++++++++++++++++ ---------------- d + - q-q capacitance c=q/v unit: farad, f. 1 f=1 coulomb/volt

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+ + + + + + + + - - - - - - - - d + - Q -Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

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Page 1: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

++++++++

--------

d

+ -

Q -Q

Capacitance C=Q/VUnit: farad, F. 1 F=1 coulomb/volt

Page 2: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

QUICK QUIZ 16.5

You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires

connected to the plates from touching each other. When you pull the plates farther apart,

do the following quantities increase, decrease, or stay the same? (a) C; (b) Q; (c) E

between the plates; (d) V.

Page 3: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

QUICK QUIZ 16.5 ANSWER

(a) C decreases

(b) Q stays the same

(c) E stays the same

(d) V increases

Page 4: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

DielectricDielectric

• Insulating materials

• Capacitance increased by a factor

Page 5: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Dielectric Constant, KDielectric Constant, KA measure of how effective it is in reducing an electric field

across the plates

Q -Q

Vo

Co=Q/Vo

Q -Q

+

+

+

V=Vo/KC=Q/V=KQ/Vo=KCo

Page 6: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Electric Field Energy

Work must be done to separate positive and negative charges against the Coulomb forces attracting them together. This work is stored as potential energy or electric field energy in a capacitor.

Page 7: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

V

+ -

e

V

++++

----

e

Vinitial

Vfinal

The average potential difference during the charge transfer is:

<V>=(Vfinal+Vinitial)/2

=(Vf+0)/2=Vf/2

The total charge transferred is Q

The work done is

W=U=Q<V>=(1/2) QV

=(1/2) CV2=(1/2) Q2/C

(C=Q/V)

-Q+Q

Page 8: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Energy Stored in a Capacitor

• Energy stored = ½ Q ΔV

• From the definition of capacitance, this can be rewritten in different forms

C2

QVC

2

1VQ

2

1Energy

22

Page 9: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

A Summary of the various formulas for charge Q, A Summary of the various formulas for charge Q, potential difference V, capacitance and energy Wpotential difference V, capacitance and energy W

Known Quantities

Unknown

QuantityC, V C, Q Q, V W, C W, V W, Q

Q= CV (2WC)1/2 2W/V

V= Q/C (2W/C)1/2 2W/Q

C= Q/V 2W/V2 Q2/2W

W= CV2/2 Q2/2C QV/2

Page 10: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

A capacitor is initially charged to 2 V. It is then connected to a 4 V battery. What is the ratio of the final to the initial energy stored in the capacitor?

(a) 2 (c) 6 (e) 10

(b) 4 (d) 8X

Questions:

Page 11: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

The figure below shows four parallel plate capacitors: A, B, C, and D. Each capacitor carries the same charge q and has the same plate area A. As suggested by the figure, the plates of capacitors A and C are separated by a distance d while those of B and D are separated by a distance 2d. Capacitors A and B are maintained in vacuum while capacitors C and D contain dielectrics with Constant = 5.

1.Which list below places the capacitors in order of increasing capacitance?

(a) A, B, C, D (b) B, A, C, D (c) A, B, D, C (d) D, C, B, A (e) B, A, D, C

X

2.Which capacitor has the largest potential difference between its plates?

(a) A (b) C (c) A and D are the same (d) B (e) D and larger than B or C.

X

Page 12: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Static ElectricityStatic Electricity

• Coulomb’s Law: F=kQ1Q2/r2

• Electric Field: E=F/Q• E=0 inside the a good conductor• Potential Difference or Voltage: V=Ed (uniform E)• Potential due to a point charge: V=kQ/r• Potential Energy:PE=QV • Capacitance: C=Q/V• Electric Field Energy: U=QV/2=CV2/2=Q2/2C

Page 13: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Flowing ElectricityFlowing Electricity

• A conducting path is needed

• A potential difference is also needed

Potential difference is analogous to water pressure

Page 14: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Chapter 17 Electric CurrentsElectric Currents

Electric Current

Ohm’s Law

Resistivity

Electric Power

Superconductivity

Page 15: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Electric CurrentElectric Current

If the net charge Q goes past in the time interval t, then the average current is

I=Q/t (Electric Current)

The unit of electric current is the Ampere (amps), A.

1 A=1coulomb/second

Electrons hop from one atom to the next, but the effects of electricity travel at the speed of light

300,000 km/s (186,000 mile/s)

Page 16: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Direct Current, DC: In DC, all the electrons move in the same direction while the

electricity flows. DC is produced by batteries and similar devices

Alternating Current, AC: In AC, the direction of electron movement changes many times

each second. AC is generated by electric generators in power plants

In the US, the frequency of AC is 60 Hz, i.e., the current changes

direction 120 times per second. This frequency is 50 Hz in much of the rest of the world

Page 17: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

For a current to occurFor a current to occur,,

• There must be a path between the two points along which charge can flow (Metals, many liquids, and plasmas (charged gases) are conductors)

• There must be a potential difference between the two points (A superconductor is an exception).

A large potential difference means a large “push” given to each charge

Page 18: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

The Role of Potential Difference

Greater height

Increased flow of water The higher voltage, the more current

Flow of Water Electric current

Page 19: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Why is Potential Difference needed for a current to flow?

Resistance

Page 20: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Ohm’s Law(German Physicist, Georg Ohm, 1787-1854)

I=V/R

Definition of Resistance R=V/I

The unit for R is Ohm, 1=1V/1A

Page 21: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Which one of the following circuits has the largest resistance?

X

Page 22: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

R=V/I

I

V0

The Ohm’s law applies for many normal conductors, but fails for other materials, and devices.

Linear

Page 23: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Non-Ohmic behavior

Page 24: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Factors that governFactors that govern R

• Material The ability to carry electric current varies drastically

• Length L The longer the conductor, the greater its resistance

• Cross sectional area A The thicker the conductor, the less its resistance

Page 25: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

R=L/A is the resistivity, and depends on the material

used reflects intrinsic properties of the material is a function of temperature

The unit of is m

Page 26: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

The potential difference across the ends of a wire is doubled in magnitude. If Ohm’s law is obeyed, which one of the following statements concerning the resistance of the wire is true?

(a) The resistance is one half of its original value.

(b) The resistance is twice its original value.

(c) The resistance is not changed.

(d) The resistance increases by a factor of four.

(e) The resistance decreases by a factor of four.

X

Question:

Page 27: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Question:The resistance of a conductor does not depend on its

(a) mass

(b) length

(c) cross-sectional area

(d) resistivity

Answer: a

Page 28: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Question: A certain wire has a resistance R. The resistance of another wire, identical with the first except for having twice its diameter, is

(a) R/4(b) R/2(c) 2R(d) 4R

Answer: a

Page 29: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Question: A certain piece of copper is to be shaped into a conductor of minimum resistance. Its length and cross-sectional area

(a) should be, respectively, L and A

(b) should be, respectively, 2L and A/2

(c) should be, respectively, L/2 and 2A

(d) do not matter, since the volume of copper remains the same.

Answer: c

Page 30: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Electric PowerElectric Power

Electric energy is readily transformed into other forms of

energy

Page 31: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

The work that must be done to take

Q through V W=QV

Since I=Q/t or Q=It,

W=IVt (Electric work)

Recall P=work done/time interval=W/t,

P=IV (Electric Power)

Using Ohm’s law I=V/R,

P=I2R=V2/R

The unit for Power is Watt, 1W=1J/s

Page 32: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Question: Which draws more current, a 100-W light bulb or a 75-W bulb? Which has the higher resistance?

Answer:

Let P1=100 W and P2=75 W. Since P=IV, V=constant

P1/P2=I1/I2=100/75=4/3 The 100-W bulb draws more current

Also, P=V2/R, where V=constant

P1/P2=R2/R1=100/75=4/3 The 75-W has higher resistance

Page 33: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Question: Electric power is transferred over large distances at very high voltages. Explain how the high voltage reduces power losses in the transmission lines.

Answer: The power transmitted P=IV. So the higher V, the smaller I.

Since the power lost P=I2R, the smaller I leads to smaller I2R loss for a given transmission line of resistance R.

Page 34: ++++++++++++++++ ---------------- d + - Q-Q Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Superconductor

T0

Conductor

Tc

The highest Tc is about –150 C or –300 F found in “high temperature” superconductors