Download PHY 1160, Ch 18 homework (WP)

PHY 1160C
Homework
Chapter 18: Electric Potential & Capacitance
Ch 18: 1, 3, 10, 11, 21, 27, 23, 40, 59, 64, 68
18.1 What is the potential difference between two points if
6 µJ of work is required to move 12 µC of charge from one
point to the other?
D PE el = W = 6 µJ
D V = D PEel = 6 µJ = 0.5 J = 0.5 V
q
12 µC
C
V = 0.5 V
18.3
The electric field in the region between two parallel
plates connected to a 12–volt power supply is 100 N/C. What is
the separation distance between the plates?
We can use Equation 18.10,
E=– D V
D r
D r=– D V
E
D r = – 12 V
100 N/C
D r = – 0.12 V
N/C
D r = – 0.12 V J/C N m
N/C V
J
r = 0.12 m = 12 cm
The minus sign just says the potential increases in the direction
opposite to the electric field.
Ch 18, p 1
18.10 Electrons in a color television are accelerated through a
potential difference of about 25,000 V. With what speed do they
strike the front of the picture tube?
KE = D PE = D V q = D V e
KE = (1/2) m v2 = (25,000 V) (e) = 25,000 eV = 25 keV
(1/2) (9.11 x 10–31 kg) v2 = 25,000 eV
–19
J N m kg m/s2
v2 = (25,000 eV) (2) 1.6 x 10
eV
J
N
(9.11 x 10–31kg)
v 2 = 8.78 x 1015 m2/s2
v = 9.37 x 107 m/s
That is nearly one-third the speed of light!
18.21
How many electrons, accelerated through a potential
difference of 20,000 V, must bombard one gram of water to
raise its temperature 1.0 C°?
First, how much energy, in joules, does each single electron
have?
E = 20,000 eV
–19
J
E = 20,000 eV 1.6 x 10
eV
E = 3.2 x 10–15 J
Now, how much energy is required to raise the temperature of
one gram of water by D T = 1 C°? That requires ideas about
specific heat from Chapter 12 from last semester. In particular,
we need Equation 12.5,
Ch 18, p 2
Q=cmD T
For water, the specific heat is
c = 4186 J/kg C°
Q = (4186 J/kg C°) (0.001 kg) (1.0 C°) = 4.186 J
How many electrons, each with energy of E = 3.2 x 10–15 J, must
deposit their energy into the water to give it 4.186 J of energy?
Q = 4.186 J = n E = n (3.2 x 10–15 J)
4.186 J
n= Q=
E
3.2 x 10–15 J
n = 1.31 x 1015 electrons
18.27 What is the electric potential at a distance of 1.00, 2.00,
3.00, 4.00, and 5.00 m from a 1.25 µC charge? What is the
electric field at each of those distances?
Make graphs of
electric potential versus distance and of electric field strength
versus distance.
The potential is calculated from
V=kQ/r
and the electric field is calculated from
E = k Q / r2
These calculations can be done readily by a spreadsheet
For Q = 1.25 µC = 1.25 x 10–6C = 0.000 001 25 C, this gives
Ch 18, p 3
distance potential E field
( m ) ( V = J/C ) ( N/C)
1.00
11,250
11,250
2.00
5,625
2,813
3.00
3,750
1,250
4.00
2,813
703
5.00
2,250
450
While you can certainly make nicer graphs by hand or by using
Graphical Analysis (as we did last semester), we can also let
ClarisWorks make this graph for us.
But before we make the graph, let us do a few more calculations
at some more distances, just to make the graph smoother and,
perhpas, more interesting or informative,
distance potential
E field
( m ) ( V = J/C )
( N/C)
0.05
225,000
4,500,000
0.10
112,500
1,125,000
0.25
45,000
180,000
0.50
22,500
45,000
1.00
11,250
11,250
2.00
5,625
2,813
3.00
3,750
1,250
4.00
2,813
703
5.00
2,250
450
10.00
1,125
113
20.00
563
28
Ch 18, p 4
V=kQ/r
50,000
P
o
t
e
n
t
i
a
l
40,000
30,000
20,000
10,000
0
0.00
2.00
4.00
6.00
distance (in meters)
8.00
10.00
Now, we do the same sort of calculations for the electric field,
A
1
2
3
4
5
6
7
8
9
10
11
12
13
B
distance
E field
(m)
( N/C)
0.05
4,500,000
0.10
1,125,000
0.25
180,000
0.50
45,000
1.00
11,250
2.00
2,813
3.00
1,250
4.00
703
5.00
450
10.00
113
20.00
28
Again, before we make the graph, let us do a few more
calculations at some more distances, just to make the graph
smoother and, perhpas, more interesting or informative,
Ch 18, p 5
E = k Q / r^2
200,000
E
•
f 100,000
i
e
l
d
0
0.00
2.00
4.00
6.00
distance (in meters)
8.00
10.00
18.34 What is the electric potential at the unfilled corner of
the square in the figure below?
V=?
- 2 µC
1.0 m
1.0 m
-4µC
+3µC
Figure 18.45 Problem 18.34
Calculate the potential due to each individual charge and then
add those together for the net potential. Potential is a scalar
so this will be ordinary, common, scalar addition
V=kQ/r
V 2 = (9 x 109) ( – 2 x 10–6) / 1.0 = – 1.8 x 104 V
Ch 18, p 6
V 3 = (9 x 109) ( + 3 x 10–6) / 1.414 = + 1.35 x 104 V
V 4 = (9 x 109) ( – 4 x 10–6) / 1.0
V net = V2 + V3 + V4
= – 3.6 x 104 V
=
V net =
– 4.05 x 104 V
– 4.05 x 104 V
18.40 A 120 µF capacitor stores ± 180 µC of charge on its
plates.
What is the voltage of the battery to which it is
connected?
C = Q/D V or D V = Q/C
D V = 180 µC/120µF
V = 1.5 V
18.59
To run a particular laser, 9.5 J of energy must be
discharged through a "flashlamp" to provide an intense flash of
light.
This energy is released by a capacitor that has been
charged by a 12 kV power supply. What is the capacitance of
this capacitor? How much charge is stored and then released?
PE cap = Q D V / 2
Q = 2 PEcap / D V
Q = (2) (9.5 J)/12,000 V = 1.58 x 10–3 C
Q = 1.58 x 10–3 C is the charge stored and then released by
the capacitor.
C = Q/D V
C = 1.58 x 10–3 C / 12,000 V = 1.32 x 10–7 F
C = 0.132 µF
Ch 18, p 7
18.64
In the series capacitor circuit shown below C1 = 1.0 µF,
C2 = 2.5 µF, C3 = 5.0 µF, and V = 12.2 V. Find the equivalent
capacitance of this circuit.
What energy is stored in the
equivalent capacitor? Calculate the voltage across each of the
capacitors and the energy stored in each capacitor.
C
C
2
1
C
3
V
Figure 18.47 Problem 18.67
The equivalent capacitance is given by
1 = 1 + 1 + 1
Ceq C1
C2
C3
1 =
1
1
1
+
+
Ceq 1.00 µF
2.50 µF
5.00 µF
1 = 1.6
Ceq
µF
Ceq = 0.625 µF
PE cap = C (D V)2 / 2 = (0.625 x 10–6) (12.2)2 / 2
PE cap = 4.65 x 10–5 J
This is the energy that would be stored in the single equivalent
capacitor or the total energy stored in the three individual
capacitors.
Ch 18, p 8
Now we must find the voltage across each individual capacitor
and the energy stored in each individual capacitor. The charge
on each capacitor is the same and is the same as the charge on
the equivalent capacitor,
C = Q/D V or D V = Q/C or Q = C D V
Q = C D V = (0.625 x 10–6) (12.2) = 7.625 x 10–6 C
D V 1 = Q/C1 = 7.625 x 10–6 C / 1.00 x 10–6 F = 7.625 V =
V1
D V 2 = Q/C2 = 7.625 x 10–6 C / 2.50 x 10–6 F = 3.05
V=
V2
D V 3 = Q/C3 = 7.625 x 10–6 C / 5.00 x 10–6 F = 1.525 V =
D V tot = D V 1 + D V 1 + D V 1 = 12.2 V
V3
Now, for the energy in each capacitor,
PE cap = Q2 / 2C
PE cap,1 =Q 2/2C1=(7.625x10–6C)2/[2(1.00x10–6F)]=2.91x10–11 J
PE cap,2 =Q 2/2C1=(7.625x10–6C)2/[2(2.50x10–6F)]=1.17x10–11 J
PE cap,3 =Q 2/2C1=(7.625x10–6C)2/[2(5.00x10–6F)]=5.81x10–12 J
PEcap,tot = PEcap,1 + PEcap,1 + PEcap,1 = 4.65x10–11 J
And, as we would expect—as we must have—this is the energy
stored in the equivalent capacitor.
Ch 18, p 9
18.68 Find the equivalent capacitor for the following circuit.
Determine the energy stored in this equivalent capacitor. Find
the voltage across each of the individual capacitors and the
charge and energy stored in each.
72 µF
54 µF
12 v
87 µF
Figure 18.49 Problem 18.68
First, find an equivalent capacitor for the two that are in series,
72 µF
54 µF
12 v
87 µF
For series capacitors, the equivalent capacitor is given by
1 = 1 + 1
Ceq C1
C2
1 =
1
1
+
Ceq 72 µF
87 µF
1 = 0.0254
Ceq
µF
Ceq = 39.4 µF
Ch 18, p 10
Now we can re-draw this circuit as
54 µF
39 µF
12 v
and find the new equivalent capacitor that replaces the 54 µF
capacitor and the 39 µF capacitor which are now connected in
series. For such series capacitors, the equivalent capacitor is
given by
Ceq = C1 + C2
Ceq = 54 µF + 39 µF = 93 µF
Ceq = 93 µF
93 µF
12 v
The energy stored in this equivalent capacitor is
PE cap = C(D V)2/2
PE cap = (93 x 10–6)(12)2 / 2
PE cap = 6.7 x 10–3 J = 6.7 mJ
PE cap = 6.7 mJ
Ch 18, p 11
Now, what is happening to each of the individual capacitors:
54 µF: The voltage across this capacitor is the full 12 V
Q = C D V = (54 x 10–6) (12) = 6.48 x 10–4 C
PE cap = C (D V)2 / 2 = (54 x 10–6) (12)2 / 2 = 3.89 x 10–3 J
72 µF and 87 µF capacitors in series: Their equivalent capacitor
of 39 x 10–6 F or 39 µF has 12 V across its ends so the charge
on that equivalent capacitor is
Q = C D V = (39 x 10–6) (12) = 4.68 x 10–4 C
and this is also the charge on the 72 µF capacitor and the
charge on the 87 µF capacitor since they are in series
72 µF: The voltage is given by
D V = Q/C = 4.68 x 10–4 / 72 x 10–6 = 6.5 V
The charge is 4.68 x 10 –4 C since that is the same as the charge
on the equivalent capacitor.
The energy stored in the capacitor is
PE cap = C (D V)2 / 2 = (72 x 10–6) (6.5)2 / 2 = 1.51 x 10–3 J
87 µF: The voltage is given by
D V = Q/C = 4.68 x 10–4 / 87 x 10–6 = 5.4 V
Notice that 6.5 V + 5.4 V = 11.9 V
12.0 V. This is due to
rounding error since the equivalent capacitor was really
39.396 µF which we wrote as merely 39 µC. If we want greater
accuracy, we can keep all the results to one or two more
significant figures.
The charge is 4.68 x 10 –4 C since that is the same as the charge
Ch 18, p 12
on the equivalent capacitor.
The energy stored in the capacitor is
PE cap = C (D V)2 / 2 = (87 x 10–6) (5.4)2 / 2 = 1.27 x 10–3 J
The energy stored in these two capacitors is
PE = 1.51 x 10–3 J + 1.27 x 10–3 J = 2.78 x 10–3 J
and that should be very close to the energy stored in this
equivalent capacitor
PE cap = C (D V)2 / 2 = (39 x 10–6) (12)2 / 2 = 2.80 x 10–3 J
Again, the small discrepency is simply a rounding error
Ch 18, p 13