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Unformatted text preview: 21.58. 7 21.60. 21.61. 21.62. IDENTIFY and SETUP: The electric field produced by an infinite sheet of charge with charge density 0' has magnitude E = ‘32! . The field is directed toward the sheet if it has negative charge and is away from the sheet if it
has positive charge. , EXECUTE: (a) The field lines are sketched in Figure 21.58a. (b) The field lines are sketched in Figure 21 .58b. EVALUATE: The spacing of the field lines indicates the strength of the field. In part (a) the two fields add
between the sheets and subtract in the regions to the left of A and to the right of B. In part (b) the opposite is true. +0" +20
IDENTIFY: The field appears like that of a point charge along way from the disk and an infinite sheet close to the disks center. The field is symmetrical on the right and left.
SET UP: For a positive point charge, E is proportional to l/rzand is directed radially outward. For an infinite sheet of positive charge, the field is uniform and is directed away from the sheet. EXECUTE: The field is sketched in Figure 21.60.
EVALUATE: Near the disk the field lines are parallel and equally spaced, which corresponds to a uniform field. Far from the disk the ﬁeld lines are getting farther apart, corresponding to the 1/r2 dependence for a point charge. ./
\ Figure 21.60 IDENTIFY: Use symmetry to deduce the nature of the ﬁeld lines.
(3) SET UP: The only distinguishable direction is toward the line or away from the line, so the electric ﬁeld lines are perpendicular to the line of charge, as shown in Figure 21 .61a. ini>l< Figure 21.6121 (b) EXECUTE and EVALUATE: The magnitude of the electric ﬁeld is inversely proportional to the spacing of the
field lines. Consider a circle of radius r with the line of charge passing through the center, as shown in
Figure 21.61b. Figure 21.61b The spacing of field lines is the same all around the circle, and in the direction perpendicular to the plane of the
circle the lines are equally spaced, so E depends only on the distance r. The number of field lines passing out
through the circle is independent of the radius of the circle, so the spacing of the ﬁeld lines is proportional to the
reciprocal of the circumference Zﬂ'r of the circle. Hence E is proportional to l/r. IDENTIFY: Field lines are directed away from a positive charge and toward a negative charge. The density of
field lines is proportional to the magnitude of the electric field. SET UP: The field lines represent the resultant field at each point, the net field that is the vector sum of the fields
due to each of the three charges. EXECUTE: (a) Since field lines pass from positive charges and toward negative charges, we can deduce that the
top charge is positive, middle is negative, and bottom is positive. (b) The electric field is the smallest on the horizontal line through the middle charge, at two positions on either
side where the field lines are least dense. Here the ycomponents of the ﬁeld are cancelled between the positive
charges and the negative charge cancels the xcomponent of the field from the two positive charges. EVALUATE: Far from all three charges the field is the same as the field of a point charge equal to the algebraic sum of the three charges. 22.6. 22.7. 22.8. 22.9. IDENTIFY: Use Eq.(22.3) to calculate the flux for each surface. SETUP: <1>=E A =EAcos¢ where/T: Aft. EXECUTE: (a) as] = —j'(1eft) . «1251 = —(4><103 N/C)(0.10 my cos(900~ 36.90) : —24 N  mz/C.
r252 = +IE(top) . cps: = —(4x103 N/C)(0.10 m)2,cos90° = 0. 1353 = +j‘(right) . $53 = +(4><103 N/C)(0.10 m)2 cos(90° — 369°) = +24 N  mZ/C. r354 = —l€ (bottom) . (D54 2 (4X 103 N/C)(0.10 m)2 cos90° = 0. n55 = +i (front) . (1255 = +(4><103 N/C)(0.10 my cos36.9° = 32 N » mz/C . r256 = 4 (back) . cosﬁ = —(4x 103 N/C)(0. 10 my cos 36.9° = —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. (a) IDENTIFY: Use Eq.(22.5) to calculate the ﬂux through the surface of the cylinder. SET UP: The line of charge and the cylinder are sketched in Figure 22.7. LBJ Figure 22.7 EXECUTE: The area of the curved part of the cylinder is A = 27W]. The electric field is parallel to the end caps of the cylinder, so E .3 = 0 for the ends and the ﬂux through the
cylinder end caps is zero. The electric field is'normal to the curved surface of the cylinder and has the same magnitude E = l/ 271'60r at all
points on this surface. Thus ¢ = 0° and i w
end we E ,u (6,00><10'6 C/m)(0.400 m)
c1> =EA =EA= 2/2 2 1— =
E cow ( my“ 7") 60 8.854X10’” CZ/Nm2 (b) In the calculation in part (a) the radius r of the cylinder divided out, so the ﬂux remains the same,
(DE = 2.71x105 Nm2/C. ,1; (6.00x10* C/m)(0.800 m) 5 2 .
(c) (DE = — — _12 2 2  5.42X10 N  m / C (twice the flux calculated in parts (b) and (0)).
60 8.854x10 C /Nm EVALUATE: The ﬂux depends on the number of field lines that pass through the surface of the cylinder.
IDENTIFY: Apply Gauss’s law to each surface. SET UP: Q3!lcl is the algebraic sum of the charges enclosed by each surface. Flux out of the volume is positive and
ﬂux into the enclosed volume is negative. EXECUTE: (3) (psi = (11/60 = (4.00x10'9 cye0 = 452 NmZ/C.
(b) (1)31 = (12/50 = (—7.80x10‘9 C)/60 = —881 NmZ/C. (c) (1)53 = (q1 + q2)/60 = ((4.00— 7.80)><10‘9 cyq, = —429 N . mZ/C.
(d) (1)34 = (q1 + q3)/60 = ((4.00 + 2.40)><10‘° eye.) = 723 N  m2/C. (e) (D35 = (q1 + q2 + q3)/os0 = ((4.00— 7.80 + 2.40)><10‘9 C)/§, = —158 N  m2/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: Apply the results in Example 21.10 for the field of a spherlcal shell of charge.
lql SET UP: Example 22.10 shows that E = 0 inside a uniform spherical shell and that E = k? outside the shell. —2.71><105 NmZ/C EXECUTE: (a) E = 0 15.0><10'6 C 7
= : . 109N 2/C2 ——=3.75><10 N/C
(b) r 0.060 m and E (8 99X 111 ) (0060 my
15.0x10'6 C 7
= = . 109N~ 2/C2 ——————=1.ll><10 N/C
(c) r 0.110mand E (8 99X m ) (0110 my EVALUATE: Outside the shell the electric field is the same as if all the charge were concentrated at. the center of
the Shell. But inside the shell the field is not the same as for a point charge at the center of the shell, 1n31de the shell the electric field is zero. 22.11. 22.38. IDENTIFY: Apply Gauss’s law.
SET UP: In each case consider a small Gaussian surface in the region of interest. EXECUTE: (3) Since E is uniform, the flux through a closed surface must be zero. That is: (I) = 43E 11/3 = g :21; deV : 0 :> fpdV = 0. But because we can choose any volume we want, p must be zero if the integral equals zero.
(b) If there is no charge in a region of space, that does NOT mean that the electric field is uniform. Consider a closed volume close to, but not including, a point charge. The field diverges there, but there is no charge in that
region. EVALUATE: The electric field within a region can depend on charges located outside the region. But the flux
through a closed surface depends only on the net charge contained within that surface. IDENTIFY: Apply Gauss’s law. SET UP: Use a Gaussian surface that is a cylinder of radius r, length l and that has the line of charge along its axis.
The charge on a length l of the line of charge or of the tube is q = 0d . EXECUTE: (a) (i) For r < a , Gauss’s law gives E(27rrl) = Q“ = a—l and E = a .
60 60 27W
(ii) The electric field is zero because these points are within the conducting material.
2 1
(iii) For r > b , Gauss’s law gives E(27rrl) =—B‘ﬂ =1 and E = —a—— .
‘ ‘0 6o 7560’” The graph of E versus r is sketched in Figure 22.38. (b) (i) The Gaussian cylinder with radius r, for a < r < b , must enclose zero net charge, so the charge per unit length
on the inner surface is —a. (ii) Since the net charge per length for the tube is +05 and there is ——a on the inner surface, the charge per unit length on the outer surface must be +2a.
EVALUATE: For r > b the electric field is due to the charge on the outer surface of the tube. £le Figure 22.38 1:4" an, 21:3: £48€w§gé£€lis¢if¢£§ﬁhyz 22.39. (a) IDENTIFY: Use Gauss’s law to calculate E(r).
(i) SET UP: r < (1: Apply Gauss’s law to a cylindrical Gaussian surface of length l and radius r, where r < a, as sketched in Figure 22.39a. —a EXECUTE: (DE 2 E(27rrl)
Qencl = 0d (the charge on the length l
of the line of charge) Figure 22.393 (DE = Q8“ gives E (27571 ) = ﬂ 50 6o
a . The enclosed charge is positive so the direction of E is radially outward. E :
27reor (ii) a < r < [7: Points in this region are within the conducting tube, so E = 0. (iii) SET UP: r > b: Apply Gauss’s law to a cylindrical Gaussian surface of length l and radius r, where r > b, as
sketched in Figure 22.3%. EXECUTE: 435 = E (anl)
Qencl 2 al (the charge on length l of the line of charge) —al (the charge on
length l of the tube) Thus Qencl = 0. Figure 223%
(DE = QM gives'E(27rrl) = 0 and E = 0. The graph of E versus r is sketched in Figure 22.39c.
60
E
a/queor
I
I
r
a
Figure 22.390 (b) IDENTIFY: Apply Gauss’s law to cylindrical surfaces that lie just outside the inner and outer surfaces of the
tube. We know E so can calculate Qe (i) SET UP: inner surface
Apply Gauss’s law to a cylindrical Gaussian surface of length l and radius r, where a < r < b. ncl ’ EXECUTE: This surface lies within the conductor of the tube, where E = 0, so (DE 2 0. Then by Gauss’s law QM = 0. The surface encloses charge 051 on the line of charge so must enclose charge —al on the inner surface of the tube. The charge per unit length on the inner surface of the tube is ~05. (ii) outer surface The net charge per unit length on the tube is —a. We have shown in part (i) that this must all reside on the inner
surface, so there is no net charge on the outer surface of the tube. EVALUATE: For r < a the electric field is due only to the line of charge. For r > b the electric field of the tube is
the same as for a line of charge along its axis. The ﬁelds of the line of charge and of the tube are equal in magnitude
and opposite in direction and sum to zero. For r < a the electric field lines originate on the line of charge and terminate on the surface charge on the inner surface of the tube. There is no electric field outside the tube and no
surface charge on the outer surface of the tube. 22.44. 22.45. IDENTIFY: Apply Gauss’s law and conservation of charge. SET UP: Use a Gaussian surface that is a sphere of radius r and that has the point charge at its center.
1 . . EXECUTE: (a) For r < a , E = T95, radlally outward, s1nce the charge enclosed is Q, the charge of the point
71'60 r charge. For a < r < b , E = 0 since these points are within the conducting material. For r > b , E = ,
71'60 r radially inward, since the total enclosed charge is —2Q. (b) Since a Gaussian surface with radius r, for a < r < b, must enclose zero net charge, the total charge on the inner Q 47m2 .
(c) Since the net charge on the shell is —3Q and there is ——Q on the inner surface, there must be —2Q on the outer 2Q
43192 '
(d) The field lines and the locations of the charges are sketched in Figure 22.44a.
(e) The graph of E versus r is sketched in Figure 22.44b. surface is —Q and the surface charge density on inner surface is 0 : — surface. The surface charge density on the outer surface is 0' z — w Q spread on
inner surface —2Q spread on outer
1 surface a  b (a) (b)
Figure 22.44 EVALUATE: For r < a the electric field is due solely to the point charge Q. For r > b the electric field is due to
the charge —2Q that is on the outer surface of the shell.
IDENTIFY: Apply Gauss’s law to a spherical Gaussian surface with radius r. Calculate the electric field at the surface of the Gaussian sphere.
(3) SET UP: (i) r < a: The Gaussian surface is sketched in Figure 22.4521. EXECUTE: (DE = EA 2 E(47[r2) Qencl = 0; no charge is enclosed 9i (DE = °1 says E(4ﬂ'r2) =0 and E=0.
60 Figure 22.453 (ii) a < r < b: Points in this region are in the conductor of the small shell, so E = 0.
(iii) SET UP: b < r < c: The Gaussian surface is sketched in Figure 22.45b.
Apply Gauss’s law to a spherical Gaussian surface with radius b < r < c. EXECUTE: (DE = EA = E(47rr2)
The Gaussian surface encloses all of the small
shell and none of the large shell, so Q = +2q. encl (D E = 92% gives E (47rr2) = a so E = 2g 2 . Since the enclosed charge is positive the electric ﬁeld is radially
50 60 “or
outward. (iv) 0 < r < d: Points in this region are in the conductor of the large shell, so E = 0. 2246 22.46. Chapter 22 (v) SET UP: r > d : Apply Gauss’s law to a spherical Gaussian surface with radius r > d, as shown in Figure 22.45c. EXECUTE: @E : EA : 5(4mz) The Gaussian surface encloses all of the small shell
and all of the large shell, so Qencl = +2q + 4q 2 6g. Figure 22.45c Q5“ gives E (475%) = :32
60 50
6q
47rq,r2
The graph of E versus r is sketched in Figure 22.45d. (1i: E E = . Since the enclosed charge is positive the electric field is radially outward. E 6q/47r60d2 2q/471'50b2 a b c d Figure 22.45d (b) IDENTIFY and SET UP: Apply Gauss’s law to a sphere that lies outside the surface of the shell for which we want to ﬁnd the surface charge.
EXECUTE: (i) charge on inner surface of the small shell: Apply Gauss’s law to a spherical Gaussian surface with radius a < r < b. This surface lies within the conductor of the small shell, where E = 0, so (DE = 0. Thus by Gauss’s law Qend = 0, so there is zero charge on the inner surface of the small shell. (ii) charge on outer surface of the small shell: The total charge on the small shell is +2q. We found in part (i) that
there is zero charge on the inner surface of the shell, so all +2q must reside on the outer surface. (iii) charge on inner surface of large shell: Apply Gauss’s law to a spherical Gaussian surface with radius c < r < d.
The surface lies within the conductor of the large shell, where E = 0, so (DE = 0. Thus by Gauss’s law le = O. The
surface encloses the +2q on the small shell so there must be charge ——2q on the inner surface of the large shell to make the total enclosed charge zero.
(iv) charge on outer surface of large shell: The total charge on the large shell is +4q. We showed in part (iii) that the charge on the inner surface is —2q, so there must be +6q on the outer surface. EVALUATE: The electric field lines for b < r < c originate from the surface charge on the outer surface of the
inner shell and all terminate on the surface charge on the inner surface of the outer shell. These surface charges have
equal magnitude and opposite sign. The electric field lines for r > d originate from the surface charge on the outer surface of the outer sphere. IDENTIFY: Apply Gauss’s law.
SET UP: Use a Gaussian surface that is a sphere of radius r and that is concentric with the charged shells. EXECUTE: (a) (i) For r < a, E = 0, since the charge enclosed is zero. (ii) For a < r < b,E = 0, since the points
are w1th1n the conducting material. (111) For b < r < c,E = Z—Ezg, outward, Since charge enclosed 1s +2q.
EEO r (iv) For c < r < d, E = 0, since the points are within the conducting material. (v) For r > d, E = 0, since the net
charge enclosed is zero. The graph of E versus r is sketched in Figure 22.46. Gauss’s Law 2247 (h) (i) small shell inner surface: Since a Gaussian surface with radius r, for a < r < b, must enclose zero net
charge, the charge on this surface is zero. (ii) small shell outer surface: +2q . (iii) large shell inner surface: Since a Gaussian surface with radius r, for c < r < d , must enclose zero net charge, the charge on this surface is —2q .
(iv) large shell outer surface: Since there is —2q on the inner surface and the total charge on this conductor is —2q t, the charge on this surface is zero.
EVALUATE: The outer shell has no effect on the electric field for r < c . For 7 > d the electric field is due only to
the charge on the outer surface of the larger shell. E a b c d
Figure 22.46 541‘ i
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This homework help was uploaded on 04/09/2008 for the course PH 1120 taught by Professor Keil during the Spring '08 term at WPI.
 Spring '08
 Keil

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