Physics 2113 Physics 2113 Lecture: 15 MON 29 SEP Lecture: 15 MON 29 SEP

Capacitance I Capacitance I

Physics 2113

Jonathan Dowling

Capacitors and CapacitanceCapacitors and Capacitance

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

+Q

–Q

Uses: storing and releasing electric charge/energy. Most electronic capacitors:micro-Farads (μF),pico-Farads (pF) — 10–12 FNew technology: compact 1 F capacitors

Uses: storing and releasing electric charge/energy. Most electronic capacitors:micro-Farads (μF),pico-Farads (pF) — 10–12 FNew technology: compact 1 F capacitors

Potential DIFFERENCE between conductors = V

Q = CCV where C = capacitanceFor fixed Q the larger the capacitance the larger the stored voltage.

Units of capacitance: Farad (F) = Coulomb/Volt

CapacitanceCapacitance

• 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. +Q

–Q

(We first focus on capacitorswhere gap is filled by AIR!)

Capacitance depends only on GEOMETRICAL factors and on the MATERIAL that separates the two conductors. It does not depend on the voltage across it or the charge on it at all.

(a) Same

(b) Same

Electrolytic (1940-70)Electrolytic (new)

Paper (1940-70)

Tantalum (1980 on) Ceramic (1930 on)Mica (1930-50)

Variableair, mica

CapacitorsCapacitors

Parallel Plate CapacitorParallel Plate Capacitor

+Q

E field between conducting plates:

Relate E to potential difference V:

What is the capacitance C ?

Area of each plate = ASeparation = dcharge/area = σ= Q/A

σ

We want capacitance: C = Q/V 

–Q

d

Capacitance and Your iPhone!Capacitance and Your iPhone!

Parallel Plate Capacitor — ExampleParallel Plate Capacitor — Example

• 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?

Lesson: difficult to get large valuesof capacitance without specialtricks!

Lesson: difficult to get large valuesof capacitance without specialtricks!

C = ε0A/d

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

= 2.21 x 10–9 F

(Very Small!!)

IsolatedIsolated Parallel Plate Capacitor: ICPP Parallel Plate Capacitor: ICPP

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

• Charge = Q, potential voltage difference = V.

• Battery is then disconnected.

If the plate separation is INCREASED, does the capacitance C:

(a) Increase?

(b) Remain the same?

(c) Decrease?

If the plate separation is INCREASED, does the capacitance C:

(a) Increase?

(b) Remain the same?

(c) Decrease?

+Q –Q

• Q is fixed!• d increases!• C decreases (= ε0A/d)• V=Q/C; V increases.

Parallel Plate Capacitor & Battery: ICPPParallel Plate Capacitor & Battery: ICPP

• 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.

+Q –Q

• V is fixed constant by battery!• C decreases (=ε0A/d)• Q=CV; Q decreases• E = σ/ε0 = Q/ε0A decreases

Does the Electric Field Inside:

(a) Increase?

(b) Remain the Same?

(c) Decrease? Battery does work on capacitor to maintain constant V!

Spherical CapacitorSpherical Capacitor

What is the electric field insidethe capacitor? (Gauss’ Law)

Radius of outer plate = bRadius of inner plate = a

Concentric spherical shells:Charge +Q on inner shell,–Q on outer shellRelate E to potential difference

between the plates:

Spherical CapacitorSpherical Capacitor

What is the capacitance?C = Q/V =

Radius of outer plate = bRadius of inner plate = a

Concentric spherical shells:Charge +Q on inner shell,–Q on outer shell

Isolated sphere: let b >> a,

Cylindrical CapacitorCylindrical Capacitor

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

Relate E to potential differencebetween the plates:

Radius of outer plate = bRadius of inner plate = a

cylindrical Gaussiansurface of radius r

Length of capacitor = L+Q on inner rod, –Q on outer shell

SummarySummary• Any two charged conductors form a capacitor.

•Capacitance : C= Q/V

•Simple Capacitors:

Parallel plates: C = ε0 A/d

Spherical: C = 4πε0 ab/(b-a)

Cylindrical: C = 2πε0 L/ln(b/a)]

•Parallel plates: C = ε0 A/d

Spherical:

•

Cylindrical: C = 2πε0 L/ln(b/a)]

Hooked to batteryV is constant.

Q=VC

(a) d increases -> C decreases -> Q decreases

(b) a inc -> separation d=b-a dec. -> C inc. -> Q increases

(a) b increases -> separation d=b-a inc.-> C inc. -> Q increases

Capacitors in Parallel: V=ConstantCapacitors 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

A B

C D

C1

C2

Q1

Q2

CeqQtotal

V = VAB = VA –VB

V = VCD = VC –VD

VA VB

VC VD

ΔV=V

PAR-V (Parallel V the Same)

PARALLEL: • V is same for all capacitors• Total charge = sum of Q

Capacitors in Series: Q=ConstantCapacitors in Series: Q=Constant

• Q1 = Q2 = Q = Constant

• VAC = VAB + VBC

A B C

C1 C2

Q1 Q2

Ceq

Q = Q1 = Q2

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)

PARALLEL: • V is same for all capacitors• Total charge = sum of Q

(a) Parallel: Voltage is same V on each but charge is q/2 on each since

q/2+q/2=q.

(b) Series: Charge is same q on each but voltage is V/2 on eachsince V/2+V/2=V.

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

““Compiling” in parallel and in seriesCompiling” in parallel and in series

• 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 Series?Example: Parallel or Series?

What is the charge on each capacitor?

C1=10 μF

C3=30 μF

C2=20 μF

120V

• Qi = CiV • V = 120V = Constant• Q1 = (10 μF)(120V) = 1200 μC • Q2 = (20 μF)(120V) = 2400 μC• Q3 = (30 μF)(120V) = 3600 μC

Note that:• Total charge (7200 μC) 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)

Example: Parallel or SeriesExample: Parallel or Series

What is the potential difference across each capacitor?

C1=10μF C3=30μFC2=20μF

120V

• Q = CserV• Q is same for all capacitors• Combined Cser is given by:

• Ceq = 5.46 μF (solve above equation)

• Q = CeqV = (5.46 μF)(120V) = 655 μC

• V1= Q/C1 = (655 μC)/(10 μF) = 65.5 V

• V2= Q/C2 = (655 μC)/(20 μF) = 32.75 V

• V3= Q/C3 = (655 μC)/(30 μF) = 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 Parallel?Example: Series or Parallel?

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

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

10 μF

10μF 10V

10μF

5μF5μF 10V

• The two 5μF capacitors are in parallel

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

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

capacitor of interest• Hence, Q = (10μF )(5V) = 50μC

Neither: Circuit Compilation Needed!