# 1 PHYS 1101 Practice problem set 9, Chapter 29: 6, 15, 28, 41, 52

PHYS 1101 Practice problem set 9, Chapter 29: 6, 15, 28, 41, 52, 63, 71, 79;
Chapter 30: 27, 33, 57, 62, 65, 67, 68, 83
29.6.
Model: The charges are point charges.
Visualize: Please refer to Figure Ex29.6.
Solve: For a system of point charges, the potential energy is the sum of the potential energies due to all pairs of charges:
U elec = ∑
i,j
Kq i q j
rij
= U 12 + U 13 + U 23 = K
q1 q 2
r12
+K
q1 q 3
r13
+K
q 2 q3
r23
(2 nC )(−1.0 nC ) (−1 nC )( −1.0 nC ) 
9
2
2  (2 nC )(− 1.0 nC )
= 9.0 × 10 N m / C 
+
+

0.03
m
0.03 m
0.03 m


(
)
= − 0.60 × 10 −6 J − 0.60 × 10 −6 J + 0.30 × 10 −6 J = −0.90 × 10 −6 J
Assess: Note that U12 = U21, U13 = U31, and U23 = U32.
29.15. Model: Energy is conserved. The potential energy is determined by the electric potential.
Visualize:
The figure shows a before-and-after pictorial representation of an electron moving through a potential difference.
Solve: (a) Because the electron is a negative charge and it slows down as it travels, it must be moving from a region of
higher potential to a region of lower potential.
(b) Using the conservation of energy equation,
Kf + Uf = Ki + Ui ⇒ Kf + qVf = Ki + qVi
⇒ Vf − V i =
⇒ ΔV = −
mv 2i
2e
=−
1
(Ki − Kf ) = 1
(− e )
q
(9.11 × 10
−31
(
(
1
2
)(
2
mv i − 0 J
5
)
kg 5.0 × 10 m / s
2 1.60 × 10 −19 C
)
)
2
= −0.712 V
Assess: The negative sign with ΔV verifies that the electron moves from a higher potential region to a lower potential
region.
29.28. Model: The net potential is the sum of the potentials due to each charge.
Visualize: Please refer to Figure Ex29.28.
Solve: From the geometry in the figure,
1.5 cm
1.5 cm
1.5 cm
1.5 cm
=
=
= cos 30° ⇒ r1 = r2 = r3 =
= 1.732 cm
r1
r2
r3
cos30°
The potential at the dot is
1
V=
1 q1
1 q2
1 q3
+
+
4πε 0 r1 4πε 0 r2
4πε 0 r3
−9
−9
−9
C 1.0 × 10 C 1.0 × 10 C 
9
2
2  2.0 × 10
= 9.0 × 10 N m / C 
−
−
 = 0V
 0.01732 m
0.01732 m
0.01732 m 
(
)
Assess: Potential is a scalar quantity, so we found the net potential by adding three scalar quantities.
29.41. Model: Mechanical energy is conserved. Metal spheres are point particles and they have point charges.
Visualize:
Solve: (a) The system could have both kinetic and potential energy, although here K = 0 J. The energy of the system is
E0 = K 0 + U0 = 0 J +
qA qB
4πε 0 r0
=
(9.0 × 10
9
2
Nm / C
2
)( 2.0 × 10
−6
)(
C 2.0 × 10
−6
C
0.05 m
) = 0.720 J
(b) In static equilibrium, the net force on sphere A is zero. Thus
T = FB on A =
qA qB
2
0 0
4πε r
=
(9.0 × 10
9
2
Nm / C
2
)(2.0 × 10
(0.05 m )
−6
)(
C 2.0 × 10
2
−6
C
) = 14.4 N
(c) The spheres move apart due to the repulsive electric force between them. The surface is frictionless, so they continue to
slide without stopping. When they are very far apart (r1 → ∞), their potential energy U1 → 0 J. Energy is conserved, so we
have
E1 = K 1 + U 1 =
1
2
2
m A v A1 +
1
2
2
m B vB1 + 0 J = E 0
Momentum is also conserved: Pafter = mAvA1 + mBvB1 = Pbefore = 0 kg m/s. Note that these are velocities and that vA1 is a
negative number. From the momentum equation,
m v
v A1 = − B B1
mA
Substituting this into the energy equation,
2
E0 =
⇒ vB1 =
1
2
 m v 
2
m A  − B B1  + 12 m B vB1 =

mA 
2 E0
m B (m B m A + 1)
=
1
2
m
 2
m B  B + 1 vB1
 mA

2 (0.720 J )
(0.004 kg )(4
g 2 g + 1)
= 10.95 m / s
Using this result, we can then find v A1 = − m B v B1 m A = −21.9 m / s . These are the velocities, so the final speeds are 21.9
m/s for the 2 g sphere and 10.95 m/s for the 4 g sphere.
2
29.52. Model: Energy is conserved.
Visualize:
The dipole initially is at the higher potential energy. As it moves to its minimum energy state, the decrease in the potential
energy appears as rotational kinetic energy.
 
Solve: The potential energy of a dipole is U dipole = − p ⋅ E = − pE cosθ , where θ is the angle between the dipole and the
electric field. The energy conservation equation Kf + Uf = Ki + Ui is
1
2
Iω 2 + U f = 0 J + U i ⇒
⇒
1
2
1
2
Iω 2 = U i − U f = − pE cos90° − (− pE cos 0°)
2
Iω = pE ⇒ ω =
2 pE
I
2
The moment of inertia of the dipole is Idipole = 2 m( 12 s ) . Substituting into the above equation,
ω =
2 (qs )E
2m ( s )
1
2
2
(
)
4 2.0 × 10 −9 C (0.10 m )(1000 V / m )
=
(1.0 × 10
−3
)
kg ( 0.10 m )
2
29.63. Model: The potential at any point is the superposition of the potentials due to all charges. Outside a uniformly
charged sphere, the electric potential is identical to that of a point charge Q at the center.
Visualize: Please refer to Figure P29.63. Sphere A is the sphere on the left and sphere B is the one on the right.
Solve: The potential at point a is the sum of the potentials due to the spheres A and B:
Va = VA at a + VB at a =
(
9
2
1 QA
1
QB
+
4π ε 0 R A 4π ε0 0.70 m
= 9.0 × 10 N m / C
2
)
100 × 10 −9 C
0.30 m
= 3000 V + 321 V = 3321 V
(
9
2
+ 9.0 × 10 N m / C
2
)
25 × 10 −9 C
0.70 m
Similarly, the potential at point b is the sum of the potentials due to the spheres A and B:
Vb = VB at b + VA at b =
1 QB
1
QA
+
4πε 0 R B 4πε 0 0.95 m
−9
C 100 × 10 −9 C 
9
2
2  25 × 10
= 9.0 × 10 N m / C 
+

 0.05 m
0.95 m 
(
)
= 4500 V + 947 V = 5447 V
Thus, the potential at point b is higher than the potential at a. The difference in potential is Vb – Va = 5447 V –
3321 V = 2126 V.
Assess: VA at a = 3000 V and the sphere B has a potential of 225 V at point a. The spherical symmetry dictates that the
potential on a sphere’s surface be the same everywhere. So, in calculating the potential at point a due to the sphere B we
used the center-to-center separation of 1.0 m rather than a separation of 100 cm – 30 cm = 70 cm from the center of sphere
B to the point a. The former choice leads to the same potential everywhere on the surface whereas the latter choice will
lead to a distribution of potentials depending upon the location of the point a. Similar reasoning also applies to the
potential at point b.
3
29.71. Model: Because the rod is thin, assume the charge lies along the semicircle of radius R.
Visualize: Please refer to Figure P29.71. The bent rod lies in the xy-plane with point P as the center of the semicircle.
Divide the semicircle into N small segments of length Δs and of charge ΔQ = (Q πR )Δs , each of which can be modeled as
a point charge. The potential V at P is the sum of the potentials due to each segment of charge.
Solve: The total potential is
V=
1
∑ V = ∑ 4πε
ΔQ
i
i
i
R
0
=
1
4πε 0
Q
1
Q
∑  πR  Δs = 4π ε
R
i
0
πR
2
∑ RΔθ
=
i
1
Q
4πε 0 πR
∑ Δθ .
i
All of the terms come to the front of the summation because these quantities did not change as far as the summation is
concerned. The summation does not have to convert to an integral because the sum of all the Δθ around the semicircle is π.
Hence, the potential at the center of a charged semicircle is
1 Qπ
1 Q
Vcenter =
=
4πε 0 πR
4π ε0 R
29.79. Model: Energy and momentum are conserved.
Visualize: Please refer to Figure CP29.79. Let the two spheres have masses mA and mB, and speeds vA and vB when they
are very far apart.
Solve: The energy conservation equation Kf + Uf = Ki + Ui is
(
⇒
1
2
1
2
2
2
1
)
m A v A + 12 m B v B + 0 J = 0 J +
(0.002 kg ) v2A + 12 ( 0.001 kg )v 2B
(
9
2
= 9.0 × 10 N m / C
2
2
qAqB
4π ε 0 rAB
⇒ v A + 0.5 v B = 0.0090 m
2
2
)
/s
(2.0 × 10
−9
)(
)
C 1.0 × 10 −9 C
2.0 × 10 −3 m
2
To solve for vA and vB, we need another equation relating vA and vB. From the momentum conservation equation Pafter =
Pbefore we get
mAvA + mBvB = 0 kg m/s ⇒ (2.0 g)vA + (1.0 g)vB = 0 kg m/s ⇒ vB = −2vA
Substituting this expression into the energy conservation equation,
2
v A + 0.5 (2v A ) = 0.009 m / s ⇒ 3v A = 0.009 m / s
2
2
2
2
2
2
Solving these equations, we get vA = –0.0548 m/s and vB = +0.110 m/s. Thus the speeds are 0.0548 m/s for A and 0.110
m/s for B.
30.27. Visualize: Please refer to Figure Ex30.27.
Solve: Any two electrodes, regardless of their shape, form a capacitor whose capacitance is defined as C = Q ΔVC . The
capacitance is
C=
20 × 10 −9 C
100 V
= 2.0 × 10
−10
F = 200 pF
4
30.33. Solve: (a) From Equation 30.33, the energy stored in the charged capacitor is
UC =
1
2
1  Aε 0 
( ΔVC )2
C( ΔVC ) =
2
2 d 
2
=
(
−12
2
2
1 π (0.01 m ) 8.85 × 10 C / N m
2
0.50 × 10 −3 m
) (200 V )
2
= 1.11 × 10
−7
J
(b) From Equation 30.35, the energy density in the electric field is
uE =
1
2
2
ε 0E =
1
2
2
ε0
1
 ΔV C 
−12
2
= 8.85 × 10
C / Nm
 d 
2
(
2
2
200 V
) 0.5 × 10
−3

3
= 0.708 J / m
m
30.57. Model: Capacitance is a geometric property of two electrodes.
Visualize:
Solve: The ratio of the charge to the potential difference is called the capacitance: C = Q ΔVC . The potential difference
across the capacitor is
1
ΔVC =
4π ε 0 R1
1
1
⇒ C = 4πε 0 
−

R
R
 1

2 
(
⇒ R1 R2 = 100 × 10
−12
Q
1
−
4πε 0 R 2
−1
= 4π ε0
)(
Q
R1 R 2
−3
Q 1
1 

−

4πε 0  R1 R2 
= 4πε 0
R2 − R1
F 1.0 × 10
=
)(
R1 R2
1.0 × 10 −3 m
9
2
m 9.0 × 10 N m / C
2
= 100 × 10
) = 900 × 10
−12
−6
F
m
2
Using R2 = R1 + 1.0 mm,
(
R1 R1 + 1.0 × 10
⇒ R1 =
−3
)
m = 900 × 10
− 1.0 × 10 −3 m ±
−6
(1.0 × 10
2
2
(
m ⇒ R1 + 1.0 × 10
−3
m
)
2
−3
)
m R1 − 900 × 10
+ 3600 × 10 −6 m 2
−6
m =0
= 0.0295 m = 2.95 cm
The outer radius is R2 = R1 + 0.001 m = 0.0305 m = 3.05 cm. So, the diameters are 5.90 cm and 6.10 cm.
30.62. Visualize:
The pictorial representation shows how to find the equivalent capacitance of the three capacitors shown in the figure.
Solve: Because C1 and C2 are in parallel, their equivalent capacitance Ceq 12 is
Ceq 12 = C1 + C2 = 20 µF + 60 µF = 80 µF
Then, Ceq 12 and C3 are in series. So,
5
1
C eq
=
1
Ceq 12
+
1
=
C3
1
80 µF
+
1
10 µ F
=
9
80
(µ F) −1
⇒ C eq =
80
9
µF = 8.9 µF
30.65. Model: Assume the battery is an ideal battery.
Visualize:
The pictorial representation shows how to find the equivalent capacitance of the three capacitors shown in the figure.
Solve: Because C2 and C3 are in series,
1
1
1
1
1
10
(µ F) −1 ⇒ Ceq 23 = 2410 µF = 2.4 µ F
=
+
=
+
=
Ceq 23
C2 C3
4 µF 6 µ F 24
Ceq 23 and C1 are in parallel, so
Ceq = Ceq 23 + C1 = 2.4 µF + 5 µF = 7.4 µF
A potential difference of ΔVC = 9 V across a capacitor of equivalent capacitance 7.4 µF produces a charge
Q = CeqΔVC = (7.4 µF)(9 V) = 66.6 µC
Because Ceq is a parallel combination of C1 and Ceq 23, these capacitors have ΔV1 = ΔVeq 23 = ΔV C = 9 V. Thus the charges
on these two capacitors are
Q1 = (5 µF)(9 V) = 45 µC
Qeq 23 = (2.4 µF)(9 V) = 21.6 µC
Because Qeq 23 is due to a series combination of C2 and C3, Q2 = Q3 = 21.6 µC. This means
ΔV2 =
Q2
C2
=
21.6 µ C
4 µF
= 5.4 V
ΔV3 =
Q3
C3
=
21.6 µC
6 µF
= 3.6 V
In summary, Q1 = 45 µC, V1 = 9 V; Q2 = 21.6 µC, V2 = 5.4 V; and Q3 = 21.6 µC, V3 = 3.6 V.
30.67. Model: Capacitance is a geometric property.
Visualize: Please refer to Figure P30.67. Shells R1 and R2 are a spherical capacitor C. Shells R2 and R3 are a spherical
capacitor C′. These two capacitors are in series.
Solve: The ratio of the charge to the potential difference is called the capacitance: C = Q ΔVC . The potential differences
across the capacitors C and C′ are
ΔVC =
ΔVC ′ =
1
Q
4π ε 0 R1
1
Q
4πε 0 R2
−
1
Q
4πε 0 R 2
−
 1
1 
⇒ C = (4πε 0 )
−

 R1 R 2 
1
Q
4πε 0 R3
−1
=
Q 1
1 
−


4πε 0  R1
R2 
=
Q 1
1 

−

4πε 0  R 2
R3 
 1
1
C ′ = (4πε 0 )
−

 R2
R3 
−1
Because these two capacitors are in series,
1
C net
=
1
C
+
1
C′
=
1  1
1
1
1
1  R3 − R1 
−
+
− =



4π ε 0  R1 R2 R 2 R 3  4πε 0  R1 R 3 
 RR 
1
 ( 0.01 m )( 0.03 m ) 
−12
C net = 4πε 0  1 3  =
 = 1.67 × 10 F = 1.67 pF
9
2
2 
 R3 − R1  9.0 × 10 N m / C  0.03 m − 0.01 m 
Assess:
Cnet depends on only the inner and outer shells, not on R2.
6
30.68. Model: Assume the battery is ideal.
Visualize:
The circuit in Figure P30.68 has been redrawn to show that the six capacitors are arranged in three parallel combinations,
each combination being a series combination of two capacitors.
Solve: (a) The equivalent capacitance of the two capacitors in series is 12 C . The equivalent capacitance of the six
capacitors is 32 C .
(b) As points a and b are midpoints of identical capacitors, Va = Vb = 6.0 V. Therefore, the potential difference between
points a and b is zero.
30.83. Visualize:
Solve: The charge density inside the cylinder is ρ. Consider a Gaussian surface with a radius r inside the cylinder. The
charge enclosed by this closed surface is Qenclosed = (πr2L)ρ. Using Gauss’s law,
(
)
  ρ 
 
πr 2 L ρ
πr 2 L ρ  ρ 
Qenclosed
⇒ E=
=
(
)
E
⋅
d
A
=
E
2
π
rL
=
=
r ⇒ E = 
 rrˆ
∫
ε 0 2πrL  2ε 0 
 2ε 0 
ε0
ε0
In the last step, we have taken into account the radial direction of the electric field pointing outward. The potential
difference between the surface and the axis is
R
R
 
ρ

ρ
− ρ r2
ΔV = − ∫ E ⋅ dr = − ∫
r ˆr ⋅ d r = − ∫
r dr =
2ε 0
2ε 0
2ε 0 2
0
0
R
=−
0
ρR 2
4ε 0
7
Mass

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