Transcript Unit 3 A Electric Potential Capacitance Part 2.

Bright Storm on Capacitors
(Start to minute 7:10)
Khan Academy on
Capacitance
Read and take notes on pgs: 546-547
Capacitance
• A capacitor is a device used in a variety of
electric circuits.
• The capacitance, C, of a capacitor is defined as
the ratio of the magnitude of the charge on
either conductor (plate) to the magnitude of
the potential difference between the
conductors (plates).
Section 16.6
Capacitance, Cont.
•
• Units: Farad (F)
– 1F=1C/V
– A Farad is very large
• Often will see µF or pF
• V is the potential
difference across a
circuit element or
device.
• V represents the actual
potential due to a given
charge at a given
location.
Section 16.6
Parallel-Plate Capacitor, Example
• The capacitor consists of
two parallel plates.
• Each has area A.
• They are separated by a
distance d.
• The plates carry equal and
opposite charges.
• When connected to the
battery, charge is pulled off
one plate and transferred to
the other plate.
• The transfer stops when
Vcap = Vbattery
Section 16.7
Parallel-Plate Capacitor
• The capacitance of a device depends on the
geometric arrangement of the conductors.
• For a parallel-plate capacitor whose plates are
separated by air:
Section 16.7
Electric Field in a Parallel-Plate
Capacitor
• The electric field between the plates is uniform.
– Near the center
– Nonuniform near the edges
• The field may be taken as constant throughout the region
between the plates.
Section 16.7
EXAMPLE 16.6 A Parallel-Plate Capacitor
Goal Calculate fundamental physical properties of a parallel-plate
capacitor.
Problem A parallel-plate capacitor has an area A = 2.00 10-4 m2 and a
plate separation d = 1.00 10-3 m. (a) Find its capacitance. (b) How much
charge is on the positive plate if the capacitor is connected to a 3.00-V
battery? Calculate (c) the charge density on the positive plate, assuming
the density is uniform, and (d) the magnitude of the electric field between
the plates.
Strategy Parts (a) and (b) can be solved by substituting into the basic
equations for capacitance. In part (c) use the definition of charge density,
and in part (d) use the fact that the voltage difference equals the electric
field times the distance.
SOLUTION
(a) Find the capacitance.
Substitute values into the definition of capacitance:
2.00 10-4 m2
A
C = ε0 = (8.85 10-12 C2/N · m2)
1.00 10-3 m
d
(
)
C = 1.77 10-12 F = 1.77 pF
(b) Find the charge on the positive plate after the capacitor is connected to
a 3.00-V battery.
Substitute into the capacitance charge equation to find the charge:
C = Q/ΔV → Q = CΔV = (1.77 10-12 F)(3.00 V)
Q = 5.31 10-12 C
(c) Calculate the charge density on the positive plate.
Charge density is charge divided by area:
Q 5.31 10-12 C
= 2.66 10-8 C/m2
σ= =
-4
2
A 2.00 10 m
(d) Calculate the magnitude of the electric field between the plates.
Apply ΔV = Ed.
E=
ΔV
d
=
3.00 V
-3
1.00 10 m
= 3.00 103 V/m
LEARN MORE
Remarks The answer to part (d) could also have been obtained from the
electric field derived for a parallel plate capacitor E = σ/ε0.
Question How do the following change if the distance between the plates
is doubled? (Select all that apply.)
The electric field remains the same.
The capacitance is doubled.
The charge is halved.
The electric field between the plates is
doubled.
The charge is doubled.
The capacitance is halved.
The electric field between the plates is halved.
Read and take notes on pgs: 548-550
Capacitors in Circuits
• A circuit is a collection of objects usually
containing a source of electrical energy (such
as a battery) connected to elements that
convert electrical energy to other forms.
• A circuit diagram can be used to show the
path of the real circuit.
Section 16.7
Bright Storm on Capacitors in
Parallel
Capacitors in Parallel
• When connected in
parallel, both have the
same potential
difference, V, across
them.
Section 16.8
Capacitors in Parallel
• When capacitors are first connected in the circuit,
electrons are transferred from the left plates through
the battery to the right plate, leaving the left plate
positively charged and the right plate negatively
charged.
• The flow of charges ceases when the voltage across
the capacitors equals that of the battery.
• The capacitors reach their maximum charge when
the flow of charge ceases.
Section 16.8
Capacitors in Parallel
• The potential difference
across the capacitors is the
same.
– And each is equal to the
voltage of the battery
• The total charge, Q, is
equal to the sum of the
charges on the capacitors.
– Q = Q1 + Q2
Section 16.8
More About Capacitors in Parallel
• The capacitors can be
replaced with one capacitor
with a capacitance of Ceq
– The equivalent capacitor
must have exactly the same
external effect on the circuit
as the original capacitors.
Section 16.8
Capacitors in Parallel, Final
• Ceq = C1 + C2 + …
• The equivalent capacitance of a parallel
combination of capacitors is greater than any
of the individual capacitors.
Section 16.8
EXAMPLE 16.7 Four Capacitors Connected in Parallel
Four capacitors connected in parallel.
Goal Analyze a circuit with several capacitors in
parallel.
Problem (a) Determine the capacitance of the
single capacitor that is equivalent to the parallel
combination of capacitors shown in the figure. Find
(b) the charge on the 12.0-µF capacitor and (c) the
total charge contained in the configuration. (d)
Derive a symbolic expression for the fraction of the
total charge contained on one of the capacitors.
Strategy For part (a), add the individual capacitances. For part (b), apply
the formula C = Q/ΔV to the 12.0-µF capacitor. The voltage difference is
the same as the difference across the battery. To find the total charge
contained in all four capacitors, use the equivalent capacitance in the same
formula
SOLUTION
(a) Find the equivalent capacitance.
Apply the equivalent capacitance equation:
Ceq = C1 + C2 + C3 + C4
= 3.00 µF + 6.00 µF + 12.0 µF + 24.0 µF = 45.0 µF
(b) Find the charge on the 12-µF capacitor (designated C3).
Solve the capacitance equation for Q and substitute.
Q = C3ΔV = (1.20 10-5 F)(18.0 V) = 2.16 10-4 C = 216 µC
(c) Find the total charge contained in the configuration.
Use the equivalent capacitance.
Ceq = Q/ΔV → Q = CeqΔV = (45.0 µF)(18.0 V)
= 8.10 102 µC
(d) Derive a symbolic expression for the fraction of the total charge
contained in one of the capacitors.
Write a symbolic expression for the fractional charge in the ith capacitor
and use the capacitor definition.
Qi CiΔV Ci
=
=
Qtot CeqΔV Ceq
LEARN MORE
Remarks The charge on any one of the parallel capacitors can be found as
in part (b) because the potential difference is the same. Notice that finding
the total charge does not require finding the charge on each individual
capacitor and adding. It's easier to use the equivalent capacitance in the
capacitance definition.
Question If all four capacitors had the same capacitance, what fraction of
the total charge would be held by each? Express the fraction in decimal
form.
.25
Bright Storm on Capacitors in
Series
Read and take notes on pgs: 551-552
and 553
Capacitors in Series
• When in series, the
capacitors are
connected end-to-end.
• The magnitude of the
charge must be the
same on all the plates.
Section 16.8
Capacitors in Series
• When a battery is connected to the circuit, electrons
are transferred from the left plate of C1 to the right
plate of C2 through the battery.
• As this negative charge accumulates on the right
plate of C2, an equivalent amount of negative charge
is removed from the left plate of C2, leaving it with an
excess positive charge.
• All of the right plates gain charges of –Q and all the
left plates have charges of +Q.
Section 16.8
More About Capacitors in Series
• An equivalent capacitor can
be found that performs the
same function as the series
combination.
• The potential differences
add up to the battery
voltage.
Section 16.8
Capacitors in Series, Final
•
• The equivalent capacitance of a series
combination is always less than any individual
capacitor in the combination.
Section 16.8
EXAMPLE 16.8 Four Capacitors Connected in Series
Four capacitors connected in series.
Goal Find an equivalent capacitance of capacitors
in series, and the charge and voltage on each
capacitor.
Problem Four capacitors are connected in series
with a battery, as in the figure (a) Calculate the
capacitance of the equivalent capacitor. (b)
Compute the charge on the 12-µF capacitor. (c) Find the voltage drop
across the 12-µF capacitor.
Strategy Combine all the capacitors into a single, equivalent capacitor.
Find the charge on this equivalent capacitor using C = Q/ΔV. This charge
is the same as on the individual capacitors. Use this same equation again
to find the voltage drop across the 12-µF capacitor.
SOLUTION
(a) Calculate the equivalent capacitance of the series.
Apply the series combination equivalent capacitance equation.
1
1
1
1
1
=
+
+
+
Ceq 3.0 µF 6.0 µF 12 µF 24 µF
Ceq = 1.6 µF
(b) Compute the charge on the 12-µF capacitor.
The desired charge equals the charge on the equivalent capacitor:
Q = CeqΔV = (1.6 10-6 F)(18 V) = 29 µC
(c) Find the voltage drop across the 12-µF capacitor.
Apply the basic capacitance equation.
Q
Q 29 µC
= 2.4 V
C=
→ ΔV = =
ΔV
C 12 µF
LEARN MORE
Remarks Notice that the equivalent capacitance is less than that of any of
the individual capacitors. The relationship C = Q/ΔV can be used to find
the voltage drops on the other capacitors, just as in part (c).
Question Over which capacitor is the voltage drop the smallest?
mc
the 3.0-µF capacitor
the 6.0-µF capacitor
the 12-µF capacitor
the 24-µF capacitor
Over which capacitor is the voltage drop the largest?
mc
the 3.0-µF capacitor
capacitor
the 6.0-µF capacitor
the 24-µF capacitor
the 12-µF
Bright Storm on Capacitors in
Circuits
(A must watch!!!!!)
Problem-Solving Strategy
• Be careful with the choice of units.
• Combine capacitors following the formulas.
– When two or more unequal capacitors are connected in
series, they carry the same charge, but the potential
differences across them are not the same.
• The capacitances add as reciprocals and the equivalent
capacitance is always less than the smallest individual capacitor.
– When two or more capacitors are connected in parallel,
the potential differences across them are the same.
• The charge on each capacitor is proportional to its capacitance.
• The capacitors add directly to give the equivalent capacitance.
Section 16.8
Problem-Solving Strategy, Final
• Redraw the circuit after every combination.
• Repeat the process until there is only one single
equivalent capacitor.
– A complicated circuit can often be reduced to one
equivalent capacitor.
• Replace capacitors in series or parallel with their equivalent.
• Redraw the circuit and continue.
• To find the charge on, or the potential difference
across, one of the capacitors, start with your final
equivalent capacitor and work back through the
circuit reductions.
Section 16.8
Problem-Solving Strategy, Equation
Summary
• Use the following equations when working through
the circuit diagrams:
– Capacitance equation: C = Q / V
– Capacitors in parallel: Ceq = C1 + C2 + …
– Capacitors in parallel all have the same voltage differences
as does the equivalent capacitance.
– Capacitors in series: 1/Ceq = 1/C1 + 1/C2 + …
– Capacitors in series all have the same charge, Q, as does
their equivalent capacitance.
Section 16.8
Circuit Reduction Example
Section 16.8
EXAMPLE 16.9 Equivalent Capacitance
To find the equivalent capacitance of the circuit in (a), use the series and parallel rules described in the
text to successively reduce the circuit as indicated in (b), (c), and (d).
Goal Solve a complex combination of series and parallel capacitors.
Problem (a) Calculate the equivalent capacitance between a and b for the
combination of capacitors shown in figure a. All capacitances are in
microfarads. (b) If a 12-V battery is connected across the system between
points a and b, find the charge on the 4.0-µF capacitor in the first diagram
and the voltage drop across it.
Strategy For part (a), reduce the combination step by step, as indicated in
the figure. For part (b), to find the charge on the 4.0-µF capacitor, start
with figure c, finding the charge on the 2.0-µF capacitor. This same charge
is on each of the 4.0-µF capacitors in the second diagram. One of these
4.0-µF capacitors in the second diagram is simply the original 4.0-µF
capacitor in the first diagram.
SOLUTION
(a) Calculate the equivalent capacitance.
Find the equivalent capacitance of the parallel 1.0-µF and 3.0-µF
capacitors in figure a:
Ceq = C1 + C2 = 1.0 µF + 3.0 µF = 4.0 µF
Find the equivalent capacitance of the parallel 2.0-µF and 6.0-µF
capacitors in figure a:
Ceq = C1 + C2 = 2.0 µF + 6.0 µF = 8.0 µF
Combine the two series 4.0-µF capacitors in figure b.
1
1
1
1
1
1
= + =
+
=
→ Ceq = 2.0 µF
Ceq C1 C2 4.0 µF 4.0 µF 2.0 µF
Combine the two series 8.0-µF capacitors in figure b.
1
1
1
1
1
1
= + =
+
=
→ Ceq = 4.0 µF
Ceq C1 C2 8.0 µF 8.0 µF 4.0 µF
Finally, combine the two parallel capacitors in figure c to find the
equivalent capacitance between a and b.
Ceq = C1 + C2 = 2.0 µF + 4.0 µF = 6.0 µF
(b) Find the charge on the 4.0 µF capacitor and the voltage drop across it.
Compute the charge on the 2.0-µF capacitor in figure c, which is the same
as the charge on the 4.0-µF capacitor in figure a:
Q
C=
→ Q = CΔV = (2.0 µF)(12 V) = 24 µC
ΔV
Use the basic capacitance equation to find the voltage drop across the 4.0µF capacitor in figure a:
Q
Q 24 µF
C=
→ ΔV = = 4.0 µF = 6.0 V
ΔV
C
LEARN MORE
Remarks To find the rest of the charges and voltage drops, it's just a
matter of using C = Q/ΔV repeatedly, together with facts 5C and 5E in the
Problem-Solving Strategy. The voltage drop across the 4.0-µF capacitor
could also have been found by noticing, in figure b, that both capacitors
had the same value and so by symmetry would split the total drop of 12
volts between them.
Question Which capacitor holds more charge, the 1.0-µF capacitor or the
3.0-µF capacitor?
mc
The 1.0-µF capacitor.
have the same charge.
The 3.0-µF capacitor.
They both
Read and take notes on pgs: 555-556
Energy Stored in a Capacitor
• Energy stored = ½ Q ΔV
• From the definition of capacitance, this can be
rewritten in different forms.
Section 16.9
Application
• Defibrillators
– When fibrillation occurs, the heart produces a rapid,
irregular pattern of beats.
– A fast discharge of electrical energy through the heart can
return the organ to its normal beat pattern.
• In general, capacitors act as energy reservoirs that
can be slowly charged and then discharged quickly to
provide large amounts of energy in a short pulse.
Section 16.9
EXAMPLE 16.10 Typical Voltage, Energy, and Discharge Time for a
Defibrillator
Goal Apply energy and power concepts to a capacitor.
Problem A fully charged defibrillator contains 1.20 kJ of energy stored in
a 1.10 10-4 F capacitor. In a discharge through a patient, 6.00 102 J of
electrical energy are delivered in 2.50 ms. (a) Find the voltage needed to
store 1.20 kJ in the unit. (b) What average power is delivered to the
patient?
Strategy Because we know the energy stored and the capacitance, we can
use the equation below to find the required voltage in part (a). For part (b),
dividing the energy delivered by the time gives the average power.
2
1
1
Q
Energy Stored = QΔV = C(ΔV)2 =
2
2
2C
SOLUTION
(a) Find the voltage needed to store 1.20 kJ in the unit.
Solve for ΔV.
Energy stored = 1/2CΔV2
ΔV =
√
2 (energy stored)
=
C
√
2(1.20 103 J)
1.10 10-4 F
= 4.67 103 V
(b) What average power is delivered to the patient?
Divide the energy delivered by the time.
av
=
energy delivered
Δt
=
6.00 102 J
-3
2.50 10 s
= 2.40 105 W
LEARN MORE
Remarks The power delivered by a draining capacitor isn't constant, as
we'll find in the study of RC circuits later on. For that reason, we were
able to find only an average power. Capacitors are necessary in
defibrillators because they can deliver energy far more quickly than
batteries. Batteries provide current through relatively slow chemical
reactions, whereas capacitors can quickly release charge that has already
been produced and stored.
Question If the voltage across the capacitor were doubled, the energy
stored would be multiplied by:
4
Advanced Explanation of
Capacitance with
Walter Lewin
(Optional)
Read and take notes on pgs: 557-559
Capacitors with Dielectrics
• A dielectric is an insulating material that, when
placed between the plates of a capacitor, increases
the capacitance.
– Dielectrics include rubber, plastic, or waxed paper.
• C = κCo = κεo(A/d)
– The capacitance is multiplied by the factor κ when the
dielectric completely fills the region between the plates.
– κ is called the dielectric constant.
Section 16.10
Capacitors with Dielectrics
Section 16.10
Dielectric Strength
• For any given plate separation, there is a
maximum electric field that can be produced
in the dielectric before it breaks down and
begins to conduct.
• This maximum electric field is called the
dielectric strength.
Section 16.10
Commercial Capacitor Designs
Section 16.10
An Atomic Description of Dielectrics
• Polarization occurs when there is a separation
between the average positions of its negative
charge and its positive charge.
• In a capacitor, the dielectric becomes
polarized because it is in an electric field that
exists between the plates.
• The field produces an induced polarization in
the dielectric material.
Section 16.10
More Atomic Description
• The presence of the positive
charge on the dielectric
effectively reduces some of
the negative charge on the
metal
• This allows more negative
charge on the plates for a
given applied voltage
• The capacitance increases
Section 16.10
Link to Webassign
Unit 3 A Electric Potential and
Capacitors
Discussion Questions
Link to Webassign complete
Unit 3 A Electric Potential and
Capacitance
Homework #7-11
Grading Rubric for Unit 3 A Electric Potential &Capacitance
Name: ______________________
Conceptual notes Lesson 1 “a” Electric Fields and Moving Charge ---------------------------------------____
Lesson 1 “b, c” Electric Potential Difference---------------------------------------------____
Advanced notes from text book:
Pgs 531-533---------------------------------------------------------------------------------------------_____
Pgs 535-537---------------------------------------------------------------------------------------------_____
Pgs 538-540---------------------------------------------------------------------------------------------_____
Pgs 542-544---------------------------------------------------------------------------------------------_____
Pgs 546-547---------------------------------------------------------------------------------------------_____
Pgs 548-550---------------------------------------------------------------------------------------------_____
Pgs 551-553---------------------------------------------------------------------------------------------_____
Pgs 555-556---------------------------------------------------------------------------------------------_____
Pgs 557-559---------------------------------------------------------------------------------------------_____
Example Problems:
16.1-------------------------------------------------------------------------------------------------------_____
16.2-------------------------------------------------------------------------------------------------------_____
16.3-------------------------------------------------------------------------------------------------------_____
16.4-------------------------------------------------------------------------------------------------------_____
16.6-------------------------------------------------------------------------------------------------------_____
16.7-------------------------------------------------------------------------------------------------------_____
16.8-------------------------------------------------------------------------------------------------------_____
16.9-------------------------------------------------------------------------------------------------------_____
16.10------------------------------------------------------------------------------------------------------_____
Web Assign 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 ------------------------------------------------------------------------____
2002B5B. (10 points) Two parallel conducting plates, each of area 0.30 m2, are separated by a
distance of 2.0 x 10-2 m of air. One plate has charge +Q; the other has charge -Q. An electric field
of 5000 N/C is directed to the left in the space between the plates, as shown in the diagram above.
(a) Indicate on the diagram which plate is positive (+) and which is negative (-).
(b) Determine the potential difference between the plates.
(c) Determine the capacitance of this arrangement of plates.
An electron is initially located at a point midway between the plates.
(d) Determine the magnitude of the electrostatic force on the electron at this location and state its
direction.
(e) If the electron is released from rest at this location midway between the plates, determine its
speed just before striking one of the plates. Assume that gravitational effects are negligible.
2001B3. Four charged particles are held fixed at the corners of a
square of side s. All the charges have the same magnitude Q, but
two are positive and two are negative. In Arrangement 1, shown
above, charges of the same sign are at opposite corners. Express
your answers to parts a. and b. in terms of the given quantities and
fundamental constants.
a. For Arrangement 1, determine the following.
i. The electrostatic potential at the center of the square
ii. The magnitude of the electric field at the center of the
square
The bottom two charged particles are now switched to form
Arrangement 2, shown above, in which the positively charged
particles are on the left and the negatively charged particles are on
the right.
b. For Arrangement 2, determine the following.
i. The electrostatic potential at the center of the square
ii. The magnitude of the electric field at the center of the square
c. In which of the two arrangements would more work be required
to remove the particle at the upper right corner from its present
position to a distance a long way away from the arrangement?
_________ Arrangement 1
Justify your answer
___________ Arrangement 2
(15 points) 1998B2. A wall has a negative charge distribution producing a uniform horizontal
electric field. A small plastic ball of mass 0.01 kg, carrying a charge of -80.0 C is suspended by
an uncharged, nonconducting thread 0.30 m long. The thread is attached to the wall and the ball
hangs in equilibrium, as shown above, in the electric and gravitational fields. The electric force
on the ball has a magnitude of 0.032 N.
a. On the diagram below, draw and label the forces acting on the ball.
b. Calculate the magnitude of the electric field at the ball's location due to the charged wall, and
state its direction relative to the coordinate axes shown.
c. Determine the perpendicular distance from the wall to the center of the
ball.
d. The string is now cut.
i. Calculate the magnitude of the resulting acceleration of the ball, and
state its direction relative to the coordinate axes shown.
ii.
Describe the resulting path of the ball.
1990B2. A pair of square parallel conducting plates, having sides of length 0.05 meter, are 0.01
meter apart and are connected to a 200-volt power supply, as shown above. An electron is
moving horizontally with a speed of 3 x 107 meters per second when it enters the region between
the plates. Neglect gravitation and the distortion of the electric field around the edges of the
plates.
a. Determine the magnitude of the electric field in the region between the plates and indicate its
direction on the figure above.
b. Determine the magnitude and direction of the acceleration of the electron in the region between
the plates.
c. Determine the magnitude of the vertical displacement of the electron for the time interval
during which it moves through the region between the plates.
d. On the diagram below, sketch the path of the electron as it moves through and after it
emerges from the region between the plates. The dashed lines in the diagram have been added
for reference only.
e.
A magnetic field could be placed in the region between the plates which would cause the
electron to continue to travel horizontally in a straight line through the region between the
plates. Determine both the magnitude and the direction of this magnetic field.