 lim 0 2/122 0 ε σ σ ε = + ?= ∞→ ∞→ r § 16.3 The distribution of electric field and calculation § 16.3 The distribution of electric field and calculation 16 σ+ E r E r σ? E r E r 0 2ε σ =E 0 2ε σ =E § 16.3 The distribution of electric field and calculation 0 0 ? ? ? ? ? =+= ?+ ε σ EEE += σσ ?+ σ+ E r E r σ? E r E r Between the plate Outside the plates § 16.3 The distribution of electric field and calculation 17 Example 3: There is a straight rod of charged λ per unit length. Find the electric field at point P of perpendicular distance y from the rod. zQ dd λ= r r Q E r r 3 0 4 d d πε = Solution: 2 0 4 d d r z E πε λ = magnitude z O 1 θ y P λ Qd E r d r r θ 2 θ y § 16.3 The distribution of electric field and calculation θ θ sindd cosdd EE EE y z = = θ πε λ θ πε λ sin 4 d d cos 4 d d 2 0 2 0 ∫∫ ∫∫ == == r z EE r z EE yy zz The components of the electric field z 1 θ y P λ Qd z Ed r r θ y Ed 2 θ E r d O y § 16.3 The distribution of electric field and calculation 18 θ θθθ 22222 2 csc d cscd ctg yzyr yzyz =+= =?= Unify the variables )cos(cos 4 dsin 4 )sin(sin 4 dcos 4 21 00 12 00 2 1 2 1 θθ πε λ θθ πε λ θθ πε λ θθ πε λ θ θ θ θ ?== ?== ∫ ∫ yy E yy E y z Then we have z y yzP E E EEE arctg 22 = += α The total electric field: The angle with respect to z axis: § 16.3 The distribution of electric field and calculation y EEE yz 0 21 2 0 . 0 πε λ πθθ === ≈≈ If P is very close to the rod, or the length of the rod is infinite, then Can you draw the electric field lines of the rod? § 16.3 The distribution of electric field and calculation z 1 θ y P λ Qd z Ed r r θ y Ed 2 θ E r d O y 19 r E 0 2πε λ = § 16.3 The distribution of electric field and calculation the length of the rod is infinite: If the length of the rod is L, what is the electric field at the point on the perpendicular bisector of the rod? )cos(cos 4 dsin 4 )sin(sin 4 dcos 4 21 00 12 00 2 1 2 1 θθ πε λ θθ πε λ θθ πε λ θθ πε λ θ θ θ θ ?== ?== ∫ ∫ yy E yy E y z recalling For the point on the perpendicular bisector of the rod θπθθθ ?== 21 , § 16.3 The distribution of electric field and calculation 20 22 0 22 0 0 )2/( 1 4 ) )2/( 2/ 2( 4 / )cos2( 4 yL y Q yL L y LQ y E y + = + = = πε πε θ πε λ 0= z E § 16.3 The distribution of electric field and calculation z O θ θπ ? Q 2 L y y Exercises Exercise 1: A large, flat, nonconducting surface has a uniform charge density σ. A small circular hole of radius R has been cut in the middle of the surface. Ignore fringing of the field lines around all edges, and calculate the electric field at point P, a distance z from the center of the hole along its axis. 21 Exercises Exercise 2: Figure shows an electric quadrupole.Find the value of E on the axis of the quadrupole for a point P a distance z from its center(assume z >>d) Solution: Think of the quadrupole as composed of two dipoles, the field produced by the right dipole of the pair is 2/3 0 ) 2 (2 d z qd ?πε the field produced by the left dipole is 2/3 0 ) 2 (2 d z qd + ? πε Exercises 22 Use the binomial expansions ) 2 (3) 2 ( ) 2 (3) 2 ( 433 433 d zz d z d zz d z ??? ??? ?≈+ ??≈? We obtain 4 0 4 0 2 4343 0 4 3 4 6 ) 2 31 2 31 ( 2 z D z qd z d zz d z qd E πεπε πε == +?+= Exercises Quadrupole moment D=3qd 2 Exercise 3: a plastic rod is uniformly distribute charge –Q. the rod has been bent in 120o circular arc of radius r. the coordinate is shown in Figure, what is the electric field due to the rod at point P? E r Exercises 23 Solution: θλλ ddd rsQ == 2 0 2 0 d 4 1 d 4 1 d r s r Q E λ πε πε = = Exercises θθ λ πε θ θθ λ πε θ dsin 4 1 sindd dcos 4 1 cosdd 2 0 2 0 r r EE r r EE y x == == The component of the electric field θθ λ πε dcos 4 1 d 0d 2 60 60 0 r r EE EE xx yy ∫∫ ∫ ? == == o o We can get Exercises 24 324 3 4 3 )]60sin(60[sin 4 1 dcos 4 1 d 00 0 2 60 60 0 r Q rr r r r EEE xx ππεπε λ πε θθ λ πε == ??= = == ∫ ∫ ? oo o o i r Q iEE x ? 4 83.0 ? 2 0 πε == r Exercises Exercise 4: P j r q E j r q E ? ? 22 0 22 0 πε πε ?= ?= r r Exercises 25 Exercise 5: Two parallel nonconducting rings arranged with their central axes along a common line as shown in figure.the rings are separated by a distance 3R. The net electric field at point P is zero, what is the ratio q 1 /q 2 ? Exercises Solution: Assume both charge are positive 51.0) 5 2 (2 ])2[(4 )2( )(4 2/3 2 1 2/322 0 2 2/322 0 1 ≈= + = + q q RR Rq RR Rq πεπε Exercises 26 Exercise 6: A nonconducting rod of length L has charge –q uniformly distributed along its length. What is the electric field at point P, a distance a from the end of the rod? Solution: O qd xd xaL ?+ Choose the origin at the left end of the rod. xq dd λ= 2 0 )( d 4 1 d xaL x E x ?+ = λ πε Exercises i aLa q i aLa L i xaL i xaL x iEE L L x ? )(4 1 ? )(4 ? )( 1 4 ? )( d 4 1 ? 0 0 0 0 0 2 0 + ?= + = ?+ = ?+ == ∫ πε πε λ πε λ λ πε r The electric field of the rod at point P Exercises 27 Exercises Exercise 7: A charged rod AB of length L and charge density λ′ is placed in the electric field produced by an infinite charged rod of charge density λ. Find the electric force on AB rod. i x E ? 2 0 πε λ = r The field at the point a distance x from the infinite rod qd i x x qEF ? 2 d dd 0 ' πε λλ == rr The force exerted on Solution: A B ' λ a L λ o xdq dd ' xq λ= x i a La i x x FF La a ? ln 2 ? 2 d d 0 ' 0 ' + === ∫∫ + πε λλ πε λλ rr A B ' λ a L λ o xdq x Exercises If λ and λ′are like , the force will be parallel to the x axis; otherwise the force will be antiparallel to the x axis. 28 § 16.4 motion of a charged particle in a electric field In a uniform electric field σ+ σ? + 0 v r θ x y d o jEjE ?? 0 0 ?=?= ε σ r jqEj q F ?? 0 0 ?=?= ε σ r j m qE j m q m F a ?? 0 0 ?=?= = ε σ r r Projectile motion! Example 1: Initial state of the charged particle θ,,, 000 vvyyxx rr === Find (1)the the velocity when the particle impact with the plate; (2)if the width of the plate is L, what is the condition that the particle fly out the region of the electric field? 2 00 00 0 0 2 1 tatvyy tvxx tavv vv yy x yyy xx ++= += += = § 16.4 motion of a charged particle in a electric field 29 22 0 00 0 0 0 0 )( 2 1 )sin( )cos( )(sin cos t m qE tvyy tvx t m qE vv vv y x ?++= = ?+= = θ θ θ θ Two special situations: 1P739 example 16.20 2P741 example 16.21 § 16.4 motion of a charged particle in a electric field Example 2: If an electron e is placed near the center of the charged ring with charge q and radius R on the axis of the ring, how does it move? § 16.4 motion of a charged particle in a electric field 30 The electric field at z axis 2/322 0 )(4 Rz qz E + = πε The force on the electron 2/322 0 )(4 Rz eqz F + ?= πε If z << R, then z R eq F ) 4 ( 3 0 πε ?= The electron moves in simple harmonic motion. e mR eq 3 0 4πε ω = The angular frequency § 16.4 motion of a charged particle in a electric field Applications: Ink-jet printing § 16.4 motion of a charged particle in a electric field 31 SvSv r r ?== )cos( θΦ 1. The flux of a vector § 16.5 Gauss’s law for electric field and its applications S r Uniform vector field § 16.5 Gauss’s law for electric field and its applications Sv r r dd ?=Φ r r GM m F rg ? )( 2 ?== r r The gravitational field can be imagined as the force per unit mass at a given point in space. It is the same as the acceleration due to gravity at that location. 2. Gravitational field: M g r ∫∫ ?=?= SareaSarea SdSd r r r r vv ΦΦ 1 S 2 S Nonuniform vector field 32 ? ? ? ? ? >< <> == ==?= 20cosd 20cosd 20cosd cosddd πθθ πθθ πθθ θΦ Sg Sg Sg SgSg v r S r d g r θ ∫∫ ∫∫ ?== ?== SS Sg Sg r r r r dd dd SS ΦΦ ΦΦ § 16.5 Gauss’s law for electric field and its applications Gauss’s law for the gravitational field: 1solid angle 222 cosd ? dd d r S r rS r A φ ? = ? == r 2 Gauss’s law The flux of the gravitational field of M through a closed surface ∫ ?= S clsd dSg r r Φ S r d AS dcosd =φ r ? r r M ?d φ § 16.5 Gauss’s law for electric field and its applications 33 Because φθ cosdcosdd 2 S r GM SgSg ?==? r r ∫∫∫ ?=?=?= ? φ Φ d cosd d 2 S clsd GM r S GMSg r r GMSg πΦ 4d S clsd ?=?= ∫ r r S r d AS dcosd =φ r ? r r M ?d φ S θ g r § 16.5 Gauss’s law for electric field and its applications GM S r GM rSr r GM Sg SSS π Φ 4 d ? d ? d 22 ?= ?=??=?= ∫∫∫ r r The flux through the spherical surface which surrounded the M The flux through the arbitrary surface which surrounded the M GMSg S πΦ 4d ?=?= ∫ r r § 16.5 Gauss’s law for electric field and its applications 34 § 16.5 Gauss’s law for electric field and its applications The flux through the arbitrary surface which don’t surrounded M 0d =?= ∫ S Sg r r Φ The flux through the arbitrary surface which surrounded several particles ∑ ?= ????= N i Si N MG GMGMGM )within( 21 4 444 π πππΦ L SgggSg N SS r s L rr r r d)(d 21 ?+++=?= ∫∫ Φ 3. The Gauss’s law for the electric field § 16.5 Gauss’s law for electric field and its applications r r GM m F rg ? )( 2 ?== r r r r Q q F E ? 4 1 2 0 πε == r r 0 4 1 πε →? → G QM If we do the substitution 35 GMSg S πΦ 4d ?=?= ∫ r r QSE S 0 1 d ε Φ =?= ∫ rr similarly, the flux of electric field of a charge Q through a closed surface S with the charge inside The flux of gravitational field of a mass M through a closed surface S with the mass inside § 16.5 Gauss’s law for electric field and its applications ∫∫ =?= SS QSE within 0 d 1 d ε Φ rr For continuous distribution of charge If there is a collection of charges, then ∑ ∫ =?= N i Si S QSE )within( 0 1 d ε Φ rr Gauss’s law is a very general relationship between the flux of the total electric field through any closed surface and the total (net)charge enclosed by the surface. § 16.5 Gauss’s law for electric field and its applications 36 § 16.5 Gauss’s law for electric field and its applications S r ? S r ? S r ? § 16.5 Gauss’s law for electric field and its applications Exercise 1: What is the flux of the electric field through this closed cylindrical surface in a uniform electric field? 37 § 16.5 Gauss’s law for electric field and its applications Exercise 2: A butterfly net is in a uniform electric field of magnitude E. The rim, a circle of radius a, is aligned perpendicular to the field. Find the electric flux through the netting. Note: 1the electric field is caused by all charges anywhere, even those not enclosed in the surface, but the flux is depended only on the charges enclosed in the surface; 2the Gaussian surface S can be chose any way that seems useful; 3the Gauss’s law is the fundamental equation, since it expresses the relationship between electric field and the electric charge. E r § 16.5 Gauss’s law for electric field and its applications 38 4. The application of Gauss’s law (i)The shape of the closed surface involved is arbitrary, this means that in some situation we can choose the surface that will assist us in finding the magnitude of the field as easy as possible choose such a shape of the surface that a constant magnitude electric field exists over all or at least part of the surface. (ii)the symmetry of the charge distribution is a clue to choosing the shape of the Gaussian surface over which to evaluate the flux integral. § 16.5 Gauss’s law for electric field and its applications (iii)Gauss’s law still applies to any closed surface; we just cannot easily use the law to find the field because the flux integral on the left hand side of the law is too complicated to evaluate with ease or in closed form. ∫∫ =?= SS QSE within 0 d 1 d ε Φ rr Note the physical meaning of every quantity in the equation! § 16.5 Gauss’s law for electric field and its applications 39 1the magnitude of the electric field of a single, point charge The symmetry implies that at a fixed distance r away from the charge, the field has constant magnitude. Choose the Gaussian surface as the spherical surface of radius r, the integral of the left hand of the Gauss’s law is § 16.5 Gauss’s law for electric field and its applications )4(d0cosdd 2 spherespheresphere rEsEsESE π===? ∫∫∫ o rr 2 0 0 2 4 4 r Q E Q rE πε ε π = = Therefore we have § 16.5 Gauss’s law for electric field and its applications 40 2the magnitude of the electric field of an infinite, uniform charged straight-line The magnitude of the electric field is constant over the lateral area of the Gaussian surface but is not constant over the ends. ∫∫ ∫∫ ?+?+ ?=? end2end1 lateralS dd dd SESE SESE rrrr rrrr 0= 0= l S r d S r d E r § 16.5 Gauss’s law for electric field and its applications )2(dddd laterallaterallateralS rlESESESESE π===?=? ∫∫∫∫ rrrr 0 insd. 0 d 1 ε λ ε l q = ∫ ∫∫ =? insd. 0 S d 1 d qSE ε rr Gauss’s law r E 0 2πε λ = If we change the line charge into an infinite, uniform charged solid cylinder with charge density ρ and radius R, how to calculate the electric field distribution? § 16.5 Gauss’s law for electric field and its applications 41 3The magnitude of the electric field inside and outside a uniformly charged solid sphere (a)Inside the solid sphere R r E r S r d Choose a spherical Gaussian surface of radius r 3 0 insd. 0 2 S 3 41 d 1 )4(dd rq rESESE S πρ εε π = ==? ∫ ∫∫ rrr r R Q rr r E 3 00 3 2 0 4 1 33 4 4 1 πεε ρ ρπ πε === § 16.5 Gauss’s law for electric field and its applications R r E rS r d (b)Outside the spherical surface Choose a spherical Gaussian surface of radius r Qq rESESE S 0 insd. 0 2 S 1 d 1 )4(dd εε π = ==? ∫ ∫∫ rrr 2 0 2 0 4 1 4 1 r Q Q r E πεπε == Same as a pointlike charge. If the charge distribution is nonuniform, ···? § 16.5 Gauss’s law for electric field and its applications 42 How about a thin spherical shell, radius R and total charge q? § 16.5 Gauss’s law for electric field and its applications How about a spherical shell, inner radius a and outer radius b, volume charge density ρ? § 16.5 Gauss’s law for electric field and its applications 43 S r 4The magnitude of the electric field of uniformly charged an infinite sheet Choose a cylindrical closed surface. The flux through the lateral area is zero. The total flux of the field is the sum of the flux through each end of the cylindrical surface. S r § 16.5 Gauss’s law for electric field and its applications S r S r ESSESESE 2ddd rightleftS =+=? ∫∫∫ rr 0 insd. 0 d 1 ε σ ε S q = ∫ 0 2ε σ =E If we change the sheet into a plate of thickness d, what will be the electric field distribution inside the plate and outside the plate? § 16.5 Gauss’s law for electric field and its applications According to Gauss’s law