ch. 24 capacitance & dielectrics

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Ch. 24 Capacitance & Dielectrics AP Physics C

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Ch. 24 Capacitance & Dielectrics. AP Physics C. Capacitance. Ability of a capacitor to store charge Ratio of charge stored to potential difference 1 C/V = 1 Farad (F). Capacitor. A device that stores electric charge - PowerPoint PPT Presentation

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Page 1: Ch. 24 Capacitance & Dielectrics

Ch. 24 Capacitance & Dielectrics

AP Physics C

Page 2: Ch. 24 Capacitance & Dielectrics

Capacitance• Ability of a capacitor to store

charge• Ratio of charge stored to potential

difference• 1 C/V = 1 Farad (F)

QCV

Page 3: Ch. 24 Capacitance & Dielectrics

Capacitor• A device that stores electric charge• Used with resistors in timing circuits

because it takes time for a capacitor to fill with charge

• Used to smooth varying DC supplies by acting as a reservoir of charge

• Used in filter circuits because capacitors easily pass AC (changing) signals but they block DC (constant) signals

Page 4: Ch. 24 Capacitance & Dielectrics

Types of Capacitors• Polarized

• Unpolarized

Page 5: Ch. 24 Capacitance & Dielectrics

Polarized Capacitors• Electrolytic

capacitors are polarized and they must be connected the correct way.

• Examples:

• Schematic Symbol

Page 6: Ch. 24 Capacitance & Dielectrics

Un-polarized Capacitors• Small value

capacitors are un-polarized and may be connected either way round.

• Examples:

• Schematic Symbol

Page 8: Ch. 24 Capacitance & Dielectrics

Factors Affecting Capacitance:• The capacitance of a capacitor is

affected by three factors:– The area of the plates – The distance between the plates – The dielectric constant of the

material between the plates

Page 9: Ch. 24 Capacitance & Dielectrics

Area of the plates:• Larger plates provide greater

capacity to store electric charge. Therefore, as the area of the plates increase, capacitance increases.

Page 10: Ch. 24 Capacitance & Dielectrics

Distance between the plates:• Capacitance is directly

proportional to the electrostatic force field between the plates. This field is stronger when the plates are closer together. Therefore, as the distance between the plates decreases, capacitance increases. As the distance between the plates increases, capacitance decreases.

Page 11: Ch. 24 Capacitance & Dielectrics

The dielectric between the plates:• The ability of the dielectric to

support electrostatic forces is directly proportional to the dielectric constant. Therefore, as the dielectric constant increases, capacitance increases.

Page 12: Ch. 24 Capacitance & Dielectrics

Capacitance:• http://micro.magnet.fsu.edu/electro

mag/java/capacitance/index.html

oACd

Page 13: Ch. 24 Capacitance & Dielectrics

Sample Problem 1• An air-filled parallel-plate capacitor

has a plate area of 2.0 cm2 and a plate separation of 1.00 mm.– Find its capacitance.– If a potential difference of 100 V is

applied across the plates, what is• The charge on each plate and• The electric field between the plates?

– What is its capacitance for a plate separation of 3.00 mm?

Page 14: Ch. 24 Capacitance & Dielectrics

Sample Problem 2• A spherical conductor consists of

a spherical conducting shell of radius b and charge – Q concentric with a smaller conducting sphere of radius a and charge Q. Find the capacitance of this device.

• What is the magnitude of the electric field outside the spherical capacitor?

Page 15: Ch. 24 Capacitance & Dielectrics

Sample Problem 3• A solid cylindrical conductor of

radius a and charge Q is coaxial with a cylindrical shell of negligible thickness, radius b > a, and charge – Q. Find the capacitance of this cylindrical capacitor if its length is l.

• What is the magnitude of the electric field outside the cylindrical capacitor?

Page 16: Ch. 24 Capacitance & Dielectrics

Energy Storage• The work it takes to charge up a

capacitor is equal to the electrical energy stored in a capacitor

• Alternate forms:

Page 17: Ch. 24 Capacitance & Dielectrics

Capacitors in Circuits:• Series Circuit-

– Same charge– Potential differences add

Page 18: Ch. 24 Capacitance & Dielectrics

Capacitors in Circuits:

• Parallel Circuit-– Charges add– Potential differences are the same

Page 19: Ch. 24 Capacitance & Dielectrics

Sample Problem 4• Two capacitors, 6 μF and 10 μF,

are connected in series across a potential difference of 24 V.– Determine:

• Equivalent capacitance,• Charge stored on each capacitor, • Potential difference across each

capacitor, and• Energy stored on each capacitor.

Page 20: Ch. 24 Capacitance & Dielectrics

Sample Problem 5• Two capacitors, 6 μF and 10 μF,

are connected in parallel across a potential difference of 24 V.– Determine:

• Equivalent capacitance,• Charge stored on each capacitor, • Potential difference across each

capacitor, and• Energy stored on each capacitor.

Page 21: Ch. 24 Capacitance & Dielectrics

Sample Problem 6• Given the following values:

– C1 = 10 pF– C2 = 20 pF– C3 = 60 pF– ΔV = 120 V

• Find the equivalent capacitance, charge stored on each capacitor, energy stored in each capacitor and the potential difference across each capacitor.

Page 22: Ch. 24 Capacitance & Dielectrics

Sample Problem 7• A capacitor with a capacitance of 8.0

μF is charged by connecting it to a source of potential difference 120 V. Once the capacitor is charged fully, it is disconnected from the source.– What is the charge stored on the 8.0 μF

capacitor?– What is the energy stored in the capacitor?

Page 23: Ch. 24 Capacitance & Dielectrics

Sample Problem 8• The charged capacitor in the previous

problem is connected to an uncharged 4.0 μF capacitor.– Explain how the charge is distributed once the

connection is made.– How much charge is stored on each capacitor?– What is the potential difference across each

capacitor?– What is the total energy stored on this

combination?– Compare this to the energy stored before the

connection was made.

Page 24: Ch. 24 Capacitance & Dielectrics

Conceptual Questions:• A parallel-plate

capacitor is fully charged and then disconnected from the source. When the plates are pulled apart, do the following quantities increase, decrease, or stay the same?

• Capacitance?

• Charge stored?

• Energy stored?

• Potential difference?

• Electric field strength?

Page 25: Ch. 24 Capacitance & Dielectrics

Conceptual Questions:• A parallel-plate

capacitor is fully charged and remains connected from the source. When the plates are pulled apart, do the following quantities increase, decrease, or stay the same?

• Capacitance?

• Charge stored?

• Energy stored?

• Potential difference?

• Electric field strength?

Page 26: Ch. 24 Capacitance & Dielectrics

Dielectrics:• What are they?

– Non-conducting materials

• What are the three functions of a dielectric?– Increases capacitance– Increases maximum

operating voltage– Mechanical support

that prevents the plates from touching; Allows the plates to be even closer together

Page 27: Ch. 24 Capacitance & Dielectrics

Dielectric Strength:• The Dielectric Strength is the

maximum Electric Field that the material can withstand before the material breaks down as an insulator and permits current to flow through the material.

Page 28: Ch. 24 Capacitance & Dielectrics

Material Dielectric Constantk = e / eo

Dielectric Strength(V/m)

Vacuum 1

Air (1 atm) 1.00054 1-3 x10 6

Air (100 atm) 1.0548

Bakelite (typical) 4.9 24x106

Glass (Pyrex) 4.5-5.5 15x106

Polystyrene (typical) 2.6 25x106

Mylar (typical) 3.5 6-13 x10 6

Paper (typical) 3.5 14x106

Porcelain (typical) 7 4x106

Teflon (typical) 2 60x106

Mineral Oil (typical) 4.5 12x106

Water (20 oC) 80.4 Conductor

Page 29: Ch. 24 Capacitance & Dielectrics

Dielectric Constant:• Depends on the permittivity-The

permittivity of a substance is a characteristic which describes how it affects any electric field set up in it. A high permittivity tends to reduce any electric field present. We can increase the capacitance of a capacitor by increasing the permittivity of the dielectric material.

o

Page 30: Ch. 24 Capacitance & Dielectrics

Voltage remains constant:• A fully-charged parallel-plate capacitor

remains connected to a battery while a dielectric is slide between the plates. Do the following quantities increase, decrease, or stay the same?– Capacitance– Charge stored– Electric field strength between the plates– Potential difference– Energy stored in the capacitor

Page 31: Ch. 24 Capacitance & Dielectrics

Charge remains constant:• A fully-charged parallel-plate capacitor is

disconnected from a battery and then a dielectric is slid between the plates. Do the following quantities increase, decrease, or stay the same?– Capacitance– Charge stored– Electric field strength between the plates– Potential difference– Energy stored in the capacitor

Page 32: Ch. 24 Capacitance & Dielectrics

Sample Problem 9:• A parallel-plate capacitor has plates of

dimensions 2.0 cm by 3.0 cm separated by a 1.0 mm thickness of paper. The dielectric constant of paper is 3.7. – Find its capacitance.– What is the maximum charge that can be

placed on the capacitor? The dielectric strength of paper is 16 x 106 V/m.

– What is the maximum energy that can be stored in the capacitor?

Page 33: Ch. 24 Capacitance & Dielectrics

Sample Problem 10• Suppose that the capacitance in the

absence of a dielectric is 8.50 pF and that the capacitor is charged to a potential difference of 12.0 V. If the battery is disconnected and a slab of polystyrene (dielectric constant of 2.56) is inserted between the plates, what is the Uo – U?

Page 34: Ch. 24 Capacitance & Dielectrics

Polarization of a Dielectric:• If a material contains polar molecules, they

will generally be in random orientations when no electric field is applied. An applied electric field will polarize the material by orienting the dipole moments of polar molecules.

Page 35: Ch. 24 Capacitance & Dielectrics

Parallel-Plate with Dielectric:• The insertion of a dielectric will cause the

effective electric field to decrease and the capacitance to increase.

Page 36: Ch. 24 Capacitance & Dielectrics

Sample Problem 11• A parallel-plate capacitor with a

plate separation of d has a capacitance Co in the absence of a dielectric. What is the capacitance when a slab of dielectric material of dielectric constant K and thickness 1/3d is inserted between the plates?

Page 37: Ch. 24 Capacitance & Dielectrics

Gauss’s law in dielectrics• The enclosed charge,

Qenclosed = (σ – σinduced)A• Applying Gauss’s law, gives

EA = (σ – σinduced)A/εo

Page 38: Ch. 24 Capacitance & Dielectrics

Gauss’s law in dielectrics• Using: E = Eo/K

• Eo = σ/εo

• E = (σ – σinduced)/εo

• You get that: (σ – σinduced) = σ/K

Page 39: Ch. 24 Capacitance & Dielectrics

• EA = (σ – σinduced)A/εo

• Becomes: EA = σA/Kεo

KEA = σA/εo

• Gauss’s law in dielectrics:

∫KE·dA = Qencl-free/ εo

Page 40: Ch. 24 Capacitance & Dielectrics

Sample Problem 12• In the spherical capacitor in

sample problem #2, the volume between the concentric spherical conducting shells is filled with an insulating oil with dielectric constant K. Use Gauss’s law to find the capacitance.