Physics Department @ Brock University

BROCK UNIVERSITY
Page 1 of 3
Final Exam: April 2010 Number of pages: 3
Course: PHYS 1P22/1P92 Number of students: 121
Examination date: 21 April 2010 Number of hours: 3
Time of Examination: 14:00 – 17:00
Instructor: S. D’Agostino
A formula sheet is included with the test paper. No other aids are permitted except for a
non-programmable, non-graphing calculator.
Communication (during the exam) with extraterrestrial beings or entities in extradimensional
hyperspaces is strictly forbidden. (Of course, such communications before the exam are
permitted and, indeed, encouraged.)
Solve all problems in the booklets provided.
Total number of marks: 50.
1. [10 marks] Clearly indicate whether each statement is TRUE or FALSE. Then provide
a BRIEF explanation (i.e. one or two sentences), including a clear and complete
correction of the statement if it is false. Your explanation may include formulas, if you
wish. Remember, brevity and clarity are courtesy.
(a) Although the total energy is conserved for all systems, the mechanical energy is
not conserved for a simple harmonic oscillator, because the oscillator sometimes
goes faster and sometimes goes slower.
(b) A wave is called a standing wave when it is a fixed pattern that does not oscillate.
(c) When car engine oil (which is black) is spilled into a puddle, the oil appears to
have many different colours; this is because mixing the oil with water causes the
oil to separate into its many different types of molecules.
(d) For a conductor in equilibrium, the electric field inside the conductor is zero,
regardless of the electric field outside the conductor.
(e) An insulator cannot be charged, because current cannot pass through an insulator.
(f) The electric field is constant along any equipotential surface.
(g) Electricity is produced at a generating station by rotating loops of wire in a
magnetic field.
(h) Light bulbs in a series circuit are not necessarily equally bright; it depends on
their resistances.
(i) Light bulbs in a parallel circuit are equally bright, regardless of their resistances.
(j) A charged particle in a magnetic field moves along a circular path.

PHYS 1P22/1P92 April 2010 Page 2 of 3
2. [5 marks] A 25 pF parallel-plate capacitor with an air gap between the plates is
connected to a 100 V battery.
(a) Calculate the charge on the positive plate.
(b) Repeat Part (a) if a Teflon slab (with a dielectric constant of 2.0) completely fills
the gap between the plates.
3. [5 marks] A 1.0 nC point charge is located 5 cm North of point A, and a −1.0 nC
point charge is located 5 cm South of point A. Calculate the strength and direction of
the electric field 5 cm East of point A.
4. [5 marks] The electric field strength is 20,000 V/m inside a parallel-plate capacitor
for which the distance between the plates is 1.0 mm. An electron is released from rest
at the negative plate. Calculate the electron’s speed when it reaches the positive plate.
5. [5 marks] Consider the circuit in the diagram.
23_FigureP59.pdf
(a) Calculate the current flowing through the 2.0 Ω resistor.
(b) Calculate the power dissipated by the 20 Ω resistor.
(c) Calculate the cost to run the circuit for 30 days if electricity costs 10 cents per
kWh.
6. [5 marks] A thin film of MgF 2 , which has an index of refraction of 1.38, coats a piece
of glass, which has an index of refraction of 1.50. Calculate the thinnest film that
results in destructive interference for the reflection of light of wavelength 500 nm.
7. [5 marks] A mass spectrometer has a constant magnetic field of 0.050 T. Particles of
charge +e are injected into the spectrometer with a speed of 2.5 ×10 5 m/s.
(a) Calculate the mass of the particles if each of them moves along a circular path of
radius 0.21 m.
(b) Calculate the strength of the electric field that must be introduced (while the magnetic
field remains the same) so that the particles injected into the spectrometer
will travel along a straight path.

PHYS 1P22/1P92 April 2010 Page 3 of 3
8. [5 marks] The circuit shown in the figure encloses an area of 25 cm 2 . The magnetic
field (indicated by the dots) increases steadily from 0 T to 0.50 T in 10 ms. Calculate
the current in the resistor during this time.
25_FigureP18.pdf
9. [5 marks] Two 50 g blocks are held 30 cm above a table, as shown in the figure. One
of them is just barely touching the top of a spring that is 30 cm long. The blocks are
released at the same time. The block on the left hits the table at exactly the same
time as the block on the right reaches the lowest point of its motion. Calculate the
value of the spring’s stiffness constant.
14_FigureP57.pdf

Potentially useful data and formulas
F = K |q 1| |q 2 |
r 2
U elec = qV
⃗ E =
⃗ F
q
E = ∆V
d
Q = C∆V parallel-plate capacitor: E = Q
ɛ 0 A
C = κɛ 0A
d
point charge: E = K |q|
r 2
point charge: V = K q r
E ′ = E κ
and
U C = 1 2 · Q2
C = 1 2 C(∆V )2 u E = U C
volume = 1 2 κɛ 0E 2
C = ɛ 0A
d
R = ρL A
I = ∆Q
∆t
∆V = IR
P = ∆U
∆t = I∆V = I2 R =
(∆V )2
R
Kirchhoff:
∑
Iin = ∑ I out
∆V loop = ∑ i
∆V i = 0
R eq = R 1 + R 2 + R 3 + · · · + R N R eq =
( 1
R 1
+ 1 R 2
+ 1 R 3
+ · · · + 1
R N
) −1
B = µ 0I
2πr
B = µ 0NI
2R
B = µ 0 I N L
(long wire)
(centre of loop with N turns)
(inside a solenoid)
⃗F = q⃗v × ⃗ B which is equivalent to F = qvB sin θ
F = ILB F = µ 0LI 1 I 2
2πd
Φ = AB cos θ
E = N
∆Φ
∣ ∆t ∣

kinetic energy = 1 2 mv2
gravitational potential energy = mgh
centripetal force = mv2
r
F = −kx x = A cos(2πft) v = −(2πf)A sin(2πft) a = −(2πf) 2 A cos(2πft)
U = 1 2 kx2
f = 1 T
f = 1
2π
√
k
m
v = fλ
n = c v
double-slit (bright):
θ m = m λ d
and
y m = m λL d
(dark): y m =
(
m + 1 ) λL
2 d
diffraction grating (bright): d sin θ m = mλ and y m = L tan θ m
thin films:
2t = m λ n
and 2t =
(
m + 1 ) λ
2 n
K = 8.99 × 10 9 N · m2
C 2
ɛ 0 = 8.85 × 10 −12 C 2
N · m 2
µ 0 = 1.26 × 10 −6 T · m
A
−11 N · m2
G = 6.67 × 10
kg 2
g = 9.80 m/s 2
mass of proton (and neutron) = 1.67 × 10 −27 kg
mass of electron = 9.11 × 10 −31 kg
fundamental unit of charge: e = 1.60 × 10 −19 C
speed of light in vacuum: c = 3.00 × 10 8 m/s

Solutions
1.(a) F. The speed changes because energy is continuously converted from kinetic to
potential and back again.
(b) F. A standing wave is a wave confined between barriers. The reflections of the wave
from the barriers interfere to create the pattern of nodes and oscillating anti-nodes.
(c) F. The oil spreads out over the water in a thin film; however, the thickness of the film
varies. Light incident on the film is reflected if it has the right wavelength, which depends on
the thickness of the film. Thus, regions of the film with different thicknesses reflect different
wavelengths of light, which means that different regions of the film appear to be a different
colour.
(d) T. Because conductors have many ”free” electrons, the free electrons in a conductor
will move in response to the applied electric field. In an extremely short time, some free
electrons will arrange themselves on the surface of the conductor so that the net electric
field inside the conductor (the superposition of the applied field and the field due to the
rearranged charges in the conductor) is zero.
(e) F. For example, rubbing your feet on a carpet on a dry winter day transfers charge
between you and the carpet (the latter is an insulator). Although charges cannot easily pass
through an insulator, charge can be transferred onto the surface of an insulator.
(f) F. The electric potential is constant along an equipotential surface. The magnitude
of the electric field depends on the rate of change of the electric potential, and so can be
different at different points along an equipotential surface.
(g) T. The idea behind this is called Faraday’s law of induction. The voltage induced in
the loop of wire (and therefore the current, as well) depends on the orientation and size of
the loop, the number of turns of wire, the strength of the magnetic field, and the rate at
which the loop turns.
(h) T. The current through each element of the series circuit is the same, but the brightness
of each bulb depends on the power dissipated, which is I 2 R. Therefore the brightness
of each bulb depends on its resistance.
(i) F. The voltage across each branch of a parallel circuit is the same, but the brightness
of each bulb depends on the power dissipated, which is V 2 /R, and therefore depends on the
resistance of each bulb.
(j) F. It depends on the initial direction of the charged particle, and it also depends on
whether the magnetic field is constant or not. Under certain circumstances the path can be
circular, but it also might be a straight line, a spiral, or a more complicated path.
2.(a) Q = CV = 25 pF × 100 V = 2500 pC = 2.5 nC
(b) The capacitance of the capacitor is now 2.0 × 25 pF = 50 pF, so the charge on each
plate is now Q = CV = 50 pF × 100 V = 5000 pC = 5.0 nC
3. 2545 V/m South
4. The electron “falls” through a potential difference of 20, 000 × 0.001 = 20 V. Thus,
the electron loses potential energy of e∆V = 1.6 × 10 −19 × 20 = 3.2 × 10 −18 J. The increase

in the electron’s kinetic energy is equal to its loss of potential energy, so:
1
2 mv2 − 1 2 mv2 0 = 3.2 × 10 −18
1
2 mv2 − 1 2 m(0)2 = 3.2 × 10 −18
mv 2 = 6.4 × 10 −18
v 2 6.4 × 10−18
=
√
9.11 × 10 −31
6.4 × 10 −18
v =
9.11 × 10 −31
= 2.651 × 10 6
Thus, the final speed of the electron is 2.7 × 10 6 m/s.
5. (a) The resistance R of the parallel part of the circuit satisfies
1
R = 1 20 + 1 5
= 1 20 + 4 20
= 5 20
= 1 4
R = 4 Ω
Thus, the total resistance of the circuit is 2 + 4 + 4 = 10 Ω. The current flowing from
the “battery,” all of which passes through the 2-Ω resistor, is therefore
I = V R
= 100
10
= 10 A
(b) The total current that flows in the parallel part of the circuit is I = 10 A. Let I 1
represent the current flowing through the 20-Ω resistor, and let I 2 represent the current
flowing through the 5-Ω resistor. Then
I 1 + I 2 = 10 A
I 2 = 10 − I 1
Also, because the voltage drop across each branch of the parallel part of the circuit is

the same,
I 1 R 1 = I 2 R 2
20I 1 = 5I 2
20I 1 = 5 (10 − I 1 )
20I 1 = 50 − 5I 1
25I 1 = 50
I 1 = 2 A
The power dissipated by the 20-Ω resistor is therefore
P = I 2 1R 1
= 2 2 × 20
= 80 W
(c) From Part (a), the total resistance of the circuit is 10 Ω and the total current leaving
the “battery” is 10 A. Therefore, the total power dissipated by the circuit is P = I 2 R =
10 2 × 10 = 1000 W.
The total energy E dissipated by the circuit in 30 days is
E = P · t
= 1000 W · 30 days
= 1 kW · 30 × 24 hours
= 720 kWh
Therefore, the total cost to run the circuit for 30 days is 0.1 × 720 = $72.
6. For destructive interference with this configuration of indices of refraction, the thickness
t of the film satisfies
(
2t = m + 1 ) λ
2 n
The thinnest film has the minimum value of m+1/2; that is, m = 0. Therefore, the thickness
of the thinnest film is
t = λ
4n
= 500
(4)(1.38)
= 90.6 nm
7. (a) Because each particle moves in a circle of radius r, the acceleration (centripetal) of
each particle is v 2 /r. Because the path of each particle is circular, the particles are injected
perpendicular to the direction of the magnetic field. Therefore, the magnitude of the force

acting on each particle is qvB. By Newton’s second law of motion,
F = ma
qvB = mv2
r
m = qvBr
v 2
= qBr
v
= 1.6 × 10−19 (0.05)(0.21)
2.5 × 10 5
= 6.7 × 10 −27 kg
(b) In this situation, if each particle is to move in a straight line, then the net force on
each must be zero. This means that the magnitude of the electric force must be the same
as the magnitude of the magnetic force (their directions are opposite, but we don’t have to
worry about that in the following calculation). Thus,
qE = qvB
E = vB
= ( 2.5 × 10 5) (0.05)
= 12, 500 V/m
8. By Faraday’s law of induction, the emf induced in the circuit by the changing magnetic
field is
E = ∆Φ
∆t = BA
∆t
= (0.5)(0.0025)
0.01
= 0.125 V
By Lenz’s law, the direction of the induced emf is the same as the emf induced by the battery.
Thus, the net emf in the circuit is 9+0.125 = 9.125 V. The current in the resistor is therefore
I = V R
= 9.125
20
= 0.456 A
9. The time t needed for the block on the left to hit the table satisfies
∆y = 1 2 gt2
t 2 = 2∆y
g
= (2)(0.3)
9.8
= 0.06122
t = 0.2474 s

This time represents half a cycle for the block on the right; thus, the period of the block
on the right is
T = 2(0.2474) = 0.495 s
Because the angular frequency of oscillation ω satisfies ω 2 = k/m, the spring’s stiffness
constant is
k = mω 2
= 4π2 m
T 2
= 4π2 (0.05)
0.495 2
= 8.1 N/m