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Lecture 7
Happy Groundhog Day!!
Punxsutawney Phil saw his shadow this morning…6 more weeks of winter!
Energy Storage in Capacitors
The electric potential energy, U, stored in a capacitor is equal to the amount of work required to charge
it. (work is required because of the force that charges exert on each other.)
Let’s define U = 0 to be the electric potential energy of an uncharged capacitor (Q = 0, V = 0).
When the capacitor is fully charged, it will have a charge Q and potential difference V = Q/C.
At some intermediate point during charging, when the capacitor has a charge q, the potential difference
will be V = q/C.
To increase the charge
on the capacitor by a
small amount dq will
then require an
amount of work, dW
q
dW = ΔVdq = dq
C
To find the total
work required to
charge the capacitor,
we need to integrate
from q = 0 to q = Q:
W =∫
Q
0
q
Q2
dq =
C
2C
Energy storage in capacitors (cont.)
Thus, the electric potential energy stored in a charged capacitor is:
1 Q2
U=
2 C
• Larger capacitance => less energy is required to store a fixed
charge Q
1
U = CV 2
2
Since Q = CV , we can also express this as:
• Larger capacitance => greater ability to store energy at
given potential difference.
• Analogous to elastic potential energy: U = 1 kx (P7E)
2
2
1
U
=
QV
Using V = Q / C , we can re-write the same expression as:
2
• This can be thought of as the total charge Q multiplied by the
average potential difference (V/2) during the process of charging.
Electric field energy
The process of charging a capacitor creates an electric field between the conductors
•Wherever there is a potential difference, there is an electric field
We can think of the energy stored in a capacitor as being stored in the electric field between
the plates.
For a parallel plate capacitor, the energy density u of the electric field is defined as the stored
energy divided by the volume between the plates. If the plate area is A and plate separation is d:
A
For a parallel plate capacitor, C = ε 0 d
1 CV 2
U
U
u=
=
= 2
Volume Ad
Ad
and V = Ed :
1
2
1 CV 2
ε 0 A d (Ed ) 1
u= 2
=2
= ε 0 E 2 [J/m 3 ]
Ad
Ad
2
(
)
Electric field energy
The energy density:
1
u = ε 0 E 2 [J/m 3 ]
2
for the special case of a parallel plate capacitor is actually much more general and
describes the energy density of any electric field in vacuum.
Thus we can think of the electric potential energy as being associated with the
arrangement of the charges, or as being stored in the electric field those charges
create.
Although we think of a vacuum as being “empty”, if it contains an electric field, it
isn’t really empty after all, since it contains energy.
Later in the course, we will use this relationship in connection with electromagnetic
waves, which can transport energy through a vacuum.
Quiz: Energy stored in capacitor
You reposition the two plates of a capacitor so that the capacitance doubles. There is vacuum
between the plates.
If the charges +Q and –Q on the two plates are kept constant in this process, the energy
stored in the capacitor…
A.becomes 4 times greater.
B.becomes twice as great.
C.remains the same.
D.becomes 1/2 as great.
E.becomes 1/4 as great.
1
U = CV 2
2
1 Q2
U=
2 C
1
U = QV
2
Quiz: Energy stored in capacitor
You reposition the two plates of a capacitor so that the capacitance doubles. There is vacuum
between the plates.
If the charges +Q and –Q on the two plates are kept constant in this process, the energy
stored in the capacitor…
A.becomes 4 times greater.
B.becomes twice as great.
C.remains the same.
D.becomes 1/2 as great.
E.becomes 1/4 as great.
1 Q2
U=
2 C
1 Q2
U=
2 C
1 Q2
U' =
2 2C
Dielectrics
What happens if we fill the region between the plates with a dielectric?
A: The potential difference between the plates decreases.
➡ Suppose the potential decreases by a factor K > 1, so V=V0/K.
With a vacuum between the plates, if the charge stored is Q and the
potential difference is V0, the capacitance is C0 = Q/V0
With a dielectric between the plates, if the charge stored is Q and the
potential difference is V < V0, the capacitance is:
C=
Dielectric const.
K=
C
C0
Q
Q
= K = KC0
V
V0
For Q=const.
V=
V0
K
C = KC0
Dielectric constants
The factor K depends on the material we insert, and is called the “dielectric constant”
Vacuum: C =1. Air: C ~ 1. For other materials, K > 1:
Induced Charge and Dielectric Polarization
If a dielectric reduces the potential difference between two
plates with fixed charge by a factor K, it must also reduce the
electric field by the same factor:
E0
E=
K
for constant Q
Why? Electric field of the charges on the plates polarizes the
dielectric material, creating an induced charge on the surfaces.
•The charges inside the material of the dielectric rearrange
themselves due to the presence of the field E0
•The induced charges on the surface of the dielectric create an
induced field opposite to E0
•The induced field partially cancels E0, so the total field
inside the dielectric is less than E0
Induced polarization is the effect
that dominates in most materials
used as dielectrics in capacitors
Dielectric Permittivity
The product ε = Kε 0 is called the permittivity of the dielectric.
Using the permittivity, the electric field in the dielectric can be expressed in the same form
as for vacuum:
σ
E=
ε
Similarly, for a parallel plate capacitor with a dielectric:
C = KC0 = Kε 0
A
A
=ε
d
d
And the energy density of the electric field in a dielectric is:
1
u = εE 2
2
Induced Charge
For most materials, and moderate field strengths, the induced surface charge density of the dielectric is proportional to the original field E0
•We will assume that is true
If the charge density on the plates is σ and the induced charge density on the surface of the dielectric is σi, the net surface charge density, σnet, on each side of the capacitor is σ – σi
Since the electric field is given by:
• E0 = σ ε
0
• E=
σ − σ i E0
=
ε0
K
Solving for σi:
E=
σ net
we have
ε0
1⎞
⎛
σ i = σ ⎜ 1 − ⎟ induced surface charge density
⎝ K⎠
If K is very large, σi≅σ, and thus the net electric field and voltage are much reduced.
Properties of Capacitor
Vacuum
Electric field
σ
E=
ε
σ
E0 =
ε0
Capacitance
A
C0 = ε 0
d
Electric energy density
1
u0 = ε 0 E02
2
With dielectric
Q
C=
Vab
A
C=ε
d
1
u = εE 2
2
Dielectric constant
K=
ε0: permittivity of free space
ε = Kε 0
In dielectric replace ε0 with ε
Permittivity
C V0
=
C0 V
Brief Announcement
— Ch 24 homework will be assigned asap, due next week.
Chapter 25
Current & Resistance
Electric Current
●
●
Electric current is the rate of flow of charge
through some region of space (or a wire).
The SI unit of current is the ampere (A)
●
●
1A=1C/s
The symbol for electric current is I
dQ
I=
dt
Direction of Current
●
●
●
The charges passing
through the area could be
positive or negative or both
It is conventional to assign
to the current the same
direction as the flow of
positive charges
Convention: direction of
current flow is opposite
the direction of the flow
of electrons
#thanksbenfranklin
Charge Motion in a Conductor
●
Electric field forces cause the electrons to move in the wire and
create a current
• Most electron motion in
a conductor is random,
caused by collisions.
• The net “drift velocity”
is very slow.
Current and Drift Speed
̣
̣
̣
̣
Wire of cross-section area A
n = charge carriers per Volume
nAΔx = number of charges that
move through A in time Δt
Total charge is ΔQ = (nAΔx)e
Drift Velocity
dQ
I=
= n q Avd
dt
Electron drift is SLOW
Current:
I ' 1 A = 1 C/s
Density of electrons in wire:
Electron charge:
Area of wire:
A ⇠ 10
Definition
of current:
v=
q ⇠ 10
6
n ⇠ 1029 /m3
19
C
m2
Drift Velocity
dQ
I=
= nqAv
dt
I
1C/s
= 29
nqA
10 · 10 19 · 10
6 C/m
= 10
4
m/s
wikipedia
slower than a snail
faster than continental drift
Why is the current flow nearly instantaneous?
If drift velocities are ~10-4 m/s, why do the lights turn on instantly?
Answer: a conductor has mobile charges all along its length.
Turning on a switch establishes an electric field through the conductor almost
instantaneously, which means all mobile charges start drifting immediately.
•A charge doesn’t need to travel from the switch to the light-bulb for the light-bulb
to turn on – there are already charges inside the light-bulb, and the light-bulb turns
on as soon as they start drifting, which happens almost instantly.
Current and Work
We just showed that the kinetic energy of the charges does not increase appreciably.
So Where does the energy go?
•Due to the frequent collisions of the moving charges, the energy is transferred to the (stationary)
ions of the material, increasing their vibrational energy, and thus their temperature.
Bulk of the work done by the electric field goes into heating the conductor.
(this is ‘wasted’ energy, lots of room to increase efficiency)
Quiz: current
These four wires are made of the same metal. Rank in order, from largest to
smallest, the electron currents ia to id:
A.id>ia>ib>ic
B.ib=id>ia=ic
C.ic>ib>ia>id
D.ic>ia=ib>id
E.ib=ic>ia=id
dQ
I=
= n q Avd
dt
Quiz: current
These four wires are made of the same metal. Rank in order, from largest to
smallest, the electron currents ia to id:
A.id>ia>ib>ic
dQ
I=
= n q Avd
dt
B.ib=id>ia=ic
C.ic>ib>ia>id
D.ic>ia=ib>id
E.ib=ic>ia=id
(πr )v
2
(πr )2v
2
(π 4r )v
2
⎛ 1 2⎞
⎜⎝ π r ⎟⎠ 2v
4
Quiz: current and drift speed
Two copper wires of different diameter are joined end to end, and a current flows in
the wire combination.
When electrons move from the larger-diameter wire into the smaller-diameter wire,
A.their drift speed increases.
B.their drift speed decreases.
C.their drift speed stays the same.
D.not enough information given to decide
Quiz: current and drift speed
Two copper wires of different diameter are joined end to end, and a current flows in
the wire combination.
When electrons move from the larger-diameter wire into the smaller-diameter wire,
A.their drift speed increases.
B.their drift speed decreases.
C.their drift speed stays the same.
D.not enough information given to decide
Analogy: water is a pipe (when radius of pipe decreases, the speed increases).
The current I is constant for the whole wire.
If A decreases, v has to decrease to compensate.
dQ
I=
= nqAv
dt
Current density
Current:
dQ
I=
= n q Avd
dt
Take out the area factor => current density: J =
I
= n q vd ⎡ A 2 ⎤
⎣ m⎦
A
In some cases, it is useful to treat the current density as a vector:
• q > 0, vd same direction as E
• q < 0, vd opposite direction as E
Total current through a surface:
J describes how charges flow at a certain point
• Direction of J changes around the circuit
• Magnitude of J = I/A can change along the circuit (e.g. when A changes)
Conductivity
●
Currents flow in response to electric fields.
●
For some materials (e.g. metals), the current density
is [nearly] directly proportional to the electric field
●
The constant of proportionality, σ, is called the
conductivity of the conductor
J=σE
“Ohm’s Law”
current density:
Ohm’s Law
●
●
●
●
I
= n q vd
A
Ohm’s law : for many materials, the
current density is proportional to the
electric field: J = σ E
Most metals obey Ohm’s law - Materials
that obey Ohm’s law are called ohmic
Not all materials follow Ohm’s law
●
●
J=σE
J=
Materials that do not obey Ohm’s law
are said to be nonohmic
Ohm’s law is not a fundamental law
of nature
Ohm’s law is an empirical
relationship valid only for certain
materials
●
Specifically when conduction properties
are determined by collisions of
electrons with atoms
●
●
●
●
Ohm (1789 -1854)
German physicist
Formulated idea of
resistance
Discovered Ohm’s Law