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.

QUICK QUIZ 16.5 ANSWER (a) C decreases (b) Q stays the same (c) E stays the same (d)  V increases

Dielectric Insulating materials Capacitance increased by a factor 

Dielectric Constant, K Dielectric Constant, K A measure of how effective it is in reducing an electric field across the plates Q-Q V o C o =Q/V o Q-Q V=V o /K C=Q/V=KQ/V o =KC o

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.

V + - e V e V initial V final The average potential difference during the charge transfer is: =(V final +V initial )/2 =(V f +0)/2=V f /2 The total charge transferred is Q The work done is W=U=Q =(1/2) QV =(1/2) CV 2 =(1/2) Q 2 /C (C=Q/V) -Q+Q

Energy Stored in a Capacitor Energy stored = ½ Q ΔV From the definition of capacitance, this can be rewritten in different forms

A Summary of the various formulas for charge Q, potential difference V, capacitance and energy W KnownQuantities Unknown Quantity C, VC, QQ, V W, CW, VW, Q Q=CV (2WC) 1/2 2W/V V=Q/C (2W/C) 1/2 2W/Q C=Q/V 2W/V 2 Q 2 /2W W= CV 2 /2Q 2 /2CQV/2

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) 8 X Questions:

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

Static Electricity Coulomb’s Law: F=kQ 1 Q 2 /r 2 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=CV 2 /2=Q 2 /2C

Flowing Electricity A conducting path is needed A potential difference is also needed Potential difference is analogous to water pressure

Electric Currents Chapter 17 Electric Currents Electric Current Ohm’s Law Resistivity Electric Power Superconductivity

Electric 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)

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

For 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

The Role of Potential Difference Greater height Increased flow of waterThe higher voltage, the more current Flow of Water Electric current

Why is Potential Difference needed for a current to flow? Resistance

Ohm’s Law (German Physicist, Georg Ohm, ) I=V/R Definition of Resistance R=V/I The unit for R is Ohm,  1  =1V/1A

Which one of the following circuits has the largest resistance? X

R=V/I I V 0 The Ohm’s law applies for many normal conductors, but fails for other materials, and devices. Linear

Non-Ohmic behavior

Factors that govern Factors 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

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 

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:

Question:The resistance of a conductor does not depend on its (a)mass (b)length (c)cross-sectional area (d)resistivity Answer: a

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

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

Electric Power Electric energy is readily transformed into other forms of energy

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=I 2 R=V 2 /R The unit for Power is Watt, 1W=1J/s

Question: Which draws more current, a 100-W light bulb or a 75-W bulb? Which has the higher resistance? Answer: Let P 1 =100 W and P 2 =75 W. Since P=IV, V=constant P 1 /P 2 =I 1 /I 2 =100/75=4/3 The 100-W bulb draws more current Also, P=V 2 /R, where V=constant P 1 /P 2 =R 2 /R 1 =100/75=4/3 The 75-W has higher resistance

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=I 2 R, the smaller I leads to smaller I 2 R loss for a given transmission line of resistance R.

Superconductor  T 0 Conductor Tc The highest Tc is about –150 C or –300 F found in “high temperature” superconductors