Physics 1B: Electricity & Magnetism
Summer Session I, 2010
Final Exam, July 30, 2010
This is version
A
!
Useful coefficients:
Mass of proton, m
p
= 1
.
67
×
10
−
27
kg; Mass of electron, m
e
= 9
.
11
×
10
−
31
kg
Unit of elementary charge
e
= 1
.
6
×
10
−
19
C
Coulomb’s constant,
k
e
= 8
.
99
×
10
9
N m
2
C
−
2
Permittivity of free space
ǫ
0
=
1
4
πk
e
= 8
.
85
×
10
−
12
C
2
N
−
1
m
−
2
1 eV = 1.6
×
10
−
19
J
g
= 9
.
8 m s
−
2
Permeability of free space
μ
o
= 4
π
×
10
−
7
T m A
−
1
Speed of light
c
= 3
×
10
8
m/s
1.
Which of the following characteristics are held in common by both gravitational and
electrostatic forces when dealing with either point masses or charges?
a. inverse square distance law applies
b. forces are conservative
c. potential energy is a function of distance of separation
d. all of the above choices are valid
Solution: d.
2. An electrical device with a resistance of 1Ω is operated at a voltage of 2.0 V. How many
electrons pass through a point inside the device in a time Δt = 0.8 seconds?
We must calculate the current via I = ΔV/R = 2V/1.0Ω = 2.0 A.
From the definition of current, I = ΔQ/Δt we can calculate the total charge which passes
through the device: ΔQ = I Δt = (2.0 C/s)(0.8 s) = 1.6 C
Finally, we divide by the unit of elementary charge:
N = ΔQ/(1.6
×
10
−
19
C/e
−
) = 1.6 C / (1.6
×
10
−
19
C/e
−
) = 1
.
0
×
10
+19
electrons
1
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3. In the core of a supergiant star, nuclear fusion is occurring: in one reaction, three helium
nuclei (
4
He; 2 protons, 2 neutrons each) fuse together to form a single carbon nucleus (
12
C;
six protons and six neutrons in each). Suppose three
4
He nuclei have just fused to create a
12
C nucleus, and there’s an extra
4
He nucleus sitting 4 nm away (1 nm = 10
−
9
m). What is
the resulting electrostatic force between the
12
C and
4
He nuclei? (Note: temperatures here
are high enough that all atoms are ionized, i.e., all electrons have been stripped, so assume
electrons are not relevant here.)
4. Two protons are initially 2
×
10
−
9
m apart. What is the CHANGE in potential energy
(P.E.
final
– P.E.
initial
) if they’re brought in to a separation distance of 1
×
10
−
9
m?
The initial potential energy is PE(initial)= (
k
e
q
1
q
2
)/r =
(8
.
99
×
10
9
N m
2
C
−
2
×
(1
.
6
×
10
−
19
C
)
2
/
2
e

9
m
) =
1
.
15
×
10
−
19
J.
The final potential energy is PE(final)= (
k
e
q
1
q
2
)/r =
( 8
.
99
×
10
9
N m
2
C
−
2
×
(1
.
6
×
10
−
19
C
)
2
/
1
e

9
m
) =
2
.
30
×
10
−
19
J.
PE(final) – PE(initial = 1
.
15
×
10
−
19
J
Using 1 eV = 1.6e19 J, this amount of energy is equal to 0.72 eV.
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 Summer '09
 Berman
 Electron, Charge, Magnetism, Mass, Magnetic Field, Electric charge

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