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Unformatted text preview: we find that
1
1
k q1 q2 + m1 m2 d=
q
q
Ex 2 − 1 m2 m1 2 1
1 9 N⋅m −7.0 ×10−6 C +18 ×10−6 C + 8.99 ×10 2
−5
−5
C 1.4 × 10 kg 2.6 ×10 kg = 6.5 m
=
−6
−6 ( +2500 N/C ) +18 ×10 C − −7.0 ×10 C −5
−5 2.6 ×10 kg 1.4 ×10 kg ______________________________________________________________________________ 51. SSM REASONING AND SOLUTION The net electric field at point P in Figure 1 is the
vector sum of the fields E+ and E–, which are due, respectively, to the charges +q and –q.
These fields are shown in Figure 2.
E+
Pα P a E– l l +q α a M
2d
Figure 1 –q
Figure 2 According to Equation 18.3, the magnitudes of the fields E+ and E– are the same, since the
triangle is an isosceles triangle with equal sides of length l. Therefore, E+ = E– = k q / l 2 .
The vertical components of these two fields cancel, while the horizontal components
reinforce, leading to a total field at point P that is horizontal and has a magnitude of
k q EP = E+ cos α +E – cos α = 2 2 cos α
l At point M in Figure 1, both E+ and E– are horizontal and point to the right. Again using
Equation 18.3, we find
k q k q 2k q
EM = E+ +E – = 2 + 2 = 2
d
d
d Chapter 18 Problems 987 Since EM/EP = 9.0, we have
EM
2k q / d 2
1
=
=
= 9.0
EP 2k q ( cos α ) / l 2 ( cos α ) d 2 / l 2 But from Figure 1, we can see that d/l = cos α. Thus, it follows that
1
= 9.0
cos3α cos α = 3 1/ 9.0 = 0.48 or The value for α is, then, α = cos –1 ( 0.48 ) = 61° . ______________________________________________________________________________
52. REASONING The net charge q carried by the particle determines the magnitude F of the
electrical force that the external electric field of magnitude E exerts on the particle, via
F = qE (Equation 18.2). The electric field is horizontal, so the electric force acting on the
particle is also horizontal. Taking east as the positive x direction, the electric force gives the
particle a horizontal component of acceleration: ax. Newton’s second law holds that the
particle’s horizontal acceleration is ax = F
(Equation 4.2a), where m is the mass of the
m particle. Combining Equation 18.2 and Equation 4.2a, we obtain
ax = F qE
=
mm (1) We will use Equation (1) to determine q. Note that the algebraic sign of q is the same as
that of ax, which we will obtain by using kinematics. The particle’s horizontal acceleration
is related to its horizontal displacement x by x = v0 xt + 1 axt 2 (Equation 3.5a), where v0x is
2
the horizontal component of its initial velocity v0 and t is the elapsed time. The particle’s
initial velocity is entirely horizontal, so that v0x = v0, and Equation 3.5a becomes
x = v0t + 1 axt 2
2 (2) In order to determine the elapsed time t, we consider the vertical component of the particle’s
projectile motion. Because the particle is launched horizontally, the vertical component v0y
of its initial velocity is zero, and the vertical component ay of its acceleration is that due to
gravity: ay = −9.80 m/s2, where we have taken upward as the positive y direction.
Therefore, from y = v0 yt + 1 ayt 2 (Equation 3.5b), we have that
2 988 ELECTRIC FORCES AND ELECTRIC FIELDS y = ( 0 m/s ) t + 1 ayt 2 = 1 ayt 2
2
2 (3) where y = −6.71×10−3 m is the particle’s vertical displacement. The value of y is negative
because the particle moves downward.
SOLUTION Solving Equation (1) for q, we obtain q= max (4) E Solving Equation (2) for ax yields
1 a t2
2x = x − v0t ax = or 2 ( x − v0t ) (5) t2 Substituting Equation (5) into Equation (4), we find that q= 2m ( x − v0t ) (6) Et 2 Solving Equation (3) for t yields
2y
t=
ay
2 or t= ( ) 2 −6.71× 10−3 m
2y
=
= 0.0370 s
ay
−9.80 m/s 2 Therefore, from Equation (6), the charge carried by the particle is q= 2 m ( x − v0t )
Et 2 = ( ) 2 1.50 × 10 −6 kg 0.160 m − ( 8.80 m/s )( 0.0370 s ) ( 925 N/C )( 0.0370 s ) 2 = −3.92 × 10 −7 C 53. REASONING AND SOLUTION Since the thread makes an angle of 30.0° with the
vertical, it can be seen that the electric force on the ball, Fe, and the gravitational force, mg,
are related by
Fe = mg tan 30.0°
The force Fe is due to the charged ball being in the electric field of the parallel plate
capacitor. That is, Chapter 18 Problems Fe = E qball 989 (1) where qball is the magnitude of the ball's charge and E is the magnitude of the field due to
the plates. According to Equation 18.4 E is E= q
ε0 A (18.4) where q is the magnitude of the charge on each plate and A is the area of each plate.
Substituting Equation 18.4 into Equation (1) gives
q qball
Fe = mg tan 30.0° =
ε0 A
Solving for q yields q= ε 0 Amg tan 30.0°
qball ( )( )( )( ) 8.85 × 10 –12 C2 / N ⋅ m 2 0.0150 m 2 6.50 × 10 –3 kg 9.80 m/s 2 tan 30.0° =
0.150 × 10 –6 C
= 3.25 × 10 –8 C
______________________________________________________________________________
54. REASONING AND SOLUTION Gauss' Law is given by text Equation 18.7: ΦE = Q ε0 , where Q is the net charge enclosed by the Gaussian surface.
a. Φ E =
b. Φ E =
c. Φ E = 3.5 × 10 – 6 C
8.85 × 10 −1...
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 Spring '13
 CHASTAIN
 Physics, The Lottery

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