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22 - GAUSS’S LAW 7 22.5.IDENTIFY The flux through the...

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Unformatted text preview: GAUSS’S LAW 7 22.5.IDENTIFY: The flux through the curved upper half of the hemisphere is the same as the flux through the flat circle 22.6. 22.8. 22.23. defined by the bottom of the hemisphere because every electric field line that passes through the flat circle also must pass through the curved surface of the hemisphere. SET UP: The electric field is perpendicular to the flat circle, so the flux is simply the product of E and the area of the flat circle of radius r. EXECUTE: ¢5=EA=E(7rr2)=7rr2E EVALUATE: The flux would be the same if the hemisphere were replaced by any other surface bounded by the flat circle. IDENTIFY: Use Eq.(22.3) to calculate the flux for each surface. SETUP: ¢=E-Z=EAcos¢whmZ-Aa. Enema: (a) as] - -}(1efi) . <r>M = —(4x 10’ N/C)(0.10 m)2 cos(90° — 369°) = -24 N -m’/c. as) = +12 (top) . es, = —(4x 10’ N1C)(0.10 m)2 cos90° = o. as, - +}(right) . cps, = +(4 x 10’ N/C)(0. 10 m)2 cos(90° — 369°) = +24 N - mz/C. a, = 42 (bottom) . o,‘ = (4x 10’ N/C)(0.10 m): cos90° = o . as, - +1“ (front) . <15, = +(4x103 N/cxo.ro m)z cos36.9° = 32 N - ml/c . a, - —:‘ (back) . o,‘ = -(4 x103 N/C)(0.10 m)’ cos36.9° = -32 N . mZ/C . EVALUATE: (b) The total flux through the cube must be zero; any flux entering the cube must also leave it, since the field is uniform. Our calculation gives the result; the sum of the fluxes calculated in part (a) is zero. IDENTIFY: Apply Gauss’s law to each surface. SET UP: Q“,el is the algebraic sum of the charges enclosed by each surface. Flux gut of the volume is positive and flux into the enclosed volume is negative. EXECUTE: (a) (PSI = qI/F; = (4.00x10" C)/R, = 452 N 'mI/C. (b) tbs2 = qzlfi, =(—7.80x10‘° C)/Po = -881 N 'mZIC. (e) (93, = (q1 ~+ qz)/P0 = ((4.00— 7.80) x 10" C)/l?, = -429 N -m‘/C. (d) 1% = (q1 + q,)/R, = ((4.00+ 2.40)x10'° C)/F; = 723 N - mZ/C. (e) tbs, = (qI + q2 + q,)/E’, = ((4.00—7.80+ 2.40)x10” C)/P0 = —158 N. mz/C. EVALUATE: (f) All that matters for Gauss’s law is the total amount of charge enclosed by the surface, not its distribution within the surface. IDENTIFY: The electric field inside the conductor is zero, and all of its initial charge lies on its outer surface. The introduction of charge into the cavity induces charge onto the surface of the cavity, which induces an equal but opposite charge on the outer surface of the conductor. The net charge on the outer surface of the conductor is the sum of the positive charge initially there and the additional negative charge due to the introduction of the negative charge into the cavity. 22.25. 22.26. (a) SET UP: First find the initial positive charge on the outer surface of the conductor using q‘ = 0.4, where A is the area of its outer surface. Then find the net charge on the surface afier the negative charge has been introduced into the cavity. Finally use the definition of surface charge density. EXECUTE: The original positive charge on the outer surface is qi = M = 0(4nr’) = (6.37 x 10.. C/m2)47t(0.250 m2) = 5.00x10“ C/m2 After the introduction of —0.500 pC into the cavity, the outer charge is now 5.00 ,uC — 0.500 pC = 4.50 pC The surface charge density is now a' — 7 — 2—”?- — m = 5.73 x10" C/m2 ‘ ’ , , . . q q (b) SET UP: Usrng Gauss s law, the electric field is E _ T _ E: = 3,47": EXECUTE: Substituting numbers gives —6 E: figf—L—z: 6.47 x105 N/C. (8.85x10 C /N -m )(47r)(0.250 m) (c) SET UP: We use Gauss’s law again to find the flux. (D5 =% . O EXECUTE: Substituting numbers gives 4) = —0.500x 10“ C 5 8.85x10'” c‘m-m’ EVALUATE: The excess charge on the conductor is still +5.00 pC, as it originally was. The introduction of the —0.500 ,uC inside the cavity merely induced equal but opposite charges (for a net of zero) on the surfaces of the conductor. IDENTIFY: The magnitude of the electric field is constant at any given distance from the center because the charge density is uniform inside the sphere. We can use Gauss’s law to relate the field to the charge causing it. a SET UP: Gauss’s 1a tells us that EA = 1 , and the char e densi is ‘ven b = l = _q__ . ( ) W F3 8 ty 8' Y P V (4 ”MR, EXECUTE: Solving for q and substituting numbers gives q = EAR, = E(4:rr’)F{, = (1750 N/C)(4z)(0.500 m)’(8.85 x 10'12 C’IN - m1) = 4.866 x 10" C . Using the formula -I for charge density we get p = 1 = ———g— = M2— = 2.60x 10'7 C/m’. = —5.65 x10‘ N - mZ/Cz. (b) SET UP: Take a Gaussian surface of radius r = 0.200 m, concentric with the insulating sphere. The charge enclosed within this surface is qml = pV = p(§nr’) , and we can treat this charge ad‘a point-charge, using Coulomb’s law E = 4;: 1‘1?- . The charge beyond r = 0.200 m makes no contribution to the electric field. 0 r EXECUTE: First find the enclosed charge: 4.... = p(;flr3) = (2.60 x10" C/m’)[-§-z(0.200m)3] = 8.70x10’9 c Now treat this charge as a point-charge and use Coulomb’s law to find the field: 8.70x10" c (0.200 m)2 EVALUATE: Outside this sphere, it behaves like a point-charge located at its center. Inside of it, at a distance r from the center, the field is due only to the charge between the center and r. IDENTIFY: Apply Gauss’s law and conservation of charge. SET UP: Use a Gaussian surface that lies wholly within he conducting material. EXECUTE: (a) Positive charge is attracted to the inner surface of the conductor by the charge in the cavity. Its magnitude is the same as the cavity charge: q," = + 6.00 nC, since E = 0 inside a conductor and a Gaussian surface that lies wholly within the conductor must enclose zero net charge. E=(9.00x10'N-m2/C2) =1.96x1o’N/c 22.30. 22.58. (b) On the outer surface the charge is a combination of the net charge on the conductor and the charge “lefi behind" when the +6.00 nC moved to the inner surface: q” =q_ + qm =>qm =qm'9m =5.00nC—6.00nC=-1.00nC. EVALUATE: The electric field outside the conductor is due to the charge on its surface. IDENTIFY: The net electric field is the vector sum of the fields due to each of the four sheets of charge. SET UP: The electric field of a large sheet of charge is E = a / ZR, . The field is directed away from a positive sheet and toward a negative sheet. _Lzl Ll 12L 12L}. ' EXECUTE. (a) At A: E 2|?) 2P0 + 2F; 2"?) 2P ——(laz|+la,[+|a‘|— lull). l A =¥(5 pC/mz + 2 yC/m’ + 4 yC/mz —6 pc/m‘) = 2.82x10’ N/C to the lefi. 0 (b) E = 0' +1131+12L E=J-(IUII‘F'UJI +|U4| ‘l’zb' " 2P 2P 2P 2P 2P E, = —(6 pC/mz + 2 pC/mz +4 yc/m2 — 5 yC/mz) = 3.95x10’ N/C to the left. «)E.=l”—4+ 11"- sz-l— m=~<lazl+lvsI-lml-lml)- 2P 2P 2P 2P 2P EC = E“ ,Ic/m2 + 6 pC/nfi —5 pC/mz — 2 pc/m’) = 1.69x10’ N/C to the lefl 0 EVALUATE: The field at C is not zero. The pieces of plastic are not conductors. IDENTIFY: Apply Gauss’s law. SET UP: Use a Gaussian surface that is a sphere of radius r and that is concentric with the spherical distribution of charge. The volume of a thin spherical shell of radius r and thickness dr is (W = 4nr2dr . l l EXECUTE: (a) Q= jp(r)dV= 4x fp(r)r‘dr= 4m IF—Jr zdr=41rpo|: Var-3% [r’dr] 0 0 3 4 9:4” [R 4R —— --———- =0 . The total charge is zero. 3 3R 4 (b)Forr2R, (IE Edi: =9§d—=o,soE=o. 0 i I" "_ 47‘ ’ I I2 I 2_47tpo ’11 I 4 ’I3 I (c)For rSR, uE-dA-%=E- °p(r)r dr . E4Irr - F3 [or dr—a—E‘Lr dr and -&__’___ _I’.» _L E Prl[3 3R]: 3P [1 R]- (d) The graph of E versus r is sketched in Figure 22.58. M P_o_ 2p°r R e Wh E maxm -—-=0. 'Ih —-—-—5‘—=0 d =—.Atth' , () ere ma 1 um, dr rsgives— 3P 332R an rm 2 isr PoR 1 PoR E=—-— l—— =———. 3P2[ 2] 12F; EVALUATE: The result in part (b) for r S R gives E = 0 at r = R ; the field is continuous at the surface of the charge distribution. 22.65. poR/Izeo 12?. R Figure 22.58 IDENTIFY: Let —dQ be the electron charge contained within a spherical shell of radius r’ and thickness dr' . Integrate r' from 0 to r to find the electron charge within a sphere of radius r. Apply Gauss’s law to a sphere of radius r to find the electric field E(r) . SET UP: The volume of the spherical shell is (W = 4m" dr' . EXECUTE: (a) Q(r) = Q— IpdV = Q_ 32? Jew/Mad, = Q— fa? Erna—2m. dr' _ Q(r) = Q — 499-126" — afir2 — 2ar — 2) = Qe'z""°[2(r/ao)2 + 2(r/ao) + 1]. 3 3 aoa Note if r —> oo, Q(r) —~) 0 ; the total net charge of the atom is zero. (b) The electric field is radially outward. Gauss’s law gives E(47rr’) = % and 0 E = "Qer: .. (2(r/ao)2 + 2(r/ao) + 1) . (c) The graph of E versus r is sketched in Figure 22.65. What is plotted is the scaled E, equal to E/ Ev,” , kQ versus scaled r, equal to r/ ao . EM,“e = ‘T is the field of a point charge. r EVALUATE: As r —> 0 , the field approaches that of the positive point charge (the proton). For increasing r the electric field falls to zero more rapidly than the 1/ r2 dependence for a point charge. Scaled E 1.00 0.80 0.20 mm“ 0.00 E 5 ' ‘ 0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 Figure 22.65 ...
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