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# Physics 2102 Lecture: 08 THU 11 FEB - PowerPoint PPT Presentation

Physics 2102 Jonathan Dowling. Physics 2102 Lecture: 08 THU 11 FEB. Capacitance I. 25.1–4. Capacitors and Capacitance. Capacitor: any two conductors, one with charge +Q , other with charge –Q. –Q. Potential DIFFERENCE between conductors = V. +Q.

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Jonathan Dowling

### Physics 2102 Lecture: 08 THU 11 FEB

Capacitance I

25.1–4

Capacitor: any two conductors, one with charge +Q, other with charge –Q

–Q

Potential DIFFERENCE between conductors = V

+Q

Uses: storing and releasing electric charge/energy.

Most electronic capacitors:

New technology:

compact 1 F capacitors

Q = CV where C = capacitance

Units of capacitance:

–Q

Capacitance

• Capacitance depends only on GEOMETRICAL factors and on the MATERIAL that separates the two conductors

• e.g. Area of conductors, separation, whether the space in between is filled with air, plastic, etc.

(We first focus on capacitors

where gap is filled by AIR!)

Electrolytic (new)

Paper (1940-70)

Capacitors

Variable

air, mica

Ceramic (1930 on)

Tantalum (1980 on)

Mica (1930-50)

We want capacitance: C = Q/V

E field between the plates: (Gauss’ Law)

Area of each plate = A

Separation = d

charge/area = 

= Q/A

Relate E to potential difference V:

+Q

-Q

What is the capacitance C ?

• A huge parallel plate capacitor consists of two square metal plates of side 50 cm, separated by an air gap of 1 mm

• What is the capacitance?

C = 0A/d

= (8.85 x 10–12 F/m)(0.25 m2)/(0.001 m)

= 2.21 x 10–9 F

(Very Small!!)

Lesson: difficult to get large values

of capacitance without special

tricks!

+Q

Isolated Parallel Plate Capacitor

• A parallel plate capacitor of capacitance C is charged using a battery.

• Charge = Q, potential difference = V.

• Battery is then disconnected.

• If the plate separation is INCREASED, does Potential DifferenceV:

(a) Increase?

(b) Remain the same?

(c) Decrease?

• Q is fixed!

• C decreases (=0A/d)

• V=Q/C; V increases.

+Q

Parallel Plate Capacitor & Battery

• A parallel plate capacitor of capacitance C is charged using a battery.

• Charge = Q, potential difference = V.

• Plate separation is INCREASED while battery remains connected.

Does the Electric Field Inside:

(a) Increase?

(b) Remain the Same?

(c) Decrease?

• V is fixed by battery!

• C decreases (=0A/d)

• Q=CV; Q decreases

• E = Q/0A decreases

What is the electric field inside

the capacitor? (Gauss’ Law)

Radius of outer plate = b

Radius of inner plate = a

Concentric spherical shells:

Charge +Q on inner shell,

–Q on outer shell

Relate E to potential difference

between the plates:

What is the capacitance?

C = Q/V =

Radius of outer plate = b

Radius of inner plate = a

Concentric spherical shells:

Charge +Q on inner shell,

–Q on outer shell

Isolated sphere: let b >> a,

What is the electric field in between the plates? Gauss’ Law!

Radius of outer plate = b

Radius of inner plate = a

Length of capacitor = L

+Q on inner rod, –Q on outer shell

Relate E to potential difference

between the plates:

cylindrical Gaussian

• Any two charged conductors form a capacitor.

• Capacitance : C= Q/V

• Simple Capacitors:Parallel plates: C = 0 A/dSpherical: C = 4e0 ab/(b-a)Cylindrical: C = 20 L/ln(b/a)]

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

PAR-V (Parallel V the Same)

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

SERI-Q (Series Q the Same)

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 C) 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