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Physics 2102 Lecture: 12 MON FEB

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Physics 2102

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

- 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

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

- 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

Parallel: Circuit Splits Cleanly in Two (Constant V)

C1=10 F

C2=20 F

C3=30 F

120V

- 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

Series: Isolated Islands (Constant Q)

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

10F

10F

10V

5F

5F

10V

Neither: Circuit Compilation Needed!

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

- 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
- Hence, Q = (10F )(5V) = 50C

- 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

General expression for any region with vacuum (or air)

volume = Ad

- 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)

- 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/dSpherical : 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