Physics 2102 Physics 2102 Lecture: 08 THU 18 FEB Lecture: 08 THU 18 FEB

Capacitance II Capacitance II

Physics 2102

Jonathan Dowling

25.4–7

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Capacitors in Parallel: Capacitors in Parallel: V=ConstantV=Constant

• An ISOLATED wire is an equipotential surface: V=Constant

• Capacitors in parallel have SAME potential difference but NOT ALWAYS same charge!

• VAB = VCD = V

• Qtotal = Q1 + Q2

• CeqV = C1V + C2V

• Ceq = C1 + C2

• Equivalent parallel capacitance = sum of capacitances

A B

C D

C1

C2

Q1

Q2

CeqQtotal

V = VAB = VA –VB

V = VCD = VC –VD

VA VB

VC VD

V=V

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Cparallel = C1 + C2

PAR-V (Parallel: V the Same)

Capacitors in Series: Capacitors in Series: Q=ConstantQ=Constant

• Q1 = Q2 = Q = Constant

• VAC = VAB + VBC

A B C

C1 C2

Q1 Q2

Ceq

Q = Q1 = Q2

21 C

Q

C

Q

C

Q

eq

+=

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1

Cseries

=1

C1

+1

C2

SERIES: • Q is same for all capacitors• Total potential difference = sum of V

Isolated Wire:Q=Q1=Q2=Constant

SERI-Q: Series Q the Same

Capacitors in Parallel and in Capacitors in Parallel and in SeriesSeries

• In series : 1/Cser = 1/C1 + 1/C2

Vser = V1  + V2

Qser= Q1 = Q2

C1 C2

Q1 Q2

C1

C2

Q1

Q2

• In parallel : Cpar = C1 + C2

Vpar = V1 = V2

Qpar = Q1 + Q2 Ceq

Qeq

Example: Parallel or Example: Parallel or Series?Series?

What is the charge on each capacitor?

C1=10 F

C3=30 F

C2=20 F

120V

• Qi = CiV • V = 120V on ALL Capacitors (PAR-V)• Q1 = (10 F)(120V) = 1200 C • Q2 = (20 F)(120V) = 2400 C• Q3 = (30 F)(120V) = 3600 CNote that:• Total charge (7200 C) is

shared between the 3 capacitors in the ratio C1:C2:C3

— i.e. 1:2:3

Parallel: Circuit Splits Cleanly in Two (Constant V)

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Cpar = C1 + C2 + C3 = 10 + 20 + 30( )μF = 60μF

Example: Parallel or Example: Parallel or SeriesSeries

What is the potential difference across each capacitor?

C1=10F C3=30FC2=20F

120V

• Q = CserV• Q is same for all capacitors (SERI-Q)• Combined Cser is given by:

€

1

Cser

=1

(10μF)+

1

(20μF)+

1

(30μF)

• Ceq = 5.46 F (solve above

equation)

• Q = CeqV = (5.46 F)(120V) = 655

C

• V1= Q/C1 = (655 C)/(10 F) = 65.5

V

• V2= Q/C2 = (655 C)/(20 F) = 32.75

V

• V3= Q/C3 = (655 C)/(30 F) = 21.8

V

Note: 120V is shared in the ratio of INVERSE capacitances i.e. (1):(1/2):(1/3)

(largest C gets smallest V)

Series: Isolated Islands (Constant Q)

Example: Series or Example: Series or Parallel?Parallel?

In the circuit shown, what is the charge on the 10F capacitor?

In the circuit shown, what is the charge on the 10F capacitor?

10 F

10F 10V

10F

5F5F 10V

• The two 5F capacitors are in parallel

• Replace by 10F • Then, we have two 10F

capacitors in series• So, there is 5V across the 10 F

capacitor of interest by symmetry

• Hence, Q = (10F )(5V) = 50C

Neither: Circuit Compilation Needed!

Energy U Stored in a Energy U Stored in a CapacitorCapacitor

• Start out with uncharged capacitor

• Transfer small amount of charge dq from one plate to the other until charge on each plate has magnitude Q

• How much work was needed? dq

∫=Q

VdqU0

∫ ==Q

C

Qdq

C

q

0

2

2 2

2CV=

Energy Stored in Electric Field of Energy Stored in Electric Field of CapacitorCapacitor

• Energy stored in capacitor: U = Q2/(2C) = CV2/2 • View the energy as stored in ELECTRIC FIELD• For example, parallel plate capacitor: Energy

DENSITY = energy/volume = u =

€

u =Q2

2CAd= =

⎟⎠⎞

⎜⎝⎛ Ad

dA

Q

0

2

2ε 2

0

2

2 A

Q

ε 22

20

2

0

0 E

A

Q εε

ε=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

volume = AdGeneral

expression for any region with vacuum

(or air)

Dielectric ConstantDielectric Constant

• If the space between capacitor

plates is filled by a dielectric, the

capacitance INCREASES by a

factor

• This is a useful, working definition

for dielectric constant.

• Typical values of are 10–200+Q –Q

DIELECTRIC

C = ε A/dC = ε A/d The and the constant εεo are both called dielectric constants. The has no units (dimensionless).

The and the constant εεo are both called dielectric constants. The has no units (dimensionless).

Atomic View

Emol

Ecap

Molecules set up counter E field Emol that somewhat cancels out capacitor field Ecap.

This avoids sparking (dielectric breakdown) by keeping field inside dielectric small.

Hence the bigger the dielectric constant the more charge you can store on the capacitor.

Example Example

• Capacitor has charge Q, voltage V• Battery remains connected while

dielectric slab is inserted.• Do the following increase,

decrease or stay the same:– Potential difference?– Capacitance?– Charge?– Electric field?

dielectric slab

ExampleExample

• Initial values: capacitance = C; charge = Q; potential difference = V; electric field = E;

• Battery remains connected

• V is FIXED; Vnew = V (same)

• Cnew = C (increases)

• Qnew = (C)V = Q (increases).

• Since Vnew = V, Enew = V/d=E (same)

dielectric slab

Energy stored? u=ε0E2/2 => u=ε0E2/2 = εE2/2

SummarySummary• Any two charged conductors form a capacitor.• Capacitance : C= Q/V

• Simple Capacitors:Parallel plates: C = ε0 A/d

Spherical : C = 4 ε0 ab/(b-a)

Cylindrical: C = 2 ε0 L/ln(b/a)

• Capacitors in series: same charge, not necessarily equal potential; equivalent capacitance 1/Ceq=1/C1+1/C2+…

• Capacitors in parallel: same potential; not necessarily same charge; equivalent capacitance Ceq=C1+C2+…

• Energy in a capacitor: U=Q2/2C=CV2/2; energy density u=ε0E2/2

• Capacitor with a dielectric: capacitance increases C’=kC