C C. κ κ A C 0. A d. κ κ. A d A 3. d A. κ κ (1)

Transcription

1 7 icture the roblem We can moel this system as two caacits in series, of thickness / an of thickness / an use the uation f the uivalent caacitance of two caacits connecte in series. xress the uivalent caacitance of the two caacits connecte in series: elate the caacitance of to its ielectric constant an thickness: elate the caacitance of to its ielectric constant an thickness: Substitute an simlify to obtain: *7 icture the roblem et the charge on the caacit with the air ga be an the charge on the caacit with the ielectric ga be. If the caacitances of the caacits were initially, then the caacitance of the caacit with the ielectric inserte is '. We can use the conservation of charge an the uivalence of the otential ifference across the caacits to obtain two uations that we can solve simultaneously f an. ly conservation of charge uring the insertion of the ielectric to obtain: () Because the caacits have the same otential ifference across ()

2 them: Solve uations () an () simultaneously to obtain: an * Determine the oncet When current flows, the charges are not in uilibrium. In that case, the electric fiel rovies the fce neee f the charge flow. *5 icture the roblem The resistance of the metal bar varies irectly with its length an inversely with its cross-sectional area. Hence, to minimize the resistance of the bar, we shoul connect to the surface f which the ratio of the length to the contact area is least. Denoting the surfaces as a, b, an c, comlete the table to the right: Surface / a 8.8 b. c.5 Because connecting to surface c minimizes : (c) is crect. 7 icture the roblem The ower issiate in the resist is given by I. We can exress the ower issiate when the current is I an, assuming that the resistance oes not change, exress the ratio of the two rates of energy issiation to fin the ower issiate when the current is I. xress the ower issiate in the resist when the current in it is I: xress the ower issiate in the resist when the current in it is I: Divie the secon of these uations by the first to obtain: I ' ' ( I ) 9I 9I 9 I ' 9 an () is crect.

3 icture the roblem Because the otential ifference across the two combinations of resists is constant, we can use to relate the ower elivere by the battery to the uivalent resistance of each combination of resists. xress the ower elivere by the battery when the resists are connecte in series: etting reresent the resistance of the ientical resists, exress : Substitute to obtain: xress the ower elivere by the battery when the resists are connecte in arallel: xress the uivalent resistance of the ientical resists connecte in arallel: s s () ( )( ) Substitute to obtain: () Divie uation () by uation () to obtain: s Solve f an evaluate : ( W) 8 W s an (e) is crect.

4 5 icture the roblem We can solve K mev f the velocity of an electron in the beam an use the relationshi between current an rift velocity to fin the beam current. (a) xress the kinetic energy of the K mev beam: Solve f v: v K m e Substitute numerical values an evaluate v: v 9 ( )(.6 J/e) ke m/s kg (b) Use the relationshi between current an rift velocity (here the velocity of an electron in the beam) to obtain: I nev xress the cross-sectional area of the beam in terms of its iameter D: π D Substitute to obtain: I πnev D Substitute numerical values an evaluate I: Substitute numerical values an evaluate I: I ( 5 cm )(.6 )( 5.9 m/s)( m) 7. µ π

5 *9 icture the roblem We can use Ohm s law to exress the ratio of the otential ifferences across the two wires an to relate the resistances of the wires to their lengths, resistivities, an cross-sectional areas. Once we ve foun the ratio of the otential ifferences across the wires, we can use to ecie which wire has the greater electric fiel. (a) ly Ohm s law to exress the otential ro across each wire: I an I Divie the first of these uations by the secon to exress the ratio of the otential ros across the wires: I I () elate the resistances of the wires to their resistivity, cross-sectional area, an length: Divie the first of these uations by the secon to exress the ratio of the resistances of the wires: an because an. Substitute in uation () to obtain: Substitute numerical values (see Table 65- f the resistivities of coer an iron) an evaluate the ratio of the otential ifferences: (b) xress the electric fiel in each conuct in terms of its length an the otential ifference across it:.7 Ω m Ω m 8 8 an.7

6 Divie the first of these uations by the secon to obtain: Because 5.88 : is greater in the iron wire. *77 icture the roblem We can use fv to fin the ower the electric mot must evelo to move the car at 8 km/h against a frictional fce of N. We can fin the total charge that can be elivere by the batteries using NI t. The total electrical energy elivere by the batteries befe recharging can be foun using the efinition of emf. We can fin the istance the car can travel from the efinition of wk an the cost er kilometer of riving the car this istance by iviing the cost of the ruire energy by the istance the car has travele. (a) xress the ower the electric mot must evelo in terms of the see of the car an the friction fce: (b) Use the efinition of current to exress the total charge that can be elivere befe charging: fv ( N)( 8 km/h) 6.7kW 5.76 M ( h) NI t 6 6s h where N is the number of batteries. (c) Use the efinition of emf to exress the total electrical energy available in the batteries: W ε 69.MJ ( 5.76M)( ) () elate the amount of wk the batteries can o to the wk ruire to overcome friction: W f Solve f an evaluate : f W 69.MJ N 57.6km (e) xress the cost er kilometer as the ratio of the ratio of the cost of the energy to the

7 istance travele befe recharging: ( $.9/kW h) εit ( $.9/kW h)( )( 6 h) ost/km 57.6km $./ km *97 icture the roblem et I be the current elivere by the left battery, I the current elivere by the right battery, an I the current through the 6-Ω resist, irecte own. We can aly Kirchhoff s rules to obtain three uations that we can solve simultaneously f I, I, an I. Knowing the currents in each branch, we can use Ohm s law to fin the otential ifference between oints a an b an the ower elivere by both the sources. (a) ly Kirchhoff s junction rule at junction a: ly Kirchhoff s loo rule to a loo aroun the outsie of the circuit to obtain: ly Kirchhoff s loo rule to a loo aroun the left-han branch of the circuit to obtain: I I I ( Ω) I ( Ω) I ( Ω ) I ( Ω ) I ( Ω) I ( 6Ω) I Solve these uations simultaneously to obtain: I.667, I.889, an I.56 (b) ly Ohm s law to fin the otential ifference between oints a an b: (c) xress the ower elivere by the - battery in the left-han branch of the circuit: ab left ( Ω) ( 6Ω)(.56 ) I ε ( )(.667 ) 8. W I xress the ower elivere by the - battery in the right-han branch of the circuit: right ε I ( )(.889 ).7 W