Phys 222 Exam Review 1 PPT

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Physics 222 Exam Review
Outline
• Overview/Mind-map
• What each equation does
• Practice Problems
Sorry about the boring theme.
• I couldn’t find a suitable theme that I liked,
that didn’t mess up my text.
• However, looking at green is said to increase
creativity and stimulate brain function. May
use this later.
Useful tip: Storing variables in
calculator
• π‘˜π‘’ = 9 ∗ 109
• πœ–0 = 8.85 ∗ 10−12
𝑬 = −𝛁V
Electric Force
Multiply by q
Electric Potential
π‘‰π‘Ž − 𝑉𝑏
Potential Energy
𝐴
=−
Divide by q
Electric Field
𝑬 ⋅ 𝑑𝒍
𝐡
Capacitors
Interacts with dipole
Dipole
Resistors
Charge
Overview
•
•
•
•
•
•
Fluids
Electric force -> Electric field
Potential energy -> Electric potential
E->V and V->E
Capacitors and energy stored inside them
Resistors
𝐹
𝑃=
𝐴
• This is the definition of Pressure: Force/Area
𝑃 = 𝑃0 + πœŒπ‘”π‘¦
• Used to calculate the pressure at a depth of y
in some medium, usually water.
• Example: 5 m deep under the water,
pressure=1.01 ∗ 105 + 103 ∗ 9.8 ∗ 5
𝑑𝑉
= 𝐴𝑣 = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘
𝑑𝑑
• This is the continuity equation for fluid flow.
• In English, it means that the amount of stuff
going through a pipe is constant, so shrinking
the pipe means that the water will go faster.
1 2
𝑃 + πœŒπ‘”π‘¦ + πœŒπ‘£ = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘
2
• Also known as Bernoulli’s Equation
• Classic problem: calculating velocity of water
shooting out of a hole in a container.
– In the case that the radius of the hole is small, the
velocity at the bottom simplifies to 𝑣 = 2𝑔𝑦 ,
where y is the distance from the water level to the
hole.
Questions?
π‘­πΆπ‘œπ‘’π‘™π‘œπ‘šπ‘
π‘ž1 π‘ž2
= π‘˜π‘’ 2 𝒓
π‘Ÿ
𝑭
𝑬=
π‘ž0
• Just like electric force, but without the test
charge q0. It’s still a vector.
π‘ž
𝑬 = π‘˜π‘’ 2 𝒓
π‘Ÿ
• Electric field of a point charge.
Electric field lines go from + to -.
• Also, line
density
indicates field
strength
𝒑 = |𝑄|𝒅
• Dipole moment, BY DEFINITION
• Notice: Dipole moment points from negative
to positive….opposite of the direction E points
𝝉=𝒑×𝑬
• Torque on a dipole by the external electric
field
– Note that E is not the E produced by the dipole
– E is external
• Torque is maximum when dipole moment and
E are perpendicular.
π‘ˆ = −𝒑 ⋅ 𝑬
• Potential energy is minimum (also called stable
equilibrium) when two things are true:
– Dipole moment is parallel to E
– Dipole moment points in the same direction as E.
• Potential energy is maximum (also called
unstable equilibrium) when two things are true:
– Dipole moment is parallel to E.
– Dipole moment points in the opposite direction as E.
Questions?
π‘žπ‘’π‘›π‘
Φ𝐸 =
𝑬 ⋅ 𝑑𝒂 =
πœ–0
π‘ π‘’π‘Ÿπ‘“π‘Žπ‘π‘’
• Also known as Gauss’s Law
• Really there are two equations here…but
they’re both equally valid…always.
Steps in solving Gauss’s Law Problems
– Draw a picture of the object. Pick a good Gaussian
surface.
– Write down the expression of Gauss’s Law that
involves the dot product between E and A. (If E is
perpendicular to A, the flux is 0 for that surface.
Otherwise, use symmetry to get rid of the integral.)
– Write down the expression of Gauss’s Law that
involves the total charge q.
– Set the two expressions equal to each other and
eliminate variables.
2πœ†
𝐸 = π‘˜π‘’ π‘Ÿ
π‘Ÿ
• Electric field a distance r away from a long
wire with charge density πœ†.
𝜎
𝑬=± π‘₯
2πœ–
• This one is especially important.
• This is the electric field anywhere away from a
large sheet of charge.
• Notice that the electric field doesn’t depend
on distance, and always points perpendicular
to the surface.
• Classic problem: What is the electric field
between two parallel plates of charge ±Q and
area A?
• Answer: Since you have two plates of opposite
charge, the E fields add, and thus…
• 𝜎=
𝑄
𝐴
𝐸=2
𝜎
2πœ–0
=
𝜎
πœ–0
=
𝑄
π΄πœ–
𝐴
𝐄 → 𝑉:
𝑉 → 𝑬:
π‘‰π‘Ž − 𝑉𝑏 = −
𝑬 ⋅ 𝑑𝒍
𝐡
𝑬 = −𝛁V
• These relations let you go from either E to V or
vice-versa. If you know one, you can calculate
the other.
π‘Š = −Δπ‘ˆ = −π‘ž0 Δ𝑉
This equation can be used to:
• If you’re given Δπ‘ˆ, you can find the work
done.
• If a point charge of charge q0 goes through a
potential difference of Δ𝑉, it tells you the
work done on the charge.
π‘ˆ = π‘ž0 𝑉
• This is the defining relation between potential
energy (U) and electric potential (V).
• Note: since q0 can be positive or negative, U
and V do not necessarily have the same sign.
• One more time: Electric potential (V) is not
the same thing as electric potential energy (U)
• But let’s rewrite it.
𝑉 = π‘ˆ/π‘ž0
Similar to:
𝑬=
𝑭
.
π‘ž0
π‘˜π‘’ π‘ž
𝑉=
π‘Ÿ
• Electric potential of a point charge q, a
distance r away, assuming V=0 at infinity.
• Potential goes up as you get closer to the
point charge.
Δ𝑉 = ±πΈπ‘‘
• Rather important: This is the potential
difference between two parallel conducting
plates, otherwise known as a capacitor.
• What is E for a capacitor again? 𝐸 = 𝜎/πœ–0
𝑄
𝐢=
𝑉
• Definition of capacitance
• C=Charge Q/ Voltage drop
𝐴
𝐢 = πœ–0
𝑑
• Special case of capacitance when you’re
looking at a parallel plate capacitor.
• Notice that the capacitance doesn’t depend
on the charge on the plates.
πΆπ‘’π‘ž = 𝐢1 + 𝐢2 + β‹―
• Adding capacitors in PARALLEL.
1
1
1
= + +β‹―
πΆπ‘’π‘ž 𝐢1 𝐢2
• Adding capacitors in SERIES.
π‘Žπ‘
𝐢 = 4πœ‹πœ–0
𝑏−π‘Ž
• Special case of capacitance when you’re
looking at a two concentric spherical
conducting shells.
• The radius of the smaller shell is a, the radius
of the larger shell is b.
𝐢 = 4πœ‹πœ–0 𝑅
• The capacitance of a single spherical shell of
radius R
2
1 2 𝑄
1
π‘ˆ = 𝐢𝑉 =
= 𝑄𝑉
2
2𝐢 2
• You use these equations to calculate the stored
energy in a capacitor
• Okay, but there’s 3 different equations, so which
one is appropriate for my problem?
– If they just ask you to calculate U, use the one that has
variables you know.
– If they ask you what happens to U if you double the
charge, halve d, etc…
• If the capacitors are stand-alone, use
𝑄2
.
2𝐢
1
2
• If the capacitors are connected to a voltage source, use 𝐢𝑉 2
πœ– = πœ…πœ–0
• This relates the permittivity of free space to
the permittivity in a medium of dielectric
constant πœ….
𝐢 = πœ…C0
• Relates capacitance without a dielectric (C0) to
capacitance with a dielectric.
1 2
𝑒 = πœ–πΈ
2
• The energy density when you have an electric
field E in a medium of permittivity πœ–.
• For example, let’s say you have a cube (L=3) filled
with water (πœ… = 78). The cube has the same E
field everywhere (E=5).
– Then
1
u=
2
∗ 78 ∗ πœ–0 ∗ 52
3
1
2
– Total stored energy=π‘ˆ = ∫ 𝑒 𝑑𝑉 = 3 ⋅ ⋅ 78 ⋅ πœ–0 ⋅ 52
𝑑𝑄
𝐼=
𝑑𝑑
• Definition of current.
𝑉 = 𝐼𝑅
• Ohm’s Law: A relationship between voltage,
current, and resistance.
• Pretty fundamental.
𝐼
𝐽 = = π‘žπ‘›π‘£π‘‘
𝐴
• 𝐽=Current density
• n = density of charge carriers
• Vd=drift velocity (average velocity of charge
carriers)
• q=charge on a charge carrier (usually e=1.6 ∗
10−19 𝐢)
𝐸 = 𝐽𝜌
• Microscopic Ohm’s Law:
• E-field (E) = Current density (J) x resistivity (𝜌)
𝐿
𝑅=𝜌
𝐴
• Resistance of a conductor of resistivity 𝜌,
length L, and cross-sectional area A
𝜌 = 𝜌0 1 + 𝛼 𝑇 − 𝑇0
• Resistivity changes as a function of
temperature.
𝑅 = 𝑅0 1 + 𝛼 𝑇 − 𝑇0
• Resistance changes as a function of
temperature
π‘…π‘’π‘ž = 𝑅1 + 𝑅2 + β‹―
• Adding resistances in series.
1
1
1
=
+
+β‹―
π‘…π‘’π‘ž 𝑅1 𝑅2
• Adding resistances in parallel.
Practice Problems
In the figure to the right, there are two charges
connected by a massless insulating rod…and remember
to use VECTORS when appropriate…
Draw the electric dipole.
Torque caused by the electric field=
Dipole moment=
Potential energy as it is right now=
Which way will the dipole begin to rotate?
(Clockwise/Counter-clockwise)
How much work is done in rotating the dipole from its
current position to the stable equilibrium position?
What does the work in question f?
A block of MagicFoam (length 10 cm, width 10 cm,
height 3 cm) sits on top of a calm body of water.
MagiFoam density=0.5 g/cm3. How much of the
block is submerged?
A 10 kg block floats in the water. What is the
buoyant force on it?
A house with a roof of area 5 m2 has winds of 50
m/s above it. What is the force on the roof caused
by the pressure difference?
How much energy does it take to bring two
electrons within .1 nπ‘š of each other?
Some water is flowing at a rate of 20 mph in a pipe.
Further on in the pipe, the pipe halves its diameter.
What is the new speed of the water?
If current is going from your hand
to your foot, which direction are
the electrons going?
Final Questions?
Thank you, and good luck!
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