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+++++++++++++. - - - - - - - - - - - - -. +++++++++++++. - - - - - - - - - - - - -. Capacitance Energy & Dielectrics. Q. V. E. 0. 0. Q. V. E. - Q. A. r. a. + + + +. + Q. d. - - - - -. b. L. - Q. a. + Q. b. Cylindrical. Spherical. Parallel Plates. - PowerPoint PPT Presentation

Capacitance Energy & Dielectrics

CapacitanceEnergy & DielectricsE0QEV0V+++++++++++++- - - - - - - - - - - - -

+++++++++++++- - - - - - - - - - - - -

QLecture 8#Summary of important capacitor geometries The definition of the capacitance relates Q to V via C:

The capacitance depends on the geometry:dA- - - - -+ + + +Parallel Plates

abL

r+Q-QCylindrical

ab+Q-QSpherical

In SI unit system: C has units of Farads or F (1F = 1C/V)e o has units of F/m

Lecture 8#(a) V = (2/3)V0(b) V = V0(c) V = (3/2)V0What is the relationship between V0 and V in the systems shown below?d(Area A)V0+Q-Qconductor(Area A)V+Q-Qd/3d/3Lecture 8# The electric field in the conductor = 0. The electric field everywhere else is: E = Q/(Ae0) To find the potential difference, integrate the electric field:

(a) V = (2/3)V0(b) V = V0(c) V = (3/2)V0What is the relationship between V0 and V in the systems shown below?d(Area A)V0+Q-Qconductor(Area A)V+Q-Qd/3d/3Lecture 8#Another Way(a) V = (2/3)V0(b) V = V0(c) V = (3/2)V0What is the relationship between V0 and V in the systems shown below?d(Area A)V0+Q-Qconductor(Area A)V+Q-Qd/3d/3 The arrangement on the right is equivalent to capacitors (each with separation = d/3) in SERIES!!

conductord/3(Area A)V+Q-Qd/3d/3+Q-Qd/3

Lecture 8#Energy of a CapacitorHow much energy is stored in a charged capacitor? Calculate the work provided (usually by a battery) to charge a capacitor to +/- Q:Calculate incremental work dW needed to add charge dq to capacitor at voltage V (there is a trick here!):

- +In terms of the voltage V:The total work W to charge to Q is then given by:

Look at this!Two ways to write WLecture 8#Capacitor VariablesIn terms of the voltage V:

The total work to charge to Q equals the energy U stored in the capacitor:

Look at this!Two ways to write U

You can do one of two things to a capacitor :hook it up to a battery specify V and Q followsput some charge on it specify Q and V followsLecture 8#Question!dA- - - - -+ + + +Suppose the capacitor shown here is charged to Q and then the battery is disconnected. Now suppose I pull the plates further apart so that the final separation is d1.How do the quantities Q, C, E, V, U change?How much do these quantities change?.. exercise for student!!Answers:

Q:C:E:V:U:remains the same.. no way for charge to leave.increases.. add energy to system by separatingdecreases.. since capacitance depends on geometryincreases.. since C , but Q remains same (or d but E the same)remains the same... depends only on charge densityLecture 8#Related QuestionSuppose the battery (V) is kept attached to the capacitor. Again pull the plates apart from d to d1.Now what changes?How much do these quantities change?.. exercise for student!!Answers:

C:V:Q:E:U:decreases (capacitance depends only on geometry)must stay the same - the battery forces it to be Vmust decrease, Q=CV charge flows off the platedA- - - - -+ + + +Vmust decrease ( )

must decrease ( , )

Lecture 8#Two identical parallel plate capacitors are connected to a battery, as shown in the figure. C1 is then disconnected from the battery, and the separation between the plates of both capacitors is doubled. 2) What is the relation between the charges on the two capacitors ?a) Q1 > Q2 b) Q1 = Q2 c) Q1 < Q23) How does the electric field between the plates of C2 change as separation between the plates is increased ? The electric field:a) increases b) decreases c) doesnt change

Preflight 8:Lecture 8#Two identical parallel plate capacitors are connected to a battery, as shown in the figure. C1 is then disconnected from the battery, and the separation between the plates of both caps is doubled.5) What is the relation between the voltages on the two capacitors? a) V1 > V2 b) V1 = V2 c) V1 < V2

Preflight 8:Lecture 8#ddV2d2dVC1C2Two identical parallel plate capacitors are connected to a battery. What is the relation between the U1, the energy stored in C1, and the U2, energy stored in C2?C1 is then disconnected from the battery and the separation between the plates of both capacitors is doubled.(a) U1 < U2(b) U1 = U2(c) U1 > U2Lecture 8#ddV2d2dVC1C2Two identical parallel plate capacitors are connected to a battery. What is the relation between the U1, the energy stored in C1, and the U2, energy stored in C2?C1 is then disconnected from the battery and the separation between the plates of both capacitors is doubled.(a) U1 < U2(b) U1 = U2(c) U1 > U2 What is the difference between the final states of the two capacitors? The charge on C1 has not changed. The voltage on C2 has not changed. The energy stored in C1 has definitely increased since work must be done to separate the plates with fixed charge, they attract each other. The energy in C2 will actually decrease since charge must leave in order to reduce the electric field so that the potential remains the same.

Initially:

Later:Lecture 8#Where is the Energy Stored?Claim: energy is stored in the electric field itself. Think of the energy needed to charge the capacitor as being the energy needed to create the field. The electric field is given by:

The energy density u in the field is given by:

Units:This is the energy density, u, of the electric field.To calculate the energy density in the field, first consider the constant field generated by a parallel plate capacitor, where

++++++++ +++++++- - - - - - - - - - - - - -- Q+QLecture 8#Energy DensityExample (and another exercise for the student!)Consider E- field between surfaces of cylindrical capacitor:Calculate the energy in the field of the capacitor by integrating the above energy density over the volume of the space between cylinders.is general and is not restricted to the special case of the constant field in a parallel plate capacitor.Claim: the expression for the energy density of the electrostatic field

Compare this value with what you expect from the general expression:

Lecture 8#Consider two cylindrical capacitors, each of length L. C1 has inner radius 1 cm and outer radius 1.1cm.C2 has inner radius 1 cm and outer radius 1.2cm.(a) U2 < U1(b) U2 = U1(c) U2 > U1If both capacitors are given the same amount of charge, what is the relation between U1, the energy stored in C1, and U2, the energy stored in C2?

1C2

1.21C11.1Lecture 8#Consider two cylindrical capacitors, each of length L. C1 has inner radius 1 cm and outer radius 1.1cm.C2 has inner radius 1 cm and outer radius 1.2cm.(a) U2 < U1(b) U2 = U1(c) U2 > U1If both capacitors are given the same amount of charge, what is the relation between U1, the energy stored in C1, and U2, the energy stored in C2?

1C2

1.21C11.1The magnitude of the electric field from r = 1 to 1.1 cm is the same for C1 and C2. But C2 also has electric energy density in the volume 1.1 to 1.2 cm. In formulas:

Lecture 8#DielectricsEmpirical observation:Inserting a non-conducting material between the plates of a capacitor changes the VALUE of the capacitance.Definition:The dielectric constant of a material is the ratio of the capacitance when filled with the dielectric to that without it:

k values are always > 1 (e.g., glass = 5.6; water = 78)They INCREASE the capacitance of a capacitor (generally good, since it is hard to make big capacitors)They permit more energy to be stored on a given capacitor than otherwise with vacuum (i.e., air):

Lecture 8#Parallel Plate ExampleDeposit a charge Q on parallel plates filled with vacuum (air)capacitance C0.Disconnect from batteryThe potential difference is V0 = Q / C0.Now insert material with dielectric constant k .Charge Q remains constantCapacitance increases C = k C0Voltage decreases from V0 to:

Q+++++++++++++++- - - - - - - - - - - - - - -E0V0

Electric field decreases also: - - - - - - - - - - - - - - -

+++++++++++++++QVE+-+-+-+-+-+-+-

Note: The field only decreases when the charge is held constant!Lecture 8#Parallel Plate Example, assuming Q constantHOW ABOUT GAUSS' LAW?How can field decrease if charge remains the same?Answer: the dielectric becomes polarized in the presence of the field due to Q. The molecules partially align with the field so that their negative charge tends toward the positive plate. The field due to this rearrangement inside the dielectric (a la dipole) opposes the original field and is therefore responsible for the reduction in the effective field.Some of the original field lines now end on charges in the dielectric

QQEV0VE0+++++++++++++- - - - - - - - - - - - -

+++++++++++++- - - - - - - - - - - - -

++-+-+-+-+-+-+-+-+-4 modify Gauss Law, see AppendixLecture 8#

Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After C2 has been charged and disconnected it is filled with a dielectric. 7) Compare the voltages of the two capacitors.a) V1 > V2 b) V1 = V2 c) V1 < V28) Compare the electric fields between the plates of bothcapacitors.a) E1 > E2 b) E1 = E2 c) E1 < E2Preflight 8:Lecture 8#10) When we insert the dielectric into the capacitor C2 we do:a) positive work

Preflight 8:Recall the meaning of negative work the energy of the system is reduced. The dielectric is sucked into the capacitor. When the charge is constant, the total energy of the capacitor decreases because the presence of the dielectric increases the capacitance . It turns out that the dielectric is pulled in even if