Name: ________________________________________ Code # _________
UCSD Physics 2B
Practice Exam 2
Summer Session 2008
ID: X
Potential, Work & Capacitors
READ ME !!!
Helpful Hints:
Do this exam a few times... Once without the answers,
Once after seeing the solutions & Once again with your formula sheet
This is a good time to add formulas/constants to your sheet.
Remember, this is a timed exam... work swiftly so you have time to check your answers.
1. The voltage between the cathode (negative plate) and the (positive) screen of a computer monitor is
12.0kV . If an electron leaves the cathode at rest, what is its speed as it hits the screen? Electron
mass is me = 9.11 × 10 −31 kg
A. 2.96 × 10 −23
m
s
B. 6.49 × 10 7
m
s
C. 2.05 × 10 6
m
s
D. 7.7 × 10 15
m
s
2. How much work is done by moving a charge q = 2.0 C between two points which potential
difference is ΔV = 6.0 V ?
A. 2.0 J
B. 3.0 J
C. 6.0 J
D. 12 J
3. You want to store 7.30 × 10 10 excess electrons on the negative plate of a capacitor at V = 12.0 V . How
large a capacitor must you use?
A. 0.166μ F
B. 6.08 × 10 9 F
C. 1.40 × 10−7 F
1
D. 973 p F
Name: ________________________
ID: A
4. The electric potential in a region of space is given by
ÁÊ V ˜ˆ
ÁÊ
V ˜ˆ
V ÊÁË x, y ˆ˜¯ = 55 V + ÁÁÁÁ 33 ˜˜˜˜ x − ÁÁÁÁ 22 2 ˜˜˜˜ y 2
Ë m¯
Ë m ¯
What is the electric field in this region?
ÁÊ
V ˜ˆ
A. ÁÁÁÁ +55 ˜˜˜˜ 8i
m¯
Ë
ÊÁ
ÊÁ V
V ˆ˜
B. ÁÁÁÁ +33 ˜˜˜˜ i8 + ÁÁÁÁ 22 2
m¯
Ë
Ë m
ÁÊÁ V
˜ˆ
ÁÁ 55 + 33 V − 22 V ˜˜˜ ÊÁÁ i8 +
Á m
m
m ˜¯ Ë
Ë
ÊÁ
ÊÁ V ˆ˜
V ˆ˜
D. ÁÁÁÁ −33 ˜˜˜˜ i8 + ÁÁÁÁ 44 2 ˜˜˜˜ y j8
m¯
Ë
Ë m ¯
C.
ˆ˜
˜˜ y j8
˜˜
¯
j8 ˆ˜˜
¯
5. You connect three capacitors as shown in the diagram below.
C1 = C2 = 5.00 μF and C3 = 10.0 μF . What is the equivalent capacitance of this network?
A. 5.00 μF
B. 10.0 μF
C. 15.0 μF
D. 3.33 μF
6. Two protons in a Helium nucleus (2 protons & 2 neutrons) are 6.0 × 10 −15 m apart. What is the
electrostatic potential energy (energy of assembly) of this pair?
A. 4.7 × 10 −34 J
B. 4.3 × 10 −24 J
C. 3.8 × 10 −14 J
2
D. 2.4 × 104 J
Name: ________________________
ID: A
7. A charge q = 5.00 μC is in a uniform electric field of intensity E = 3.50 × 10 5 N / C . How much work
is required to move this charge a distance r = 50.0 cm along a straight path making an angle θ = 33.0°
with the electric field vector? Assume positive work.
A. 277 J
B. 0.477 J
C. 87.5 J
D. 0.734 J
8. The magnitude of the electric field between two parallel plates separated by a distance d = 333 mm is
E = 150 V / m. What is the potential difference between the plates?
A. 150 V
B. 450 V
C. 50.0 V
D. zero
9. You charge a 30 pF capacitor by connecting it across a 150 V battery. How much energy is stored
in the capacitor when it is fully charged?
A. 3.4 x 10-11 J
B. 4.5 x 10-11 J
C. 6.7 x 10-11 J
3
D. 3.4 x 10-8 J
Name: ________________________
ID: A
10. A coaxial cable (cylindrical symmetry) consists of a wire of radius a = 0.30 mm and an outer
conducting shell of radius b = 1.0 mm as shown in the cross-section drawing below. The shaded area
between the cylinders is non-conducting.
The capacitance per unit length is approximately
A. 46 nF/m
B. 17 nF/m
C. 46 pF/m
4
D. 92 pF/m
Name: ________________________
ID: A
11. A capacitor consists of two concentric conducting spherical shells, an inner shell of radius a = 0.95
mm and an outer shell of radius b = 1.0 mm as shown in the cross-section drawing below.
The capacitance of this arrangement is approximately
A.
B.
C.
D.
E.
1.7 nF
46 pF
2.1 pF
23 nF
zero
12. A very large parallel plate capacitor has plates of Area A = 4200 m2 and capacitance C = 17 μ F . What is the
separation distance d between the plates?
A. 4.0 nm
B. 71 mm
C. 0.25 Gm
5
D. 2.2 mm
Name: ________________________
ID: A
13. In the network below C 1 = 13μF and C 2 = 26μF . What is the equivalent capacitance?
A. 39 μ F
B. 8.7 F
C. 39 F
D. 8.7 μ F
14. In the network below C 1 = 13μF and C 2 = 26μF . What is the equivalent capacitance?
A. 39 μ F
B. 8.7 F
C. 39 F
6
D. 8.7 μ F
ID: A
UCSD Physics 2B
Answer Section
Practice Exam 2
Potential, Work & Capacitors
MULTIPLE CHOICE
1. ANS: B
We solve this by equating the Energy gain through Work done on the electron by the field W = qΔV with the
1
final kinetic energy of the moving electron KE = mv 2
2
W = KE ⇒ v =
2qΔV
=
m
Ê
ˆÊ
ˆ
2 ÁÁ 1.60 × 10 −19 C ˜˜ ÁÁ 12.0 × 10 3 V ˜˜
m
Ë
¯Ë
¯
= 6.49 × 10 7
s
ÁÊÁ 9.11 × 10−31 kg ˜ˆ˜
Ë
¯
PTS: 1
2. ANS: D
The Work on a charge by an Electric Field is the Charge times Potential Difference:
ÊÁ
J ˆ˜
W = q (ΔV ) = (2.0C ) ÁÁÁÁ 6.0 ˜˜˜˜ = 12J
C¯
Ë
You might recall that in units, 1Volt = 1
Joule
Coulomb
PTS: 1
3. ANS: D
Here we use the Capacitor Formula but remember to express the charge in Coulombs:
Q = CV ⇒
ˆ ÁÊ
10 ˜
−19
ÁÊ
˜ˆ
Q ÁË 7.30 × 10 e ˜¯ ÁË 1.60 × 10 C / e ˜¯
=
= 9.73 × 10 −10 F = 973 × 10 −12 F = 973 p F
C=
V
12.0V
PTS: 1
4. ANS: D
Ê
ˆ
ÊÁ ∂V ∂V ∂V ˆ˜
˜˜ = − ÁÁÁ 33 V ,− 44 V y,0 ˜˜˜ = −33 V i8 + 44 V y j8
E = − ÁÁÁÁ
,
,
Á
˜˜
˜
2
Á m
˜
m
m
m2
Ë
¯
Ë ∂x ∂y ∂z ¯
PTS: 1
5. ANS: A
C1 and C2 are in parallel so the capacitances ADD: C 12 = C 1 + C 2 = 10.0μF .
The network C12 is in series with C3, so
1
1
1
1
1
2
=
+
=
+
=
C series C 12 C 3 10.0μF 10.0μF 5.00μF
Hence the equivalent capacitance of the network is C series = 5.00μF .
PTS: 1
1
ID: A
6. ANS: C
Since there are only two charges, the energy to assemble them is simply the energy required to bring the
second one (take your pick)
ÊÁ
ˆÊ
ˆ2
Á 9.0 × 109 Nm2 / C 2 ˜˜ ÁÁ 1.6 × 10 −19 C ˜˜
ÊÁ kq ˆ˜ ke 2
Ë
¯Ë
¯
Á ˜
PE = qV = q ÁÁÁ ˜˜˜ =
=
= 3.8 × 10 −14 J
Ê
ˆ
−15
Á r ˜
r
Á
˜
Á 6.0 × 10 m˜
Ë ¯
Ë
¯
PTS: 1
7. ANS: D
Work is the line-integral of force * distance along the path. Since the Electric Field is a constant vector, the
force is constant F = qE and since the path is a straight line (constant angle) everything comes out of the
integral, and were left with just the path length, r:
b
b
b
r
W = ∫ F ⋅ dL = ∫ qE ⋅ dL = ∫ qE cos θ dL = qE cos θ ∫ dL
a
a
a
0
N ˆ˜ Ê
Ê
ˆ ÊÁ
ˆ
= qEr cos θ = ÁÁ 5.00 × 10 −6 C ˜˜ ÁÁÁÁ 3.50 × 10 5 ˜˜˜˜ ÁÁ 50.0 × 10−2 m˜˜ cos (33°) = 0.734 J
Ë
¯Ë
¯
C ¯Ë
PTS: 1
8. ANS: C
Of course you can just plug in the formula, but the idea behind it is this. Work and energy are the same. The
Potential Energy difference between two places is the work it takes to get a charge from one place to another.
The Potential Difference is the work it takes to move a unit charge. Since the field between the plates is
uniform, let’s just take the shortest path from one to another - right along the field line. The answer is just
field times distance.
|
|ΔV | = | −
||
∫
PTS: 1
9. ANS: A
1
1
W = CV 2 =
2
2
d
0
|
ÊÁ
V ˆ˜ Ê
ˆ
E ⋅ dL | = Ed = ÁÁÁ 150 ˜˜˜˜ ÁÁ 333 × 10 −3 m˜˜ = 50.0V
Á
¯
Ë
m
||
¯
Ë
ÊÁ
2
ˆ
Á 30 × 10 −12 F ˜˜ (150V ) = 3.4 × 10 −7 J
Ë
¯
PTS: 1
2
ID: A
10. ANS: C
b
V = ∫ E ⋅ dr = ∫
a
C=
b
a
ˆ
Ê
2k λ
⋅ dr = 2k λ lnÁÁÁ b ˜˜˜
a
r
¯
Ë
Q
λL
=
V 2k λ ln ÁÊÁ b ˜ˆ˜
Á a˜
¯
Ë
1
1
C
−12
=
=
ˆ
Ê
ÊÁ
ˆ˜ = 46 × 10 F / m
−3
L 2k ln ÁÁ b ˜˜
Á a ˜ 2 ÊÁÁ 9.0 × 10 9 Nm2 / C 2 ˆ˜˜ ln ÁÁÁ 1.0 × 10 m ˜˜˜
¯
Ë
Ë
¯ ÁÁÁ 0.30 × 10 −3 m ˜˜˜
Ë
¯
PTS: 1
11. ANS: C
1
1
C= Ê
= 2.1 × 10 −12 F
ˆ
ˆ = Ê
Ê
˜
˜
Á
Á
1
1
1
1
ˆÁ
9
2
2˜
˜˜
˜ Á
−
k ÁÁ
−
˜˜
b ˜¯ ÁË 9.0 × 10 Nm / C ˜¯ ÁÁÁ
−3
−3
Ë a
m
m
1.0
×
10
0.95
×
10
¯
Ë
PTS: 1
12. ANS: D
ÊÁ
ˆ
Á 4200m2 ˜˜
Ë
¯
A
A ÊÁ
ˆ
C = ε0 ⇒ d = ε0
= Á 8.85 × 10 −12 C 2 / Nm2 ˜˜ Ê
= 2.2mm
−6
d
C Ë
¯ ÁÁ 17 × 10 F ˆ˜˜
Ë
¯
PTS: 1
13. ANS: A
Ê
ˆ Ê
ˆ
For a parallel network, C parallel = C 1 + C 2 = ÁÁ 13 × 10−6 F ˜˜ + ÁÁ 26 × 10 −6 F ˜˜ = 39μF
Ë
¯ Ë
¯
or, more directly,
C parallel = C 1 + C 2 = 13μF + 26μF = 39μF
Note that, when all the devices have the same units - microfarads, for example - you can just add the numbers
without the exponents, and your answer will also be in microfarads!
PTS: 1
14. ANS: D
For a series network,
1
1
1
1
1
1
=
+
=
+
=
⇒ C series = 8.7μF
C series C 1 C 2 13 × 10 −6 F 26 × 10 −6 F 8.7 × 10−6 F
Here it's easier to just write
1
1
1
1
1
1
=
+
=
+
=
⇒ C series = 8.7μF
C series C 1 C 2 13μF 26μF 8.7μF
PTS: 1
3