Transcript Storing Electrical Energy

Storing Electrical Energy
Capacitors
Overview
• Storing electrical charge
• Defining capacitance
• Applications
• Relationships
Storing electrical potential energy
Squeeze a spring  stored elastic
potential energy
Hold magnets together  stored
magnetic potential energy
Hold electric charges together 
stored electrical potential
energy
Holding charges
Voltage source (e.g., battery)
Two conductive plates,
separated by a
non-conductor called a
dielectric
Charging a capacitor
When one plate is connected
to the voltage source (left
plate in this example), an
electric field is created,
causing electrons to flow from
left plate towards positive
terminal.
Electrons are pulled toward
other plate.
Charged capacitor
After some time, the potential
difference between the
capacitor plates is equal to the
potential difference from the
battery. When
Vcapacitor = Vbattery, the electrons
stop flowing.
The capacitor is considered
fully charged.
Some applications
Storing large amounts of charge for later release
e.g., camera flash, defibrillator
Computer interface components
e.g., touch screen, keyboards
Protecting components from surges in direct current
e.g., adapters, surge protectors
Uninterrupted power supply
e.g., power for computers and other electronic
devices with changing load requirements
In conjunction with resistors, timing circuits
e.g., pacemakers or intermittent windshield wipers
Etc.
Virtually every piece of modern electronics
contains capacitors.
Read more here: http://electronics.howstuffworks.com/capacitor2.htm
Quantifying capacitance
1st: consider the electric field
created by two parallel
plates
• Strong and uniform electric field
between the plates
• 0 N/C outside the plates
Recall
2nd: remember the relationship
between strength of the electric
field, voltage and distance
Strength of Electric Field
V
𝐸= −
𝑑
Depends on voltage, e.g., battery
Depends on separation of charges
So,
• 2V  2E
• 2d  ½E
Quantifying capacitance
3rd: draw the connection
between charge stored
and voltage
Electric field lines are the result of
electric charges.
Q  E
And we know that E  V
So, Q  E  V
Or, Q  V
Or, 𝑄  𝑉
Quantifying capacitance
Charge is proportional to
voltage,
𝑄𝑉
Named in honor of Michael Faraday, an English
scientist (1791 – 1867) who connected fields of
electricity and magnetism.
4th:
add a proportionality
constant,
𝑄 = 𝐶𝑉
Capacitance
Measured in units of coulombs per volt,
abbreviated as farads
1 farad = the capacitance that can hold 1
coulomb of charge with 1 volt
potential difference.
1F=1C/1V
Typical capacitance
Since a coulomb is a ridiculously
large amount of charge and a
volt is relatively small, a farad
𝑄
(𝐶 = ) is ridiculously large.
𝑉
Typical capacitors store
10-12 F (picofarads) to
10-3 F (1000 microfarads).
Example
The figure at right shows the ratio of
charge to voltage of three different
capacitors. Which capacitor has the
greatest capacitance: A, B, or C?
Justify your answer.
Q = CV, so C= Q/V.
Steeper the slope, bigger the value of
Think
C. it through first.
Check the solution by moving this box.
Line A has the steepest slope and
therefore the largest capacitance.
Example
Two plates of a capacitor hold +2500 C and -2500 C of charge respectively
when the potential difference is 960 V. What is the capacitance?
𝑄 = 2500 𝐶
𝑉 = 960 𝑉
Try it first.
𝑄
2500 x 10−6 C
𝐶 =the solution
= by moving this box.=
Check
𝑉
𝐶 = 2.6μ𝐹
960 V
2.6x10−6
𝐶
𝑉
Example
An 8500-pF capacitor holds plus and minus charges of 1.65 x 10-7C. What
voltage is required?
𝐶 = 8500 x 10−12 F
𝑄
𝑄
𝐶𝑉, so 𝑉 =
Try =
it first.
𝐶
−7
Check the
solution
by moving
1.65
x 10
C this box.
𝑉=
8500 x 10−12 F
𝑉 = 19 V
𝑄 = 1.65 x 10−7 C
Factors that affect capacitance
Area of plates
Rationale:
𝑄
𝐶 =
𝑉
Charges can spread across
wider area on each plate, i.e.,
capacitor is able to maintain
more larger charge separation
per volt
AQC
AC
Factors that affect capacitance
Distance between plates
Rationale:
Recall V = Ed; so, at constant V,
dE
Smaller electric field leads to less
ability to hold charges apart,
E Q
𝑄
Since, 𝐶 =
𝑉
d EQ C
dC
Factors that affect capacitance
Material between plates
Chemistry nerds: polar molecules!
Rationale:
The structure of some materials minimizes the
strength of the electric field as the field passes
through.
We use the Greek letter kappa, , to represent
dielectric constant
𝐸0
≝
𝐸0 − 𝐸1
where 𝐸0 is strength of field produced
by environment
𝐸1 is the strength of the field in
the material
The positive side of the newly-aligned polar
molecules attract more negative charges to the
negative-charged plate, allowing more charges to
accumulate.
QC
Dielectrics
Maximum strength of field before dielectric
breaks down and charges start flowing
Factors that affect capacitance
Area of plates
AC
Distance between plates
dC
Material between plates
C
Quantifying capacitance, part 2
𝐶
𝐴
𝑑
𝐴
𝐶 = 𝜖0 
𝑑
A quick check on units:
Proportionality constant
𝐶2
−12
Permittivity of free space, 8.85 𝑥 10
𝑁𝑚2
1
Recall Coulomb’s constant, 𝑘 = 4𝜖
0
𝐶2 ∙ 𝑚 2
𝑁𝑚2 𝑚
=
𝐶 𝐶
𝑁𝑚
𝐶
1
𝐶
= 𝐽 𝐶 =𝑉 𝐶= 𝑉 = 𝐹
Example
A parallel plate capacitor has an area of 1.00 m2 and a spacing of
0.500 mm. If the insulator has a dielectric constant of 4.9, what is the
capacitance?
𝐴 = 1.00 𝑚2
𝑑 = 0.500 x 10−3 𝑚
 = 4.9
2
2
𝐴
C
1.00
𝑚
−12
Try it first.
𝐶
=
𝜖

=
(8.85
𝑥
10
)(4.9)
0
2
Check the solution
by
moving
this
box.
𝑑
𝑁𝑚
0.500 x 10−3 𝑚
𝐶 = 8.8 x 10−8 𝐹
Example
In one kind of computer keyboard, each key is attached to
one plate of a parallel plate capacitor. The other plate is
fixed. The capacitor is maintained at 5.0 V.
When the key is pressed down, the top plate moves closer,
changing the capacitance and allowing charge to flow again.
The circuit detects the change and sends a signal to the
computer screen.
If each plate is a square of 36.0 mm2 and the plate separation
changes from 4.0 mm to 1.2 mm when a key is pressed, how
much charge flows through the circuit? Assume there is air
between the plates instead of a flexible insulator.
Example
𝑉 = 5.0 V
𝑑𝑓 = 1.2 x 10−3 𝑚
∆𝑄 = ∆𝐶𝑉
𝐴 = 36 𝑚𝑚2 = 36 x 10−6 𝑚2
𝑑𝑖 = 4.0 x 10−3 𝑚  = 1.00
where ∆𝐶 =
𝐴
𝜖0 
𝑑𝑓
−
𝐴
𝜖0 
𝑑𝑖
=
1
𝜖0 𝐴(
𝑑𝑓
1
− )
𝑑𝑖
Try it first.
1 how1the capacitance changes when the key is up and when it is pressed down.
Here’s a hint: consider
Check
the𝜖solution
by moving
∆𝑄 =
− this
𝑉 box.
0 𝐴
𝑑𝑓 𝑑𝑖
= 8.85
2
−12 C
x 10
Nm2
∆𝑄 =0.93 x 10-12 C
1
0.0012 𝑚
1.00)(36 x 10−6 𝑚 (
−
1
) 5V
0.004 𝑚
Example
An electric field of 8.50 x 105 V/m is desired between two parallel plates,
each of area 45.0 cm2 and separated by 2.45 mm of air. What charge must
be on each plate?
𝑉
𝑚
x 10−3 𝑚
𝐸 = 8.50 x 105
A = 45.0 𝑐𝑚2 = 45 x 10−4 𝑚2
𝑑 = 2.45
=1.00
Q=?
𝐴
Try it first.
𝑄 = 𝐶𝑉 where 𝐶 = 𝜖0  and 𝑉 = −𝐸𝑑
𝑑 this box.
Check the solution by moving
So,
𝑄 = 𝜖0 𝐴𝐸
Q=(8.85
𝑄 = 3.4x10−8 C
2
−12 C
x 10
)(1.00)(45
𝑁𝑚2
x 10−4 𝑚2 )(8.50 x 105 V/𝑚)
Investigate
Spend some time with the PHeT
simulation. Click on the diagram at
right.
Confirm the relationships
𝑄 = 𝐶𝑉
and
𝐴
𝐶 = 𝜖0 
𝑑
Energy stored in a spring
k = spring constant, measure of spring’s stiffness
Energy stored in a capacitor
Storing energy in a capacitor is like storing energy in a
spring, easy at first then increasingly harder.
To calculate the amount of work (which also equals
the system’s potential energy), find the average work
per charge.
𝑉𝑓 − 𝑉𝑖
𝑃𝐸 = 𝑊 = 𝑄
2
@ 𝑉𝑖 = 0
1
𝑃𝐸 = 𝑄𝑉
2
Energy stored in a capacitor
1
𝑈 = 𝑄𝑉
2
The energy stored in a capacitor is the ½ the
product of the capacitance and the square of
the voltage.
Where 𝑄 = 𝐶𝑉,
So, 𝑈 =
1
𝑄𝑉
2
=
1 𝑄2
2 𝐶
=
1
𝐶𝑉 2
2
Example
650 V is stored in a 2800-pF capacitor. How much energy is stored?
𝐶 = 2800 x 10−12 𝐹
𝑉 = 650 V
𝑈 =?
Try it first.
1
1
2
𝑈
=the𝐶𝑉
(2800
x 10−12
Check
solution=
by moving
this box.
2
2
𝑈 = 5.9 x 10−4 𝐽
𝐹)(650 V)2
Example
A cardiac defibrillator is used to shock a heart that is beating erratically.
A capacitor stores 5.0 kV and stores 1200 J. What is its capacitance?
𝑉 = 5.0 x 103 V
𝐶 =?
𝑈 = 1.2 x 103 𝐽
2𝑈
Try it first.
2
𝑈
= ½ 𝐶𝑉
so, 𝐶 = 2
Check the solution by moving this box.
𝑉
= 2 (1.2 𝑥 103 𝐽) / (5.0 𝑥 103 V)
𝐶 = 9.6 x 10−5 𝐹
Bonus! Interesting insight from these relationships
𝑈=
1
𝐶𝑉 2
2
where 𝐶 =
𝐴
𝜖0 
𝑑
and 𝑉 2 = 𝐸 2 𝑑 2
1
𝐴 2 2
𝑈 = 𝜖0  𝐸 𝑑
2
𝑑
𝑈=
1
𝜖0 𝐸 2 𝐴𝑑
2
where 𝐴𝑑 = 𝑣𝑜𝑙𝑢𝑚𝑒
Energy density in space is proportional to the
square of the electric field strength.
So,
𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦
𝑔𝑖𝑣𝑒𝑛 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑝𝑎𝑐𝑒
 𝐸2
Not just in capacitors, but everywhere: light, radio
waves, and every other type of electromagnetism!