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)– 1 F = 1 C / 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:

C = ε0 A

= (8.85 10-12 C2/N · m2) ( 2.00 10-4 m2 ) d 1.00 10-3 m

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:

σ = Q

= 5.31 10-12 C

= 2.66 10-8 C/m2 A 2.00 10-4 m2

(d) Calculate the magnitude of the electric field between the plates.

Apply ΔV = Ed.

E = ΔV

= 3.00 V

= 3.00 103 V/m d 1.00 10-3 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 charge is halved.

The capacitance is doubled. 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.

C = Q → ΔV =

Q =

29 µC = 2.4 V

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 the 6.0-µF capacitor the 12-µF

capacitor the 24-µF capacitor

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:

C = Q → Q = CΔV = (2.0 µF)(12 V) = 24 µC ΔV

Use the basic capacitance equation to find the voltage drop across the 4.0-µF capacitor in figure a:

C = Q → ΔV =

Q =

24 µF = 6.0 V

ΔV C 4.0 µF

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. The 3.0-µF capacitor. They both have the same charge.

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.

Energy Stored = 1 QΔV =

1 C(ΔV)2 =

Q2 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)

= √ 2(1.20 103 J)

C 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

= 6.00 102 J

= 2.40 105 W Δt 2.50 10-3 s

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 ___________ Arrangement 2 Justify your answer

(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.