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# Physics 2102 Lecture: 12 MON FEB - PowerPoint PPT Presentation

Physics 2102 Jonathan Dowling. Physics 2102 Lecture: 12 MON FEB. Capacitance II. 25.4–5. V = V AB = V A –V B. C 1. Q 1. V A. V B. A. B. C 2. Q 2. V C. V D. D. C. V = V CD = V C –V D. C eq. Q total.  V=V. Capacitors in Parallel: V=Constant.

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Jonathan Dowling

### Physics 2102 Lecture: 12 MON FEB

Capacitance II

25.4–5

V = VAB = VA –VB

C1

Q1

VA

VB

A

B

C2

Q2

VC

VD

D

C

V = VCD = VC –VD

Ceq

Qtotal

V=V

Capacitors in Parallel: V=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

Q1

Q2

B

C

A

C1

C2

Capacitors in Series: Q=Constant

Isolated Wire:

Q=Q1=Q2=Constant

• Q1 = Q2 = Q = Constant

• VAC = VAB + VBC

Q = Q1 = Q2

• SERIES:

• Q is same for all capacitors

• Total potential difference = sum of V

Ceq

C1

Q1

C2

Q2

Qeq

Q1

Q2

Ceq

C1

C2

Capacitors in parallel and in series

• In parallel :

• Cpar = C1 + C2

• Vpar= V1 = V2

• Qpar= Q1 + Q2

• In series :

1/Cser = 1/C1 + 1/C2

Vser= V1  + V2

Qser= Q1 = Q2

C1=10 F

C2=20 F

C3=30 F

120V

Example: Parallel or Series?

• Qi = CiV

• V = 120V = Constant

• Q1 = (10 F)(120V) = 1200 C

• Q2 = (20 F)(120V) = 2400 C

• Q3 = (30 F)(120V) = 3600 C

What is the charge on each capacitor?

• Note that:

• Total charge (7200 mC) is shared between the 3 capacitors in the ratio C1:C2:C3— i.e. 1:2:3

Example: Parallel or Series

C2=20mF

C3=30mF

C1=10mF

What is the potential difference across each capacitor?

• Q = CserV

• Q is same for all capacitors

• Combined Cser is given by:

120V

• 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)

10 F

10F

10F

10V

5F

5F

10V

Example: Series or Parallel?

Neither: Circuit Compilation Needed!

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

• The two 5F capacitors are in parallel

• Replace by 10F

• Then, we have two 10F capacitors in series

• So, there is 5V across the 10 F capacitor of interest

• Hence, Q = (10F )(5V) = 50C

• 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

Energy Stored in Electric Field of Capacitor

• 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 =

10mF (C1)

20mF (C2)

Example

• 10mFcapacitor is initially charged to 120V.

• 20mF capacitor is initially uncharged.

• Switch is closed, equilibrium is reached.

• How much energy is dissipated in the process?

Initial charge on 10mF = (10mF)(120V)= 1200mC

After switch is closed, let charges = Q1 and Q2.

Charge is conserved: Q1 + Q2 = 1200mC

• Q1 = 400mC

• Q2 = 800mC

• Vfinal= Q1/C1 = 40 V

Also, Vfinal is same:

Initial energy stored = (1/2)C1Vinitial2 = (0.5)(10mF)(120)2 = 72mJ

Final energy stored = (1/2)(C1 + C2)Vfinal2= (0.5)(30mF)(40)2 = 24mJ

Energy lost (dissipated) = 48mJ

• Any two charged conductors form a capacitor.

• Capacitance : C= Q/V

• Simple Capacitors:Parallel plates: C = e0 A/d Spherical : C = 4pe0 ab/(b-a)Cylindrical: C = 2pe0 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 densityu=e0E2/2

• Capacitor with a dielectric: capacitance increasesC’=kC

Summary