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:
Kq i
q j
U elec
= ∑ = U 12
+ U 13
+ U 23
= K q q 1 2
+ K q 1q 3
+ K q 2
q 3
r 12
r 13
i,j
r ij
= ( 9.0 × 10 9 N m 2 / C 2 ⎡ ( 2 nC )( −1.0 nC ) (
)
+ 2 nC ) ( −1.0 nC ) ( −1 nC) + ( −1.0 nC ) ⎤
⎢
⎥
⎣ 0.03 m
0.03 m
0.03 m ⎦
r 23
= −0.60 × 10 −6 J − 0.60 × 10 −6 J + 0.30 × 10 −6 J = −0.90 ×10 −6 J
Assess: Note that U 12 = U 21 , U 13 = U 31 , and U 23 = U 32 .
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,
⇒ ΔV = − mv 2
i
2e
K f + U f = K i + U i ⇒ K f + qV f = K i + qV i
⇒ V f − V i = 1 (
q K i − K f
) =
1
( −e)
= − 9.11 × 10
−31
kg
2 1.60 × 10 −19 C
1
( 2 mv 2 i − 0 J )
( )( 5.0 ×10 5 m / s) 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° ⇒ r 1 = r 2 = r 3 =
r 3 cos30°
The potential at the dot is
r 1
r 2
= 1.732 cm
1

V =
1
4πε 0
q 1
r 1
+
1
4πε 0
q 2
r 2
+
1
4πε 0
q 3
r 3
( ) 2.0 × 10 −9 C
= 9.0 × 10 9 N m 2 / C 2 ⎡
0.01732 m − 1.0 × 10 −9 C
0.01732 m − 1.0 ×10 −9 C ⎤
⎣ ⎢
0.01732 m ⎦ ⎥ = 0 V
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
E 0
= K 0
+ U 0
= 0 J + q A q B
4πε 0
r 0
= 9.0 ×10 9 N m 2 / C 2
(b) In static equilibrium, the net force on sphere A is zero. Thus
( )( 2.0 ×10 −6 C) ( 2.0 ×10 −6 C)
0.05 m
( )( 2.0 × 10 −6 C) ( 2.0 × 10 −6 C)
= 0.720 J
T = F B on A = q q A B
4πε 0 r = 9.0 × 10 9 N m 2 / C 2
= 14.4 N
2
0 ( 0.05 m ) 2
(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 (r 1 → ∞), their potential energy U 1 → 0 J. Energy is conserved, so we
have
E 1
= K 1
+ U 1
= 1 2
2
m A
v A1
+ 1 2
2
m B
v B1
+ 0 J = E 0
Momentum is also conserved: P after = m A v A1 + m B v B1 = P before = 0 kg m/s. Note that these are velocities and that v A1 is a
negative number. From the momentum equation,
Substituting this into the energy equation,
⇒ v B1
=
⎛
E 0 = 1 2 m A − m v B B1
⎜
⎝ m A
2 E 0
m B m B m A +1
v A1
⎞
⎟
⎠
2
= − m B v B1
m A
+ 1 2
2 m B v B1
⎛ m
= 1 2 m B
⎞ 2
B ⎜ + 1⎟ v B1
⎝ m A ⎠
( ) = 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
= − pEcosθ , where θ is the angle between the dipole and the
electric field. The energy conservation equation K f + U f = K i + U i is
The moment of inertia of the dipole is I dipole
( )
1
2
Iω 2 + U f
= 0 J + U i
⇒ 1 2
Iω 2 = U i
− U f
= −pE cos90° − − pEcos 0°
ω =
2( qs)E
( ) 2 =
2m 1 2
s
⇒ 1 2
Iω 2 = pE ⇒ ω =
2 pE
I
= 2 m( 1 2 s) 2 . Substituting into the above equation,
( ) 0.10 m
( 1.0 ×10 −3 kg) 0.10 m
4 2.0 ×10 −9 C
( )( 1000 V / m)
= 0.283 rad / s
( ) 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:
V a
= V A at a
+ V B at a
=
1
4πε 0
Q A
R A
+
1
4πε 0
= ( 9.0 × 10 9 N m 2 / C 2
) 100 × 10 −9 C
0.30 m
= 3000 V + 321 V = 3321 V
Q B
0.70 m
+ ( 9.0 × 10 9 N m 2 / 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:
V b = V B at b + V A at b =
1
4πε 0
( )
= 9.0 × 10 9 N m 2 / C 2
Q B
R B
+
= 4500 V + 947 V = 5447 V
1
4πε 0
Q A
0.95 m
⎛ 25 ×10 −9 C
0.05 m + 100 ×10 −9 C⎞
⎜
⎟
⎝
0.95 m ⎠
Thus, the potential at point b is higher than the potential at a. The difference in potential is V b – V a = 5447 V –
3321 V = 2126 V.
Assess: V A 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 = ∑ V i =
i
1
i
∑ 4πε 0
ΔQ
R = 1
4πε 0
R
⎛ Q ⎞
⎝ πR ⎠ Δs 1 Q
1 Q
∑ =
RΔθ
i
4πε 0
∑ =
πR 2
i
4πε 0
∑ Δθ .
πR
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
V center =
4πε 0
πR = 1 Q
4πε 0
R
29.79. Model: Energy and momentum are conserved.
Visualize: Please refer to Figure CP29.79. Let the two spheres have masses m A and m B , and speeds v A and v B when they
are very far apart.
Solve: The energy conservation equation K f + U f = K i + U i is
⇒ 1 2
2
( 0.002 kg )v A
1 2
2
m A
v A
+ 1 2
( 2
m B
v B ) + 0 J = 0 J +
+ 1 2
2
( 0.001 kg )v B
1
4πε 0
= 9.0 ×10 9 N m 2 / C 2
q A
q B
r AB
(
( ) 2.0 × 10 −9 C) ( 1.0 ×10 −9 C)
⇒ v A 2 + 0.5 v B 2 = 0.0090 m 2 / s 2
2.0 ×10 −3 m
To solve for v A and v B , we need another equation relating v A and v B . From the momentum conservation equation P after =
P before we get
m A v A + m B v B = 0 kg m/s ⇒ (2.0 g)v A + (1.0 g)v B = 0 kg m/s ⇒ v B = −2v A
Substituting this expression into the energy conservation equation,
v A
2
+ 0.5( 2v A
) 2 = 0.009 m 2 / s 2 2
⇒ 3v A
= 0.009 m 2 / s 2
Solving these equations, we get v A = –0.0548 m/s and v B = +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 ΔV C
. The
capacitance is
−9
20 ×10 C
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
U C
= 1 2 C ΔV C
= 1 2
( ) 2 = 1 2
⎛
⎝
Aε 0
d
⎞
⎠ ΔV C
( ) 2
( )
0.50 ×10 −3 m
π( 0.01 m ) 2 8.85 ×10 −12 C 2 / N m 2
(b) From Equation 30.35, the energy density in the electric field is
u E = 1 2 ε 0E 2 = 1 2 ε ⎛
0
⎝
ΔV C
d
⎞
⎠
2
( )
= 1 2 8.85 ×10 −12 C 2 / N m 2
( 200 V ) 2 = 1.11 × 10 −7 J
⎛ 200 V ⎞
⎝ 0.5 ×10 −3 m ⎠
2
= 0.708 J / m 3
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 ΔV C
. The potential difference
across the capacitor is
Using R 2 = R 1 + 1.0 mm,
ΔV C
=
⎡ 1
⇒ C = 4πε 0 − 1 ⎤
⎢ ⎥
⎣ ⎢ R 1 R 2 ⎦ ⎥
1
4πε 0
−1
Q
R 1
−
1
4πε 0
Q
R 2
=
Q ⎡ 1
− 1 ⎤
⎢ ⎥
4πε 0 ⎣ ⎢ R 1 R 2 ⎦ ⎥
= 4πε 0
R 1
R 2
R 2 − R 1
= 4πε 0
R 1
R 2
1.0 ×10 −3 m = 100 ×10 −12 F
⇒ R 1
R 2
= ( 100 × 10 −12 F) ( 1.0 ×10 −3 m) ( 9.0 ×10 9 N m 2 / C 2
) = 900 × 10 −6 m 2
( ) = 900 ×10 −6 m ⇒ R 1 2 + 1.0 × 10 −3 m
R 1
R 1
+1.0 × 10 −3 m
⇒ R 1
=
−1.0 ×10 −3 m ± ( 1.0 × 10 −3 m) 2 + 3600 ×10 −6 m 2
2
( )R 1
− 900 ×10 −6 m = 0
= 0.0295 m = 2.95 cm
The outer radius is R 2 = R 1 + 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 C 1 and C 2 are in parallel, their equivalent capacitance C eq 12 is
Then, C eq 12 and C 3 are in series. So,
C eq 12 = C 1 + C 2 = 20 µF + 60 µF = 80 µF
5

1
C eq
=
1
C eq 12
+ 1 C 3
=
30.65. Model: Assume the battery is an ideal battery.
Visualize:
1
80 µF + 1
10 µ F = 9 ( 80 µF)
−1 ⇒ C eq = 80
9 µF = 8.9 µF
The pictorial representation shows how to find the equivalent capacitance of the three capacitors shown in the figure.
Solve: Because C 2 and C 3 are in series,
C eq 23 and C 1 are in parallel, so
1
C eq 23
= 1 C 2
+ 1 C 3
=
1
4 µF + 1
6 µF = 10 ( 24 µF)
−1 ⇒ C eq 23
= 24
10
µF = 2.4 µ F
C eq = C eq 23 + C 1 = 2.4 µF + 5 µF = 7.4 µF
A potential difference of ΔV C = 9 V across a capacitor of equivalent capacitance 7.4 µF produces a charge
Q = C eq ΔV C = (7.4 µF)(9 V) = 66.6 µC
Because C eq is a parallel combination of C 1 and C eq 23 , these capacitors have ΔV 1
= ΔV eq 23
= ΔV C
= 9 V. Thus the charges
on these two capacitors are
Q 1 = (5 µF)(9 V) = 45 µC Q eq 23 = (2.4 µF)(9 V) = 21.6 µC
Because Q eq 23 is due to a series combination of C 2 and C 3 , Q 2 = Q 3 = 21.6 µC. This means
ΔV 2
= Q 21.6 µC
2
= = 5.4 V ΔV 3
= Q 21.6 µC
3
=
C 2
4 µF
C 3
6 µF = 3.6 V
In summary, Q 1 = 45 µC, V 1 = 9 V; Q 2 = 21.6 µC, V 2 = 5.4 V; and Q 3 = 21.6 µC, V 3 = 3.6 V.
30.67. Model: Capacitance is a geometric property.
Visualize: Please refer to Figure P30.67. Shells R 1 and R 2 are a spherical capacitor C. Shells R 2 and R 3 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 ΔV C
. The potential differences
across the capacitors C and C′ are
ΔV C
=
ΔV C ′ =
1
4πε 0
1
4πε 0
Q
R 1
−
Q
R 2
1
4πε 0
1
−
4πε 0
Q
R 2
=
Q
R 3
=
Q ⎡ 1
− 1 ⎤
⎢ ⎥
4πε 0 ⎣ ⎢ R 1 R 2 ⎦ ⎥
Q ⎡ 1
− 1 ⎤
⎢ ⎥
4πε 0 ⎣ ⎢ R 2 R 3 ⎦ ⎥
( )
⇒ C = 4πε 0
⎛ 1
− 1 ⎞
⎜ ⎟
⎝ R 1
R 2
⎠
−1
( )
C ′ = 4πε 0
⎛ 1
− 1 ⎞
⎜ ⎟
⎝ R 2
R 3
⎠
−1
Because these two capacitors are in series,
1
= 1 C net
C + 1 C ′
1 ⎛ 1
= − 1 + 1 − 1 ⎞ 1 ⎛ R
⎜
⎟ =
3
− R 1
⎞
⎜ ⎟
4πε 0
⎝ R 1
R 2
R 2
R 3
⎠ 4πε 0
⎝ R 1
R 3
⎠
( )( 0.03 m )
⎛ R
C net
= 4πε
1
R 3
⎞
1 ⎡ 0.01 m ⎤
0 ⎜ ⎟ =
⎢
⎥ = 1.67 ×10 −12
⎝ R 3
− R 1
⎠ 9.0 × 10 9 N m 2 / C 2 ⎣ 0.03 m − 0.01 m ⎦
Assess: C net depends on only the inner and outer shells, not on R 2 .
F = 1.67 pF
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 1 2 C . The equivalent capacitance of the six
capacitors is 3 2 C .
(b) As points a and b are midpoints of identical capacitors, V a = V b = 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 Q enclosed = (πr 2 L)ρ. Using Gauss’s law,
∫ E ⋅ dA
= E( 2πrL ) = Q (
enclosed
= πr 2 L)ρ
⇒ E = πr 2 Lρ
ε 0
ε 0 2πrL = ⎛ ρ ⎞
⎜ ⎟ r ⇒ ⎛ ρ ⎞
E = ⎜ ⎟ rˆ r
⎝ 2ε 0 ⎠ ⎝ 2ε 0 ⎠
ε 0
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
ΔV = −
∫
E
⋅ dr
= −
R
∫
0
ρ
r ˆ r ⋅ dr = −
2ε 0
R
∫
0
ρ
2ε 0
r dr
= −ρ r 2 R
= − ρR2
2ε 0 2 0
4ε 0
7