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Unformatted text preview: 21.33. IDENTIFY: Eq. (21.3) gives the force on the particle in terms of its charge and the electric field between the
plates. The force is constant and produces a constant acceleration. The motion is similar to projectile motion' use
constant acceleration equations for the horizontal and vertical components of the motion. ,
(3) SET UP: The motion is sketched in Figure 21.33a. 2.00 cm
e———.—_—_> 050ml v0 1 },1’ Foranelectron q=—e.
H——’
E Figure 21.3321 F 2 (1E and q negative gives that ii" and E are in opposite directions, so 17' is upward. The free—body diagram
for the electron is given in Figure 21.33b.
y EXECUTE: F =
T a 2 ma F=€E eE=ma
x Figure 21.33b Solve the kinematics to find the acceleration of the electron: Just misses upper plate says that x — x0 = 2.00 cm
when y — y0 = +0.500 cm. xcomponent
VOX =v0 =1.60><106 m/s, a, =0, x—xo =0.0200 m, t=? .. _ L 2
x xO—v0xt+2axt .x—xo=.9£2@gn_=msxm—ss
v0, 1.60x10 m/s In this same time t the electron travels 0.0050 In vertically: ycomponent
t =1.25x10r‘s, voy = 0, y — yo = +0.0050 m, ay = '2 2
y — yo = voyt+%ayt = ___2(y ‘ yo) _ m = 6.40><10‘3 m/Sz a __
y t2 (1.25x10‘8 s)2 (This analysis is very similar to that used in Chapter 3 for projectile motion, except that here the acceleration is upward rather than downward.) This acceleration must be produced by the electricfield force: eE = ma E _ £13 _ (9.109x10‘31 kg)(6.40><10l3 m/sz) e 1.602X10“19 C Note that the acceleration produced by the electric field is much larger than g, the acceleration produced by gravity, so it is perfectly ok to neglect the gravity force on the elctron in this problem. —19
(b, 0=£=W=34Mw mz
m 1.673X 10 kg p
This is much less than the acceleration of the electron in part (a) so the vertical deﬂection is less and the proton won’t hit the plates. The proton has the same initial speed, so the proton takes the same time
t =1.25><10‘8 s to travel horizontally the length of the plates. The force on the proton is downward (in the same direction as E, since q is positive), so the acceleration is downward and ay = ~3.49x10‘° m/82.
y —— yo = voyt +gayt2 = %(—3.49><1010 m/sz)(1.25x10'8 s)2 = ~2.73X10_6 m. The displacement is 2.73x10’l5 m, = 364 N/C downward. _
(c) EVALUATE: The displacements are in opposite directions because the electron has negative charge and the proton has positive charge. The electron and proton have the same magnitude of charge, so the force the electric
ﬁeld exerts has the same magnitude for each charge. But the proton has a mass larger by a factor of 1836 so its
acceleration and its vertical displacement are smaller by this factor. II ,I,,, a r__l_ JAM” MAJ ,1: AI . r"  21.35. IDENTIFY: Apply constant acceleration equations to the ‘motion of the‘electron. SET UP: Let +x be to the right and let + y be downward. The electron moves 2.00 cm to the right and 0.50 cm
downward. EXECUTE: Use the horizontal motion to find the time when the electron emerges from the field.
x— x0 = 0.0200 m, ax =0, v0x =1.60><106 m/s . x— x0 = volt +3th2 gives t = l.25><10'8 s . Since a =0, v0y+v 2 vx=l.60X106 m/s. y—yo=0.0050m,v0y=0,t=l.25x10_ss. y—y0=[ y} gives vy=8.00><105 m/s. Then v: vi+v§ =1.79x106 m/s. EVALUATE: vy = v0y + ayt gives ay = 6.4><lOl3 rn/s2 . The electric field between the plates is E = e 1_60X10—19 C = 364 V/m . This is not a very large field.
21.44. IDENTIFY: For a point charge, E 2 kl—LZZl . For the net electric field to be zero, E1 and E2 must have equal
_ r
magnitudes and opposite directions.
‘SET UP: Let q1 = +0.500 HC and q2 = +8.00 nC. E is toward a negative charge and away from a positive charge.
EXECUTE: The two charges and the directions of their electric fields in three regions are shown in Figure 21.44.
Only in region II are the two electric fields in opposite directions. Consider a point a distance x from ql so a
distance 1.20 m—x from q2 . E1 = E2 gives kES—Oizn—C— = kﬁéogn—C—z. 16x2 = (1.20 m—x)2. 4x = i(1.20 m—x)
x (1.20 — x)
and x = 0.24 m is the positive solution. The electric field is zero at a point between the two charges, 0.24 m from
the 0.500 nC charge and 0.96 m from the 8.00 nC charge.
EVALUATE: There is only one point along the line connecting the two charges where the net electric ﬁeld is zero.
This point is closer to the charge that has the smaller magnitude.
I II III I H m
E’ ‘11 E2 E1 q? E‘ E2 £11 E1 42 E1
:3 ° " ' ’ ‘ % W—H—
+ +
E2 < x ——> E2 E1 + E2 _ E2
(8) (b)
Figure 21.44
21.45. IDENTIFY: Eq.(21.7) gives the electric field of each point charge. Use the principle of superposition and add the electric field vectors. In part (b) use Eq.(21.3) to calculate the force, using the electric field calculated in part (a).
(3) SET UP: The placement of charges is sketched in Figure 21 .45a. Figure 21.453 The electric field of a point charge is directed away from the point charge if the charge is positive and toward the
1 la! point charge if the charge is negative. The magnitude of the electric field is E = where r is the distance between the point where the field is calculated and the point charge. (i) At point a the fields E1 of ql and 1732 of q2 are directed as shown in Figure 21.45b. y
E2 q2<0
x
qx>0 a___ .
E1 Figure 21.45b +7....W .mmwwawmnmmmmwmw , iw q,l % 2100x104) C EXECUTE: EI : —2 — (8.988x109 N  mz/CZ)»—~—2— = 449.4 N/C
47560 r] (0.200 m)
49
E2 : “Li—q?! : (8.988X10° N . mZ/CZ)M§ :1248 N/C
47Z'60 r2 (0.600 m) E” = 449.4 N/C, EU 2 0 E21 21248 N/C, E2y :0 Ex : E” + Eu 2 +4494 N/C + 124.8 N/C : +5742 N/C Ey :Ely +E2y =0 The resultant field at point a has magnitude 574 N/C and is in the +x—direction. (ii) SET UP: At point b the fields E1 of (11 and E2 of q2 are directed as shown in Figure 21.450. y
E2 E1
r b5: x
q1 > 0 612 < 0
Figure 21.45c
‘ —9
EXECUTE: E1 = i—@ = (8.988><109 N . mZ/CZ)M2£ = 12.5 N/C
47Z'€0 rl (1.20 In)
—9
E2 '= —1—lq—§i = (8.988x109 N  1112/c2)§'00——X—19—2C = 280.9 N/C
4m.) r2 (0.400 m) E” = 12.5 N/C, EU 2 0 En = —280.9 N/C, Ezy = 0 Ex = EIx +E2x 2 +125 N/C — 280.9 N/C = —268.4 N/C Ey =E1y +E2y =0 The resultant field at point b has magnitude 268 N/C and is in the —x—direction. (iii) SET UP: At point e the fields E1 of £11 and E2 of q2 are directed as shown in Figure 21.45d. E1 Ez y
<————,—> .___ x
c ‘11 > 0 q2 < 0 Figure 21.45d —9
EXECUTE: E1 = —1—@ = (8.988x109 N  mZ/C2)—2ﬂ)—ﬂ2—C = 449.4 N/C
4ﬂ'e0 r1 (0.200 In)
—9
E2 2—1—@ = (8.988x109 N . tie/Cam = 44.9 N/C EIx = —449.4 N/C, E1y = 0 E“ = +449 N/C, E2y = 0 E = Eu + EN = —449.4 N/C + 44.9 N/C = —404.5 N/C Ey =E1y +E2y =0 The resultant field at point b has magnitude 404 N/C and is in the —xdirection. (b) SET UP: Since we have calculated E at each point the simplest way to get the force is to use F‘ = —eE'.
EXECUTE: (i) F = (1.602x10—‘9 C)(574.2 N/C) = 9.20x10‘” N, — xdirection (ii) F = (1.602x10‘19 C)(268.4 N/C) = 4.30x10‘17N, + x—direction (iii) F = (1.602x10”’9 C)(404.5 N/C) = 6.48x10‘17 N, +xdirection EVALUATE: The general rule for electric field direction is away from positive charge and toward negative
charge. Whether the field is in the +x— or — xdirection depends on where the field point is relative to the charge
that produces the field. In part (a) the field magnitudes were added because the fields were in the same direction
and in (b) and (c) the field magnitudes were subtracted because the two fields were in opposite directions. In
part (b) we could have used Coulomb's law to find the forces on the electron due to the two charges and then
added these force vectors, but using the resultant electric field is much easier. 21.50. IDENTIFY: Apply Eq.(21.7) to calculate the ﬁeld due to each charge and then calculate the vector sum of those ﬁelds:
SET UP: The ﬁelds due to q1 and to q2 are sketched in Figure 21.50. —9
EXECUTE: E2 = LWH) = —150i N/C .
47E0 (0.6 m) X
a 1 1 A 1 . . .
E =— 4.00x10'9 C —————— 0.600 i+~—————— 0.800 ‘ = 21.6i +28.8' N C.
1 47reo( {(1.00 m)2( ) (1.00 m)2( )1) ( J) / E = El + E = (—1284 N/C)i + (28.8 N/C)j. E=./(128.4 N/C)2 +(28.8 N/C)2 = 131.6 N/C at 9 = tan‘1 =12.6° above the —x axis and therefore 196.2° counterclockwise from the +x axis. 128.4
EVALUATE: E1 is directed toward q1 because q1 is negative and E2 is directed away from (12 because q2 is positive.
y
o 91
E1
___._. x
E2 ‘11
Figure 21.50
21.79. IDEI‘TTIFY: Use Coulomb's law to calculate the forces between pairs of charges and sum these forces as vectors to
find the net charge. (3) SET UP: The forces are sketched in Figure 21.7921. EXECUTE: 17‘] + F3 = 0, so the net force is F = 132. 1 q(3q) _ 642 F = —~ _ ,
47l'60 (LA/2f 47:60L2 away from the vacant corner. Figure 21.7%
(b) SET UP: The forces are sketched in Figure 21.7%. 1 2
EXECUTE: F2 =—1— q(3q) 3‘1 4mg (Jay W 1 q(3q) 3q2 F=F=
‘ 3 47:60 L2 475503 The vector sum of E and F3 is E3 = «(Ff + F32. Figure 21.7%
2
E3 = JEFI = é—[Zq—z; 1—?"l3 and 152 are in the same direction.
4ﬂ'eoL
3q2 1 . .
F = F13 + F2 = 4 L2 $2 + E , and 1s directed toward the center of the square.
75% EVALUATE: By symmetry the net force is along the diagonal of the square. The net force is only slightly larger
when the —3q charge is at the center. Here it is closer to the charge at point 2 but the other two forces cancel. ...
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 Spring '08
 Keil

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