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PHYSICS 222 SI FINAL
EXAM REVIEW
WITH SI LEADER: ROSALIE 
ELECTROSTATICS THEORY
• Two point charges, +q and (unknown) Q produce the net electric field shown
at point p in the figure. What is the charge Q in terms of q?
SOLUTION
• Q = -q
If the magnitude of Q were any greater or less than q, the vector E would have
some force in the x direction. It doesn’t, so we know the magnitude of Q must be
the same as q. We also know that the charge of Q must be opposite, to push
the particle the same way that q pulls it.
ELECTROSTATICS THEORY
• Two uncharged metal spheres are touching. A third, positively charged sphere
is placed to the left of sphere 1. With this third sphere still in place, spheres 1
and 2 are moved away from each other so that they are no longer touching,
and the third sphere is removed. What are the charges on spheres 1 and 2?
1
3
2
SOLUTION
• Sphere 1 is negatively charged and sphere 2 is positively charged. (Try to
draw a diagram, it helps!)
CONDUCTORS
• An uncharged conducto has a hollow cavity inside of it. Within this cavity,
there is a charge of +10 C that does not touch the conductor. There are no
other charges in the vicinity. What are the charges of the inner and outer
surfaces of the conductor?
SOLUTION
• The charge on the inner surface is -10 C and the charge on the outer surface
is 10 C (A diagram helps on this one as well)
ELECTROSTATICS - THEORY
• The figure shows three chargers labeled 𝑄1 , 𝑄2 , 𝑎𝑛𝑑 𝑄3, and electric field
lines in the region surrounding the charges. What are the signs of these
charges?
SOLUTION
• 𝑄1 is positive
• 𝑄2 is negative
• 𝑄3 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒
• Electric field lines start at positive charges and end at negative charges.
ELECTROSTATICS -1
• A solid metal ball of radius 2.0 cm carries a total charge of -0.1 μC. What is
the magnitude of the electric field at a distance from the ball center of
1.0cm?
ELECTROSTATICS – 1 SOLUTION
•0
• Conductors carry charge on their surface. A charge inside of a conductor is 0.
ELECTROSTATICS
When two point charges are a distance d part, the electric force that each one
feels from the other has magnitude F. In order to make this force twice as strong,
what would the distance have to be changed to?
ELECTROSTATICS SOLUTION
𝑑
2
If 𝐹 =
𝑘𝑞2
,
𝑑2
then in order to make 𝑑 2 one half of its original value, it is
necessary to divide d by 2
FLUX AND GAUSS LAW
• A point charge 𝑞1 = 3.8 𝑛𝐶 is located on the x-axis at x = 2.1 m, and a
second point charge 𝑞2 = −6.2 𝑛𝐶 is on the y- axis at y = 1.15 m. What is
the total electric flux due to these two point charges through a spherical
surface centered at the origin with radius r = 1.55 m?
SOLUTION
• Charge 1 isn’t enclosed in the sphere, but charge 2 is. The equation for flux is:
Φ=
𝑞
𝜀0
=
−6.2 × 10−9
8.85 × 10−12
=
𝑁𝑚2
700
𝐶
MAGNETIC FIELDS
• A small metal rod has a mass of 0.1 kg and it lies in a magnetic field of 1 T. If
the rod is 1 m long and carries a current, what current is necessary so that the
net force on the rod is zero?
SOLUTION
• F = mg = IL x B  I x (1m X 1T) = .1kg x 9.8 m/s^2 solve for I = 98 A
MAGNETIC FIELD
• A 1 m by 1 m circuit containing only a 50 Ω
resistor lies perpendicular in a
magnetic field. The magnetic field changes from 5 T to 1 T in 30 ms. What is
the magnitude of the induced current?
SOLUTION
• Induced emf is ε = -dΦ/dt = -AdB/dt = (1m2)(4T) / (30 ms) = 133.33 V
which is equal to IR so current is 2.66 A
FLUX AND GAUSS LAW
• What is the net flux through a cube of side 10 cm if a + 2 C
placed 20 cm from the center of the cube?
point charge is
SOLUTION
•0
There is no charge inside of the cube, so there is no flux.
MAGNETIC FIELD
•
An electron moving with a velocity = 5.0 × 107 m/s enters a region of space
where perpendicular electric and magnetic fields are present. The electric
field is E = 1j. What magnetic field will allow the electron to go through the
region without being deflected?
SOLUTION
•
If the electron is to keep the same velocity and path, the force from the magnetic
field must be the same as the force from the electric field so:
F = qE = qv x B
-1.6 x 10−19 N j = 8 x 10−12 i x B
-1.6 x 10−19 N j = (0 – 0)i + (0 – (8 x 10−12 )(𝐵𝑧 ))j + ((8 x 10−12 ) 𝐵𝑦 – 0)k
-1.6 x 10−19 N j = - (8 x 10−12 )(𝐵𝑧 )j
B = 2 x 10−8 T k
CIRCUITRY -1
Consider a fully-charged parallel plate capacitor. The capacitor is far from other
objects and is not connected to a battery. The plates are now being pulled apart.
What happens to the potential difference between the plates as they are being
separated?
a)
b)
c)
d)
e)
It decreases
It increases
It remains constant
It is zero
Cannot be determined from given information
CIRCUITRY 1 – SOLUTION
• B) It increases!
• ∆V = Ed
Where d is the distance apart. As d increases, so does ∆V!
CIRCUITRY 2
Two resistors of 15 and 30 Ω are connected in parallel. If the combination is
connected in series with a 9.0-V battery and a 20-Ω resistor, what is the current
through the 15-Ω resistor?
CIRCUITRY 2 - SOLUTION
Use the loop rule (the sum of potential drops
around any loop is 0) around the left loop and
the outer loop
• I = .2 A
20 Ω
15 Ω
9v
30 Ω
Use the junction rule (the sum of all currents
entering a junction must be the same as he sum
of all current exiting the junction) on the upper
junction
9𝑉 = 20Ω𝐼1 + 15Ω𝐼2
9𝑉 = 20Ω𝐼1 + 30Ω𝐼3
𝐼1 = 𝐼2 + 𝐼3
Solve these as a system of equations for 𝐼1 , then
use 𝐼1 to find 𝐼2 = .2 A
LR CIRCUIT
• A series LR circuit consists of a 2.0-H inductor with negligible internal
resistance, a 100-ohm resistor, an open switch, and a 9.0-V ideal power
source. After the switch is closed, what is the maximum power delivered by the
power supply?
SOLUTION
𝑃 = 𝐼𝑉 =
𝑉2
𝑅
=
81
𝑊
100
= .81𝑊. Since the question asks about the maximum
power, the current will be maximal (and constant) and the inductor will just act like
a piece of wire.
LRC CIRCUIT
In a series LRC circuit, the frequency at which the circuit is at resonance is 𝑓0 . If
you double the resistance, the inductance, the capacitance, and the voltage
amplitude of the ac source, what is the new resonance frequency?
SOLUTION
• The circuit is at resonance when 𝜔 =
1
𝐿𝐶

1
1

2𝐿2𝐶
2 𝐿𝐶
=
𝜔
2
POWER
• If you have 2 circuits each connected to a 120 V power source and each
containing 3 bulbs of resistance 20 Ω (one circuit has the three bulbs in
parallel and the other has the bulbs in series), which bulbs will glow the
brightest? What is the power delivered to each bulb?
• 𝑃 = 𝐼𝑅2
First, solve for the current going through each bulb
in the top circuit (Kirchoff’s laws are handy for this):
120 V
SOLUTION
120V = (I)(20Ω) I = 6 A
Plug this into out power equation: 𝑃𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 =
(6𝐴)(20)2 = 2400 W
V = IR  120 = (I)(60)  I = 2 A
𝑃𝑠𝑒𝑟𝑖𝑒𝑠 = (2𝐴)(20)2 = 800 W
So: 𝑃𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 > 𝑃𝑠𝑒𝑟𝑖𝑒𝑠 and the series bulbs glow
the brightest!
120 V
Now find the power in the second loop: find the
current through the circuit: 𝑅 = 𝑅1 + 𝑅2 + 𝑅3 =
60 Ω
AC CIRCUITS
• Which one of the phasor diagrams shown below best represents a series LRC
circuit driven at resonance?
AC CIRCUIT SOLUTION
• At resonance, 𝑣𝐿 = −𝑣𝑐 so it must be #3
ELECTROMAGNETIC WAVES
• An electromagnetic wave with frequency 64 Hz travels in an insulating
magnetic material that has dielectric constant 3.64 and relative permeability
5.18. At this frequency, the electric field has amplitude 7.2 V/m. Determine
the speed of propagation of the wave, the wavelength, and the amplitude of
the magnetic field.
SOLUTION
𝑐
•𝑣=
𝑘𝑘𝑚
•λ=
𝑣
𝑓
•𝐵=
𝐸
𝑐
= 9.91(107 )
= 1.06 106
=
𝐸
λ𝑓
= 1.04 10−6 𝑇
LIGHT WAVES
• A laser emits 1.00 mW of average light power in a cylindrical beam of
radius 5.00 mm. The intensity of the light at a point in the beam is ____
W/m2
SOLUTION
•
𝐴 = 𝜋𝑟 2 = 7.85 × 10−5 𝑚2
• 𝐼𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑃𝑜𝑤𝑒𝑟
𝑎𝑟𝑒𝑎
=
0.001 𝑊
7.85×10−5 𝑚2
= 12.7
𝑊
𝑚2
POLARIZATION
A beam of unpolarized light of intensity I0 passes through a series of ideal
polarizing filters with their polarizing directions turned to various angles as
shown in the figure. What are the intensities at points a and b?
SOLUTION
•𝐼=
𝐼0 𝑐𝑜𝑠 2 𝜃 and natural light going through a single filter is reduced to
one half its original value
• 𝐼𝑎 =
𝐼0
2
• 𝐼𝑏 =
𝐼0
𝑐𝑜𝑠 2 60
2
=
𝐼
8
LAB QUESTION
•
In experiment RR – Reflection and refraction, you used the set up shown in the figure
below. A monochromatic laser beam is incident onto the flat side of a semi-circular
prism, right at its center. The orientation angles of the flat side of the prism (α) and
of the refracted ray (β) are measured relative to the direction of the incident ray.
For the first orientation you try, α = 37° and β = 155°. For a good measurement of
the index of refraction, we will have to get many more measurements, but we can use
the first one to obtain a rough estimate. Based on this first measurement, what is the
index of refraction of the material?
SOLUTION
• Snells law: 𝑛1 𝑠𝑖𝑛θ1 = 𝑛2𝑠𝑖𝑛θ2
• Since the medium for this system is air, 𝑛1 = 1.
θ1
•
Angles 1 and 2 are the angles
θ2
between the incoming and refracted
ray and the normal to the surface
So:
θ1 + α = 90°  θ1 = 53°
θ2 + α + 90° = β  θ2 = 28°
(1)𝑠𝑖𝑛53 = 𝑛2 𝑠𝑖𝑛28
𝑛2 = 1.7
MIRROR
A dentist uses a curved mirror to view teeth on the upper side of the mouth.
Suppose she wants an erect image with a magnification of 2.00 when the mirror
is 1.25 cm from a tooth. (Treat this problem as though the object and image lie
along a straight line.) Determine what kind of mirror is needed using a ray
diagram, and determine the focal length and radius of curvature of the mirror
SOLUTION
•𝑚=
• 2=
•
1
1.25
𝑠′ 1
− ,
𝑠 𝑠
𝑠′
−
1.25
1
−
2.5
+
1
𝑠′
=
1
𝑓
→ 𝑠 ′ = −2.5
=
1
𝑓
→ 𝑓 = 2.5
LENSES
• Draw a lens diagram for the following lenses and define the images:
Converging lens
Diverging lens
LENSES SOLUTION
• Draw a lens diagram for the following lenses and define the images:
Real, inverted, diminished
Imaginary, upright, diminished
SIGN CONVENTIONS FOR LENSES
• Object distance (s) is positive if it is on the same side as incoming light –
otherwise negative (the left side)
• Image distance (s’) is positive if on the same side as outgoing light – otherwise
negative (the right side)
• Radius of curvature (R) is positive if the center of curvature is on the same side
as outgoing light, otherwise negative. (the center of the circle is on the right
side)
LENS PROBLEM
Consider an object with s = 12 cm that produces an image with s’ = 15 cm.
• Find the focal length
• Tell whether the image is real or virtual
SOLUTION
• 1𝑠 + 𝑠′1 = 𝑓1
→
1
12
+
1
15
=
1
𝑓
 f = 6.67 cm
• The image is real because s’ is positive
LENS
A converging lens has a focal length f. An object is placed at a distance between f and
2f on a line perpendicular to the center of the lens. The image formed is located at
what distance from the lens?
•
•
•
•
•
A) between the lens and f
f
2f
Farther than 2f
Between f and 2f
SOLUTION
• Larger than 2f
• Draw a lens diagram
LENSES
• A glass converging lens has one flat side and one side with a radius of
curvature of 20.0 cm. The focal length of the lens is ____ cm. (The index of
refraction of the glass in the lens is 1.50)
SOLUTION
• 𝑓1 = (𝑛 − 1)(𝑅1
1
• f = 40 cm
−
1
)
𝑅2
→ (.5)(
1
20
1
∞
− )
2 SLIT
• In a double slit experiment, if the slit separation is increased, which of the
following happens to the interference pattern?
SOLUTION
• Minima locations are given by 𝑦𝑚 =
𝑚𝑅λ
𝑑
so if d increases, y decreases
meaning that the minima get closer together.
RELATIVITY
• A man stands in the exact center of a glass freight car moving at very high
speed. He holds a flashlight in each hand and directs them at opposite walls,
switching them on at the same time. You stand still outside of the freight train
and watch it go by. In the mans frame of reference, which beam of light
travels faster? In your frame of reference which beam of light travels faster?
In the mans frame of reference which wall is struck by light first? In your fram
of reference which wall is struck by light first?
SOLUTION
• Light always moves at the same speed, c, no matter what the frame of
reference is. So we see the light moving at the same speed as the man does.
• The man sees the light hitting both walls at the same time. However, we see the
light hit the rear wall first, because the rear wall moves “towards” the beam
of light, while the front wall moves “away” from the light. Isn’t relativity fun?
RELATIVITY
• A star is moving towards the earth with a speed at 90% of the speed of light.
It emits light, which moves away from the star in every direction at the speed
of light, c. Relative to us on earth, what is the speed of light moving toward us
from the star?
SOLUTION
• C. Light always moves at the speed of light, no matter what the reference
frame is.
HEISENBERG UNCERTAINTY
• If the accuracy in measuring the position of a particle increases, the accuracy
in measuring its velocity will do what?
• A) increase
• B) remain the same
• C) decrease
• D) it is impossible to say since the two measurements are unrelated and do not
affect each other
SOLUTION
• It decreases
Heisenberg’s uncertainty principle tells us that ∆𝑥∆𝑝 ≥
ħ
2
This means that if the accuracy in x increases, the opposite must be true of p.
QUESTIONS?