Unformatted text preview: . Chapter 19 Problems 1047 SOLUTION a. Using Equation 19.1 and the fact that WAB = +Fs, we find
WAB = + Fs = EPE A − EPE B
F= EPE A − EPE B
s = 9.0 × 10−4 J
= 4.5 × 10−3 N
0.20 m As discussed in the reasoning, the direction of the force is from A toward B .
b. From Equation 18.2, we find that the electric field has a magnitude of
F 4.5 ×10−3 N
E=
=
= 3.0 ×103 N/C
−6
q0 1.5 ×10 C The direction is the same as that of the force on the positive charge, namely
from A toward B . 59. REASONING The potential V at a distance r from a proton is V = k(+e)/r (see
Equation 19.6), where +e is the charge of the proton. When an electron (q = −e) is placed at
a distance r from the proton, the electric potential energy is EPE = −eV, as per
Equation 19.3.
SOLUTION The difference in the electric potential energies when the electron and proton
are separated by rfinal = 5.29 × 10−11 m and when they are very far apart (rinitial = ∞) is EPE final − EPE initial = (−e)ke (−e)ke
−
rfinal
rinitial ( = − (8.99 × 109 N ⋅ m 2 / C2 ) 1.60 × 10−19 C 1
1
×
− =
−11 5.29 × 10 m ∞ ) 2 −4.35 × 10−18 J 60. REASONING The drawing shows a set of equipotential surfaces, and position D is halfway
∆V
between the +450V equipotential and the +550V equipotential. We can use E =
∆s
(Equation 19.7a, without the minus sign) to determine the magnitude E of the electric field,
where ∆V is the potential difference (a positive number) between the two equipotential
surfaces and ∆s is the distance between them. 1048 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL SOLUTION The distance between the equipotentials is equal to the width of two squares
on the grid, each of which is 2.0 cm wide. Therefore, ∆s = 2(2.0 cm) = 4.0 cm. In SI base
units, this is 1m −2
∆s = 4.0 cm = 4.0 × 10 m 100 cm ∆V
Now, according to E =
(Equation 19.7a, without the minus sign), the magnitude of the
∆s
electric field at position D is ( E= ) ∆V ( 550.0 V − 450.0 V )
=
= 2500 V/m
∆s
4.0 × 10 −2 m Because the electric field points from high potential to low potential, at position D the
electric field must point from the +550.0V equipotential to the +450.0V equipotential.
Therefore, the direction of the electric field at D is towards the bottom of the drawing . 61. SSM REASONING AND SOLUTION Combining Equations 19.1 and 19.3, we have
WAB = EPE A – EPE B = q0 (V A – VB ) = ( +1.6 × 10−19 C)(0.070 V) = 1.1× 10−20 J 62. REASONING The potential at a distance r from a point charge q is given by Equation 19.6
as V = kq / r . Therefore, the potential difference between the locations B and A can be
written as
kq kq
VB − VA =
−
rB rA
We can use this relation to find the charge q.
SOLUTION Solving the equation above for q yields
q= VB − VA
1
1
k − r B rA = 9 8.99 × 10 45.0 V
= −6.0 × 10−8 C
2
N⋅m 1
1
− 2 4.00 m 3.00 m C Chapter 19 Problems 1049 63. REASONING AND SOLUTION The capacitance is C = q0/V0 = q/V. The new charge q is,
therefore, ( ) 5.3 × 10−5 C ( 9.0 V )
q0V
=
= 8.0 × 10−5 C
q=
V0
6.0 V 64. REASONING AND SOLUTION The charge on the empty capacitor is q0 = C0V0. With the dielectric in place, the charge remains the same. However, the new capacitance is C = κC0
and the new voltage is V. Thus,
q0 = CV = κC0 V = C0V0
Solving for the new voltage yields
V = V0/κ = (12.0 V)/ 2.8 = 4.3 V The potential difference is 12.0 − 4.3 = 7.7 V . The change in potential is a decrease . 65. REASONING The electric field E that exists between two points in space is, according to
Equation 19.7a, proportional to the electric potential difference ∆V between the points
divided by the distance ∆x between them: E = − ∆V/ ∆x. SOLUTION
a. The electric field in the region from A to B is E= − 5.0 V − 5.0 V
∆V
=−
=
∆x
0.20 m − 0 m 0V/m b. The electric field in the region from B to C is
E= − 3.0 V − 5.0 V
∆V
=−
= 1.0 × 10 1 V/ m
∆x
0.40 m − 0.20 m c. The electric field in the region from C to D is
E= − 1.0 V − 3.0 V
∆V
=−
= 5.0 V / m
∆x
0.80 m − 0.40 m 1050 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL 66. REASONING AND SOLUTION
a. Let d be the distance between the charges. The potential at the point x1 = 4.00 cm to the
left of the negative charge is V = 0= k q1
k q2
−
d − x1
x1 which gives q1
q2 = d
−1
x1 (1) Similarly, at the point x2 = 7.00 cm to the right of the negative charge we have
V= 0 = k q1
k q2
−
x2 + d
x2 which gives
q1
d
=
+1
q2
x2 (2) Equating Equations (1) and (2) and solving for d gives d = 0.187 m . b. Using the above value for d in Equation (1) yields q1
= 3.67 .
q2 67. SSM WWW REASONING If we assume that the motion of the proton and the electron
is horizontal in the +x direction, the motion of the proton is determined by Equation 2.8,
x = v0t + 1 apt 2 , where x is the distance traveled by the proton, v0 is its initial speed, and ap
2
is its acceleration. If the distance between the capacitor places is d, then this relation
becomes 1 d = v0t + 1 apt 2...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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