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)Q

CV

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

Types of Capacitors

• Polarized

• Unpolarized

Polarized Capacitors• Electrolytic

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

• Examples:

• Schematic Symbol

Un-polarized Capacitors• Small value

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

• Examples:

• Schematic Symbol

Charging/Discharging• http://micro.magn

et.fsu.edu/electromag/java/capacitor/index.html

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

Area of the plates:

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

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.

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.

Capacitance:

• http://micro.magnet.fsu.edu/electromag/java/capacitance/index.html

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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?

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?

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?

Energy Storage

• The work it takes to charge up a capacitor is equal to the electrical energy stored in a capacitor

• Alternate forms:

Capacitors in Circuits:

• Series Circuit-– Same charge– Potential differences add

Capacitors in Circuits:

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

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.

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.

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.

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?

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.

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?

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?

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

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.

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

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.

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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

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

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?

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?

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.

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

effective electric field to decrease and the capacitance to increase.

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?

Gauss’s law in dielectrics

• The enclosed charge,

Qenclosed = (σ – σinduced)A

• Applying Gauss’s law, gives

EA = (σ – σinduced)A/εo

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

• Eo = σ/εo

• E = (σ – σinduced)/εo

• You get that: (σ – σinduced) = σ/K

• EA = (σ – σinduced)A/εo

• Becomes: EA = σA/Kεo

KEA = σA/εo

• Gauss’s law in dielectrics:

∫KE·dA = Qencl-free/ εo

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.

Works Cited:• http://www.kpsec.freeuk.com/capacit.htm• http://micro.magnet.fsu.edu/electromag/electri

city/capacitance.html• http://www.ac.wwu.edu/~vawter/PhysicsNet/T

opics/TopicsMainTemplate.html• http://www.matter.org.uk/schools/SchoolsGlos

sary/permittivity.html• http://hyperphysics.phy-astr.gsu.edu/hbase/el

ectric/dielec.html