51 ch24 capacitance

7/30/2019 51 Ch24 Capacitance

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Capacitance and dielectrics (sec. 24.1)

Capacitors in series and parallel (sec. 24.2)

Energy storage in capacitorsand electric field energy (sec. 24.3)

Dielectrics (sec. 24.4)

Molecular model / polarization (sec. 24.5)

Chapter 24 Capacitance andDielectrics

C 2012 J. Becker


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Learning Goals - we will learn: Ch 24

The nature of capacitors, and how tocalculate their ability to store charge.

How to analyze capacitors connectedin a network.

How to calculate the amount of energystored in a capacitor.

What dielectrics are, and how they make

capacitors more effective.


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Any two conductors insulated from oneanother form a CAPACITOR.

A "charged" capacitorcan store charge.When a capacitor isbeing charged, negativecharge is removed fromone side of the

capacitor and placedonto the other, leavingone side with a negative

charge (-q) and theother side with apositive charge (+q).


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A charged parallelplate capacitor.

Q = C Vwhere C = eo A / d

for a parallel plate capacitor,where eo is the permittivity of

the insulating material(dielectric) between plates.

Recall that we used Gauss's Lawto calculate the electric field (E)between the plates of a charged

capacitor:E = s / eo where there is avacuum between the plates.

Vab

= E d, so E = Vab

/dThe unit of capacitance is called the Farad (F).


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1 / Ceq= 1 / C1 + 1 / C2

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Two capacitors in series andthe equivalent capacitor.

V = V1 + V2 and Q = C V


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Ceq = C1 + C2

Two capacitors in parallel and

the equivalent capacitor.

Q = Q1 + Q2 and Q = C V


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Capacitors can store charge and ENERGYDU = q DV and the potential V increases as the

charge is placed on the plates (V = Q / C).

Since the V changes as the Q is increased, wehave to integrate over all the little charges

dq being added to a plate:DU = q DVleads toU = V dq = q/C dq = 1/CoQ q dq = Q2 / 2C.And using Q = C V, we get

U = Q2 / 2C = C V2/ 2 = Q V / 2

ENERGY STORED IN A CAPACITORThe potential energy stored in the system ofpositive charges that are separated from the

negative charges is like a stretched springthat has otential ener associated with it.


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ELECTRIC FIELD ENERGY

Here's another way to think of the energy

stored in a charged capacitor: If we considerthe space between the plates to contain theenergy (equal to 1/2 C V2) we can calculate an

energy DENSITY (Joules per volume).The volume between the plates is area x plateseparation (A d). Then the energy density u is

u = 1/2 C V2

/ A d =eoE2 / 2Substituting C =eoA / d and V = E d

u = 1/2 (eo A/d)(Ed)2

/Ad = o E2

/2C 2012 J. F. Becker


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Energy density: u = eoE2 / 2This is an important result because it tells usthat empty space contains energy if there is

an electric field (E) in the "empty" space.

If we can get an electric field to travel(or propagate) we can send or transmit

energy and information throughempty space!!!

C 201 2 J. F. Becker


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Effect of a dielectric between the plates of aparallel plate capacitor.

Note the charge is constant !

DIELECTRICCONSTANT:

K= C / Co= ratio of thecapacitances

V = Vo / K


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A dielectric is added between the plates of acharged capacitor (battery not connected):Q = Qo, therefore Q = C V and Qo = Co Vo

Co Vo = C V,and if Vo decreases to V, Co must increase to

C to keep equation balanced, and

V = Vo Co/C

Definition of DIELECTRIC CONSTANT:

K= C / Co = ratio of the capacitances

V = Vo / K

C 2012 J. F. Becker


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You slide a slab of dielectric between the plates of a parallel-plate

capacitor. As you do this, the charges on the plates remain

constant.

What effect does adding the dielectric have on the energy stored

in the capacitor?

A. The stored energy increases.

B. The stored energy remains the same.

C. The stored energy decreases.

D. not enough information given to decide

Q24.18


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The charges induced on the surface of thedielectric (insulator) reduce the electric field.


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Polarization of a

dielectric in anelectric field E givesrise to thin layers of

bound charges onthe dielectrics

surfaces, creating

surface chargedensities

+si and si.


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A neutral sphere

B in the electricfield of a charged

sphere A is

attracted to thecharged spherebecause ofpolarization.


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For each problem:

Draw a clear and carefully labeled diagram.Write the necessary equations in terms ofvariables.

Explicitly show all steps in calculations, including units.


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Polarization of a dielectric inan electric field E

51 ch24 capacitance

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