ELECTRIC CHARGE AND ELECTRIC FIELD

LCTRIC CHARG AND LCTRIC FILD (a) IDNTIFY and ST UP: Use the chage of one electon ( 6 C) to find the numbe of electons euied to poduce the net chage XCUT: The numbe of ecess electons needed to poduce net chage is C = = electons e 6 C/electon (b) IDNTIFY and ST UP: Use the atomic mass of lead to find the numbe of lead atoms in 8 kg of lead Fom this and the total numbe of ecess electons, find the numbe of ecess electons pe lead atom XCUT: The atomic mass of lead is 7 kg/mol, so the numbe of moles in 8 kg is mtot 8 kg n = = = 865 mol N A (Avogado s numbe) is the numbe of atoms in mole, so the M 7 kg/mol numbe of lead atoms is N = nn A = (865 mol)(6 atoms/mol) = 8 atoms The numbe of electons ecess electons pe lead atom is = 859 8 atoms VALUAT: ven this small net chage coesponds to a lage numbe of ecess electons But the numbe of atoms in the sphee is much lage still, so the numbe of ecess electons pe lead atom is ve small IDNTIFY: The chage that flows is the ate of chage flow times the duation of the time inteval ST UP: The chage of one electon has magnitude e = 6 C 4 XCUT: The ate of chage flow is, C/s and t = μs = s 4 Q 9 Q = (, C/s)( s) = C The numbe of electons is ne = = 5 6 C VALUAT: This is a ve lage amount of chage and a lage numbe of electons IDNTIFY: Fom ou mass estimate the numbe of potons in ou bod You have an eual numbe of electons ST UP: Assume a bod mass of 7 kg The chage of one electon is 6 C 7 XCUT: The mass is pimail potons and neutons of m = 67 kg The total numbe of potons and 7 kg 8 8 neutons is np and n = = 4 About one-half ae potons, so n 7 p = = ne The numbe of 67 kg 8 electons is about The total chage of these electons is 8 9 Q = ( 6 C/electon)( electons) = 5 C VALUAT: This is a huge amount of negative chage But ou bod contains an eual numbe of potons and ou net chage is zeo If ou ca a net chage, the numbe of ecess o missing electons is a ve small faction of the total numbe of electons in ou bod 4 IDNTIFY: Use the mass m of the ing and the atomic mass M of gold to calculate the numbe of gold atoms ach atom has 79 potons and an eual numbe of electons ST UP: N = 6 atoms/mol A poton has chage +e A XCUT: The mass of gold is 77 g and the atomic weight of gold is 97 g mol So the numbe of atoms 77 g is NAn= (6 atoms/mol) = 54 atoms 97 g mol The numbe of potons is 4 5 n p = (79 potons/atom)(54 atoms) = 47 potons Q= ( np )(6 C/poton) = 68 C 4 (b) The numbe of electons is ne = np = 47 VALUAT: The total amount of positive chage in the ing is ve lage, but thee is an eual amount of negative chage -

- Chapte k 5 IDNTIFY: Appl F = and solve fo ST UP: F = 65 N 9 k (899 N m /C )( C) XCUT: = = = 7 m = 7 km F 65 N VALUAT: Chaged objects tpicall have net chages much less than C 6 IDNTIFY: Appl Coulomb's law and calculate the net chage on each sphee ST UP: The magnitude of the chage of an electon is e = 6 C 6 XCUT: F = This gives = P F = 4 πp (457 N)( m) = 4 C And P 6 theefoe, the total numbe of electons euied is n= / e= (4 C)/(6 C/electon) = 89 electons VALUAT: ach sphee has 89 ecess electons and each sphee has a net negative chage The two like chages epel 7 IDNTIFY: Appl Coulomb s law ST UP: Conside the foce on one of the sphees (a) XCUT: = = F N 7 F = = so = = 5 m = 74 C (on each) 9 P P (/4 πp ) 8988 N m /C (b) = 4 4 F F 7 7 F = = so = = = (74 C) = 7 C P P 4(/4 πp) (/4 πp) 6 And then = 4 = 48 C VALUAT: The foce on one sphee is the same magnitude as the foce on the othe sphee, whethe the sphee have eual chages o not 8 IDNTIFY: Use the mass of a sphee and the atomic mass of aluminum to find the numbe of aluminum atoms in one sphee ach atom has electons Appl Coulomb's law and calculate the magnitude of chage on each sphee ST UP: N A = 6 atoms/mol = n ee, whee n e is the numbe of electons emoved fom one sphee and added to the othe XCUT: (a) The total numbe of electons on each sphee euals the numbe of potons 5 kg 4 ne = np = ()( NA) = 75 electons 698 kg mol (b) Fo a foce of 4 N to act between the sphees, 4 N 4 πp F = = This gives 4 4 = 4 π P ( N)(8 m) = 84 C The numbe of electons emoved fom one sphee and 5 added to the othe is n e = / e= 57 electons (c) n e/ ne = 77 VALUAT: When odina objects eceive a net chage the factional change in the total numbe of electons in the object is ve small 9 IDNTIFY: Appl F = ma, with F = k ST UP: XCUT: a = 5g = 45 m/s An electon has chage F e = 6 C = ma= (855 kg)(45 m/s ) = 9 N The sphees have eual chages, so F = k and 6 F 9 N 6 9 C = = (5 m) = 9 C N = = = 9 9 4 electons The k 899 N m /C e 6 C chages on the sphees have the same sign so the electical foce is epulsive and the sphees acceleate awa fom each othe

lectic Chage and lectic Field - VALUAT: As the sphees move apat the epulsive foce the eet on each othe deceases and thei acceleation deceases (a) IDNTIFY: The electical attaction of the poton gives the electon an acceleation eual to the acceleation due to gavit on eath ST UP: Coulomb s law gives the foce and Newton s second law gives the acceleation this foce poduces e ma = and = P P ma ( 9 )( 9 N m /C 6 C ) XCUT: = ( 9 kg)( 98 m/s ) = 58 m VALUAT: The electon needs to be about 5 m fom a single poton to have the same acceleation as it eceives fom the gavit of the entie eath (b) IDNTIFY: The foce on the electon comes fom the electical attaction of all the potons in the eath ST UP: Fist find the numbe n of potons in the eath, and then find the acceleation of the electon using Newton s second law, as in pat (a) n = m /m p = (597 4 kg)/(67 7 kg) = 57 5 potons a = F/m = P 4 p e ne R πp = me mer XCUT: a = (9 9 N m /C )(57 5 9 )(6 C) /[(9 kg)(68 6 m) ] = 4 m/s One can ignoe the gavitation foce since it poduces an acceleation of onl 98 m/s and hence is much much less than the electical foce VALUAT: With the electical foce, the acceleation of the electon would neal 4 times geate than with gavit, which shows how stong the electical foce is IDNTIFY: In a space satellite, the onl foce acceleating the fee poton is the electical epulsion of the othe poton ST UP: Coulomb s law gives the foce, and Newton s second law gives the acceleation: a = F/m = (/ 4 πp ) (e / )/m XCUT: (a) a = (9 9 N m /C )(6-9 C) /[(5 m) (67-7 kg)] = 4 m/s (b) The gaphs ae sketched in Figue VALUAT: The electical foce of a single stationa poton gives the moving poton an initial acceleation about, times as geat as the acceleation caused b the gavit of the entie eath As the potons move fathe apat, the electical foce gets weake, so the acceleation deceases Since the potons continue to epel, the velocit keeps inceasing, but at a deceasing ate Figue IDNTIFY: Appl Coulomb s law ST UP: Like chages epel and unlike chages attact 6 (55 C) XCUT: (a) F = This gives N = P 4 πp (m) 6 foce is attactive and <, so =+ 64 C (b) F = N The foce is attactive, so is downwad VALUAT: The foces between the two chages obe Newton's thid law and 6 64 C =+ The

-4 Chapte IDNTIFY: Appl Coulomb s law The two foces on must have eual magnitudes and opposite diections ST UP: Like chages epel and unlike chages attact XCUT: The foce F that eets on has magnitude the diection, so must be positive F = F gives cm = = = 4 cm F = k and is in the + diection F must be in k = k ( nc) 75 nc VALUAT: The esult fo the magnitude of doesn t depend on the magnitude of 4 IDNTIFY: Appl Coulomb s law and find the vecto sum of the two foces on Q ST UP: The foce that eets on Q is epulsive, as in ample 4, but now the foce that eets is attactive XCUT: The -components cancel We onl need the -components, and each chage contibutes euall 6 6 ( C) (4 C) = = sin = 7 N (since sin = 6) Theefoe, the total foce is F F α α 4 πp (5 m) F = 5N, in the -diection VALUAT: If is μc and is + μc, then the net foce is in the +-diection 5 IDNTIFY: Appl Coulomb s law and find the vecto sum of the two foces on ST UP: Like chages epel and unlike chages attact, so F and F ae both in the +-diection 5 4 4 XCUT: F = k = 6749 N, F = k = 4 N F = F + F = 8 N 4 F = 8 N and is in the +-diection VALUAT: Compaing ou esults to those in ample, we see that F on = F on, as euied b Newton s thid law 6 IDNTIFY: Appl Coulomb s law and find the vecto sum of the two foces on ST UP: F on is in the +-diection XCUT: F 9 6 6 (9 N m C ) ( C) ( C) on = = N ( ) ( 6 m) ( F on ) =+ N F Q on is eual and opposite to F on Q (ample 4), so ( Q ) ( F Q on ) = 7 N F = ( F on ) + ( FQ on ) = N F ( F ) ( FQ ) F = and on F on = N and = on + on = N + 7 N = 7 N The magnitude of the total foce is F = ( N) + ( 7 N) = 5 N tan = 4, so F is 7 4 counteclockwise fom the + ais, o counteclockwise fom the + ais VALUAT: Both foces on ae epulsive and ae diected awa fom the chages that eet them 7 IDNTIFY and ST UP: Appl Coulomb s law to calculate the foce eeted b and on Add these foces as vectos to get the net foce The taget vaiable is the -coodinate of XCUT: F is in the -diection F = k = 7 N, so F =+ 7 N F = F + F and F = 7 N F = F F = 7 N 7 N = 7 N Fo F to be negative, must be on the F -ais k F = k, so = = 44 m, so = 44 m VALUAT: attacts in the + -diection so must attact in the -diection, and is at negative

lectic Chage and lectic Field -5 8 IDNTIFY: Appl Coulomb s law ST UP: Like chages epel and unlike chages attact Let F be the foce that eets on and let F be the foce that eets on XCUT: The chage must be to the ight of the oigin; othewise both and would eet foces in the + diection Calculating the two foces: 9 6 6 (9 N m C )( C)(5 C) F = = = 75 N, in the + diection 4 πp ( m) 9 6 6 (9 N m C ) ( C) (8 C) 6 N m F = =, in the diection 6 N m 6 N m We need F = F F = 7 N, so 75 N = 7 N = = 44 m 75 N + 7 N is at = 44 m VALUAT: F = 4 N F is lage than F, because is lage than and also because is less than 9 IDNTIFY: Appl Coulomb s law to calculate the foce each of the two chages eets on the thid chage Add these foces as vectos ST UP: The thee chages ae placed as shown in Figue 9a Figue 9a XCUT: Like chages epel and unlike attact, so the fee-bod diagam fo is as shown in Figue 9b Figue 9b F = P F = P 9 (5 C)(5 C) 6 F = (8988 N m /C ) = 685 N ( m) 9 ( C)(5 C) 7 F = (8988 N m /C ) = 8988 N (4 m) The esultant foce is R F F = + R = 6 7 6 R = F+ F = 685 N +8988 N = 58 N 6 The esultant foce has magnitude 58 N and is in the -diection VALUAT: The foce between and is attactive and the foce between and is eplusive IDNTIFY: Appl F = k to each pai of chages The net foce is the vecto sum of the foces due to and ST UP: Like chages epel and unlike chages attact The chages and thei foces on ae shown in Figue

-6 Chapte XCUT: (4 C)(6 C) F = k = (899 N m /C ) = 594 N 9 9 9 7 ( m) (5 C)(6 C) F = k = (899 N m /C ) = 997 N 9 9 9 7 ( m) 7 7 = + =+ = 4 N The net foce has magnitude F F F F F 4 N and is in the + diection VALUAT: ach foce is attactive, but the foces ae in opposite diections because of the placement of the chages Since the foces ae in opposite diections, the net foce is obtained b subtacting thei magnitudes Figue IDNTIFY: Appl Coulomb s law to calculate each foce on Q ST UP: Let F be the foce eeted b the chage at = a and let F be the foce eeted b the chage at = a a / XCUT: (a) The two foces on Q ae shown in Figue a sinθ = and = ( a + ) is the / ( a + ) distance between and Q and between and Q Q Qa (b) F = F + F = F = F + F = sinθ = 4 πp ( a + ) 4 πp ( a + ) Q (c) At =, F =, in the + diection P a (d) The gaph of VALUAT: F vesus is given in Figue b F = fo all values of and F > fo all Figue IDNTIFY: Appl Coulomb s law to calculate each foce on Q ST UP: Let F be the foce eeted b the chage at = a and let F be the foce eeted b the chage at a = a + cosθ = = The distance between each chage and Q is ( ) / XCUT: (a) The two foces on Q ae shown in Figue a (b) When >, F and <, F and (c) At =, F = (d) The gaph of ( a + ) / Q Q F = F + F = cos = F ae negative θ / 4 πp ( a + ) 4 πp ( a + ) F ae positive and the same epession fo F vesus is sketched in Figue b F applies F = F + F = When

VALUAT: The diection of the net foce on Q is alwas towad the oigin lectic Chage and lectic Field -7 Figue IDNTIFY: Appl Coulomb s law to calculate the foce eeted on one of the chages b each of the othe thee and then add these foces as vectos (a) ST UP: The chages ae placed as shown in Figue a = = = 4 = Figue a Conside foces on 4 The fee-bod diagam is given in Figue b Take the -ais to be paallel to the diagonal between and 4 and let (b) R = F + F + F = Figue b P L P L 8πPL + be in the diection awa fom Then F is in the + -diection XCUT: F = F P L = F = P L F = F sin 45 = F/ F =+ F cos 45 =+ F/ F =+ F sin 45 =+ F/ F =+ F cos 45 =+ F/ F =, F = F R = F + F + F = (/ ) + = ( + ) R = ( + ) Same fo all fou chages 8π P L

-8 Chapte VALUAT: In geneal the esultant foce on one of the chages is diected awa fom the opposite cone The foces ae all epulsive since the chages ae all the same B smmet the net foce on one chage can have no component pependicula to the diagonal of the suae k 4 IDNTIFY: Appl F = to find the foce of each chage on + The net foce is the vecto sum of the individual foces ST UP: Let =+ 5 μc and = 5 μc The chage + must be to the left of o to the ight of in ode fo the two foces to be in opposite diections But fo the two foces to have eual magnitudes, + must be close to the chage, since this chage has the smalle magnitude Theefoe, the two foces can combine to give zeo net foce onl in the egion to the left of Let d + 6 m fom k XCUT: F = F gives d (845)(6 m) be positive, so d = = 7 m 845 VALUAT: When + be a distance d to the left of, so it is a distance k = d =± ( d + 6 m) =± (845)( d + 6 m) d must ( d + 6 m) + is at = 7 m, F is in the The net foce would be zeo when + is at = 7 m diection and F is in the + diection 5 IDNTIFY: F = Since the field is unifom, the foce and acceleation ae constant and we can use a constant acceleation euation to find the final speed 7 ST UP: A poton has chage +e and mass 67 kg XCUT: (a) F = = 6 F 44 N (b) a = = = 7 6 m/s m 67 kg 6 (6 C)(75 N/C) 44 N 6 5 (c) v = v + at gives v = (6 m/s )( s) = 6 m/s VALUAT: The acceleation is ve lage and the gavit foce on the poton can be ignoed 6 IDNTIFY: Fo a point chage, = k ST UP: is towad a negative chage and awa fom a positive chage XCUT: (a) The field is towad the negative chage so is downwad 9 C = (899 N m /C ) = 4 N/C (5 m) (b) = 9 k (899 N m /C )( C) = = 5 m N/C VALUAT: At diffeent points the electic field has diffeent diections, but it is alwas diected towad the negative point chage 7 IDNTIFY: The acceleation that stops the chage is poduced b the foce that the electic field eets on it Since the field and the acceleation ae constant, we can use the standad kinematics fomulas to find acceleation and time (a) ST UP: Fist use kinematics to find the poton s acceleation v = when it stops Then find the electic field needed to cause this acceleation using the fact that F = XCUT: v = v + a( ) = (45 6 m/s) + a( m) and a = 6 4 m/s Now find the 7 electic field, with = e e = ma and = ma/e = (67 kg)(6 4 m/s 9 )/(6 C) = 6 N/C, to the left (b) ST UP: Kinematics gives v = v + at, and v = when the electon stops, so t = v /a XCUT: t = v /a = (45 6 m/s)/(6 4 m/s 8 ) = 4 s = 4 ns (c) ST UP: In pat (a) we saw that the electic field is popotional to m, so we can use the atio of the electic fields e/ p me/ mp = m / m = and ( ) e e p p XCUT: e = [(9 7 kg)/(67 kg)]( 6 N/C) = 8 N/C, to the ight VALUAT: ven a modest electic field, such as the ones in this situation, can poduce enomous acceleations fo electons and potons

lectic Chage and lectic Field -9 8 IDNTIFY: Use constant acceleation euations to calculate the upwad acceleation a and then appl F = to calculate the electic field ST UP: Let + be upwad An electon has chage = e XCUT: (a) v = and a = a, so = vt+ a t gives = at Then ( ) (45 m) F ma (9 kg) ( m s ) a = = = m s = = = = 569 N C 6 t ( s) 6 C The foce is up, so the electic field must be downwad since the electon has negative chage (b) The electon s acceleation is ~ g, so gavit must be negligibl small compaed to the electical foce VALUAT: Since the electic field is unifom, the foce it eets is constant and the electon moves with constant acceleation 9 (a) IDNTIFY: (4) elates the electic field, chage of the paticle, and the foce on the paticle If the paticle is to emain stationa the net foce on it must be zeo ST UP: The fee-bod diagam fo the paticle is sketched in Figue 9 The weight is mg, downwad Fo the net foce to be zeo the foce eeted b the electic field must be upwad The electic field is downwad Since the electic field and the electic foce ae in opposite diections the chage of the paticle is negative mg = Figue 9 mg (45 kg)(98 m/s ) 5 XCUT: = = = 9 C and = 9 μc 65 N/C (b) ST UP: The electical foce has magnitude F = = e The weight of a poton is w= mg F = w so e = mg 7 mg (67 kg)(98 m/s ) 7 XCUT: = = = N/C e 6 C This is a ve small electic field 8 VALUAT: In both cases = mgand = ( m/ ) g In pat (b) the m/ atio is much smalle ( ) than in pat (a) ( ) so is much smalle in (b) Fo subatomic paticles gavit can usuall be ignoed compaed to electic foces IDNTIFY: Appl = P ST UP: The ion nucleus has chage + 6 e A poton has chage + e (6)(6 C) XCUT: (a) = = 4 N/C 4 πp (6 m) (6 C) (b) poton = = 55 N/C 4 πp (59 m) VALUAT: These electic fields ae ve lage In each case the chage is positive and the electic fields ae diected awa fom the nucleus o poton IDNTIFY: Fo a point chage, = k The net field is the vecto sum of the fields poduced b each chage A chage in an electic field epeiences a foce F = ST UP: The electic field of a negative chage is diected towad the chage Point A is m fom and 5 m fom Point B is m fom and 5 m fom XCUT: (a) The electic fields due to the chages at point A ae shown in Figue a 9 65 C = k = (899 N m /C ) = 5 N/C A (5 m) 9 5 C 4 = k = (899 N m /C ) = 4 N/C ( m) A

- Chapte Since the two fields ae in opposite diections, we subtact thei magnitudes to find the net field = = 874 N/C, to the ight (b) The electic fields at points B ae shown in Figue b 9 65 C = k = (899 N m /C ) = 569 N/C B ( m) 9 5 C = k = (899 N m /C ) = 97 N/C B (5 m) Since the fields ae in the same diection, we add thei magnitudes to find the net field = + = 654 N/C, to the ight (c) At A, = 874 N/C, to the ight The foce on a poton placed at this point would be F 5 = = (6 C)(874 N/C) = 4 N, to the ight VALUAT: A poton has positive chage so the foce that an electic field eets on it is in the same diection as the field IDNTIFY: The electic foce is F = Figue ST UP: The gavit foce (weight) has magnitude w= mg and is downwad XCUT: (a) To balance the weight the electic foce must be upwad The electic field is downwad, so fo an upwad foce the chage of the peson must be negative w= F gives mg = and mg (6 kg)(98 m/s ) = = = 9 C 5 N/C 9 (9 C) 7 (b) F = k = (899 N m /C ) = 4 N The epulsive foce is immense and this is not a ( m) feasible means of flight VALUAT: The net chage of chaged objects is tpicall much less than C IDNTIFY: () gives the foce on the paticle in tems of its chage and the electic field between the plates The foce is constant and poduces a constant acceleation The motion is simila to pojectile motion; use constant acceleation euations fo the hoizontal and vetical components of the motion (a) ST UP: The motion is sketched in Figue a Fo an electon = e Figue a F = and negative gives that F and ae in opposite diections, so F is upwad The fee-bod diagam fo the electon is given in Figue b Figue b F = ma XCUT: e = ma Solve the kinematics to find the acceleation of the electon: Just misses uppe plate sas that = cm when =+ 5 cm -component 6 v = v = 6 m/s, a =, = m, t =? = v t+ a t

t m 8 = = = 5 s 6 v 6 m/s In this same time t the electon tavels 5 m veticall: -component 8 t = 5 s, v =, =+ 5 m, a =? = v t+ a t lectic Chage and lectic Field - ( ) (5 m) a = = = 64 m/s 8 t (5 s) (This analsis is ve simila to that used in Chapte fo pojectile motion, ecept that hee the acceleation is upwad athe than downwad) This acceleation must be poduced b the electic-field foce: e = ma ma (99 kg)(64 m/s ) = = = 64 N/C e 6 C Note that the acceleation poduced b the electic field is much lage than g, the acceleation poduced b gavit, so it is pefectl ok to neglect the gavit foce on the elcton in this poblem e (6 C)(64 N/C) (b) a = = = 49 m/s 7 mp 67 kg This is much less than the acceleation of the electon in pat (a) so the vetical deflection is less and the poton won t hit the plates The poton has the same initial speed, so the poton takes the same time 8 t = 5 s to tavel hoizontall the length of the plates The foce on the poton is downwad (in the same diection as, since is positive), so the acceleation is downwad and a = 49 m/s 6 = v t+ a t = ( 49 m/s )(5 s) = 7 m The displacement is 7 m, 8 6 downwad (c) VALUAT: The displacements ae in opposite diections because the electon has negative chage and the poton has positive chage The electon and poton have the same magnitude of chage, so the foce the electic field eets has the same magnitude fo each chage But the poton has a mass lage b a facto of 86 so its acceleation and its vetical displacement ae smalle b this facto 4 IDNTIFY: Appl (7) to calculate the electic field due to each chage and add the two field vectos to find the esultant field ST UP: Fo, ˆ = ˆj Fo, = ˆ cosθi+ ˆ sinθ ˆj, whee θ is the angle between and the +-ais 9 (9 N m /C )( 5 C) 4 XCUT: (a) ˆ ˆ = j = = ( 8 N/C) j P 4 m ( ) ( ) = = = 9 (9 N m /C )( C) P m + (4 m) 4 8 N/C The angle of, -ais, is 8 tan 4 cm ( ) = 69 Thus cm 4 ˆ ˆ ˆ ˆ = (8 N/C)( i cos69 + jsin69 ) = ( 6485 N/C) i + (864 N/C) j 4 (b) The esultant field is ˆ ˆ + = ( 6485 N/C) i+ ( 8 N/C + 864 N/C) j ˆ 4 + = ( 6485 N/C) i (95 N/C) ˆj measued fom the VALUAT: is towad since is negative is diected awa fom, since is positive 5 IDNTIFY: Appl constant acceleation euations to the motion of the electon ST UP: Let + be to the ight and let + be downwad The electon moves cm to the ight and 5 cm downwad XCUT: Use the hoizontal motion to find the time when the electon emeges fom the field 6 8 = m, a =, v = 6 m s = vt + a t gives t = 5 s Since a =, 6 8 v + v 5 v = 6 m s = 5 m, v =, t = 5 s = t gives v = 8 m s 6 Then v= v + v = 79 m s VALUAT: v = v + at gives a = 64 m/s The electic field between the plates is ma (9 kg)(64 m/s ) = = = 64 V/m This is not a ve lage field e 6 C

- Chapte 6 IDNTIFY: Use the components of fom ample 6 to calculate the magnitude and diection of Use F = to calculate the foce on the 5 nc chage and use Newton's thid law fo the foce on the 8 nc chage ST UP: Fom ample 6, = ( N/C) i+ ˆ (4 N/C) ˆj XCUT: (a) = + = ( N/C) + (4 N/C) = 78 N/C tan = tan (4 ) = 58, so θ = 8 counteclockwise fom the +-ais (b) (i) F = so 8 F = (78 N C)(5 C) = 445 N, at 5 below the +-ais 8 (ii) 445 N at 8 counteclockwise fom the +-ais VALUAT: The foces in pat (b) ae epulsive so the ae along the line connecting the two chages and in each case the foce is diected awa fom the chage that eets it 7 IDNTIFY and ST UP: The electic foce is given b () The gavitational foce is we = meg Compae these foces (a) XCUT: w e = (99 kg)(98 m/s ) = 89 N 4 In amples 7 and 8, = N/C, so the electic foce on the electon has magnitude 4 5 F = = e = (6 C)( N/C) = 6 N we 89 N 5 5 557 = = F 6 N The gavitational foce is much smalle than the electic foce and can be neglected (b) mg = m= g = = 6 m 6 kg 4 4 = = 79 ; m= 79 m e m 99 kg e 4 6 / (6 C)( N/C)/(98 m/s ) 6 kg VALUAT: m is much lage than m e We found in pat (a) that if m= me the gavitational foce is much smalle than the electic foce is the same so the electic foce emains the same To get w lage enough to eual F, the mass must be made much lage (c) The electic field in the egion between the plates is unifom so the foce it eets on the chaged object is independent of whee between the plates the object is placed 8 IDNTIFY: Appl constant acceleation euations to the motion of the poton = F/ 7 ST UP: A poton has mass m p = 67 kg and chage + e Let + be in the diection of motion of the poton e XCUT: (a) v = a = = vt+ a t gives mp 7 (6 m)(67 kg) = = 48 N C 6 (6 C)(5 s) e 4 (b) v = v + at = t = m s m p e = a t = t Solving fo gives mp VALUAT: The electic field is diected fom the positivel chaged plate towad the negativel chaged plate and the foce on the poton is also in this diection 9 IDNTIFY: Find the angle θ that ˆ makes with the +-ais Then = ˆ (cos θ ) i+ ˆ (sin θ ) ˆj ST UP: tan θ = / 5 π XCUT: (a) tan = ad = ˆ ˆj π (b) tan = ad ˆ = i ˆ+ ˆj 4 6 (c) tan = 97 ad = 9 = ˆ i+ ˆ 9 ˆj (Second uadant) + VALUAT: In each case we can veif that ˆ is a unit vecto, because ˆ ˆ=

lectic Chage and lectic Field - 4 IDNTIFY: The net foce on each chage must be zeo ST UP: The foce diagam fo the 65 μc chage is given in Figue 4 F is the foce eeted on the chage b the unifom electic field The chage is negative and the field is to the ight, so the foce eeted b the field is to the left F is the foce eeted b the othe point chage The two chages have opposite signs, so the foce is attactive Take the + ais to be to the ight, as shown in the figue 6 8 XCUT: (a) F = = (65 C)(85 N/C) = N 6 6 9 (65 C)(875 C) F = k = (899 N m /C ) = 88 N (5 m) F = gives T + F F = and T = F F = 8 N (b) Now F is to the left, since like chages epel F = gives T F F = and T = F + F = N VALUAT: The tension is much lage when both chages have the same sign, so the foce one chage eets on the othe is epulsive Figue 4 4 IDNTIFY and ST UP: Use in () to calculate F, F = ma to calculate a, and a constant acceleation euation to calculate the final velocit Let + be east (a) XCUT: F = = (6 C)(5 N/C) = 4 N a = F / m= (4 N)/(99 kg) = +68 m/s 5 v =+ 45 m/s, a =+ 68 m/s, = 75 m, v =? 5 v = v + a ( ) gives v = 6 m/s VALUAT: is west and is negative, so F is east and the electon speeds up (b) XCUT: F = = (6 C)(5 N/C) = 4 N a = F / m= ( 4 N)/(67 kg) = 46 m/s 7 8 4 8 v =+ 9 m/s, a = 46 m/s, = 75 m, v =? 4 v = v + a ( ) gives v = 59 m/s VALUAT: > so F is west and the poton slows down 4 IDNTIFY: Coulomb s law fo a single point-chage gives the electic field (a) ST UP: Coulomb s law fo a point-chage is = (/ 4 πp ) / XCUT: = (9 9 N m /C )(6 (b) Taking the atio of the electic fields gives 9 C)/(5 5 m) = 64 N/C / plates = (64 N/C)/( 4 N/C) = 64 6 times as stong VALUAT: The electic field within the nucleus is huge compaed to tpical laboato fields! 4 IDNTIFY: Calculate the electic field due to each chage and find the vecto sum of these two fields ST UP: At points on the -ais onl the component of each field is nonzeo The electic field of a point chage points awa fom the chage if it is positive and towad it if it is negative XCUT: (a) Halfwa between the two chages, = 4 a (b) Fo < a, = = 4 πp ( a + ) ( a ) 4 πp ( a ) + a Fo > a, = + = 4 πp ( a+ ) ( a ) 4 πp ( a ) + a Fo < a, = + = 4 πp ( a+ ) ( a ) 4 πp ( a ) The gaph of vesus is sketched in Figue 4

-4 Chapte VALUAT: The magnitude of the field appoaches infinit at the location of one of the point chages Figue 4 44 IDNTIFY: Fo a point chage, = k Fo the net electic field to be zeo, and must have eual magnitudes and opposite diections ST UP: Let =+ 5 nc and =+ 8 nc is towad a negative chage and awa fom a positive chage XCUT: The two chages and the diections of thei electic fields in thee egions ae shown in Figue 44 Onl in egion II ae the two electic fields in opposite diections Conside a point a distance fom so a 5 nc 8 nc distance m fom = gives k = k 6 = ( m ) 4 =± ( m ) ( ) and = 4 m is the positive solution The electic field is zeo at a point between the two chages, 4 m fom the 5 nc chage and 96 m fom the 8 nc chage VALUAT: Thee is onl one point along the line connecting the two chages whee the net electic field is zeo This point is close to the chage that has the smalle magnitude Figue 44 45 IDNTIFY: (7) gives the electic field of each point chage Use the pinciple of supeposition and add the electic field vectos In pat (b) use () to calculate the foce, using the electic field calculated in pat (a) (a) ST UP: The placement of chages is sketched in Figue 45a Figue 45a The electic field of a point chage is diected awa fom the point chage if the chage is positive and towad the point chage if the chage is negative The magnitude of the electic field is =, whee is the distance P between the point whee the field is calculated and the point chage (i) At point a the fields of and of ae diected as shown in Figue 45b Figue 45b

9 C 4 πp ( m) 9 5 C = = (8988 N m /C ) = 48 N/C 4 πp (6 m) XCUT: = 4494 N/C, = = 48 N/C, = = = (8988 N m /C ) = 4494 N/C = + =+ 4494 N/C + 48 N/C =+ 574 N/C = + = lectic Chage and lectic Field -5 The esultant field at point a has magnitude 574 N/C and is in the +-diection (ii) ST UP: At point b the fields of and of ae diected as shown in Figue 45c XCUT: Figue 45c = = (8988 N m /C ) = 5 N/C 9 C P m 9 5 C P 4 m ( ) ( ) ( ) = = 8988 N m /C = 89 N/C = 5 N/C, = = 89 N/C, = = + =+ 5 N/C 89 N/C = 684 N/C = + = The esultant field at point b has magnitude 68 N/C and is in the -diection (iii) ST UP: At point c the fields of and of ae diected as shown in Figue 45d XCUT: Figue 45d = = (8988 N m /C ) = 4494 N/C 9 C P m 9 5 C ( ) 4 πp ( m) ( ) = = 8988 N m /C = 449 N/C = 4494 N/C, = =+ 449 N/C, = = + = 4494 N/C + 449 N/C = 445 N/C = + = The esultant field at point b has magnitude 44 N/C and is in the -diection (b) ST UP: Since we have calculated at each point the simplest wa to get the foce is to use F = e XCUT: (i) F (ii) F = (6 C)(684 N/C) = 4 N, + -diection 7 (iii) F = (6 C)(445 N/C) = 648 N, + -diection 7 = (6 C)(574 N/C) = 9 N, -diection 7 VALUAT: The geneal ule fo electic field diection is awa fom positive chage and towad negative chage Whethe the field is in the + - o -diection depends on whee the field point is elative to the chage that poduces the field In pat (a) the field magnitudes wee added because the fields wee in the same diection and in (b) and (c) the field magnitudes wee subtacted because the two fields wee in opposite diections In pat (b) we could have used Coulomb's law to find the foces on the electon due to the two chages and then added these foce vectos, but using the esultant electic field is much easie

-6 Chapte 46 IDNTIFY: Appl (7) to calculate the field due to each chage and then euie that the vecto sum of the two fields to be zeo ST UP: The field of each chage is diected towad the chage if it is negative and awa fom the chage if it is positive XCUT: The point whee the two fields cancel each othe will have to be close to the negative chage, because it is smalle Also, it can t be between the two chages, since the two fields would then act in the same diection We could use Coulomb s law to calculate the actual values, but a simple wa is to note that the 8 nc chage is twice as lage as the 4 nc chage The zeo point will theefoe have to be a facto of fathe fom the 8 nc chage fo the two fields to have eual magnitude Calling the distance fom the 4 nc chage: + = and = 9 m VALUAT: This point is 4 m fom the 8 nc chage The two fields at this point ae in opposite diections and have eual magnitudes 47 IDNTIFY: = k The net field is the vecto sum of the fields due to each chage ST UP: The electic field of a negative chage is diected towad the chage Label the chages, and, as shown in Figue 47a This figue also shows additional distances and angles The electic fields at point P ae shown in Figue 47b This figue also shows the coodinates we will use and the and components of the fields, and XCUT: 5 C (899 N m /C ) 449 N/C ( m) 6 9 6 = = = 6 9 C 6 = (899 N m /C ) = 499 N/C (6 m) 7 = + + = and = + + = + cos5 = 4 N/C 7 = 4 N/C, towad the μc chage VALUAT: The -components of the fields of all thee chages ae in the same diection Figue 47 48 IDNTIFY: A positive and negative chage, of eual magnitude, ae on the -ais, a distance a fom the oigin Appl (7) to calculate the field due to each chage and then calculate the vecto sum of these fields ST UP: due to a point chage is diected awa fom the chage if it is positive and diected towad the chage if it is negative XCUT: (a) Halfwa between the chages, both fields ae in the -diection and =, in the P a -diection (b) = fo < a = + fo > a 4 πp ( a+ ) ( a ) 4 πp ( a+ ) ( a ) = fo < a is gaphed in Figue 48 4 πp ( a+ ) ( a )

lectic Chage and lectic Field -7 VALUAT: At points on the ais and between the chages, is in the -diection because the fields fom both chages ae in this diection Fo < a and >+ a, the fields fom the two chages ae in opposite diections and the field fom the close chage is lage in magnitude Figue 48 49 IDNTIFY: The electic field of a positive chage is diected adiall outwad fom the chage and has magnitude = The esultant electic field is the vecto sum of the fields of the individual chages P ST UP: The placement of the chages is shown in Figue 49a Figue 49a XCUT: (a) The diections of the two fields ae shown in Figue 49b = = with = 5 m P Figue 49b (b) The two fields have the diections shown in Figue 49c = = ; =, = = +, in the + -diection Figue 49c = = (8988 N m /C ) = 968 N/C 9 6 C 4 πp (5 m) = = (8988 N m /C ) = 66 N/C 9 6 C 4 πp (45 m) = + = 968 N/C + 66 N/C = 66 N/C; =+ 6 N/C, =

-8 Chapte (c) The two fields have the diections shown in Figue 49d 4 m sinθ = = 8 5 m m cosθ = = 6 5 m = Figue 49d P 9 6 C = (8988 N m /C ) = 7 N/C (4 m) = P 9 6 C = (8988 N m /C ) = 57 N/C (5 m) =, = = 7 N/C =+ cos θ =+ (57 N/C)(6) =+ 94 N/C = sin θ = (57 N/C)(8) = 76 N/C = + =+ 9 N/C = + = 7 N/C 76 N/C = 5 N/C = + = (9 N/C) + ( 5 N/C) = 56 N/C and its components ae shown in Figue 49e tanα = 5 N/C tanα = =5 + 9 N/C α = 84 C, counteclockwise fom + -ais Figue 49e (d) The two fields have the diections shown in Figue 49f m sinθ = = 8 5 m Figue 49f

The components of the two fields ae shown in Figue 49g Figue 49g = cos θ, =+ cosθ = + = =+ sin θ, =+ sinθ = = lectic Chage and lectic Field -9 P 9 6 C = ( )( ) (8988 N m /C ) (5 m) = = 868 N/C = + = = sinθ = 868 N/C 8 = 8 N/C = 8 N/C, in the + -diection VALUAT: Point a is smmeticall placed between identical chages, so smmet tells us the electic field must be zeo Point b is to the ight of both chages and both electic fields ae in the +-diection and the esultant field is in this diection At point c both fields have a downwad component and the field of has a component to the ight, so the net is in the 4th uadant At point d both fields have an upwad component but b smmet the have eual and opposite -components so the net field is in the +-diection We can use this sot of easoning to deduce the geneal diection of the net field befoe doing an calculations 5 IDNTIFY: Appl (7) to calculate the field due to each chage and then calculate the vecto sum of those fields ST UP: The fields due to and to ae sketched in Figue 5 (6 C) XCUT: ˆ ˆ = ( i) = 5 i N/C 4 πp (6 m) ˆ ˆ ˆ ˆ = (4 C) (6) + (8) (6 88 ) N C = 4 π ( m) i ( m) j i + j P ˆ ˆ = + = ( 84 N/C) i+ (88 N/C) j = (84 N/C) + (88 N/C) = 6 N/C at 88 θ = tan = 6 above the ais and theefoe 96 counteclockwise fom the + ais 84 VALUAT: is diected towad because is negative and is diected awa fom because is positive Figue 5 5 IDNTIFY: The esultant electic field is the vecto sum of the field of and of ST UP: The placement of the chages is shown in Figue 5a Figue 5a

- Chapte XCUT: (a) The diections of the two fields ae shown in Figue 5b = = P Figue 5b = 97 N/C, = = 97 N/C, = 9 6 C = (8988 N m /C ) (5 m) = = 97 N/C = + = ( 97 N/C) = 479 N/C = + = The esultant electic field at point a in the sketch has magnitude 479 N/C and is in the (b) The diections of the two fields ae shown in Figue 5c -diection 9 6 C 4 πp (5 m) 9 4 πp (45 m) Figue 5c = = (8988 N m /C ) = 97 N/C 6 C = = (8988 N m /C ) = 66 N/C =+ 97 N/C, = = 66 N/C, = = + =+ 97 N/C 66 N/C =+ N/C = + = The esultant electic field at point b in the sketch has magnitude N/C and is in the (c) The placement of the chages is shown in Figue 5d + -diection m sinθ = = 6 5 m 4 m cosθ = = 8 5 m Figue 5d The diections of the two fields ae shown in Figue 5e = P Figue 5e =, = = 7 N/C = sin θ = (57 N/C)(6) = 94 N/C =+ cos θ =+ (57 N/C)(8) =+ 76 N/C 9 6 C = (8988 N m /C ) (4 m) = 7 N/C = P 9 6 C = (8988 N m /C ) (5 m) = 57 N/C

lectic Chage and lectic Field - = + = 9 N/C = + = 7 N/C + 76 N/C = 64 N/C = + = 9 N/C The field and its components ae shown in Figue 5f tanα = 64 N/C tanα = =+ 7 N/C α =, counteclockwise fom + -ais Figue 5f (d) The placement of the chages is shown in Figue 5g m sinθ = = 8 5 m 5 m cosθ = = 6 5 m Figue 5g The diections of the two fields ae shown in Figue 5h = = P 9 6 C = (8988 N m /C ) (5 m) = 868 N/C = = 868 N/C Figue 5h = cos θ, = cosθ = + = ( 868 N/C)( 6) = 4 N/C =+ sin θ, = sinθ = + = = 4 N/C, in the -diection VALUAT: The electic field poduced b a chage is towad a negative chage and awa fom a positive chage As in ecise 45, we can use this ule to deduce the diection of the esultant field at each point befoe doing an calculations λ 5 IDNTIFY: Fo a long staight wie, = π P 9 ST UP: = 449 N m /C π P XCUT: 5 C m 8 m = = πp (5 N C) VALUAT: Fo a point chage, is popotional to / Fo a long staight line of chage, is popotional to /

- Chapte λ 5 IDNTIFY: Appl () fo the finite line of chage and = fo the infinite line of chage π P ST UP: Fo the infinite line of positive chage, is in the + diection XCUT: (a) Fo a line of chage of length a centeed at the oigin and ling along the -ais, the electic field λ is given b (): = iˆ π P a + λ (b) Fo an infinite line of chage: = iˆ Gaphs of electic field vesus position fo both distibutions of π P chage ae shown in Figue 5 VALUAT: Fo small, close to the line of chage, the field due to the finite line appoaches that of the infinite line of chage As inceases, the field due to the infinite line falls off moe slowl and is lage than the field of the finite line Figue 5 54 (a) IDNTIFY: The field is caused b a finite unifoml chaged wie ST UP: The field fo such a wie a distance fom its midpoint is λ λ = = πp ( / a) + P ( / a) + ( 9 )( 8 N m / C 75 9 C/m ) XCUT: = = 4 N/C, diected upwad 6 cm (6 m) + 45 cm (b) IDNTIFY: The field is caused b a unifoml chaged cicula wie Q ST UP: The field fo such a wie a distance fom its midpoint is = We fist find the adius / 4 πp ( + a ) of the cicle using π = l XCUT: Solving fo gives = l/π = (85 cm)/π = 5 cm The chage on this cicle is Q = λl = (75 nc/m)(85 m) = 488 nc The electic field is = P Q ( + a ) / = ( 9 9 N m /C )( 488 9 C/m)( 6 m) (6 m) + (5 m) = 45 4 N/C, upwad VALUAT: In both cases, the fields ae of the same ode of magnitude, but the values ae diffeent because the chage has been bent into diffeent shapes 55 IDNTIFY: Fo a ing of chage, the electic field is given b (8) F = In pat (b) use Newton's thid law to elate the foce on the ing to the foce eeted b the ing ST UP: Q = 5 C, a = 5 m and = 4 m XCUT: (a) = Q ˆ (7 N/C) ˆ / 4 π P ( + a ) i = i 6 5 (b) ( 5 C) (7 N/C) ˆ F = F = = i = (75 N) iˆ on ing on VALUAT: Chages and Q have opposite sign, so the foce that eets on the ing is attactive /

lectic Chage and lectic Field - 56 IDNTIFY: We must use the appopiate electic field fomula: a unifom disk in (a), a ing in (b) because all the chage is along the im of the disk, and a point-chage in (c) (a) ST UP: Fist find the suface chage densit (Q/A), then use the fomula fo the field due to a disk of chage, σ = P ( R/ ) + Q Q 65 C 5 XCUT: The suface chage densit is σ = = = = 4 C/m A π π(5 m) The electic field is σ = P ( R/ ) + (b) ST UP: Fo a ing of chage, the field is = P 5 4 C/m = (885 C /N m ) 5 cm + cm = 4 5 N/C, towad the cente of the disk XCUT: Substituting into the electic field fomula gives Q ( + a ) / = P Q ( + a ) / = 9 (9 N m /C )(65 C)( m) ( m) + (5 m) / (c) ST UP: Fo a point chage, ( ) = 89 4 N/C, towad the cente of the disk = /4 πp / XCUT: = (9 9 N m /C 9 )(65 C)/( m) = 46 5 N/C (d) VALUAT: With the ing, moe of the chage is fathe fom P than with the disk Also with the ing the component of the electic field paallel to the plane of the ing is geate than with the disk, and this component cancels With the point chage in (c), all the field vectos add with no cancellation, and all the chage is close to point P than in the othe two cases 57 IDNTIFY: B supeposition we can add the electic fields fom two paallel sheets of chage ST UP: The field due to each sheet of chage has magnitude σ /P and is diected towad a sheet of negative chage and awa fom a sheet of positive chage (a) The two fields ae in opposite diections and = (b) The two fields ae in opposite diections and = σ σ (c) The fields of both sheets ae downwad and = =, diected downwad P P VALUAT: The field poduced b an infinite sheet of chage is unifom, independent of distance fom the sheet 58 IDNTIFY and ST UP: The electic field poduced b an infinite sheet of chage with chage densit σ has σ magnitude = The field is diected towad the sheet if it has negative chage and is awa fom the sheet if it P has positive chage XCUT: (a) The field lines ae sketched in Figue 58a (b) The field lines ae sketched in Figue 58b VALUAT: The spacing of the field lines indicates the stength of the field In pat (a) the two fields add between the sheets and subtact in the egions to the left of A and to the ight of B In pat (b) the opposite is tue Figue 58

-4 Chapte 59 IDNTIFY: The foce on the paticle at an point is alwas tangent to the electic field line at that point ST UP: The instantaneous velocit detemines the path of the paticle XCUT: In Fig9a the field lines ae staight lines so the foce is alwas in a staight line and velocit and acceleation ae alwas in the same diection The paticle moves in a staight line along a field line, with inceasing speed In Fig9b the field lines ae cuved As the paticle moves its velocit and acceleation ae not in the same diection and the tajecto does not follow a field line VALUAT: In two-dimensional motion the velocit is alwas tangent to the tajecto but the velocit is not alwas in the diection of the net foce on the paticle 6 IDNTIFY: The field appeas like that of a point chage a long wa fom the disk and an infinite sheet close to the disk s cente The field is smmetical on the ight and left ST UP: Fo a positive point chage, is popotional to / and is diected adiall outwad Fo an infinite sheet of positive chage, the field is unifom and is diected awa fom the sheet XCUT: The field is sketched in Figue 6 VALUAT: Nea the disk the field lines ae paallel and euall spaced, which coesponds to a unifom field Fa fom the disk the field lines ae getting fathe apat, coesponding to the / dependence fo a point chage Figue 6 6 IDNTIFY: Use smmet to deduce the natue of the field lines (a) ST UP: The onl distinguishable diection is towad the line o awa fom the line, so the electic field lines ae pependicula to the line of chage, as shown in Figue 6a Figue 6a (b) XCUT and VALUAT: The magnitude of the electic field is invesel popotional to the spacing of the field lines Conside a cicle of adius with the line of chage passing though the cente, as shown in Figue 6b Figue 6b The spacing of field lines is the same all aound the cicle, and in the diection pependicula to the plane of the cicle the lines ae euall spaced, so depends onl on the distance The numbe of field lines passing out though the cicle is independent of the adius of the cicle, so the spacing of the field lines is popotional to the ecipocal of the cicumfeence π of the cicle Hence is popotional to / 6 IDNTIFY: Field lines ae diected awa fom a positive chage and towad a negative chage The densit of field lines is popotional to the magnitude of the electic field ST UP: The field lines epesent the esultant field at each point, the net field that is the vecto sum of the fields due to each of the thee chages XCUT: (a) Since field lines pass fom positive chages and towad negative chages, we can deduce that the top chage is positive, middle is negative, and bottom is positive (b) The electic field is the smallest on the hoizontal line though the middle chage, at two positions on eithe side whee the field lines ae least dense Hee the -components of the field ae cancelled between the positive chages and the negative chage cancels the -component of the field fom the two positive chages VALUAT: Fa fom all thee chages the field is the same as the field of a point chage eual to the algebaic sum of the thee chages

lectic Chage and lectic Field -5 6 (a) IDNTIFY and ST UP: Use (4) to elate the dipole moment to the chage magnitude and the sepaation d of the two chages The diection is fom the negative chage towad the positive chage XCUT: p = d = (45 C)( m) = 4 C m; The diection of p is fom towad (b) IDNTIFY and ST UP: Use (5) to elate the magnitudes of the toue and field XCUT: τ = p sin φ, with φ as defined in Figue 6, so Figue 6 τ = psinφ 7 N m = = 86 N/C (4 C m)sin 69 VALUAT: (5) gives the toue about an ais though the cente of the dipole But the foces on the two chages fom a couple (Poblem 5) and the toue is the same fo an ais paallel to this one The foce on each chage is and the maimum moment am fo an ais at the cente is d /, so the maimum toue is 8 ( )( d/) = N m The toue fo the oientation of the dipole in the poblem is less than this maimum 64 (a) IDNTIFY: The potential eneg is given b (7) ST UP: U( φ) = p = pcos φ, whee φ is the angle between p and XCUT: paallel: and U( ) pependicula: φ = 9 and U ( 9 ) = φ = = p ( ) ( ) ( )( ) 4 ΔU 8 ( J) =Δ so = = = 9 K k ( 8 J/K) Δ U = U U = p = = (b) 6 4 9 5 C m 6 N/C 8 J kt U T VALUAT: Onl at ve low tempeatues ae the dipoles of the molecules aligned b a field of this stength A much lage field would be euied fo alignment at oom tempeatue 65 IDNTIFY: Follow the pocedue specified in pat (a) of the poblem ST UP: Use that >> d ( + d ) ( d ) d XCUT: (a) = = This gives ( d ) ( + d ) ( d 4) ( d 4) d d p = = Since >> d /4, 4 πp ( d 4) πp ( d 4) π P (b) Fo points on the -ais, is in the + diection and + is in the p, so the net field is upwad A simila deivation gives π P at points whee >> d as whee << d VALUAT: falls off like the dipole is zeo / fo a dipole, which is faste than the 66 IDNTIFY: Calculate the electic field due to the dipole and then appl F = p ST UP: Fom ample 5, dipole( ) = π P 67 C m 6 XCUT: dipole = = 4 N C The electic foce is F = = πp ( m) diection The field point is close to has the same magnitude and diection / fo a point chage The total chage of 6 (6 C)(4 N/C) = 658 N and is towad the wate molecule (negative -diection) VALUAT: dipole is in the diection of p so is in the + diection The chage of the ion is negative, so F is diected opposite to and is theefoe in the diection 67 IDNTIFY: Like chages epel and unlike chages attact The foce inceases as the distance between the chages deceases ST UP: The foces on the dipole that is between the slanted dipoles ae sketched in Figue 67a

-6 Chapte XCUT: The foces ae attactive because the + and chages of the two dipoles ae closest The foces ae towad the slanted dipoles so have a net upwad component In Figue 67b, adjacent dipoles chages of opposite sign ae close than chages of the same sign so the attactive foces ae lage than the epulsive foces and the dipoles attact VALUAT: ach dipole has zeo net chage, but because of the chage sepaation thee is a non-zeo foce between dipoles Figue 67 68 IDNTIFY: Find the vecto sum of the fields due to each chage in the dipole ST UP: A point on the -ais with coodinate is a distance = ( d/) + fom each chage XCUT: (a) The magnitude of the field the due to each chage is = =, P 4 πp ( d ) + whee d is the distance between the two chages The -components of the foces due to the two chages ae eual and oppositel diected and so cancel each othe The two fields have eual -components, d so = = sinθ, whee θ is the angle below the -ais fo both fields sinθ = 4 πp ( d ) + ( d ) + d d and dipole = = The field is the diection 4 πp ( d ) + ( d ) + 4 πp(( d ) + ) d (b) At lage, >> ( d ), so the epession in pat (a) educes to the appoimation dipole P d VALUAT: ample 5 shows that at points on the + ais fa fom the dipole, dipole The π P epession in pat (b) fo points on the ais has a simila fom 69 IDNTIFY: The toue on a dipole in an electic field is given b τ =p ST UP: τ = p sinφ, whee φ is the angle between the diection of p and the diection of XCUT: (a) The toue is zeo when p is aligned eithe in the same diection as o in the opposite diection, as shown in Figue 69a (b) The stable oientation is when p is aligned in the same diection as In this case a small otation of the dipole esults in a toue diected so as to bing p back into alignment with When p is diected opposite to, a small displacement esults in a toue that takes p fathe fom alignment with (c) Field lines fo dipole in the stable oientation ae sketched in Figue 69b VALUAT: The field of the dipole is diected fom the + chage towad the chage Figue 69 7 IDNTIFY: The plates poduce a unifom electic field in the space between them This field eets toue on a dipole and gives it potential eneg ST UP: The electic field between the plates is given b = σ / P, and the dipole moment is p = ed The potential eneg of the dipole due to the field is U = p = pcosφ, and the toue the field eets on it is τ = p sin φ

lectic Chage and lectic Field -7 XCUT: (a) The potential eneg, U = p = pcosφ, is a maimum when φ = 8 The field between the plates is = σ / P, giving U ma = (6 9 C)( 9 m)(5 6 C/m )/(885 C /N m ) = 497 9 J The oientation is paallel to the electic field (pependicula to the plates) with the positive chage of the dipole towad the positive plate (b) The toue, τ = p sin φ, is a maimum when φ = 9 o 7 In this case τ = p = pσ / P = edσ / P ma 6 ( )( )( ) ( ) τ ma = 6 C m 5 C/m 885 C / N m τ ma = 497 N m The dipole is oiented pependicula to the electic field (paallel to the plates) (c) F = VALUAT: When the potential eneg is a maimum, the toue is zeo In both cases, the net foce on the dipole is zeo because the foces on the chages ae eual but opposite (which would not be tue in a nonunifom electic field) 7 (a) IDNTIFY: Use Coulomb's law to calculate each foce and then add them as vectos to obtain the net foce Toue is foce times moment am ST UP: The two foces on each chage in the dipole ae shown in Figue 7a sinθ = 5/ so θ = 486 Opposite chages attact and like chages epel F = F + F = XCUT: Figue 7a 6 6 (5 C)( C) F = k = k = ( m) F = F sinθ = 846 N 4 N F = 846 N so F = F + F = 68 N (in the diection fom the + 5-μC chage towad the 5-μC chage) VALUAT: The -components cancel and the -components add (b) ST UP: Refe to Figue 7b The -components have zeo moment am and theefoe zeo toue F and F both poduce clockwise toues XCUT: X F Figue 7b F = F cosθ = 74 N τ = ( )(5 m) = N m, clockwise VALUAT: The electic field poduced b the μc chage is not unifom so (5) does not appl