23_InstructorSolutions

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EXECUTE: Wa→b = −1.9 ×10−8 J = Ua − Ub. Ub = Ua − Wa→b = 1.9 ×10−8 J − (−5.4 ×10−8 J) = 7.3×10−8 J. EVALUATE: When the electric force does negative work the electrical potential energy increases. IDENTIFY: The work needed to assemble the nucleus is the sum of the electrical potential energies of the protons in the nucleus, relative to infinity. SET UP: The total potential energy is the scalar sum of all the individual potential energies, where each potential energy is U = (1/ 4π P0 )(qq0 / r). Each charge is e and the charges are equidistant from each other, so the total
Ub
=
1 4π P0
qQ r
=
(8.988 ×109
Figure 23.5b
Kc = 0 (at distance of closest approach the speed is zero)
Uc
=
1 4π P0
q1q2 rc
Thus conservation of energy
Ka +Ua =Uc
gives
1 q1q2 4π P0 rc
= +0.3630 J + 0.2454 J = 0.6084 J
EXECUTE: Ka = 0 (released from rest)
Ua = +0.198 J (from part (a))
Kb
=
1 2
mvb2
Figure 23.7
Only the electric force does work, so
Wother
=0
and
U
=
1 4π P0
qQ . r
(i) rb = 0.500 m
Ka + Ua + Wother = Kb + Ub
U for the pair of point charges is given by Eq.(23.9). SET UP:
Let point a be where q2 is 0.800 m from q1 and point b be where q2 is 0.400 m from q1, as shown in Figure 23.5a.
rc
=
1 4π P0
q1q2 0.6084
J
=
(8.988 × 109
N ⋅ m2/C2 ) (−2.80 ×10−6 C)(−7.80 ×10−6 +0.6084 J
C)
=
0.323
m.
EVALUATE: U → ∞ as r → 0 so q2 will stop no matter what its initial speed is.
J
The conservation of energy equation then gives Kb = Ka + (Ua − Ub )
1 2
mvb2
=
+0.3630
J
+
(0.2454
J

0.4907
J)
=
0.1177
J
vb =
2(0.1177 1.50 ×10−3
J) kg
= 12.5
m/s
EVALUATE: The potential energy increases when the two positively charged spheres get closer together, so the kinetic energy and speed decrease. (b) IDENTIFY: Let point c be where q2 has its speed momentarily reduced to zero. Apply conservation of energy to points a and c: Ka + Ua + Wother = Kc + Uc.
U
−0.400 J
EVALUATE: The potential energy U is a scalar and can take positive and negative values.
(a) IDENTIFY and SET UP: U is given by Eq.(23.9).
EXECUTE: U = 1 qq′ 4επ 0 r
Figure 23.5a
EXECUTE:
Only the electric force does work, so
Wother
=0
and
U
=
1 4π P0
q1q2 . r
Ka
=
1 2
mva2
=
1 2
(1.50
×
10 −3
kg)(22.0
m/s)2
=
0.3630
J
Ua
=
1 4π P0
q1q2 ra
= (8.988×109
=
7.68
×
10 −14
J
(b) SET UP: The protons have equal momentum, and since they have equal masses, they will have equal speeds
and hence
equal
kinetic
energy.
ΔU
= K1 + K2
23-1
23-2 23.4.
23.5.
Chapter 23
IDENTIFY: The work required is the change in electrical potential energy. The protons gain speed after being
released because their potential energy is converted into kinetic energy.
SET UP: Points a and c are shown in Figure 23.5b.
Electric Potential 23-3
EXECUTE: Ka = +0.3630 J (from part (a)) Ua = +0.2454 J (from part (a))
23.6. 23.7.
= 2K
=
2
⎛ ⎜⎝
1 2
mv2
⎞ ⎟⎠
=
mv
2
.
EXECUTE:
Solving for v gives v =
ΔU = m
7.68 ×10−14 J 1.67 ×10−27 kg
= 6.78 × 106 m/s
EVALUATE: The potential energy may seem small (compared to macroscopic energies), but it is enough to give each proton a speed of nearly 7 million m/s. (a) IDENTIFY: Use conservation of energy:
potential
energy
is
U
=
1 4π P0
⎛ ⎜ ⎝Leabharlann e2 r+
e2 r
+
e2 r
⎞ ⎟ ⎠
=
3e2 4π P0r
.
EXECUTE: Adding the potential energies gives
U
=
3e2 4π P0r
=
3(1.60 ×10−19 C)2 (9.00 ×109 2.00 ×10−15 m
ra = 0.150 m rb = (0.250 m)2 + (0.250 m)2 rb = 0.3536 m
23.2. 23.3.
Figure 23.1
EXECUTE:
Wa→b = Ua − Ub
Ua
=
1 4π P0
q1q2 ra
=
(8.988 ×109
N ⋅ m2 / C2 ) (+2.40 ×10−6 C)(−4.30 ×10−6 0.150 m
U = (8.988 ×109 N ⋅ m2/C2 ) (+4.60 ×10−6 C)(+1.20 ×10−6 C) = +0.198 J 0.250 m
EVALUATE: The two charges are both of the same sign so their electric potential energy is positive. (b) IDENTIFY: Use conservation of energy: Ka + Ua + Wother = Kb + Ub SET UP: Let point a be where q is released and point b be at its final position, as shown in Figure 23.7.