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Unformatted text preview: 1. What is the net charge shown in Fig. 2439? The
magnitude of the middle charge is 3 ac. Solution The number of lines of force emanating from (or
terminating on) the positive (or negative) charges is
the same (14 in Fig. 24—39), so the middle charge is
I—3 ac and the outer ones are +3 ,uC. The net charge
shown is therefore 3 + 3 — 3 = 3 ac. This is reﬂected
by the fact that 14 lines emerge from the boundary of the ﬁgure. FIGURE 2439 Problem 1 Solution. 3. Two charges +q and a charge —q are at the vertices
of an equilateral triangle. Sketch some ﬁeld lines for
this charge distribution. Solution (The sketch shown follows the text‘s convention of
eight lines of force per charge magnitude q.) Problem 3 Solution. 5. A ﬂat surface with area 2.0 in2 is in a uniform
electric ﬁeld of 850 NEC. What is the electric ﬂux
through the surface when it is (a) at right angles to
the ﬁeld, (b) at 45” to the ﬁeld, and (c) parallel to
the ﬁeld? Solution (3.) When the surface is perpendicular to the ﬁeld, its
normal is either parallel or anti~parallel to E. Then
Equation 24—1 gives ‘3 = EA : EA cos(0° or 180°) 2:
:l:(35l] N/C)[2 m2) 2: i170 hNm2/C. (b) (I) EA = EA cos(45c or 135°) emu kNmQ/CM
(0.866) 2 i120 kNn1?,I’C. (c) <1: = EA cos 90° = 0. i’. A ﬂat surface with area 0.14 In2 lies in the
33; plane, in a uniform electric field given by E =
5.11 + 2.13 + 3.51} M30. Find the ﬂux through this
surface. Solution The surface can be represented by a vector area A = (0.14 m2)[il}). (Since the surface is Open, we
have a choice of normal to the x—y plane.) Then <1: : EA = :hEfcxwii n12) = $1340.14 m2) =
has kN/C)(o.14 m2) e490 xhﬁjc. (Only the
2 component of the ﬁeld contributes to the ﬂux
through the rug plane.) 11. What is the electric ﬂux through each closed
surface shown in Fig. 2443? FIGURE 2&43 Problem 11. Solution From Gauss’s law, ‘13 2 qenclosed/Eo. For the surfaces
shown, this is (a) (q — 290/50 2 —q/sg, (b) —2q/eo,
(c) and (d) l]. 13. A 2,311.0 charge is at the center of a cube 7.5 cm
on each side. What is the electric flux through one
face of the cube? Hint: Think about symmetry,
and don’t do an integral. Solution The symmetry of the situation guarantees that the
ﬂux through one face is % the ﬂux through the whole
cubical surface, so Guam = % fcubeE  (LA 2
cameraman (2.6 hC)/6(s.85x1012 Cg/Nrng) =
49.0 kNmEg’C. 1?. The electric ﬁeld at the surface of a unifornilv
charged sphere of radius 5.0 cm is 90 kNg'C. What would he the ﬁeld strength 10 cm from the
surface? The electric ﬁeld due to a uniformly charged sphere is
like the field of a point charge for points outside the
sphere, i.e., E(r) ~ Uri for r 2 R. Thus, at 10 cm
from the surface, it" : = 15 cm and Ef15 cm) :
l5;’l§)2Ex(5 cm) :: (90 kl\'/C),x'9 :: 10 kNg'Cl. 25. A spherical shell 30 cm in diameter carries a total
charge 85 ,uC distributed uniformly ever its surface. A 1.0uC point charge is located at the
center of the shell. What is the electric ﬁeld
strength (a) 5.0 cm from the center and 45cm
from the center? (c) How would your answers
change if the charge on the shell were doubled? Solution (a) The ﬁeld due to the shell is zero inside, so at r:
5 cm, the ﬁeld is due to the point charge only. Thus,
E = heir/7'2 = (9x109 Nm2/C2)(1 ,uC)f'/(U.05 mp:
(3.60x105 i\’/C)i". (b) Outside the shell, its ﬁeld isle,
that of a point charge, so at r = 45 cm, E = k(q+
one»? = (9x109 Nm2/Cz)(86 ,uC)f'/(0.45 m)2 =
(3.82x105 N/C)f. (c) If the charge on the shell mere
doubled, the ﬁeld inside would be unaffected, whiletlie
ﬁeld outside would approximately double, E :
k(1.0 so + 2x85 MC)/(45 m)2 = 7.60 MN/C. 26. The thick, spherical shell of inner radius a. and
outer radius .5 shown in Fig. 2445 carries a
uniform volume charge density p. Find an
expression for the electric ﬁeld strength in the
region tr < r < b, and show that your result is
consistent with Equation 244’ when a = 0. Solution Use the result of Gauss’s law applied to a spherically
symmetric distribution, E = qenclosed/4rr50r2. For
a < r < b in a spherical shell with charge densityr p,
genclosed : §W(T3 _ a3)pi 50 E = P(?‘3 _' G3);l350r2=
(p,"3ao)(r — (ta/T2). If a ——> 0, Equation 24? for a
uniformly charged spherical volume is recovered. Gaussian
surface FIGURE 2445 Problem 26 Solution. 31. An inﬁnitely long rod of radius R carries a
uniform volume charge density ,0. Show that the
electric ﬁeld strengths outside and inside the rod
are given, respectively, by E =: pR2/250r and
E = pr/2sg, where r is the distance from the rod
3X15. Solution The charge distribution has line symmetry (as in
Problem 29) so the flux through a coaxial cylindrical
surface of radius in (Equation 24—8) equals qcnclﬂsed/so,
from Gauss’s law. For T > R (outside the rod),
genclosed = {’31sz3 hence Eout : WR2£f27W£80 :
pita/2501'. For 1' < R (inside the rod), qenclosed =
piTT2g, hence Eu 2 pirr2£f2irr€sg == {JTfr’IQ'LTg. (The ﬁeld
direction is radially away from the symmetry axis if
p > D, and radially inward if p < O.) 35. If you “painted” positive charge on the flour, what:
surface charge density would be necessary in order"
to suspend a 15pC, 5.0g particle above the lime? Solution
A positive surface charge density or, on the ﬂoor, would produce an approximately uniform electric ﬁeld upward of E 2 0/250, at points near the ﬂoor andiiet' near an edge. The ﬁeld needed to balance the weight I of a particle, of mass m. and charge 9, is given by my = qE, therefore 0 : 250E = 250mgg'q = 2(835}: 1012 C2/Nm2)(5x10_3 kg)(9.8 m/sglﬂlhx
in—6 C) = 57.8 nC/mz. 39. A nonconducting square plate 75 cm on a side
carries a uniform surface charge density. The
electric ﬁeld strength 1 cm from the plate, not
near an edge, is ‘15 kN/C. What is the
approximate ﬁeld strength 15 m from the plate? Solution The electric ﬁeld strength close to the plate (1 cm << 75 cm) has approximate plane symmetry (E : : 0,5250), so the charge on the plate is g = 011 =
2591314 = 2(8.85x10"'12 C2/’Nn12}[45 kN;’C)x (0.75 m}2 :2: 448 nC. Very far from the plate (15 In 3) 0.75 m), the ﬁeld strength is like that from a
point charge, E = kqfr‘? 2 (9X109 Ema/(72V {448 nC)(15 Inf—Q = 17.9 N30. 45. A net charge of 5.0 ,uC is applied on one side of a
solid metal sphere 2.0 cm in diameter. After
electrostatic equilibrium is reached, what are
(a) the volume charge density inside the sphere
and (b) the surface charge density on the sphere:J
Assume there are no other charges or conductors
nearby. (c) Which of your answers depends on this
assumption, and why? Solution (a) The electric ﬁeld within a conducting medium, in
electrostatic equilibrium, is zero. Therefore, Gauss’s
law implies that the net charge contained in any clcsed
surface, lying within the metal, is zero. (b) If the
volume charge density is zero within the metal, all of
the net charge must reside on the surface of the sphere
If the sphere is electrically isolated, the charge will be
uniformly distributed (i.e., spherically symmetric}, so
or = Q/xirrR2 = (5 nC)/Llrr(1 cm}2 2 3.98x 10—3 Cfml
(c) Spherical symmetry for or depends on the
proximity of other charges and conductors. iii'. A 25U—nC point charge is placed at the center of
an uncharged spherical conducting shell 20 cm in
radius. (a) What is the surface charge density on
the outer surface of the shell? (b) What is the
electric ﬁeld strength at the shell’s outer surface? Solution is) There is a nonzero ﬁeld outside the shell. because
the net charge within is not zero. Therefore, there is a
surface charge density 0' : eoE on the outer surface of
the shell, which is uniform, if we ignore the possible
presence of other charges and conducting surfaces
outside the shell. Gauss‘s law (with reasoning similar
to Example 24—7) requires that the charge on the
Shell’s outer surface is equal to the point charge within,
50 o = (3/41ng = 250 nC/4?r(0.20 m)"2 s 497 Dogma.  {in} Then the ﬁeld strength at the outer surface is 5‘: 0/50 = 56.2 kN/C. Apoint charge q is at the center of a spherical
shellef radius R carrying charge 2:} spread Solution
 lR <. R.   ' for T 2: As explained in Example 24 5; (a) g 2 _
qencmscd : q and E = fig/CERF ;: 4M“? . gulf}; {M2 :2: 2R > Rs qenclosed :: q + 2‘; and E :I: Skq/I‘LR} " 3 {cg/4R2. ...
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 Summer '08
 schuller

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