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Material Type: Exam; Professor: Murdock; Class: Algebra-based Physics II; Subject: PHYS Physics; University: Tennessee Tech University; Term: Spring 2008;
Typology: Exams
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Name Feb. 21, 2008
Phys 2020, NSCC Exam #1 — Spring 2008
Total (100)
Multiple Choice
Choose the best answer from among the four! (2) each.
-q +q
A B C^ D
a) A b) B c) C d) D
a) The electric field is zero, but the electric potential is not zero.
b) The electric field is not zero, but the electric potential is zero. c) Neither the field nor the potential is zero. d) Both the field and potential are zero.
a) C V b) V C c) N V d) V N
a) 13 R b) 23 R c) 32 R d) 3 R
a) Into the page. b) Out of the page. c) Up. d) Down.
a) The charge that collects on each plate. (8)
The capacitance of the plates is C = 0
d Since the area is A = (20. 0 × 10 −^4 m)^2 = 2. 00 × 10 −^3 m^2
we get
C = (8. 85 × 10 −^12 )
(2. 00 × 10 −^3 m^2 ) (5. 00 × 10 −^3 m)
= 3. 54 × 10 −^12 F = 3.54 pF
Then we can use Q = CV to get the charge:
Q = (3. 54 × 10 −^12 F)(100 V) = 3. 54 × 10 −^10 C
b) The magnitude of the electric field between the plates. (2)
Use E = ∆ ∆Vs to get the magnitude of E:
(5. 00 × 10 −^3 m)
c) The magnitude of the force on a particle of charge +e if it is in the space between the plates. (4)
The magnitude of the force (on charge q = e) is
F = qE = (1. 60 × 10 −^19 C)(2. 00 × 10 4 N C ) = 3. 2 × 10 −^15 N
V = 0 V
p
V = 30.0 kV
p
v
a) Find the loss in electric potential energy the proton undergoes. (4)
The change in potential is − 30 .0 kV, so the change in po- tential energy is
∆EPE = q∆V = (1. 60 × 10 −^19 C)(− 30. 0 × 103 V) = − 4. 8 × 1015 J
that is, it loses 4. 8 × 1015 J of potential energy.
b) Find the final speed of the proton. (The mass of the proton is 1. 67 × 10 −^27 kg.) (6)
From (a), the proton must gain 4. 8 × 1015 J of kinetic energy. Then:
KE = 12 mv^2 =⇒ v^2 = 2KE m
(1. 67 × 10 −^27 kg)
= 5. 75 × 10 12 m s 22
Then: v = 2. 40 × 10 6 m s
2.0 x 10-15m
With the first proton fixed at the origin, we find the electrical potential of the second proton when it is at infinity and when it is near the other proton. In both cases, use V = k q r = +k e r. At infinity, the potential is zero. At the given distance it is
V 2 = k
e r
= (8. 99 × 10 9 N·m
2 C^2 )
(2. 0 × 10 −^15 m)
The work required is the change in potential energy, so
W = q∆V = (1. 60 × 10 −^19 C)(7. 19 × 105 V − 0) = 1. 15 × 10 −^13 J
The cross-sectional area of the wire is A = πr^2 = π(0. 200 × 10 −^3 m)^2 = 1. 26 × 10 −^7 m^2
We are given R and the resistivity of copper, so we use
R = ρ
ρ
(5.00 Ω)(1. 26 × 10 −^7 m^2 ) (1. 72 × 10 −^8 Ω · m)
= 36.6 m
40.0 V
300 kW
500 kW
400 kW
30.0 V
20.0 V
(^) - +
i 1^ i^2
i 3
Write down (clearly, please) three equations that would allow us to solve for i 1 , i 2 and i 3. You don’t need to solve these equations for these currents. (8)
Using the currents as defined in the diagram, the junction rule (at either the top or bottom junction point) gives: i 1 + i 3 = i 2
Choosing a clockwise path round the left loop, the loop rule gives us
+40.0 V + 20.0 V − i 1 (300 kΩ) = 0
Choosing a counterclockwise path around the right loop, we get
+20.0 V − i 3 (400 kΩ) + 30.0 V − i 3 (500 kΩ) = 0
Actually these equations are quite easy to solve the for three currents. We get:
i 1 = 0.200 mA i 3 = 0.055 mA i 2 = 0.255 mA
Since the proton has a positive charge, we find from the right-hand rule for magnetic forces that the field must be pointing up (in the plane of the page) in order for there to be a force out of the page. To get the magnitude of the field (since v is perpendicular to B), use
F = qvB =⇒ B =
qv
(1. 6 × 10 −^19 C)(6. 5 × 10 6 m s )
You must show all your work and include the right units with your answers!
F = ma Fc =
mv^2 r
KE = 12 mv^2
F = k
|q 1 q 2 | r^2
k =
4 π 0
F = qE E = k
|q| r^2
Eplates =
σ 0
∆EPE = ∆Uelec = q∆V Es = −
∆s
V = k
q r
1 eV = 1. 602 × 10 −^19 J
q = CV C = 0
d
Energy = 12 CV 2 Cdiel = κCvac
V = IR R = ρ
Rser = R 1 +R 2...
Rpar
∑ Iin =
∑ Iout
∑ loop
V = 0 F = qvB sin θ
r =
mv |q|B
m =
( qr^2 2 V
) B^2 F = ILB sin θ τ = NIAB sin φ
k = 8. 99 × 10 9 N C·m 2 2 0 = 8. 854 × 10 −^12 NC·m^22 e = 1. 602 × 10 −^19 C A = πr^2