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UF PHY 2049 - Lecture notes

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Slide 1Slide 2Slide 3HITTSlide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16hittIf the areas are A1 and A-A1. 2112( AAdCoC123 2.4 µF, q = 28.8 µCC2 C24 = 12 µFC1234 = 3 µF q =36 µC 212dACoThe force on a filling dielectric as it is inserted between the parallel plates of a capacitor.  LCdxdCLxAdCCCo11121xLWith the battery connected, U1 = ½CV2With the battery disconnected, U2 = Q2/2C LUdxdUF111 LUdxdUF122With the battery connected, since x is increasing downwards, a negative force is upwards, pushing the dielectric away.With the battery disconnected, the force is positive and pointed downwards, pulling in the dielectric.The force is proportional to (κ-1) and inversely to L.Chapter 26 Current and Resistance In this chapter we will introduce the following new concepts:-Electric current ( symbol i ) -Electric current density vector (symbol ) -Drift speed (symbol vd ) -Resistance (symbol R ) and resistivity (symbol ρ ) of a conductor -Ohmic and non-Ohmic conductors We will also cover the following topics:-Ohm’s law -Power in electric circuits Jr(26 - 1)HITTThe plate areas and plate separations of five parallel plate capacitors arecapacitor 1: area A0, separation d0capacitor 2: area 2A0, separation 2d0capacitor 3: area 2A0, separation d0/2capacitor 4: area A0/2, separation 2d0capacitor 5: area A0, separation d0/2Rank these according to their capacitances, least to greatest.a. 1,2,3,4,5 b. 5,4,3,2,1 c. (524) ,(13) d. 4, (12), 5,3 e. None of theseConsider the conductor shown in fig.a. All the points inside the conductor and on its surface are at the same potential. The free electrons inside the conductor move at random directiElectric currentons and thus there is not net charge transport. We now make a break in the conductorand insert a battery as shown in fig.b.Points A and B are now at potentials and , respectively.( theA BA BV VV V V- = voltage of the battery)The situation is not static any more but charges move inside the conductor so thatthere is a net charge flow in a particulardirection. This net flow of electric charge we define as "electric currrent"AB(26 - 2)i+ qconductorvri- qconductorvrConsider the conductor shown in the figureIt is connected to a battery (not shown) andthus charges move through the conductorConsider one of the cross sections through the conductor ( aa or bb or c� � �Current = rate at which charge flowc )The electric current is definedCurrent SI Unit: C/s known as the "Amperes s:" a dqitid=An electric current is represented by an arrow which has the same direction as the charge velocity. The sense of thecurrent arrow is defined as follows: If the current is due to tCurrent direction :1. he motion of chargesthe current arrow is to the charge velocity If the current is due to the motion of chargesthe current arrow is to the charge veloci tvparallel 2.antnegaiparallelposittive ivery vr(26 - 3)dqidt=i- qconductorvrAJr Current density is a vector that is defined as follows:Its magnitude The direction of is the same as that of the currentThe current through a condu iJAJ=Current densityr2SI unit for J : A/mctor of cross sectionalarea is given by the equation: if the current density is constant. If is not constant then: i JAi J dAJ A== ��rrri+ qconductorvrAJr1We note that even though the current density is a vectorthe electric current is not. This is illustrated in the figure to the left. An incoming current branches atpoint a into two currents, ,oii1 22 and . Current This equation expresses theconservation of charge at point a. Please note that we have not used vector additionoi i ii= + iJA=(26 - 4)When a current flows through a conductorthe electric field causes the charges to movewith a constant drift speed . This drift speedis superimposed on the random motion of the charges. dvDrift speedConsider the conductor of cross sectional area shown in the figure. We assumethat the current in the conductor consists of positive charges. The total charge within a length is given by: Aq L q nA=( ). This charge moves through area in a time . The current /The current density In vector form: dd ddddL e AL q nALet i nAv ev t L vnAv eiJ nv eA AJ nev= = = == = ==rrdJ nev=rrdJ nv e=(26 - 5)+-iVIf we apply a voltage across a conductor (see figure)a current will flow through the conductor. We define the conductor resistance as the ratio V the Ohm (symbolA VRiVi==SI RUnit for R :esist anceA conductor across which we apply a voltage = 1 Voltand results in a current = 1 Ampere is defined as having resistance of 1 Why not use the symbol "O" instead of " " Suppose we ha d )ViWWWQ :A : a 1000 resistor. We would then write: 1000 O which can easily be mistaken read as 10000 .A conductor whose function is to provide a specified resistance is known as a "resistor" The symbol is WWgiven to the left. VRi=(26 - 6)REr+-iVErUnlike the electrostatic case, the electric field in theconductor of the figure is not zero. We define as resistivity of the conductor the ratio In vector form: EJJEr rr==ResiSI ustivitnit r y for r2The conductivity is defined as: Using the previouV/m V m mA/ms equation takes the for1mA: J Esrsrs== =W�=ρ : r rConsider the conductor shown in the figure above. The electric field inside the conductor . The current density We substitute and into / equation and get: /V iE J E JL AE V L V A ARJ i A i L Lr r= == = = =LRAr� =LRAr= E Jr=r rJ Es=r r(26 -7)In the figure we plot the resistivity ofcopper as function of temperature . The dependence of on is almost linear. Similar dependence is observedTTrrVariation of resistivity with temperaturein many conductors. ( )The following empirical equation is used for many practical applications: The constant is known as the"temperature coefficient of resistivity". The constant is a reference tempo o ooT TTr r u aa- = -( )o8 oerature usually taken to be room temperature ( 293 K ) and is the resistivityat . For copper 1.69 10 m temperature enters the equation above as a


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