# Midterm review_

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```					                                Midterm review,
(there are only 6 more classes till the midterm)
How large a mass needs to be added to a pendulum to increase its period from
2s to 4s?

Solution: The period of a pendulum is independent of mass so it is impossible to
increase it's period by adding mass.

Let's try that again... If the pendulum has a length of L and a period of T, what
multiple of L will make a pendulum with a period of 2T?

To find T:     T =2 
   L
g
To find 2T: 2T= 2 2
   L
g

Rearrange 2T... bring the 2 inside the root sign (square it on the way in)

To find 2T: 2T=2 
   4L 
g
Now you can see that you would need a length
of 4L to a period of 2T.
Remember....Ray tracing for mirrors.

First we can use ray tracing to find out what the image will look like
Ray tracing is a relatively simple process if you can draw straight lines.

We need two lines from the object to determine the image location,
(a third one is useful as a check on the first two though)

The lines we can draw are:
1. From the object, through the center of curvature of the mirror,
then back along the same line.

2. From the object, through the focus of the mirror, then out
parallel to the axis at the point it touched the mirror.

3. From the object parallel to the axis then back out through the
focus.

4. From the object to the center of the mirror then out on a line
with an equal angle to the angle of incidence.
Now let's look at ray tracing for lenses

Ray tracing for lenses is only slightly different

We need two lines from the object to determine the image location.
A third acts as a check.

The three lines we can draw are:

1. From the object, through the center of the len and straight out

2. From the object, through the focal point to the lens then out
parallel to the axis.

3. From the object to the lens parallel to the axis, then out through
the focal point.
Important measures for lenses and mirrors*

Object distance
(s or p or do)
Image distance
(s' or q or di)

Object
height      object         f
(h or ho)
Image
Height
(h' or hi )
image
Focal length
(f)

* remember that f=C/2 … or radius of curvature /2
A few more definitions
(Sign conventions really)
Object distance (p) – distance from object to lens.
+ve if object is in front
-ve if object is behind
For mirrors this is always +ve since if the object is behind
the mirror there is no image formed

Image distance (q) – distance from image to lens.
For mirrors: +ve if image is in front of the mirror
-ve if image is behind the mirror
Watch out here!
For lenses: +ve if image is behind the lens
-ve if image is in front of the lens

Object height (h) – always +ve

Image height (h') – +ve if image is erect
-ve if object is inverted
Now that we've all agreed on the signs, we can use some
mathematics and get the same answers

R

c

Note that for a concave mirror R is always +ve
while for a convex mirror R is always -ve
Now that we've all agreed on the signs, we can use some
mathematics and get the same answers

R

c

The Mirror Equation   Magnification

1+1=2             M = h' = -q
p q R                 h     p
Example 1.

A convex spherical mirror with a radius of curvature equal to 10 cm is placed
so that the image is 1/3 the size of the object and behind the mirror.
Find the object and image distances.

What we know:
R= -10cm (Convex Mirror)
M=1/3= (-q)/(p)
So p = -3q also q is -ve
c
and
1+1=2
p q R

Putting in some numbers:         Algebra:             Answer:
1 - 1=      2                     2 = -1              q = (-10/3) cm
q 3q -10                         3q     5

Convex mirrors will always produce images that are virtual, erect, and smaller.
So the image distance will always be negative for a convex mirror.
Example 2 - The real image of a concave mirror is observed to be 4 times larger than
the object when the object is 30 cm from the front of the mirror.
What is the radius of curvature of the mirror?

What we know: q> 0; p =30cm (Since they are both in front of the mirror)
M = -q = -4
p

Using the mirror equation       Put the numbers in:        Do the math:
1+1=2                           1+ 1      =2               5R=240cm
p q R                           30 4(30)   R               R=48cm
Midterm review,
(there are only 6 more classes till the midterm)
A piano string needs to be tuned to a fundamental of 375 Hz. It currently has a
fundamental frequency of 370Hz, an effective length of 2m and a tension of 1kN.
How much should the piano tuner change the tension?
For the fundamental note the wavelength is 2L, now we know a frequency and a
wavelength, we can find velocities.(V=fλ)

Current velocity:    Needed velocity:          There are many ways to solve
this from here (I like this one
vo = (370)(4)        v = (375)(4)              'cause it's pretty...)
= 1480m/s            = 1500m/s

v=
FT
  
1500
1480
=

1000
1000

 
2
F T 
 
1500
=
1000
v
v0
=
FT


    =
F T 
FT
1480

2
1000

                            =1000  
1500
1480
−1000=27.2N increase
Equations for both mirrors and lenses                      Notice the
negative sign !
'
1 1 1 2
 = =
p q f  R                                      h −q
M= =
h  p
Object height (h) is always positive   Image height (h') is +ve if the image is upright
Image height (h') is -ve if the image is inverted
(fancy way of saying upside-down)

Signs for Mirrors                    Object distance (p) -
Signs for Lenses
R is -ve for convex mirrors
+ve if object is in front of lens
R is +ve for concave mirrors                   - ve if object is behind the lens

Object distance (p) -                          Image distance (q) -
Always +ve
Image distance (q) -                           +ve if image is behind the lens
(or if image is real)
+ve if image is in front of mirror         - ve if image is in front of the lens
(or if image is real)                      (or if image is virtual)
- ve if image is behind the mirror
(or if image is virtual)
What do we know about electric charges?

We know they come in two flavors positive and negative.

We know that like charges repel and opposite charges attract

The SI unit of charge is a Coulomb abbreviated by C

There is a fundamental unit of charge denoted e = 1.60219x10-19 C
Charge only comes in integer multiples of e: ie +/- e, +/- 2e, +/-3e

*Every electron has a charge of exactly -1e and every
proton has a charge of exactly +1e.

*This must have some great cosmic relevance but no one has figured it out to date
Copper has one valence electron per atom, so how many Coulombs of charge
does 1g (~ 1/63.5mol) of Cu have in its valence electrons?

Charge in 1g of copper = 1523.97 C

What can we compare this to?

An average lightning strike carries around 5 C,
Very large bolts can carry up to 350 C

A lightning bolt can heat the air it passes through to
~20,000 °C (36,000 °F) - about three times the
temperature at the surface of the Sun.

Moral of the story... A Coulomb is a VERY large amount of charge.
Matter contains a mind boggling amount of charge.
Since like charges repel each other (and unlike charges attract) we would like a
way to describe the how well they do this. Fortunately people have already
worked this out for us.

Coulomb's Law
⋅
∣q1∣ ∣q2∣
F 1,2=k                2
r
k= Coulomb's constant= 8.9875 x 10 9 Nm2/C2 (in SI units)

r= the length of the line joining the two charges.

The force is repulsive if the two charges have the same sign and attractive if
the charges have opposite signs. (pictures are VERY important here)
An (unrealistic) example of Coulomb's Law.

Two opposite charges of |1C| each are placed on separate towers 1 km apart (one at the
Ian Stewart building and one over our heads). What is the force exerted on one of the
charges by the other?

F= 8987 N ( this is ~ equivalent to the weight of a car pulling the charge )

Forces are forces... no matter what causes them.
... These means we are back to vectors again.

Now for the good news...

If there is more than one charge acting on a point we can calculate the
force from each charge and just add up the resultant.
A +5nC charge is placed at the origin, a +6nC charge is place 0.3m horizontally, and
a -3nC charge is placed 0.1m vertically down. What is the net force (magnitude and
direction) on the charge at the origin?

Magnitude    F = 1.38x10-5 N
Direction    θ = 77.5o below horizontal to the left

Similarly we can find the force on the 6nC charge

Magnitude     F = 1.55x10-6 N
Direction     θ = 19.3o below horizontal to the right
A copper rod stuck into the ground and a copper rod held by
a glass post are both struck by lightning. Which one can you
go over and touch right away?

The one stuck in the ground (assuming it isn't a smoking pile of
slag).
We would call this rod grounded and because copper is a
conductor the excess electrons from the lightning bolt can
quickly find their way to the ground.

Since the Earth is huge we can take it to be an infinite sink for (or a
source of) electrons. No amount of electrons we put in (or take out)
are going to make the slightest difference to it.
So what about the rod held by the glass post?

The rod is still a conductor but glass is an insulator. Insulators
make it hard for the electrons to move. If you were to grab this
rod your body, being mostly salt water which is an excellent
conductor by comparison to glass, would instantly become the
express way for the excess electrons.

(This is only good if you're the electron.)
We can classify materials according to their ability to conduct charges

Conductors – any material in which electrons are relatively free to move about.
(metals are the classic example of a conductor particularly Cu, Ag, Al, Au)

Insulators – any material in which electrons are not really free to move about.
(typical examples are glass, plastic, ceramics, or wood)

Semiconductors – any material in which electrons are somewhat free to move. These
materials can have their properties drastically altered by the addition
of small amounts of impurities (which is why we love them)
(typical examples are Si, GaAs, Ge)

Superconductors – any material in which electrons are completely free to move.
Being completely free means these materials exhibit no resistance.
To date no one has developed a superconductor that can work at room
temperature, the record is HgBa2CuO4 (134K) (room temp ~300K).
(typical examples are Nb (9.26K), Pb (7.19K), and YBCO (93K) )
How can we charge something (without the lightning)?

Polarization – When a charged object is brought close to a neutral object the charges
in the neutral object will rearrange themselves so that the opposite charge
is brought closer to the side with the charged object.
The neutral object is now said to be polarized. It works for both insulators
and conductors but the effect lasts only as long as the charged object
remains nearby.

Induction –   If a conductor is polarized by bringing a charge close to the left and the right
side is connected to ground, electrons will leave (or come) to neutralize the
charge on the right. If the grounding wire is removed, and the charge is
removed the conductor will have a resulting net charge. The net charge is
an induced charge.
Example-1 A Hydrogen atom consists of an electron orbiting a proton, (Bohr model), at a
What is the force of electrostatic attraction between the electron and proton.

We can use the force equation again.

F= kqq where q is the charge on the electron (or proton).
r2
= (9x109)(1.6x10-19)2
(5.3x10-11)2

By the way... the force of gravity between the electron and proton is ~3x10-47N

Another good question... since this force is attractive,
what keeps the electron from crashing into the proton?
Example 2 A charged sphere is connected to a spring with a 3kN/m spring constant. A
plate with an equal but opposite charge is brought close to the sphere. In the new
equilibrium the spring is stretched 2cm and the final distance between the plate and the
sphere is 1cm. What is the charge on the sphere?

First we need to find the force pulling the sphere back.

F = k*x
=3000N/m*0.02m
=60N

Second we need to equate that force to the attractive force between the plate and sphere.

60N = kq1q2 = 8.99x109Nm2/C2 (q)(q)
r2            ( 0.01m)2
q2 = 6.66x10-13
`
q = 8.16x10-7C
q = 0.82μC
Imagine you are studying a mountain field ecosystem. One thing you might be interested
would be the temperature at various points around the field.

10          12           12           11           9

12          10           11           13           12

13          14           12           14           13

15          16           14           15           15

21          19           18           18           20

Sampling points like this makes a scalar field
(it is scalar because each point has a number (the temperature), it is a field because
we have sampled the temperatrue at a number of different points in space.)
Another thing you might be interested in would be the wind at various points around the field.
Since the wind is a velocity it has a direction. Here the arrows show the direction of the wind.

3            2            2            1            4

2            1            1            3            2

3            4            2            4            3

5            6            4            2            5

1            1            1            1            0

Sampling vectors like this makes a vector field. (Now we we have a
vector associated with each point in the space we sampled)
Now replace the wind measurements with electric force measurements.

3            2             2            1             4

2            1             1            3             2

3            4             2            4             3

5            6             4            2             5

1            1             1            1             0

This makes a vector field of electric forces. This is called the electric field.
Now the formal definition:

We define an electric field, E, at a point by looking at the force on a small positive test
charge at that point.

If q0 is a small positive charge then the force on that charge is
F= kq0Q where Q is the total charge seen by the test charge.
r2

Then: E=   F = kQ         Note: E is a vector with units of N/C
q0 r2                and directed the same way as the force.

Small in this sense means that the test charge is insignificant compared to the field.
(When we measure the temperature of something, we don't worry about how much heat
Field lines
As an aid to visualizing the electric field we can draw field lines.
(these are imaginary lines that show us how a charge will move, the same as
an ocean current on a map. There are no lines in the ocean telling the water
which way to go)

But just to be annoying I am going to make up rules about it
Rules for drawing electric field lines

1. Field lines go away from positive charges and toward negative charges.

2. The number of lines leaving a +ve charge or going to a -ve charge is proportional
to the magnitude of the charge.

3. Field lines go straight to (from) infinity unless another field line deflects it.

4. Field lines can never cross.
Some examples of field line sketches

+                                -

+                  -
How about an example using the electric field

An electron, in a uniform electric field, accelerates from rest to 3x106m/s after 2mm.
What is the strength of the electric field? (e- mass = 9.11x10-31kg... charge= 1.6x10-19C)

So a=v2/2d
a= (3x106m/s)2
2(2x10-3m)
= 2.25x1015m/s2
Next
F= ma
= (9.11x10-31kg)(2.25x1015m/s2)
= 2.05x10-15N

Finally
E=F/q
= (9.11x10-31kg)(2.25x1015m/s2)
= 2.05x10-15N/1.6x10-19C
= 1.28x104N/C
Example - 4 A small 10g with a charge of 0.001C is traveling with a velocity of 300m/s.
A uniform electric field brings the mass to rest after 10m.
What is the strength of the electric field?

First we need to find the deceleration of the mass.

a=-v2/2d
=-90000/20
=4500m/s2
Second we need to find the force on the mass.

F= (0.010kg)(4500m/s2) = 45N

Finally we can find the electric field.

E= F/q = (45N)/(0.001C) = 45 000N/C
Example-5 Three charges (in μC) are placed at the three corners
of a rectangle as shown. Find the value of the electric field at the
point A.

By virtue of the superposition principle
-4      the electric field at any point is the sum
4
of all the component fields so...

E= kq
20cm                                            r2
E+ = (9x109(4x10-6)2 = 900N/C
(0.20)2
A                         4
20cm
E- = (9x109(-4x10-6)2 = -450N/C
(0.20)2+(0.20)2

E= 8.2x105N/C @ 225o (away from the negative charge)
If one charge (+5μC) is placed at the origin and a second charge of (+7μC) at x=100cm,
where can a third charge be placed so that it has no net electrostatic force on it?

100cm

x                     100cm-x
+5μC                                                 +7μC

We can use the force equation.

F= kqq
r2
0 = (8.99x109)(5x10-6)q - (8.99x109)(7x10-6)q
(x)2                 (1-x)2

Do some fancy algebra and...         5 (1-x)2 - 7 (x)2 = 0
2 x2+10x – 5 = 0
x= 0.46m, -5.46m

Note that we never specified how big q was...
So this question is the same as asking where is the E-field =0
Example- 4 Two small balls of 10g each, both with a charge of 55nC are suspended
by strings. If the balls are pushed 5cm apart at equilibrium what angle does each ball
make with the vertical?

A sketch here is definitely useful Looking at the left hand ball we can identify 3 forces
F of gravity; F of tension; F of charges
In equilibrium the forces must all balance.

In y direction: mg = T Cos(θ)
θ                                      (0.01)(9.81)= 0.0981N = T Cos(θ)

In x direction: Fch = kqq   =T Sin(θ)
T                                                   r2
-9 2
=(9x109)(55x10 ) =T Sin(θ)
0.052
charge                                       T Sin(θ)=0.0109N
mg

Finally: T Sin (θ)=0.0109N so Tan(θ)= 0.111 θ=6.34 degrees
T Cos(θ) 0.0981N
With gravity                         With Electrostatics

F = mg                               F = qE
PE = mgd                              PE = qEd

Both of these forces are called “conservative forces.”
This means that work done is equal to the negative change in potential energy
W = -ΔPE = PEf- PEi
W = qEdf- qEdi = -qEΔd

If work is done on a system then W is -ve, if the system does the work then W is +ve

A useful trick for electrostatics is to divide everything by q.

W = qEdf- qEdi = -qEΔd
q       q           q
Which leaves:
W = Edf- Edi = -EΔd         This is the work per unit charge
q
This describes the amount of work (energy) used or gained in moving a unit of charge
from one point to another in a uniform electric field.

W = Edf- Edi = -EΔd
q

ΔV = W = Edf- Edi = -EΔd. This is called a potential difference.
q

This is another kind of potential energy and so it must be relative to something
If we assume that at di=∞, Ed=0 then.... V = Ed = kq/r
f

The potential difference is more commonly called a potential. It is given the letter V
and from above we can see that it must have the units of Joules/Coulomb.

In fact the potential comes up so often that it had to be given it's own SI unit …
any guesses??
The Volt. So yes.... V is measured in V's
1V=1J/C
Note that we now have another way to express the electric field
E=N/C=V/m
+5000V

E

+                  -

0V

So a positive charge will “fall down” in an electric field by following the field lines
Potential energy is converted to kinetic energy, so it loses potential as it moves to
the area of lower potential.

A negative charge will “fall up” in an electric field by opposing the field lines
Potential energy is still converted to kinetic energy, so it loses potential energy as it moves
toward the higher potential (higher voltage).
Example 1 Find the potential 3cm away from a 50nC point charge.

Since we are asked only to find a potential take V=0 to be at infinity.

Then V=kq/r
= (8.99 x109)(50 x10-9)/(0.03)
= 15 000 V

Example 2 Given a 9V battery how much work is done to move 3mC of electrons
from the negative to the positive terminal.

9V battery implies ΔV=9V and the system is doing work, (converting
potential E to kinetic E) since the electrons are negative they want to go
to the +ve terminal. (W is +ve)

W=-qEΔd =-q(V)
W=-(-3mC)9V= 27x10-3 J
Example 3 A charge of 40μC is placed at the origin. Calculate the potential at a point
3cm along the x-axis and 4 cm along the y-axis.

V=kq/r
3cm                               =(8.99x109 )(40x10-6)/(0.032+0.042)1/2
A
=7.192 MV
r
4cm

Second find the potential energy of a -50nC charge placed at A

PE= V*q
= 7.192*106V * 50x10-9C
= 0.3596J
Example 5 How large a potential difference do you need to accelerate a proton to a
velocity of 3x106m/s if it starts at rest?

First find out how much energy is needed
KE = 1/2mv2
=1/2 (1.67x10-27kg)(3x106)2
= 7.52x10-15J
Second use the energy to find the potential. (note KE=PE)

V = PE/q
= (7.52x10-15J)/(1.602x10-19C)
= 46 941 V

We accelerate particles so often in physics that we use a special unit of energy just for them:

1eV = the energy gained by an electron going through a potential difference of 1V

1eV = 1.602 x10-19J
Example 4 You are walking along when suddenly a power line drops to the ground near
you. What is the best way to leave the area and why?

Shuffle or hop away. Don't RUN or take large steps.

Why?

The power line has a voltage relative to the ground. This will generate an electric
field around the power line. Let's say the field is 1000N/C then across a 1m stride
you have a potential of 1000V between each foot.

If you shuffle you can reduce your stride to say 0.05m. Then the same electric field
of 1000N/C produces only a 50V between your feet.

Which would you rather deal with? A 1000V or 50V?
Midterm Review

A container of flint glass (n=1.66) is filled with benzene (n=1.501). What is
the critical angle for total internal reflection when a ray travels from the
glass into the benzene?

Snell's Law... n1sinθ1=n2sinθ2

For the critical angle :θ2=90

So θc=sin-1(n2/n1) = sin-1(1.501/1.66) = 64.7o

What is the velocity of a light ray in a Lucite block (n=1.49) if the ray makes
an angle of incidence of 30o to the block.

n=c/v → v=c/n = (3x108m/s) / 1.49= 2.01x108m/s
How large a potential difference do you need to accelerate a proton to a
velocity of 4x108m/s ?

As far as we know this can't be done since this
would mean going faster than c
Of course physics has been wrong before...

How much energy is required to remove an electron from a proton if the orbital

If (as usual) we take the potential energy to be 0 at infinity then :

W =q  V =q V f −V i 
kq
Here we are interested in the potential generated by the proton V =
r
9x109⋅1.6x 10-19                       9x10 9⋅1.6x 10 -19
V f=                  =0               V i=            - 11
=27V
∞                                  5.3x10
W =1.6 x 10 -19⋅27=4.35x10−19 J
Of course if we measure this instead in electron volts then we get
J⋅1eV
W =4.35x10−19        -19
=27eV ' s
1.6x10 J
Mid term practice.

A 120 Hz oscillator vibrates a string causing waves of 31cm to
be produced. If the tension in the string is 1.2N what is the linear
density of the string?

a. 8.00x10-4 kg/m              Hint: v=√T/μ
b. 8.35x10-4 kg/m
c. 8.76x10-4 kg/m
d. 9.46x10-4 kg/m
Solution: C
μ=T/v2 and v=fλ

v=.31*120=37m/s

μ =1.2/372

μ=8.76x10-4 kg/m

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