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Physics 202, Lecture 6Today’s Topics

Capacitance (Ch. 24.1-24.3)Calculating capacitanceCombinations of capacitors: series and parallel

Energy stored in capacitors (24.4)

Capacitance

Capacitance:

stores electrical energy (by storing charge)

Capacitor: two (spatially separated) conductors, charged to +Q and -Q, with constant potential difference ΔV

C !Q

"V

Units: Farad (F)1 F= 1 C/VC

a bOur first exampleof a circuit element:

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Charging a Capacitor

UnchargedCharging

flow

of e

-

ΔV=V+-V-

Battery: source of ΔV

CC

Text problems: 24.2, 24.6, 24.8

Charge flows until Q=CΔV

Charging/discharging capacitor:

Calculating CapacitanceCapacitance is independent of charge, voltage oncapacitor: it depends on geometry of the conductors.

Examples: Parallel conducting platesConcentric spherical conductors

Concentric cylindrical conductors

Procedure: determine as a function of !V Q

Simple example: capacitance of sphere (imagine second conductor at infinity)

!V = V =kQ

RC =

Q

!V=

Q

kQ R=R

k= 4"#

0R

Text: 24.13

3

Example: Parallel Plate CapacitorConsider two metallic parallel plates with area A, separation d:

d

A

- - - - -+ + + +!

+Q

-Q

Step 1. Use Gauss’s law to get E: A ! d

!"

! = Q A

!A

!Eid!A = E

in!A ="" # !A $

0

Ein= ! "

0

!E

Step 2. Get potential difference:

!V = "!Eid!l =#

$d

%0

=Qd

%0A

Step 3. Get Capacitance: C =Q

!V="0A

d

Parallel Plate Capacitor

Realistic case

C =!0A

d

fringe field

Text: 24.15,…

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Spherical Capacitor

C =ab

k(b ! a)

-Q

+Q

Consider two concentric conducting spherical shells, radii a,ba < b

Step 1. Get electric field:

Gauss’s Law: ba

!E =

kQ

r2r̂

!V = "!Eid#!l = "

kQ

r2dr

b

a

# = kQ1

a"1

b

$%&

'()=kQ(b " a)

ab

(a < r < b)

Step 2. Get potential difference:

Step 3:

C =!

2k ln(b / a)

Cylindrical Capacitor

Consider two concentric cylindrical conducting shells, radii a,b

a < b

Step 1. Get electric field:

Gauss’s Law:

!E =

2k!

rr̂

!V = "!Eid#!l = "

2k$

rdr

b

a

# = 2k$ ln b a( )

(a < r < b)

Step 2. Get potential difference:

Step 3:

Q = !!

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Calculating Capacitance• Given 4 concentric cylinders

of radii a,b,c,d and charges+Q, -Q, +Q, -Q

• Question: What is thecapacitance between a and d?

• Note: E=0 between b and c! WHY??• A cylinder of radius r1: b < r1< c

encloses zero charge!

-Q+Q

-Q+Qa

b

c

d

Note: result of 2 cylindrical capacitors“in series” (next slides)

Vad =Q

2!"0rL

a

b

# dr + 0 +Q

2!"0rL

c

d

# dr =Q

2!"0Lln

b

a

$%&

'()+ ln

d

c

$%&

'()

$%&

'()

C =Q

Vad=

2!"0L

lnb

a

#$%

&'(+ ln

d

c

#$%

&'(

#$%

&'(

Combinations of Capacitors

ΔV1

ΔV2

ΔV

C1

C2

Q1

Q2

ΔV

ΔV

CP

Q=Q1+Q2

Equivalent Capacitance CP=Q/ΔV C = C1+ C2

C1ΔV1 =Q1C2ΔV2 =Q2

ΔV1 =ΔV2 =ΔV(why?)

CP= C1+C2+C3+…

effectively

Parallel combination: ΔV same

CP always > Ci!

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Combinations of Capacitors

effectivelyΔV1 ΔV2

+Q1

ΔV=ΔV1+ΔV2

Q

ΔV

Equivalent CapacitanceCS=Q/ΔV 1/CS = 1/C1+1/C2

C1C2

CS

Charge conservation: Q1=Q2(=Q)

-Q1 +Q2-Q2

1/CS= 1/C1+1/C2+1/C3+…

C1ΔV1 =QC2ΔV2 =Q

Series combination:

Cs=

C1C2

C1+ C

2

CS always < Ci!

Example: Connection of Charged Capacitors

Two capacitors, C1=1µF and C2=2µF are initiallycharged to ΔV1=1V and ΔV2=2V, respectively.

What are the charges in each capacitor? If the capacitors are connected in parallel as shown by

the dashed lines, what is the charge in each capacitorafter the connection?

C1

C2

- +

- +

More examples: 24.29-31

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Energy of a Capacitor• How much energy is stored in a charged capacitor?

– Calculate the work provided (usually by a battery) to charge acapacitor to +/- Q:

Incremental work dW needed to add charge dq to capacitor at voltage V:- +

• In terms of the voltage V:

• The total work W to charge to Q is then given by:

Two ways to write W

dW = V (q)idq =q

C

!"#

$%&idq

W =1

Cqdq =

0

Q

!1

2

Q2

C

W =1

2CV

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Capacitor Variables

• In terms of the voltage V:

• The total work to charge capacitor to Q equals the energy Ustored in the capacitor:

You can do one of two things to a capacitor :

• hook it up to a battery specify V and Q follows

• put some charge on it specify Q and V follows

U =1

2CV

2

U =1

Cqdq =

0

Q

!1

2

Q2

C

Q = CV

V =Q

C

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Example (I)

d

A

- - - - -+ + + +• Suppose the capacitor shown here is charged

to Q. The battery is then disconnected.

• Now suppose the plates are pulled further apart to a finalseparation d1.

• How do the quantities Q, C, E, V, U change?

• How much do these quantities change?.. See board.

• Q:• C:• E:• V:• U:

remains the same.. no way for charge to leave.

increases.. add energy to system by separating

decreases.. capacitance depends on geometry

increases.. since C ↓, but Q remains same (or d ↑ but E the same)remains the same... depends only on charge density

Answers: C1=d

d1

C V1=d1

dV

U1=

d1

dU

• Suppose the battery (V) is keptattached to the capacitor.

• Again pull the plates apart from d to d1.

• Now what changes?

• 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 plate

d

A

- - - - -+ + + +V

• How much do these quantities change?.. See board.

Answers:

U1=

d

d1

U

C1=

d

d1

C

E1=

d

d1

E

must decrease ( , )

E =!

E0

E =V

D

must decrease ( ) U =

1

2CV

2

Example (II)

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Where is the Energy stored?• Claim: energy is stored in the electric field itself.

• The electric field is given by:

⇒• The energy density u in the field is given by:

Units:

• Consider the example of a constant field generated by a parallelplate capacitor:

++++++++ +++++++

- - - - - - - - - - - - - -- Q

+Q

u =U

volume=

U

Ad=

1

2!

0E

2

E =!

"0

=Q

"0A

U =1

2!

0E

2Ad

U =1

2

Q2

C=

1

2

Q2

( A!0

/ d)

J

m3

Examples (today or next lecture): 24.42, 24.48,…