Capacitors

Phy 1161: PreLecture 06

• Today’s lecture will cover Textbook Sections 20-5 – 20-6

Comparison:Electric Potential Energy vs. Electric Potential

VAB : the difference in electric potential between points B and A

UAB : the change in electric potential energy of a charge q when moved from A to B

UAB = q VAB

q

A B

Electric Potential: Summary• E field lines point from higher to lower potential• For positive charges, going from higher to lower

potential is “downhill”• For a battery, the (+) terminal is at a higher potential

than the (–) terminalPositive charges tend to go “downhill”, from + to -Negative charges go in the opposite direction, from - to +

UAB = q VAB

Important Special CaseUniform Electric Field

+

+

+

+

+

-

-

-

-

-

Two large parallel conducting plates of area A

+Q on one plate

-Q on other plate

Then E is

uniform between the two plates: E=4kQ/A

zero everywhere else

This result is independent of plate separation

This is call a parallel plate capacitor

Parallel Plate CapacitorPotential Difference

A B

E=E0

d

V = VA – VB = +E0 d

+

+

+

+

+

-

-

-

-

-

A B

d

+

+

+

+

+

-

-

-

-

-

+

+

+

+

+

-

-

-

-

-

E=

V = VA – VB = +2E0d

Potential difference is proportional to charge: Double Q Double V

E0 = 4πkQ/A

Charge Q on plates Charge 2Q on plates

Capacitance

• The ability to store separated charge

• Definition:

• Units: Farad (F) – named in honor of Michael Faraday– 1 F = 1C/V

QC

V

From Faraday’s notebook

Capacitor

• Any pair of conductors separated by a small distance. (e.g. two metal plates)

• Capacitor stores separated charge – Positive Q on one conductor, negative Q on other– Net charge is zero

E

d

+

+

+

+

+

-

-

-

-

-

Q = CV

U = (½) Q V • Stores Energy

Capacitance of Parallel Plate Capacitor

V = Ed AND E = Q/(0A) (Between two large plates)

So: V = Qd/ /(0A) Remember: CQ/V

So:

Equation based on geometry of capacitor

A

d

AE+

-

V

0= 8.85x10-12 C2/Nm2

If there is adielectric (κ>1) between plates C = κ C0

0 AC

d

Dielectric• Placing a dielectric between

the plates increases the capacitance.

C = C0

Capacitance with dielectric

Dielectric constant ( > 1)

Capacitance without dielectric

Material Constant Material ConstantVacuum 1 Germanium 16

Polyvinyl chloride

3.18 Strontium titanate

310

Mica 3 - 6 Water 80.4

Mylar 3.1 Glycerin 42.5

Neoprene 6.70 Benzene 2.284

Plexiglass 3.40 Glass 5 – 10

Polyethylene 2.25 Air (1 atm) 1

Liquid ammonia

(-78oC)

25 Titanium dioxide (rutile)

173 perp

86 para

Dielectrics

Voltage in Circuits• Elements are connected by wires. • Any connected region of wire has the same

potential.

Vwire 1= 0 V Vwire 2= 5 V Vwire 3= 12 V Vwire 4= 15 V

VC1= _____ V VC3

= _____ VVC2= _____ V

C1 C2 C3

• The potential difference across an element is the element’s “voltage.”

Voltage in Circuits• Elements are connected by wires. • Any connected region of wire has the same

potential.

Vwire 1= 0 V Vwire 2= 5 V Vwire 3= 12 V Vwire 4= 15 V

VC1= 5 V VC3

= 3 VVC2= 7 V

C1 C2 C3

• The potential difference across an element is the element’s “voltage.”

Capacitors in Parallel• Both ends connected together by wire

C1 C2 Ceq

• Share Charge: Qeq = Q1 + Q2

• Total Cap: Ceq = (Q1 + Q2)/V = C1 + C2

= Veq • Same voltage: V1 = V2

Capacitors in Parallel• Both ends connected together by wire

C1 C2 Ceq

15 V

10 V

15 V

10 V

15 V

10 V

• Share Charge: Qeq = Q1+ Q2

• Total Cap: Ceq = (Q1+ Q2)/V = C1+ C2

= Veq • Same voltage: V1 = V2

Capacitors in Series• Connected end-to-end with NO other exits

• Same Charge: Q1 = Q2 = Qeq

Ceq

C1

C2

++++

++++

++++

+

-+-

+

-+-

+

-

+Q

-Q +Q

-Q

+Q

-Q

21

111CCCeq

• Share Voltage: V1+V2=Veq

Electromotive Force

• Battery– Maintains potential difference V – Not constant power– Not constant current– Does NOT produce or supply charges, just

“pushes” them.

+

-