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Chap 22-24 Electric field First circular accelerator in the world Easy chair Charged hair Electricity and Magnetism Electric forces hold atoms and molecules together. Electricity controls our thinking, feeling, muscles and metabolic processes. Electricity and magnetism underpin much of our current technology (e.g. computers). Electricity and magnetism are linked on a fundamental level. Key terms: electric charge electrostatics electron proton neutron nucleus positive ion negative ion ionization conductor insulator point charge electric dipole closed surface electric flux surface integral Gauss’s Law Gaussian surface Faraday’s icepail experiment electric potential energy electric potential voltage electron volt equipotential surface gradient cathode-ray tube §1 charges and couloumb’s law 1. 1 charges 1) Negative and positive 2) quantified (millikan) Q ne 19 e 1.6 10 C Fraction charge 1 e 3 3) conservation (Franklin) 4) Invariance of charges 2 e 3 1.2 Couloumb’s law 1) Model:point charge 2) Couloumb’s law Q1Q2 Q1Q2 1 F k 2 r0 r0 2 4 r r 0 k=9×109Nm2/c2, 0 =8.85×10-12c2/Nm2 q1 r21 r q2 F21 Caution: 1)valid for point charge in air 2) obey newton’s law 3) in general Henry Cavendish Experiment Fe Fg Quiz 1. Even though electric forces are very much stronger than gravitational forces, gravitational forces determine the motions in the solar system. Why? 2. When we approach another person, we are not aware of the gravitational and electric forces between us. What are the reason in each case? §2 Electric field E 2.1 Electric field Electric source F E q0 Electric field Electric field for point charge F 1 qq0 0 r 2 4 0 r F 1 q 0 E r 2 q0 4 0 r Collection of point charge E Fk k q0 1 qk 0 Ek r 2 k k k 4 0 rk superposition of electric field Electric charge 1 dq 0 Charge distribution dE r 2 4 0 r r dE dl dq dS P dq dV E dq 0 r 2 4 0 r Step to solve E: dq dE dE x dE y E x dE x E y dE y Positive electric charge Q is distributed uniformly along a line with length 2a, lying along the y-axis between y=-a and y=+a. Find the electric field at point P on the x-axis at a distance x from the origin. y a r x2 y2 O Q -a x P x dE Solution: 1 dQ Q dy Q dE dQ dy dy 2 2 2 4 r 4 2 a x y 0 0 2a Q dy x Q xdy dE x dE cos 2 2 2 2 32 2 2 4 2 a x y 4 0 0 2a x y x y Q dy y Q ydy dE y dE sin 2 2 2 2 40 2a x 2 y 2 40 2a x y x y 1 Qx a dy Q 1 Ex 32 2 2 a 40 2a 40 x x 2 a 2 x y 1 Q a ydy Ey 40 2a a x 2 y 2 Q 1 E iˆ 40 x x 2 a 2 32 0 32 Discussion: When x is much larger than a (that is, x>>a), a) Q 1 1 Qˆ ˆ E iE i 2 2 2 40 x x a 40 x This is same as the field of a point charge. b) c) When a is much larger than x (that is, a>>x), 1 ˆ Q 2a ˆ E i i 2 20 x x a 1 20 x Infinite line of charge E 20 r Field due to a Power Line A ring conductor with radius a carries a total charge Q uniformly distributed around it. Find the electric field at a point P that lies on the axis of the ring at a distance x from its center. y dQ r x2 a2 a O Q x P x dE Solution: 1 dQ dE 40 x 2 a 2 1 dQ dEx dE cos 40 x 2 a 2 Ex 1 xdQ 40 x a 2 2 32 1 x x a 2 Qx 40 x 2 a 2 Because the symmetry , Ey=0. 1 Qx ˆ E Exi 40 x 2 a 2 ˆ i 32 2 32 1 1 40 x 2 a 2 32 x 0 ,E 0 Discussion: 1 q 2). x R ,E 2 4 0 x 3). 2 x R ,E Emax 2 z dl E 2 R 2 O 2 2 R x R O y r dE x x P q dE Find the electric field caused by a disk of radius R with a uniform position surface charge density , at a point along the axis of the disk a distance x from its center. R O dQ x r dr Q x Solution: dQ dA 2rdr 2rdr dE x Ex 1 40 x 2 r 2 R 0 2rdr x 1 32 2rdr x 40 x 2 r 2 32 x 2 0 rdr R x 0 2 r 2 32 x 2 0 1 2 2 R x 1 1 dE y dE z 0 E y Ez 0 Discussion: If R is much larger than x while cons 1 1 E (an infinite sheet of charge) 2 2 2 0 R x 1 The electric field produced by an infinite plane sheet of charge is independent of the distance from the sheet. Thus the field is uniform. Its direction is everywhere perpendicular to the sheet, away from it. x Ex 2 0 Sheet 2 Example x d Sheet 1 Two infinite plane sheets are placed parallel to each other, separated by distance d. The lower sheet has a uniform positive surface charge density , and the upper sheet has a uniform negative surface charge density with the same magnitude. Find the electric field between the two sheets, above the upper sheet, and below the lower sheet. Solution: E1 E2 2 0 At all points, the direction of E1 is away from the positive charge of Sheet 1, and the direction of E2 is towards the negative charge of sheet 2. 0 E E1 E2 0 ˆj 0 above the upper sheet between the sheets below the lower sheet Example: Find the intensity of electric field of dipole in P Solution: E q i E E l 4 0 ( x l 2) 2 q q O q P x E i 2 4 0 ( x l 2) Electric dipole moment q 2 xl i E E E 2 2 2 p ql 4 0 ( x l 4) 2 xp E 2 2 2 4 0 ( x l 4) P E E Along perpendicular bisector q E E r 2 2 4 0 (r l 4) P q q l E E 2 E cos 4 0 r 3 Example Find the torque exert on dipole Solution:to pivot 0 1 1 M F l sin θ F l sin θ 2 2 qlE sin θ M ql E p E q O F l θ q Discussion (1) θ 2 (2) 0 (3) Maximum torque Torque=0 stable Torque=0,not stable equilibrum P E F § 3 Gauss law Mathematics & Physics & astronomy Qenclosed E dA 0 3.1 Electric field line 1) rule: The tangent to the field line express the direction of field the density of the line measure the intensity of the field 2) Essence of Electric field d E ds 1) originating from the positive to negative; (2) never intercross; (3) no close line. +q -q Field Lines 3.2 flux 1) In uniform field n d e EndS E cosdS EdS dS dSn de E dS En E dS dS n 2) Non uniform field de E dS dS e d e E dS S E E To closed surface e d e E dS S Discussion (1) Direction of S Open surface Closed surface n E Concave:positive, Outward: positive 3.3 Guass’s law 1) suppose a point charge is in the center of a sphere surface Q Q Q E ds k 2 ds k 2 4a 2 S S 0 a a 2) suppose a point charge is in any interior place of a sphere surface S Q E ds 0 +q q outside the surface e e1 e 2 0 +q many charges E E1 E2 ... E5 e E dS ( E1 E2 ... E5 ) dS E1 dS E2 dS ... E5 dS q1 q2 q3 0 0 0 Caution: E produced by all charges e only related to internal charge S1 S2 S q3 q1 q2 q4 q5 Guass law e S 1 E dS 0 qi i 1 e E dS S Meaning: (separated charges) dV 0 V Source field (distribution charge) Caution: (1) E in Guass’s law is the E in Guassian surface,it is produced by all the charges in the space (2) the flux only depends on the charges inside (3) Zero flux does’t mean the zero field. 4)to moving charge,coulumb’s law don’t work,but guass’s law works 3.4 Application of guass law: find E A.symmetry distribution of charges B.intensity in guass surface is uniform or piecewise uniform Example:What is the electric flux through a cylindrical surface? The electric field, E, is uniform and perpendicular to the surface. The cylinder has radius r and length L A) E 4/3 p r3 L B) E r L C) E p r2 L D) E 2 p r L E) 0 Example:In a model of the atom the nucleus is a uniform ball of +e charge of radius R. At what distance is the E field strongest? A) r = 0 B) r = R/2 C) r = R D) r = 2 R E) r = 1.5 R Conclusion: Step to find E with guass law: (1) analysis symmetry (2) from above guass surface must be closed one the field point must in guass surface E is easy to take out from integral function Find the E of the following charged ball with uniform density Solution: + ( r R) r + R r' + 3 1 q 0 R 0 E r r 2 2 4 0 r 3 0 r + rR 14 3 1 2 SE dS E 4r 0 3 r 0 q' E E r 3 0 R O r Find E of infinite long line dS Solution: r From symmetry: e E dS S s E dS t E dS b E dS s EdS E s dS E 2r l E P dS E l E E 2r l 1 0 E 2 0 r l O r Infinite plane or Sheet Charge per unit area Cm-2. Mirror symmetry about and perpendicular to plane,find E Solution: From symmetry e E dS S s E dS ls E dS rs E dS 0 ES ES 2 ES 2 ES 1 0 S E 2 0 Ex O x k r Example: find E rR Solution: R 1 k 4 3 2 E ds E 4r 0 r 3 r k E 3r 0 ? Q E ds k Q dV 4r 2 dr 2kr 2 V V r 0 2 2 kr k 2 No relation with r E 4r E 0 2 0 r R q dq 4kr dr 2kR 2 R 0 4r 2 E 2kR 2 0 kR 2 E 2 0 r 2 Inversely proportion to r2 dr r r Example:find E in p b Solution: (1) fill in the hollow ball E1 a Q1 r 2 0 4 0a (2) E produced by filling parts: r0 Q1 Q2 E E1 E 2 ( ) 4 0 a 2 b 2 E2 Q2 r 2 0 4 0b Q1 Q2 4 ( R3 r 3 ) 3 P Problem to deal with guass’s law • sphere symmetry • mirror symmetry • cylindrical symmetry R § 4 potential 4.1 work done by electric force Aa b b a b b F d l q0 E dl q qQ 4 0 a b a a Q cosdl 2 4 0 r dr qQ 1 1 Wa Wb 2 r 4 0 ra rb kQq wa r a q Q Potential energy l b F l b Discussion: a.conservative force F dl 0 C b.potential of point charge kQ kQ A U a 2 dr E dl a r a q ra c.potential difference U ab U a U b b a kQ kQ E dl ra rb a q Q F Caution: (1) belong to q0 and field source (2) with related value (3) rule for choose zero potential energy • when sources is limited,choose infinite place • in contrary, a specific place is choose • in practice,earth and the out shell of equipment 3、the calculation of potential 4.2 calculation of potential 1)potential of charged particle q q U p p E dl r p dr 4 0 r 2 4 0 rp 2)collection of charge particles: U ( p) i 4 0 ri 3)continuous charge distribution dU dq 40 r U (r ) V dq 40 r qi dV dq ds dl Find the potential of dipole Solution: r r q Ua U U k 1 k kq r r r r r>>l, 2 0 r r r 2 Ua 0 U a U max r r l cos q l pe cos Ua r2 -q Example:find potential of rod with l=15.0cm, =2.0 10-7c/m Solution: (1) dx al 3 U1 ln 2 . 5 10 V 0 4 ( l a x ) 4 a 0 0 l y (2) in perpendicular bisector l/2 dx l / 2 4 0 ( x 2 b 2 )1 / 2 U 2 dU b2 l 2 / 4 l / 2 ln 2 2 4.3 10 3V 4 0 b l /4l/2 dx P x l a l x Example:Find potential in axis z dl solution:1) dq dl dq 1 dl 4 0 r 4 0 r 2R dl 1 q U 0 4 0 r 4 0 x 2 R 2 R 2 0 x 2 R 2 q Uo In center x 0 4 0 R dU 1 R O y r dE x x P q dE q 4 0 R U O x §5 relationship between field and potential 5.1 basic concept Equivalent potential surface: U ( x, y, z ) C Character: (1) E dl (2)the density of equal potential surface reflect the intensity of electric field. (3)the direction of the line of force is the descended direction of potential 5.2 the relation between electric field and potential U ˆ U ˆ U ˆ E U { i j k) x y z u u u E ( er e ) r r z Negative of the gradient of U Example:find the E and U of the following disk x Ex P Solution: take a circular strip dq 2rdr dU dq 4 0 r x 2 U dU R2 R1 R2 R1 2 1/ 2 rdr 2 0 r x 2 rdr 2 0 r 2 x 2 1/ 2 R1 r 2 1/ 2 2 0 R2 R 2 x 2 R12 x 2 2 U r x x E Ex 1 / 2 1 / 2 x 2 0 R 2 x 2 R 2 x 2 1 2 dr