This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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: ¢=EZ=EAcos¢whmZAa. 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‘mm’
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 pointcharge, 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 pointcharge and use Coulomb’s law to ﬁnd the ﬁeld: 8.70x10" c
(0.200 m)2 EVALUATE: Outside this sphere, it behaves like a pointcharge 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'Nm2/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" szl— m=~<lazl+lvsIlmllml) 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: Var3% [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, uEdA%=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 — 499126" — 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 ...
View
Full Document
 Spring '08
 HKMIET

Click to edit the document details