1. THINK Ampere is the SI unit for current. An ampere is one coulomb per second.

Size: px
Start display at page:

Download "1. THINK Ampere is the SI unit for current. An ampere is one coulomb per second."

Transcription

1 Chapte 4 THINK Ampee is the SI unit fo cuent An ampee is one coulomb pe second EXPRESS To calculate the total chage though the cicuit, we note that A C/s and h 6 s ANALYZE (a) Thus, F I HG K J F H G I K J C h s 84 A h C s h (b) The change in potential enegy is U = q = ( 5 C)( ) = 6 6 J LEARN Potential diffeence is the change of potential enegy pe unit chage Unlike electic field, potential diffeence is a scala quantity The magnitude is U = e = 9 e = Ge (a) The change in enegy of the tansfeed chage is U = q = ( C)( 9 ) = J (b) If all this enegy is used to acceleate a -kg ca fom est, then and we find the ca s final speed to be U K mv, K U ( J) v m m kg 4 (a) E F e 77 m/s N 6 C 4 N C 4 /m c hb g 4 (b) Es 4 N C m 9 5 THINK The electic field poduced by an infinite sheet of chage is nomal to the sheet and is unifom EXPRESS The magnitude of the electic field poduced by the infinite sheet of chage is E = /, whee is the suface chage density Place the oigin of a coodinate system at the sheet and take the x axis to be paallel to the field and positive in the diection of the field Then the electic potential is 7

2 74 CHAPTER 4 E dx Ex, s z x whee s is the potential at the sheet The equipotential sufaces ae sufaces of constant x; that is, they ae planes that ae paallel to the plane of chage If two sufaces ae sepaated by x then thei potentials diffe in magnitude by = Ex = (/ )x 6 ANALYZE Thus, fo C m and 5, we have c hb g 885 C N m 5 x 6 C m s 88 m LEARN Equipotential sufaces ae always pependicula to the electic field lines Figue 4-5(a) depicts the electic field lines and equipotential sufaces fo a unifom electic field 6 (a) B A = U/q = W/( e) = (94 9 J)/( 6 9 C) = 46 (b) C A = B A = 46 (c) C B = (since C and B ae on the same equipotential line) 7 We connect A to the oigin with a line along the y axis, along which thee is no change of potential (Eq 4-8: E ds ) Then, we connect the oigin to B with a line along the x axis, along which the change in potential is which yields B A = z z x E ds 4 x dx F H G 8 (a) By Eq 4-8, the change in potential is the negative of the aea unde the cuve Thus, using the aea-of-a-tiangle fomula, we have which yields = zx E ds z b gb g (b) Fo any egion within x m, E ds is positive, but fo any egion fo which x > m it is negative Theefoe, = max occus at x = m I KJ

3 75 which yields max = 4 zx E ds b gb g (c) In view of ou esult in pat (b), we see that now (to find = ) we ae looking fo some X > m such that the aea fom x = m to x = X is 4 Using the fomula fo a tiangle ( < x < 4) and a ectangle (4 < x < X), we equie Theefoe, X = 55 m b gb g b 9 (a) The wok done by the electic field is gb g 4 4 X f q q d W q E ds dz i 9 d (6 C)(58 C/m )(56 m) (885 C /Nm ) 87 J (b) Since = W/q = z/, with set to be zeo on the sheet, the electic potential at P is z (58 C/m )(56 m) (885 C /Nm ) 7 In the inside egion between the plates, the individual fields (given by Eq 4-) ae in the same diection ( i ): E 5 9 C/m 5 9 in C/m ˆ i (4 N/C)i ˆ (885 C /Nm ) (885 C /Nm ) In the outside egion whee x > 5 m, the individual fields point in opposite diections: E 5 C/m 5 C/m i i (4 N/C)i (885 C /Nm ) (885 C /Nm ) 9 9 out ˆ ˆ ˆ Theefoe, by Eq 4-8, we have in 5 5 out E ds E dx E dx

4 76 CHAPTER 4 (a) The potential as a function of is q q E d d 4 R 8 R 9 5 (899 Nm C )(5 C)(45 m) ( m) (b) Since = () (R) = q/8 R, we have R 9 5 q (899 Nm C )(5 C) 8 R ( m) The chage is (m)( ) q N m /C 9 4 R C (a) The chage on the sphee is ( )(5 m) q 899 Nm C 9 4 R C 9 (b) The (unifom) suface chage density (chage divided by the aea of the sphee) is 4 (a) The potential diffeence is A 9 q C 8 C/m 4 R 4 5 m A B 45 q q 6 C899 9 N m C B 4 4 m m (b) Since () depends only on the magnitude of, the esult is unchanged 5 THINK The electic potential fo a spheically symmetic chage distibution falls off as /, whee is the adial distance fom the cente of the chage distibution EXPRESS The electic potential at the suface of a dop of chage q and adius R is given by = q/4 R

5 77 ANALYZE (a) With = 5 and q C, we find the adius to be Nm / C C q R m (b) Afte the two dops combine to fom one big dop, the total volume is twice the volume of an oiginal dop, so the adius R' of the combined dop is given by (R') = R and R' = / R The chage is twice the chage of the oiginal dop: q' = q Thus, q q R R / 4 4 / ( 5 ) 79 / LEARN A positively chaged configuation poduces a positive electic potential, and a negatively chaged configuation poduces a negative electic potential Adding moe chage inceases the electic potential 6 In applying Eq 4-7, we ae assuming as All cone paticles ae equidistant fom the cente, and since thei total chage is q q + q q =, then thei contibution to Eq 4-7 vanishes The net potential is due, then, to the two +4q paticles, each of which is a distance of a/ fom the cente: 4q 4q 6q 4 a / 4 a / 4 a 9 m 9 6(899 Nm C )(6 C) 7 A chage 5q is a distance d fom P, a chage 5q is a distance d fom P, and two chages +5q ae each a distance d fom P, so the electic potential at P is 9 5 q q (899 Nm C )(5 C) 4 d d d d 8 d (4 m) 4 56 The zeo of the electic potential was taken to be at infinity 8 When the chage q is infinitely fa away, the potential at the oigin is due only to the chage q : q = = d

6 78 CHAPTER 4 Thus, q /d = 64 7 C/m Next, we note that when q is located at x = 8 m, the net potential vanishes ( + = ) Theefoe, kq kq 8 m d Thus, we find q = ( q / d)(8 m) = 5 8 C = e 9 Fist, we obseve that (x) cannot be equal to zeo fo x > d In fact (x) is always negative fo x > d Now we conside the two emaining egions on the x axis: x < and < x < d (a) Fo < x < d we have d = x and d = d x Let F HG x k q q ( ) d d I F HG q KJ 4 x d x and solve: x = d/4 With d = 4 cm, we have x = 6 cm (b) Similaly, fo x < the sepaation between q and a point on the x axis whose coodinate is x is given by d = x; while the coesponding sepaation fo q is d = d x We set F x k q q I q ( ) d d x F I d x K J 4 HG KJ to obtain x = d/ With d = 4 cm, we have x = cm Since accoding to the poblem statement thee is a point in between the two chages on the x axis whee the net electic field is zeo, the fields at that point due to q and q must be diected opposite to each othe This means that q and q must have the same sign (ie, eithe both ae positive o both negative) Thus, the potentials due to eithe of them must be of the same sign Theefoe, the net electic potential cannot possibly be zeo anywhee except at infinity We use Eq 4-: HG I K J Nm C 47 4 Cm 9 p 4 5 m 5 6 Fom Eq 4- and Eq 4-4, we have (fo i = º)

7 79 p cos p cos i ep cos Wa q e cos with = 9 m Fo = 8º the gaph indicates W a = 4 J, fom which we can detemine p The magnitude of the dipole moment is theefoe 56 7 Cm (a) Fom Eq 4-5, we find the potential to be L / ( L / 4) d ln 4 d 9 (899 Nm C )(68 C/m) ln 4 (b) The potential at P is = due to supeposition 4 The potential is (6 m / ) (6 m) / 4 (8 m) 8 m P 9 dq Q (899 Nm C )(56 C) dq od od 4 R 4 R 4 R 7 m 6 We note that the esult is exactly what one would expect fo a point-chage Q at a distance R This coincidence is due, in pat, to the fact that is a scala quantity 5 (a) All the chage is the same distance R fom C, so the electic potential at C is Q 6Q 5Q, m 9 5(899 Nm C )(4 C) R R R whee the zeo was taken to be at infinity (b) All the chage is the same distance fom P That distance is R D, so the electic potential at P is Q 6Q 5Q 4 R D R D 4 R D 9 5(899 N m C )(4 C) 78 (8 m) (67 m)

8 8 CHAPTER 4 6 The deivation is shown in the book (Eq 4- though Eq 4-5) except fo the change in the lowe limit of integation (which is now x = D instead of x = ) The esult is theefoe (cf Eq 4-5) = ln 4 o L + L + d D + D + d = ln 4 = Letting d denote m, we have Q Q Q Q 4 d d d d ( m) 9 9 (899 Nm C )( C) 4 8 Conside an infinitesimal segment of the od, located between x and x + dx It has length dx and contains chage dq = dx, whee = Q/L is the linea chage density of the od Its distance fom P is d + x and the potential it ceates at P is dq dx d 4 d x 4 d x To find the total potential at P, we integate ove the length of the od and obtain: L L dx ln( ) Q d x ln L 4 d x 4 4 L d 9 5 (899 N m C )(56 C) m ln m 5 m 79 9 Since the chage distibution on the ac is equidistant fom the point whee is evaluated, its contibution is identical to that of a point chage at that distance We assume as and apply Eq 4-7: Q 4Q Q Q 4 R 4 R 4 R 4 R 9 (899 Nm C )(7 C) 4 m The dipole potential is given by Eq 4- (with = 9º in this case)

9 8 p cos p cos since cos(9º) = The potential due to the shot ac is q /4 and that caused by the long ac is q /4 Since q = + C, = 4 cm, q = C, and = 6 cm, the potentials of the acs cancel The esult is zeo THINK Since the disk is unifomly chaged, when the full disk is pesent each quadant contibutes equally to the electic potential at P EXPRESS Electical potential is a scala quantity The potential at P due to a single quadant is one-fouth the potential due to the entie disk We fist find an expession fo the potential at P due to the entie disk To do so, conside a ing of chage with adius and (infinitesimal) width d Its aea is d and it contains chage dq = d All the chage in it is at a distance D fom P, so the potential it poduces at P is pd d d 4p D D ANALYZE Integating ove, the total potential at P is R R d D R D D D Theefoe, the potential sq at P due to a single quadant is 5 (77 C/m ) sq R D D (64 m) (59 m) 59 m 4 8 8(885 C /Nm ) 5 47 LEARN Conside the limit D R The potential becomes sq R D D D D 8 8 D R R /4 qsq 8 D 4 D 4 D R whee qsq R /4 is the chage on the quadant In this limit, we see that the potential esembles that due to a point chage q sq Equation 4- applies with dq = dx = bx dx (along x m)

10 8 CHAPTER 4 (a) Hee = x >, so that 4 z bx dx x b 4 b g = 6 (b) Now x d whee d = 5 m, so that bxdx b x d 4 x d 4 = 8 Conside an infinitesimal segment of the od, located between x and x + dx It has length dx and contains chage dq = dx = cx dx Its distance fom P is d + x and the potential it ceates at P is d dq cx dx d x 4 4 d x To find the total potential at P, we integate ove the length of the od and obtain L c L xdx c c L [ x d ln( x d)] L d ln 4 d x 4 4 d 9 m (899 Nm C )(89 C/m ) m ( m)ln m 86 4 The magnitude of the electic field is given by x (5) 5m E 67 m At any point in the egion between the plates, E points away fom the positively chaged plate, diectly towad the negatively chaged one 5 We use Eq 4-4: c Ex ( x, y) ( ) x ) y ( ) x; x / m / m / m x Ey ( x, y) ( ) x ) y ( ) y y c / m / m h / m y We evaluate at x = m and y = m to obtain h

11 8 E ( /m)i ˆ ( /m)j ˆ 6 We use Eq 4-4 This is an odinay deivative since the potential is a function of only one vaiable d d E i (5 x )i ( x)i ( /m )( m)i dx dx ( 9 /m)i ˆ ˆ ˆ ˆ ˆ (a) Thus, the magnitude of the electic field is E = 9 /m (b) The diection of E is î, o towad plate 7 THINK The component of the electic field E in a given diection is the negative of the ate at which potential changes with distance in that diection EXPRESS With xyz, we apply Eq 4-4 to calculate the x, y, and z components of the electic field: Ex yz x Ey xz y Ez 4xyz z which, at (x, y, z) = ( m, m, 4 m), gives (E x, E y, E z ) = (64 /m, 96 /m, 96 /m) ANALYZE The magnitude of the field is theefoe E E E E (64 /m) ( 96 /m) (96 /m) x y z 5 m 5 N C LEARN If the electic potential inceases along some diection, say x, with / x, then thee is a coesponding nonvanishing component of E in the opposite diection ( ) E x 8 (a) Fom the esult of Poblem 4-8, the electic potential at a point with coodinate x is given by

12 84 CHAPTER 4 At x = d we obtain Q x L ln 4 L x 9 5 Q d L (899 N m C )(46 C) 5 m ln ln 4 L d 5 m d 5 m d (9 ) ln (b) We diffeentiate the potential with espect to x to find the x component of the electic field: o E x Q x L Q x x L Q ln x 4 L x x 4 L x L x x 4 x ( x L) (899 N m C )(46 C) (9 N m C), x( x 5 m) x( x 5 m) 4 (9 Nm C) E x xx ( 5 m) (c) Since Ex, its diection elative to the positive x axis is 8 (d) At x = d = 6 cm, we obtain 4 (9 Nm C) E x N/C (6 m)(6 m5 m) (e) Conside two points an equal infinitesimal distance on eithe side of P, along a line that is pependicula to the x axis The diffeence in the electic potential divided by thei sepaation gives the tansvese component of the electic field Since the two points ae situated symmetically with espect to the od, thei potentials ae the same and the potential diffeence is zeo Thus, the tansvese component of the electic field E y is zeo 9 The electic field (along some axis) is the (negative of the) deivative of the potential with espect to the coesponding coodinate In this case, the deivatives can be ead off of the gaphs as slopes (since the gaphs ae of staight lines) Thus, E E x y 5 5 /m 5 N/C x m /m N/C y m

13 85 These components imply the electic field has a magnitude of 69 N/C and a diection of 8º (with espect to the positive x axis) The foce on the electon is given by F qe whee q = e The minus sign associated with the value of q has the implication that F points in the opposite diection fom E (which is to say that its angle is found by adding 8º to that of E ) With e = 6 9 C, we obtain F ˆ ˆ ˆ ˆ ( 6 C)[(5 N/C)i ( N/C)j] ( 4 N)i (6 N)j 4 (a) Conside an infinitesimal segment of the od fom x to x + dx Its contibution to the potential at point P is Thus, d 4 ( x) dx x y 4 L d od P dx L y y 4 4 cx x y dx 9 (899 N m C )(499 C/m ) ( m) (56 m) 56 m 6 c x c x y (b) The y component of the field thee is P Ey L y y 4 4 c d c y y dy L y 9 (899 N m C )(499 C/m ) 98 N/C 56 m ( m) (56 m) (c) We obtained above the value of the potential at any point P stictly on the y-axis In ode to obtain E x (x, y) we need to fist calculate (x, y) That is, we must find the potential fo an abitay point located at (x, y) Then E x (x, y) can be obtained fom E (, ) (, ) / x x y x y x 4 We apply consevation of enegy fo the paticle with q = 75 6 C (which has zeo initial kinetic enegy): U = K f + U f, q Q whee U = 4 o

14 86 CHAPTER 4 (a) The initial value of is 6 m and the final value is (6 + 4) m = m (since the paticles epel each othe) Consevation of enegy, then, leads to K f = 9 J (b) Now the paticles attact each othe so that the final value of is 6 4 = m Use of enegy consevation yields K f = 45 J in this case 4 (a) We use Eq 4-4 with q = q = e and = nm: U k k 9 9 qq e 899 Nm C (6 C) 9 m 9 5 J (b) Since U > and U the potential enegy U deceases as inceases 4 THINK The wok equied to set up the aangement is equal to the potential enegy of the system EXPRESS We choose the zeo of electic potential to be at infinity The initial electic potential enegy U i of the system befoe the paticles ae bought togethe is theefoe zeo Afte the system is set up the final potential enegy is U f q q 4p a a a a a a 4 a p Thus the amount of wok equied to set up the system is given by 9 J 9 q (899 Nm C )( C) W U U f Ui U f 4 a 64 m LEARN The wok done in assembling the system is negative This means that an extenal agent would have to supply W ext 9 J in ode to take apat the aangement completely 44 The wok done must equal the change in the electic potential enegy Fom Eq 4-4 and Eq 4-6, we find (with = m) W 9 9 (e e e)(6 e) 5 J 899 Nm C (8)(6 C) 4 m 45 We use the consevation of enegy pinciple The initial potential enegy is U i = q /4, the initial kinetic enegy is K i =, the final potential enegy is U f = q /4,

15 87 and the final kinetic enegy is K Consevation of enegy yields The solution fo v is f mv, whee v is the final speed of the paticle q q mv q (899 Nm C )()( C) v 6 4pm kg 9 m 5 m 5 m s 46 Let = 5 m, x = m, q = 9 nc, and q = 6 pc The wok done by an extenal agent is given by F I HG KJ c hc h F HG W U q q 4 x 9 9 C 6 C 899 N m 9 KJ M C 5 m 5 m m 8 J I L M N b g b g 47 The escape speed may be calculated fom the equiement that the initial kinetic enegy (of launch) be equal to the absolute value of the initial potential enegy (compae with the gavitational case in Chapte 4) Thus, eq mv 4 O QP whee m = 9 kg, e = 6 9 C, q = e, and = m This yields 4 v = 49 m/s m/s 48 The change in electic potential enegy of the electon-shell system as the electon stats fom its initial position and just eaches the shell is U = ( e)( ) = e Thus fom U K m e v i we find the initial electon speed to be v i 9 U e (6 C)(5 ) me me 9 kg 6 66 m/s 49 We use consevation of enegy, taking the potential enegy to be zeo when the moving electon is fa away fom the fixed electons The final potential enegy is then U e / 4 d, whee d is half the distance between the fixed electons The initial f

16 88 CHAPTER 4 kinetic enegy is Ki mv, whee m is the mass of an electon and v is the initial speed of the moving electon The final kinetic enegy is zeo Thus, Hence, v 4e 4 dm K i = U f / 4 mv e d c8 99 N m C hb4gc6 Ch m 9 kg 9 9 b gc h m s 5 The wok equied is W q Q q Q q Q ( q / ) Q 4 d d 4 d d U 5 (a) Let 5 m be the length of the ectangle and w = 5 m be its width Chage q is a distance fom point A and chage q is a distance w, so the electic potential at A is 6 6 q q 9 5 C C A (899 Nm / C ) 4 w 5 m 5 m 4 6 (b) Chage q is a distance w fom point b and chage q is a distance, so the electic potential at B is B q q 4 5 m 5 m C C (899 N m / C ) w 5 78 (c) Since the kinetic enegy is zeo at the beginning and end of the tip, the wok done by an extenal agent equals the change in the potential enegy of the system The potential enegy is the poduct of the chage q and the electic potential If U A is the potential enegy when q is at A and U B is the potential enegy when q is at B, then the wok done in moving the chage fom B to A is W = U A U B = q ( A B ) = ( 6 C)( ) = 5 J (d) The wok done by the extenal agent is positive, so the enegy of the thee-chage system inceases (e) and (f) The electostatic foce is consevative, so the wok is the same no matte which path is used

17 89 5 Fom Eq 4- and Eq 4-7, we have (fo = 8º) p cos ep U q e 4 4 whee = m Using enegy consevation, we set this expession equal to e and solve fo p The magnitude of the dipole moment is theefoe p = 45 C m 5 (a) The potential enegy is q U 4 d c8 99 N m C hc5 Ch m J elative to the potential enegy at infinite sepaation (b) Each sphee epels the othe with a foce that has magnitude q F 4 d 9 6 c8 99 N m C hc5 Ch m b g 5 N Accoding to Newton s second law the acceleation of each sphee is the foce divided by the mass of the sphee Let m A and m B be the masses of the sphees The acceleation of sphee A is F 5 N aa 45 m s m 5 kg and the acceleation of sphee B is A a B F m B 5 N 5 m s kg (c) Enegy is conseved The initial potential enegy is U = 5 J, as calculated in pat (a) The initial kinetic enegy is zeo since the sphees stat fom est The final potential enegy is zeo since the sphees ae then fa apat The final kinetic enegy is m v m v A A B B, whee v A and v B ae the final velocities Thus, Momentum is also conseved, so U m v m v A A B B m v m v A A B B

18 9 CHAPTER 4 These equations may be solved simultaneously fo v A and v B Substituting vb ( ma / mb ) va, fom the momentum equation into the enegy equation, and collecting tems, we obtain U ( ma / mb )( ma mb ) va Thus, v A UmB (5 J)( kg) A A B 775 m/s m ( m m ) (5 kg)(5 kg kg) We thus obtain v B m A va mb kg 5 kg (775 m/s) 87 m/s, o vb 87 m/s 54 (a) Using U = q we can tanslate the gaph of voltage into a potential enegy gaph (in e units) Fom the infomation in the poblem, we can calculate its kinetic enegy (which is its total enegy at x = ) in those units: K i = 84 e This is less than the height of the potential enegy baie (5 e high once we ve tanslated the gaph as indicated above) Thus, it must each a tuning point and then evese its motion (b) Its final velocity, then, is in the negative x diection with a magnitude equal to that of its initial velocity That is, its speed (upon leaving this egion) is 7 m/s 55 Let the distance in question be The initial kinetic enegy of the electon is K m v, whee v i = 5 m/s As the speed doubles, K becomes 4K i Thus i e i o e U K ( 4Ki Ki ) Ki mevi, e 6 C 899 N m C 4 mv e i 9 kg m s 9 6 m 56 When paticle is at x = m, the total potential enegy vanishes Using Eq 4-4, we have (with metes undestood at the length unit) This leads to qq q q q q 4 d 4 ( d m) 4 ( m)

19 9 q q qq q d m m d which yields q = 57 C 57 THINK Mechanical enegy is conseved in the pocess EXPRESS The electic potential at (, y) due to the two chages Q held fixed at ( x,) is Q 4 x y Thus, the potential enegy of the paticle of chage q at (, y) is U q Qq 4 x y Consevation of mechanical enegy (K i + U i = K f + U f ) gives Qq K f Ki Ui U f Ki 4 x y x y i f, whee y i and axis ANALYZE (a) With we obtain K f y f ae the initial and final coodinates of the moving chage along the y q 6 5 C, Q = 5 6 C, x = m, y i = 4 m, and y f, 6 6 (5 C)( 5 C) J 4 (885 C /N m ) ( m) (4 m) ( m) J (b) We set K f = and solve fo y f (choosing the negative oot, as indicated in the poblem statement): Qq Qq Ki Ui U f J 4 x y 4 x y i f Substituting the values given, we have Ui 7 J, and y f = 85 m

20 9 CHAPTER 4 LEARN The dependence of the final kinetic enegy of the paticle on y is plotted below Fom the plot, we see that K J at y =, and K at y = 85 m The paticle oscillates between the two end-points y f 85 m f f 58 (a) When the poton is eleased, its enegy is K + U = 4 e + e (the latte value is infeed fom the gaph) This implies that if we daw a hoizontal line at the 7 volt height in the gaph and find whee it intesects the voltage plot, then we can detemine the tuning point Intepolating in the egion between cm and cm, we find the tuning point is at oughly x = 7 cm (b) Thee is no tuning point towad the ight, so the speed thee is nonzeo, and is given by enegy consevation: v = (7 e) m = (7 e)(6 x -9 J/e) 67 x -7 kg = km/s (c) The electic field at any point P is the (negative of the) slope of the voltage gaph evaluated at P Once we know the electic field, the foce on the poton follows immediately fom F = q E, whee q = +e fo the poton In the egion just to the left of x = cm, the field is E = (+ /m) î and the foce is F = N (d) The foce F points in the +x diection, as the electic field E (e) In the egion just to the ight of x = 5 cm, the field is E =( /m) î and the magnitude of the foce is F = 7 N (f) The foce F points in the x diection, as the electic field E 59 (a) The electic field between the plates is leftwad in Fig, 4-59 since it points towad lowe values of potential The foce (associated with the field, by Eq -8) is evidently leftwad, fom the poblem desciption (indicating deceleation of the ightwad moving paticle), so that q > (ensuing that F is paallel to E ); it is a poton (b) We use consevation of enegy:

21 9 K + U = K + U m pv + q = m pv + q Using q = +6 9 C, m p = 67 7 kg, v = 9 m/s, = 7, and 5, we obtain the final speed v = 65 4 m/s We note that the value of d is not used in the solution 6 (a) The wok done esults in a potential enegy gain: W = q = (e) Q 4 o R = + 6 J With R = 8 m, we find Q = 5 C (b) The wok is the same, so the incease in the potential enegy is U = + 6 J 6 We note that fo two points on a cicle, sepaated by angle (in adians), the diectline distance between them is = R sin(/) Using this fact, distinguishing between the cases whee N = odd and N = even, and counting the pai-wise inteactions vey caefully, we aive at the following esults fo the total potential enegies We use k 4 Fo configuation (whee all N electons ae on the cicle), we have U N N Nke Nke, U N R sin j R j sin j, Neven, odd j whee Fo configuation, we find N U N N N ke N ke 5, U R sin j R sin j, Neven, Nodd j j whee N The esults ae all of the fom U o ke R a pue numbe In ou table below we have the esults fo those pue numbes as they depend on N and on which configuation we ae consideing The values listed in the U ows ae the potential enegies divided by ke /R

22 94 CHAPTER 4 N U U We see that the potential enegy fo configuation is geate than that fo configuation fo N <, but fo N it is configuation that has the geatest potential enegy (a) N = is the smallest value such that U < U (b) Fo N =, configuation consists of electons distibuted at equal distances aound the cicle, and one electon at the cente A specific electon e on the cicle is R distance fom the one in the cente, and is Rsin 56R distance away fom its neaest neighbos on the cicle (of which thee ae two one on each side) Beyond the neaest neighbos, the next neaest electon on the cicle is Rsin R distance away fom e Thus, we see that thee ae only two electons close to e than the one in the cente 6 (a) Since the two conductos ae connected and must be equal to each othe Let = q /4 R = = q /4 R and note that q + q = q and R = R We solve fo q and q : q = q/, q = q/, o (b) q /q = / = (c) Similaly, q /q = / = 667 (d) The atio of suface chage densities is q 4pR q R q 4pR q R 6 THINK The electic potential is the sum of the contibutions of the individual sphees EXPRESS Let q be the chage on one, q be the chage on the othe, and d be thei sepaation The point halfway between them is the same distance d/ (= m) fom the cente of each sphee

23 95 Fo pats (b) and (c), we note that the distance fom the cente of one sphee to the suface of the othe is d R, whee R is the adius of eithe sphee The potential of eithe one of the sphees is due to the chage on that sphee as well as the chage on the othe sphee ANALYZE (a) The potential at the halfway point is Nm C C 8 C q q 4p d m 8 (b) The potential at the suface of sphee is q q C C 899 N m C R d R m m m (c) Similaly, the potential at the suface of sphee is q q C C 899 N m C d R R m m m LEARN In the limit whee d, the sphees ae isolated fom each othe and the electic potentials at the suface of each individual sphee become and q 9 8 (899 Nm C )( C), 4 R m q 9 8 (899 Nm C )( C) R m 64 Since the electic potential thoughout the entie conducto is a constant, the electic potential at its cente is also THINK If the electic potential is zeo at infinity, then the potential at the suface of the sphee is given by = Q/4 R, whee Q is the chage on the sphee and R is its adius EXPRESS Fom = Q/4 R, we find the chage to be Q 4 R ANALYZE With R 5 mand 5, we have 5 m5 8 Q 4 R 5 C Nm C

24 96 CHAPTER 4 LEARN A plot of the electic potential as a function of is shown to the ight with k / 4 Note that the potential is constant inside the conducting sphee 66 Since the chage distibution is spheically symmetic we may wite E q 4 enc ( ), whee q enc is the chage enclosed in a sphee of adius centeed at the oigin (a) Fo = 4 m, R = m, and R = 5 m, with > R > R we have E q q 4 (4 m) (899 N m C )( C C) (b) Fo R > = 7 m > R, E q 69 /m 9 6 (899 Nm C )( C) 4 67 /m 4 (7 m) (c) Fo R > R >, the enclosed chage is zeo Thus, E = The electic potential may be obtained using Eq 4-8: bg b g z Ebgd (d) Fo = 4 m > R > R, we have q q 4 (4 m) (899 N m C )( C C) 674 (e) Fo = m = R > R, we have q q 4 ( m) (899 N m C )( C C) 4 7 (f) Fo R > = 7 m > R,

25 q q 9 C C (899 Nm C ) 4 R 7 m m 4 47 (g) Fo R > = 5 m = R, 6 6 q q 9 C C (899 Nm C ) 4 R 5 m m (h) Fo R > R >, q q 9 C C (899 Nm C ) 4 R R 5 m m 4 45 (i) At =, the potential emains constant, 4 45 (j) The electic field and the potential as a function of ae depicted below: 67 (a) The magnitude of the electic field is 8 9 C899 Nm C q E 4 R 5m (b) = RE = (5 m)( 4 N/C) = 8 (c) Let the distance be x Then b g 4 x q F HG R x R I K J 5, 4 N C

26 98 CHAPTER 4 which gives R x b5 mgb5g 58 m The potential enegy of the two-chage system is U Nm C C4 C 4 x x y y cm 9 J qq Thus, 9 J of wok is needed 69 THINK To calculate the potential, we fist apply Gauss law to calculate the electic field of the chaged cylinde of adius R The Gaussian suface is a cylindical suface that is concentic with the cylinde EXPRESS We imagine a cylindical Gaussian suface A of adius and length h concentic with the cylinde Then, by Gauss law, A q E da enc he, whee qenc is the amount of chage enclosed by the Gaussian cylinde Inside the chaged cylinde ( R), qenc, so the electic field is zeo On the othe hand, outside the cylinde ( > R), qenc hso the magnitude of the electic field is q/ h E whee is the linea chage density and is the distance fom the line to the point whee the field is measued The potential diffeence between two points and is E d ANALYZE (a) The adius of the cylinde ( m, the same as R B ) is denoted R, and the field magnitude thee (6 N/C) is denoted E B Fom the equation above, we see that the electic field beyond the suface of the cylinde is invesely popotional with : RB E EB, RB

27 99 Thus, if = R C = 5 m, we obtain R 6 N/C m B EC EB 64 N C R 5 m (b) The potential diffeence between B and C is C RB ER B B R C 5 m B C d EBRB ln (6 N/C)( m)ln R C RB m 9 (c) The electic field thoughout the conducting volume is zeo, which implies that the potential thee is constant and equal to the value it has on the suface of the chaged cylinde: A B = LEARN The electic potential at a distance R can be witten as B ( ) B EBRB ln RB We see that () deceases logaithmically with 7 (a) We use Eq 4-8 to find the potential: wall Ed, o Consequently, = (R )/4 (b) The value at = is R R 4 cente C m 4 5m 7 8 4c885 C mhe b g j R Thus, the diffeence is 4 cente 78 7 THINK The component of the electic field E in any diection is the negative of the ate at which potential changes with distance in that diection EXPRESS Fom Eq 4-, the electic potential of a dipole at a point a distance away is

28 CHAPTER 4 pcos 4 whee p is the magnitude of the dipole moment p and is the angle between p and the position vecto of the point The potential at infinity is taken to be zeo ANALYZE On the dipole axis = o, so cos = Theefoe, magnitude of the electic field is p d p Ebg F d H G I K J 4 LEARN Take the z axis to be the dipole axis Fo z ( ), the othe hand, fo z ( ), E p z / E p z On / 7 Using Eq 4-8, we have Δ = A d = 4 A = A(9/m ) 7 (a) The potential on the suface is 4 6 C899 9 Nm C q 4 R m 5 6 (b) The field just outside the sphee would be 5 q E R R m, 4 m which would have exceeded M/m So this situation cannot occu 74 The wok done is equal to the change in the (total) electic potential enegy U of the system, whee qq qq qq U and the notation indicates the distance between q and q (simila definitions apply to and )

29 (a) We conside the diffeence in U whee initially = b and = a, and finally = a and = b ( doesn t change) Conveting the values given in the poblem to SI units (C to C, cm to m), we obtain U = 4 J (b) Now we conside the diffeence in U whee initially = a and = a, and finally is again equal to a and is also again equal to a (and of couse, doesn t change in this case) Thus, we obtain U = 75 Assume the chage on Eath is distibuted with spheical symmety If the electic potential is zeo at infinity then at the suface of Eath it is = q/4 R, whee q is the chage on Eath and R = 67 6 m is the adius of Eath The magnitude of the electic field at the suface is E = q/4 R, so = ER = ( /m) (67 6 m) = Using Gauss law, q = nc Consequently, 9 7 q (899 Nm C )(4958 C) 4 m The potential diffeence is = Es = (9 5 N/C)(5 m) = 9 78 The chages ae equidistant fom the point whee we ae evaluating the potential which is computed using Eq 4-7 (o its integal equivalent) Equation 4-7 implicitly assumes as Thus, we have Q Q Q Q 4 R 4 R 4 R 4 R 9 (899 Nm C )(45 C) 85 m 79 The electic potential enegy in the pesence of the dipole is 956 U qp cos ( e)( ed)cos q dipole 4 4 Noting that i = f = º, consevation of enegy leads to K f + U f = K i + U i v = e 4 o m d 5 49 = 7 5 m/s

30 CHAPTER 4 8 We teat the system as a supeposition of a disk of suface chage density and adius R and a smalle, oppositely chaged, disk of suface chage density and adius Fo each of these, Eq 4-7 applies (fo z > ) z R z z z e j e j This expession does vanish as, as the poblem equies Substituting = R and z = R and simplifying, we obtain R 5 5 (6 C/m )( m) C /Nm 8 (a) When the electon is eleased, its enegy is K + U = e 6 e (the latte value is infeed fom the gaph along with the fact that U = q and q = e) Because of the minus sign (of the chage) it is convenient to imagine the gaph multiplied by a minus sign so that it epesents potential enegy in e Thus, the value shown at x = would become e, and the 6 value at x = 45 cm becomes 6 e, and so on The total enegy ( e) is constant and can then be epesented on ou (imagined) gaph as a hoizontal line at This intesects the potential enegy plot at a point we ecognize as the tuning point Intepolating in the egion between cm and 4 cm, we find the tuning point is at x = 75 cm 8 cm (b) Thee is no tuning point towad the ight, so the speed thee is nonzeo Noting that the kinetic enegy at x = 7 cm is K = e ( 5 e) = e, we find the speed using enegy consevation: 9 K e 6 J/e v m 9 kg e 5 84 m/s (c) The electic field at any point P is the (negative of the) slope of the voltage gaph evaluated at P Once we know the electic field, the foce on the electon follows immediately fom F qe, whee q = e fo the electon In the egion just to the left of x 4 cm, the electic field is E ( /m)i ˆ and the magnitude of the foce is F 7 N

31 (d) The foce points in the +x diection (e) In the egion just to the ight of x = 5 cm, the field is E = + /m î and the foce is F = ( 6 x 7 N) î Thus, the magnitude of the foce is 7 F 6 N (f) The minus sign indicates that F points in the x diection 8 (a) The potential would be Q 4R R k e e e e 4 ee 4 Re 4 Re m electon m 6 C electon 899 N m C (b) The electic field is o 8 E 8 N C e e E 8 8 N C, 6 R 67 m e (c) The minus sign in E indicates that E is adially inwad 8 (a) Using d = m, we find the potential at P: P 9 9 e e e (899 Nm C )(6 C) 4 d 4 ( d) 4 d m 79 Note that we ae implicitly assuming that as (b) Since U = q, then the movable paticle's contibution of the potential enegy when it is at = is zeo, and its contibution to U system when it is at P is U q P 9 8 (6 C)(79 ) J Thus, the wok done is appoximately equal to W app = 8 J (c) Now, combining the contibution to U system fom pat (b) and fom the oiginal pai of fixed chages

32 4 CHAPTER 4 U fixed ( e)( e) (899 N m C )(4)(6 C) (4 m) ( m) m 8 58 J we obtain U system = W app + U fixed = 4 9 J 84 The electic field thoughout the conducting volume is zeo, which implies that the potential thee is constant and equal to the value it has on the suface of the chaged sphee: q A S 4 R whee q = 9 C and R = m Fo points beyond the suface of the sphee, the potential follows Eq 4-6: q B 4 whee = 5 m (a) We see that q S B = 4 R = 6 (b) Similaly, A B = q 4 R = 6 85 We note that the net potential (due to the "fixed" chages) is zeo at the fist location ("at ") being consideed fo the movable chage q (whee q = +e) Thus, with D = 4 m and e = 6 9 C, we obtain 4 ( D) 4 D 4 D 4 m 9 9 e e e (899 N m C )()(6 C) 79 The wok equied is equal to the potential enegy in the final configuation: W app = q = (e)(79 ) = 8 J 86 Since the electic potential is a scala quantity, this calculation is fa simple than it would be fo the electic field We ae able to simply take half the contibution that

33 5 would be obtained fom a complete (whole) sphee If it wee a whole sphee (of the same density) then its chage would be q whole = 8 C Then = whole = q whole 4 o = 8 x -6 C 4 o (5 m) = THINK The wok done is equal to the change in potential enegy EXPRESS The initial potential enegy of the system is U i q U 4 L whee q is the chage on each paticle, L is the length of the tiangle side, and U is the potential enegy associated with the inteaction of the two fixed chages Afte moving to the midpoint of the line joining the two fixed chages, the final enegy of the configuation is q U f U 4 ( L / ) Thus, the wok done by the extenal agent is q q W U U f Ui 4 L L 4 L ANALYZE Substituting the values given, we have W 9 q (899 Nm C )( C) 4 L 7 m 8 5 J At a ate of P = 8 joules pe second, it would take W/P = 8 5 seconds o about days to do this amount of wok LEARN Since all thee paticles ae positively chaged, positive wok is equied by the extenal agent in ode to bing them close 88 (a) The chages ae equal and ae the same distance fom C We use the Pythagoean theoem to find the distance b g b g d d d The electic potential at C is the sum of the potential due to the individual chages but since they poduce the same potential, it is twice that of eithe one:

34 6 CHAPTER N m C 6 C q q 4 d 4 d m (b) As you move the chage into position fom fa away the potential enegy changes fom zeo to q, whee is the electic potential at the final location of the chage The change in the potential enegy equals the wok you must do to bing the chage in: W 6 6 q C 54 5 J (c) The wok calculated in pat (b) epesents the potential enegy of the inteactions between the chage bought in fom infinity and the othe two chages To find the total potential enegy of the thee-chage system you must add the potential enegy of the inteaction between the fixed chages Thei sepaation is d so this potential enegy is q d The total potential enegy is 4 U 9 6 q 899 N m C C W 5 J 69 J 4 d m 89 The net potential at point P (the place whee we ae to place the thid electon) due to the fixed chages is computed using Eq 4-7 (which assumes as ): P e e e 4 d 4 d 4 d Thus, with d = 6 m and e = 6 9 C, we find P 9 9 e (899 Nm C )()(6 C) 6 4 d m 48 Then the equied applied wok is, by Eq 4-4, W app = (e) P = J 9 The paticle with chage q has both potential and kinetic enegy, and both of these change when the adius of the obit is changed We fist find an expession fo the total enegy in tems of the obit adius The chage Q povides the centipetal foce equied fo q to move in unifom cicula motion The magnitude of the foce is F = Qq/4 The acceleation of q is v /, whee v is its speed Newton s second law yields

35 7 and the kinetic enegy is Qq mv Qq mv, 4 4 K Qq 8 mv The potential enegy is U = Qq/4, and the total enegy is Qq Qq Qq E K U When the obit adius is the enegy is E = Qq/8 and when it is the enegy is E = Qq/8 The diffeence E E is the wok W done by an extenal agent to change the adius: Qq F I Qq F I W E E The initial speed v i of the electon satisfies HG KJ HG KJ which gives v i e K m v e, i e i 9 e c6 Jhb65 g m 9 kg 48 7 m s 9 The net electic potential at point P is the sum of those due to the six chages: P q d d d d i 5 Pi i i 4 i 4 d/ / / 5 94 d d / d d / 6 d / 4 (54 ) THINK To calculate the potential at point B due to the chaged ing, we note that all points on the ing ae at the same distance fom B EXPRESS Let point B be at (,, z) The electic potential at B is given by

36 8 CHAPTER 4 4 q z R whee q is the chage on the ing The potential at infinity is taken to be zeo ANALYZE With q = 6 6 C, z = 4 m, and R = m, we find the potential diffeence between points A (located at the oigin) and B to be B q A 4 z R R 9 6 (899 N m C )(6 C) 6 9 ( m) (4 m) m LEARN In the limit z R, the potential appoaches its point-chage limit: q 4 z 94 (a) Using Eq 4-6, we calculate the adius of the sphee epesenting the equipotential suface: 9 8 q (899 Nm C )(5 C) 45m 4 (b) If the potential wee a linea function of then it would have equally spaced equipotentials, but since they ae spaced moe and moe widely apat as inceases 95 THINK To calculate the electic potential, we fist apply Gauss law to calculate the electic field of the spheical shell The Gaussian suface is a sphee that is concentic with the shell EXPRESS At all points whee thee is an electic field, it is adially outwad Fo each pat of the poblem, use a Gaussian suface in the fom of a sphee that is concentic with the sphee of chage and passes though the point whee the electic field is to be found The field is unifom on the suface, so the flux though the suface is given by E da 4 E q enc /, whee is the adius of the Gaussian suface and q enc is the chage enclosed (i) In the egion, the enclosed chage is qenc and theefoe,

37 9 b gc h E = (ii) In the egion < <, the volume of the shell is 4 chage density is Q c h, 4, so the whee Q is the total chage on the spheical shell Thus, the chage enclosed by the Gaussian suface is 4 qenc Q Gauss law yields Q 4 E Q E 4 (iii) In the egion, the chage enclosed is qenc Q, and the electic field is like that of a point chage: Q E 4 ANALYZE (a) Fo > the field is like that of a point chage, and so is the potential: 4 Q, whee the potential was taken to be zeo at infinity (b) In the egion < <, we have Q E 4 If s is the electic potential at the oute suface of the shell ( = ) then the potential a distance fom the cente is given by Q s E d s 4 s z z Q 4 F HG F HG I KJ The potential at the oute suface is found by placing = in the expession found in pat (a) It is s = Q/4 We make this substitution and collect tems to find I KJ d

38 CHAPTER 4 c Q 4 F HG Since Q 4 h this can also be witten as I KJ ( ) (c) Fo, the electic field vanishes in the cavity, so the potential is eveywhee the same inside and has the same value as at a point on the inside suface of the shell We put = in the esult of pat (b) Afte collecting tems the esult is Q 4 c h c h, o in tems of the chage density c h (d) Using the expession fo () found in (b), we have and ( ) Q/ 4 Q ( ) 4 So the solutions agee at = and at = LEARN Electic potential must be continuous at the boundaies at = and = In the egion whee the electic field is zeo, no wok is equied to move the chage aound Thus, thee s no change in potential enegy and the electic potential is constant 96 (a) We use Gauss law to find expessions fo the electic field inside and outside the spheical chage distibution Since the field is adial the electic potential can be witten as an integal of the field along a sphee adius, extended to infinity Since diffeent expessions fo the field apply in diffeent egions the integal must be split into two pats, one fom infinity to the suface of the distibution and one fom the suface to a point inside

39 Outside the chage distibution the magnitude of the field is E = q/4 and the potential is = q/4, whee is the distance fom the cente of the distibution This is the same as the field and potential of a point chage at the cente of the spheical distibution To find an expession fo the magnitude of the field inside the chage distibution, we use a Gaussian suface in the fom of a sphee with adius, concentic with the distibution The field is nomal to the Gaussian suface and its magnitude is unifom ove it, so the electic flux though the suface is 4 E The chage enclosed is q /R Gauss law becomes q q 4 E E R 4 R If s is the potential at the suface of the distibution ( = R) then the potential at a point inside, a distance fom the cente, is z z E d s R s q q q d s 4 R R 8 R 8 R The potential at the suface can be found by eplacing with R in the expession fo the potential at points outside the distibution It is s = q/4 R Thus, L NM O QP q q R 4 R R R 8 R c h (b) The potential diffeence is q q q s c, 8 R 8 R 8 R o q /8 R 97 THINK The incease in electic potential at the suface of the coppe sphee is popotional to the incease in electic chage EXPRESS The electic potential at the suface of a sphee of adius R is given by q / 4 R, whee q is the chage on the sphee Thus, q 4 R The numbe of electons enteing the coppe sphee is N q / e, but this must be equal to ( / ) t, whee is the decay ate of the nickel ANALYZE (a) With R = m, when =, the net chage on the sphee is ( m)( ) q 899 Nm C 9 4 R C 9 Dividing q by e yields

40 CHAPTER 4 N ( C)/(6 C) electons that enteed the coppe sphee So the time equied is 9 N 695 t 8 s 8 / (7 / s) / (b) The enegy deposited by each electon that entes the sphee is E ke = 6 4 J Using the given heat capacity, we note that a tempeatue incease of T = 5 K = 5 ºC equied E CT (4 J/K)(5 K) 7 J of enegy Dividing this by E gives the numbe of electons needed to ente the sphee (in ode to achieve that tempeatue change): E 7 J N E 6 J Thus, the time needed is 5 N 475 t 8 / (7 / s) / o oughly 7 days s LEARN As moe electons get into coppe, moe enegy is deposited, and the coppe sample gets hotte 98 (a) The potential diffeence is (899 Nm C ) 4 R 4 6 m m 6 6 Q q 9 5 C 5 C (b) By connecting the two metal sphees with a wie, we now have one conducto, and any excess chage must eside on the suface of the conducto Theefoe, the chage on the small sphee is zeo (c) Since all the chages eside on the suface of the lage sphee, we have Q Q q 5 C5 C C 99 (a) The chage on evey pat of the ing is the same distance fom any point P on the axis This distance is z R, whee R is the adius of the ing and z is the distance fom the cente of the ing to P The electic potential at P is

41 dq dq 4 4 z R 4 z R q 4 z R (b) The electic field is along the axis and its component is given by dq q q E z R z R z z 4 z 4 q z 4 ( z R ) This agees with Eq -6 / / / ( ) ( ) ( ) The distance being looked fo is that whee the alpha paticle has (momentaily) zeo kinetic enegy Thus, enegy consevation leads to K + U = K + U (48 J) + (e)(9e) 4 = + (e)(9e) 4 If we set = (so U = ) then we obtain = 88 4 m (a) Let the quak-quak sepaation be To natually obtain the e unit, we only plug in fo one of the e values involved in the computation: 9 9 e/ e/ ke 4899 Nm C 6 C 4 U e e m upup e 484 Me (b) The total consists of all pai-wise tems: U e / e / e / e / e / e / 4 We imagine moving all the chages on the suface of the sphee to the cente of the the sphee Using Gauss law, we see that this would not change the electic field outside the sphee The magnitude of the electic field E of the unifomly chaged sphee as a function of, the distance fom the cente of the sphee, is thus given by E() = q/(4 ) fo > R

42 4 CHAPTER 4 Hee R is the adius of the sphee Thus, the potential at the suface of the sphee (whee = R) is given by q q R E d d 84 R R 9 Nm C C 4 4 R 6m Since the electic potential enegy is not changed by the intoduction of the thid paticle, we conclude that the net electic potential evaluated at P caused by the oiginal two paticles must be zeo: q q 4 4 Setting = 5d/ and = d / we obtain q = 5q /, o q/ q 5/ 7

Physics 2212 GH Quiz #2 Solutions Spring 2016

Physics 2212 GH Quiz #2 Solutions Spring 2016 Physics 2212 GH Quiz #2 Solutions Sping 216 I. 17 points) Thee point chages, each caying a chage Q = +6. nc, ae placed on an equilateal tiangle of side length = 3. mm. An additional point chage, caying

More information

Electrostatics (Electric Charges and Field) #2 2010

Electrostatics (Electric Charges and Field) #2 2010 Electic Field: The concept of electic field explains the action at a distance foce between two chaged paticles. Evey chage poduces a field aound it so that any othe chaged paticle expeiences a foce when

More information

Flux. Area Vector. Flux of Electric Field. Gauss s Law

Flux. Area Vector. Flux of Electric Field. Gauss s Law Gauss s Law Flux Flux in Physics is used to two distinct ways. The fist meaning is the ate of flow, such as the amount of wate flowing in a ive, i.e. volume pe unit aea pe unit time. O, fo light, it is

More information

CHAPTER 25 ELECTRIC POTENTIAL

CHAPTER 25 ELECTRIC POTENTIAL CHPTE 5 ELECTIC POTENTIL Potential Diffeence and Electic Potential Conside a chaged paticle of chage in a egion of an electic field E. This filed exets an electic foce on the paticle given by F=E. When

More information

Physics 107 TUTORIAL ASSIGNMENT #8

Physics 107 TUTORIAL ASSIGNMENT #8 Physics 07 TUTORIAL ASSIGNMENT #8 Cutnell & Johnson, 7 th edition Chapte 8: Poblems 5,, 3, 39, 76 Chapte 9: Poblems 9, 0, 4, 5, 6 Chapte 8 5 Inteactive Solution 8.5 povides a model fo solving this type

More information

2. Electrostatics. Dr. Rakhesh Singh Kshetrimayum 8/11/ Electromagnetic Field Theory by R. S. Kshetrimayum

2. Electrostatics. Dr. Rakhesh Singh Kshetrimayum 8/11/ Electromagnetic Field Theory by R. S. Kshetrimayum 2. Electostatics D. Rakhesh Singh Kshetimayum 1 2.1 Intoduction In this chapte, we will study how to find the electostatic fields fo vaious cases? fo symmetic known chage distibution fo un-symmetic known

More information

Phys102 Second Major-182 Zero Version Monday, March 25, 2019 Page: 1

Phys102 Second Major-182 Zero Version Monday, March 25, 2019 Page: 1 Monday, Mach 5, 019 Page: 1 Q1. Figue 1 shows thee pais of identical conducting sphees that ae to be touched togethe and then sepaated. The initial chages on them befoe the touch ae indicated. Rank the

More information

(Sample 3) Exam 1 - Physics Patel SPRING 1998 FORM CODE - A (solution key at end of exam)

(Sample 3) Exam 1 - Physics Patel SPRING 1998 FORM CODE - A (solution key at end of exam) (Sample 3) Exam 1 - Physics 202 - Patel SPRING 1998 FORM CODE - A (solution key at end of exam) Be sue to fill in you student numbe and FORM lette (A, B, C) on you answe sheet. If you foget to include

More information

EM-2. 1 Coulomb s law, electric field, potential field, superposition q. Electric field of a point charge (1)

EM-2. 1 Coulomb s law, electric field, potential field, superposition q. Electric field of a point charge (1) EM- Coulomb s law, electic field, potential field, supeposition q ' Electic field of a point chage ( ') E( ) kq, whee k / 4 () ' Foce of q on a test chage e at position is ee( ) Electic potential O kq

More information

17.1 Electric Potential Energy. Equipotential Lines. PE = energy associated with an arrangement of objects that exert forces on each other

17.1 Electric Potential Energy. Equipotential Lines. PE = energy associated with an arrangement of objects that exert forces on each other Electic Potential Enegy, PE Units: Joules Electic Potential, Units: olts 17.1 Electic Potential Enegy Electic foce is a consevative foce and so we can assign an electic potential enegy (PE) to the system

More information

$ i. !((( dv vol. Physics 8.02 Quiz One Equations Fall q 1 q 2 r 2 C = 2 C! V 2 = Q 2 2C F = 4!" or. r ˆ = points from source q to observer

$ i. !((( dv vol. Physics 8.02 Quiz One Equations Fall q 1 q 2 r 2 C = 2 C! V 2 = Q 2 2C F = 4! or. r ˆ = points from source q to observer Physics 8.0 Quiz One Equations Fall 006 F = 1 4" o q 1 q = q q ˆ 3 4" o = E 4" o ˆ = points fom souce q to obseve 1 dq E = # ˆ 4" 0 V "## E "d A = Q inside closed suface o d A points fom inside to V =

More information

Physics 2A Chapter 10 - Moment of Inertia Fall 2018

Physics 2A Chapter 10 - Moment of Inertia Fall 2018 Physics Chapte 0 - oment of netia Fall 08 The moment of inetia of a otating object is a measue of its otational inetia in the same way that the mass of an object is a measue of its inetia fo linea motion.

More information

Voltage ( = Electric Potential )

Voltage ( = Electric Potential ) V-1 of 10 Voltage ( = lectic Potential ) An electic chage altes the space aound it. Thoughout the space aound evey chage is a vecto thing called the electic field. Also filling the space aound evey chage

More information

! E da = 4πkQ enc, has E under the integral sign, so it is not ordinarily an

! E da = 4πkQ enc, has E under the integral sign, so it is not ordinarily an Physics 142 Electostatics 2 Page 1 Electostatics 2 Electicity is just oganized lightning. Geoge Calin A tick that sometimes woks: calculating E fom Gauss s law Gauss s law,! E da = 4πkQ enc, has E unde

More information

Potential Energy. The change U in the potential energy. is defined to equal to the negative of the work. done by a conservative force

Potential Energy. The change U in the potential energy. is defined to equal to the negative of the work. done by a conservative force Potential negy The change U in the potential enegy is defined to equal to the negative of the wok done by a consevative foce duing the shift fom an initial to a final state. U = U U = W F c = F c d Potential

More information

7.2. Coulomb s Law. The Electric Force

7.2. Coulomb s Law. The Electric Force Coulomb s aw Recall that chaged objects attact some objects and epel othes at a distance, without making any contact with those objects Electic foce,, o the foce acting between two chaged objects, is somewhat

More information

Electrostatics. 3) positive object: lack of electrons negative object: excess of electrons

Electrostatics. 3) positive object: lack of electrons negative object: excess of electrons Electostatics IB 12 1) electic chage: 2 types of electic chage: positive and negative 2) chaging by fiction: tansfe of electons fom one object to anothe 3) positive object: lack of electons negative object:

More information

OSCILLATIONS AND GRAVITATION

OSCILLATIONS AND GRAVITATION 1. SIMPLE HARMONIC MOTION Simple hamonic motion is any motion that is equivalent to a single component of unifom cicula motion. In this situation the velocity is always geatest in the middle of the motion,

More information

Chapter 23: GAUSS LAW 343

Chapter 23: GAUSS LAW 343 Chapte 23: GAUSS LAW 1 A total chage of 63 10 8 C is distibuted unifomly thoughout a 27-cm adius sphee The volume chage density is: A 37 10 7 C/m 3 B 69 10 6 C/m 3 C 69 10 6 C/m 2 D 25 10 4 C/m 3 76 10

More information

Gauss Law. Physics 231 Lecture 2-1

Gauss Law. Physics 231 Lecture 2-1 Gauss Law Physics 31 Lectue -1 lectic Field Lines The numbe of field lines, also known as lines of foce, ae elated to stength of the electic field Moe appopiately it is the numbe of field lines cossing

More information

Objects usually are charged up through the transfer of electrons from one object to the other.

Objects usually are charged up through the transfer of electrons from one object to the other. 1 Pat 1: Electic Foce 1.1: Review of Vectos Review you vectos! You should know how to convet fom pola fom to component fom and vice vesa add and subtact vectos multiply vectos by scalas Find the esultant

More information

Algebra-based Physics II

Algebra-based Physics II lgebabased Physics II Chapte 19 Electic potential enegy & The Electic potential Why enegy is stoed in an electic field? How to descibe an field fom enegetic point of view? Class Website: Natual way of

More information

2 E. on each of these two surfaces. r r r r. Q E E ε. 2 2 Qencl encl right left 0

2 E. on each of these two surfaces. r r r r. Q E E ε. 2 2 Qencl encl right left 0 Ch : 4, 9,, 9,,, 4, 9,, 4, 8 4 (a) Fom the diagam in the textbook, we see that the flux outwad though the hemispheical suface is the same as the flux inwad though the cicula suface base of the hemisphee

More information

21 MAGNETIC FORCES AND MAGNETIC FIELDS

21 MAGNETIC FORCES AND MAGNETIC FIELDS CHAPTER 1 MAGNETIC ORCES AND MAGNETIC IELDS ANSWERS TO OCUS ON CONCEPTS QUESTIONS 1. (d) Right-Hand Rule No. 1 gives the diection of the magnetic foce as x fo both dawings A and. In dawing C, the velocity

More information

Page 1 of 6 Physics II Exam 1 155 points Name Discussion day/time Pat I. Questions 110. 8 points each. Multiple choice: Fo full cedit, cicle only the coect answe. Fo half cedit, cicle the coect answe and

More information

Chapter 22: Electric Fields. 22-1: What is physics? General physics II (22102) Dr. Iyad SAADEDDIN. 22-2: The Electric Field (E)

Chapter 22: Electric Fields. 22-1: What is physics? General physics II (22102) Dr. Iyad SAADEDDIN. 22-2: The Electric Field (E) Geneal physics II (10) D. Iyad D. Iyad Chapte : lectic Fields In this chapte we will cove The lectic Field lectic Field Lines -: The lectic Field () lectic field exists in a egion of space suounding a

More information

20-9 ELECTRIC FIELD LINES 20-9 ELECTRIC POTENTIAL. Answers to the Conceptual Questions. Chapter 20 Electricity 241

20-9 ELECTRIC FIELD LINES 20-9 ELECTRIC POTENTIAL. Answers to the Conceptual Questions. Chapter 20 Electricity 241 Chapte 0 Electicity 41 0-9 ELECTRIC IELD LINES Goals Illustate the concept of electic field lines. Content The electic field can be symbolized by lines of foce thoughout space. The electic field is stonge

More information

Chapter 22 The Electric Field II: Continuous Charge Distributions

Chapter 22 The Electric Field II: Continuous Charge Distributions Chapte The lectic Field II: Continuous Chage Distibutions A ing of adius a has a chage distibution on it that vaies as l(q) l sin q, as shown in Figue -9. (a) What is the diection of the electic field

More information

Chapter 13 Gravitation

Chapter 13 Gravitation Chapte 13 Gavitation In this chapte we will exploe the following topics: -Newton s law of gavitation, which descibes the attactive foce between two point masses and its application to extended objects

More information

PHYS 1444 Lecture #5

PHYS 1444 Lecture #5 Shot eview Chapte 24 PHYS 1444 Lectue #5 Tuesday June 19, 212 D. Andew Bandt Capacitos and Capacitance 1 Coulom s Law The Fomula QQ Q Q F 1 2 1 2 Fomula 2 2 F k A vecto quantity. Newtons Diection of electic

More information

PHYS 1444 Section 501 Lecture #7

PHYS 1444 Section 501 Lecture #7 PHYS 1444 Section 51 Lectue #7 Wednesday, Feb. 8, 26 Equi-potential Lines and Sufaces Electic Potential Due to Electic Dipole E detemined fom V Electostatic Potential Enegy of a System of Chages Capacitos

More information

Lecture 8 - Gauss s Law

Lecture 8 - Gauss s Law Lectue 8 - Gauss s Law A Puzzle... Example Calculate the potential enegy, pe ion, fo an infinite 1D ionic cystal with sepaation a; that is, a ow of equally spaced chages of magnitude e and altenating sign.

More information

Physics 181. Assignment 4

Physics 181. Assignment 4 Physics 181 Assignment 4 Solutions 1. A sphee has within it a gavitational field given by g = g, whee g is constant and is the position vecto of the field point elative to the cente of the sphee. This

More information

Physics 235 Chapter 5. Chapter 5 Gravitation

Physics 235 Chapter 5. Chapter 5 Gravitation Chapte 5 Gavitation In this Chapte we will eview the popeties of the gavitational foce. The gavitational foce has been discussed in geat detail in you intoductoy physics couses, and we will pimaily focus

More information

MAGNETIC FIELD INTRODUCTION

MAGNETIC FIELD INTRODUCTION MAGNETIC FIELD INTRODUCTION It was found when a magnet suspended fom its cente, it tends to line itself up in a noth-south diection (the compass needle). The noth end is called the Noth Pole (N-pole),

More information

Chapter 25. Electric Potential

Chapter 25. Electric Potential Chapte 25 Electic Potential C H P T E R O U T L I N E 251 Potential Diffeence and Electic Potential 252 Potential Diffeences in a Unifom Electic Field 253 Electic Potential and Potential Enegy Due to Point

More information

Physics 2B Chapter 22 Notes - Magnetic Field Spring 2018

Physics 2B Chapter 22 Notes - Magnetic Field Spring 2018 Physics B Chapte Notes - Magnetic Field Sping 018 Magnetic Field fom a Long Staight Cuent-Caying Wie In Chapte 11 we looked at Isaac Newton s Law of Gavitation, which established that a gavitational field

More information

PHYSICS NOTES GRAVITATION

PHYSICS NOTES GRAVITATION GRAVITATION Newton s law of gavitation The law states that evey paticle of matte in the univese attacts evey othe paticle with a foce which is diectly popotional to the poduct of thei masses and invesely

More information

CHAPTER 10 ELECTRIC POTENTIAL AND CAPACITANCE

CHAPTER 10 ELECTRIC POTENTIAL AND CAPACITANCE CHAPTER 0 ELECTRIC POTENTIAL AND CAPACITANCE ELECTRIC POTENTIAL AND CAPACITANCE 7 0. ELECTRIC POTENTIAL ENERGY Conside a chaged paticle of chage in a egion of an electic field E. This filed exets an electic

More information

DEVIL PHYSICS THE BADDEST CLASS ON CAMPUS IB PHYSICS

DEVIL PHYSICS THE BADDEST CLASS ON CAMPUS IB PHYSICS DEVIL PHYSICS THE BADDEST CLASS ON CAMPUS IB PHYSICS TSOKOS LESSON 10-1 DESCRIBING FIELDS Essential Idea: Electic chages and masses each influence the space aound them and that influence can be epesented

More information

University Physics (PHY 2326)

University Physics (PHY 2326) Chapte Univesity Physics (PHY 6) Lectue lectostatics lectic field (cont.) Conductos in electostatic euilibium The oscilloscope lectic flux and Gauss s law /6/5 Discuss a techniue intoduced by Kal F. Gauss

More information

Objectives: After finishing this unit you should be able to:

Objectives: After finishing this unit you should be able to: lectic Field 7 Objectives: Afte finishing this unit you should be able to: Define the electic field and explain what detemines its magnitude and diection. Wite and apply fomulas fo the electic field intensity

More information

ELECTROSTATICS::BHSEC MCQ 1. A. B. C. D.

ELECTROSTATICS::BHSEC MCQ 1. A. B. C. D. ELETROSTATIS::BHSE 9-4 MQ. A moving electic chage poduces A. electic field only. B. magnetic field only.. both electic field and magnetic field. D. neithe of these two fields.. both electic field and magnetic

More information

ELECTRIC FIELD. decos. 1 dq x.. Example:

ELECTRIC FIELD. decos. 1 dq x.. Example: ELECTRIC FIELD Example: Solution: A ing-shaped conducto with adius a caies a total positive chage Q unifomly distibuted on it. Find the electic field at a point P that lies on the axis of the ing at a

More information

Look over Chapter 22 sections 1-8 Examples 2, 4, 5, Look over Chapter 16 sections 7-9 examples 6, 7, 8, 9. Things To Know 1/22/2008 PHYS 2212

Look over Chapter 22 sections 1-8 Examples 2, 4, 5, Look over Chapter 16 sections 7-9 examples 6, 7, 8, 9. Things To Know 1/22/2008 PHYS 2212 PHYS 1 Look ove Chapte sections 1-8 xamples, 4, 5, PHYS 111 Look ove Chapte 16 sections 7-9 examples 6, 7, 8, 9 Things To Know 1) What is an lectic field. ) How to calculate the electic field fo a point

More information

Physics 11 Chapter 20: Electric Fields and Forces

Physics 11 Chapter 20: Electric Fields and Forces Physics Chapte 0: Electic Fields and Foces Yesteday is not ous to ecove, but tomoow is ous to win o lose. Lyndon B. Johnson When I am anxious it is because I am living in the futue. When I am depessed

More information

From last times. MTE1 results. Quiz 1. GAUSS LAW for any closed surface. What is the Electric Flux? How to calculate Electric Flux?

From last times. MTE1 results. Quiz 1. GAUSS LAW for any closed surface. What is the Electric Flux? How to calculate Electric Flux? om last times MTE1 esults Mean 75% = 90/120 Electic chages and foce Electic ield and was to calculate it Motion of chages in E-field Gauss Law Toda: Moe on Gauss law and conductos in electostatic equilibium

More information

F Q E v B MAGNETOSTATICS. Creation of magnetic field B. Effect of B on a moving charge. On moving charges only. Stationary and moving charges

F Q E v B MAGNETOSTATICS. Creation of magnetic field B. Effect of B on a moving charge. On moving charges only. Stationary and moving charges MAGNETOSTATICS Ceation of magnetic field. Effect of on a moving chage. Take the second case: F Q v mag On moving chages only F QE v Stationay and moving chages dw F dl Analysis on F mag : mag mag Qv. vdt

More information

Physics 122, Fall September 2012

Physics 122, Fall September 2012 Physics 1, Fall 1 7 Septembe 1 Today in Physics 1: getting V fom E When it s best to get V fom E, athe than vice vesa V within continuous chage distibutions Potential enegy of continuous chage distibutions

More information

Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 1 -

Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 1 - Pepaed by: M. S. KumaSwamy, TGT(Maths) Page - - ELECTROSTATICS MARKS WEIGHTAGE 8 maks QUICK REVISION (Impotant Concepts & Fomulas) Chage Quantization: Chage is always in the fom of an integal multiple

More information

Ch 30 - Sources of Magnetic Field! The Biot-Savart Law! = k m. r 2. Example 1! Example 2!

Ch 30 - Sources of Magnetic Field! The Biot-Savart Law! = k m. r 2. Example 1! Example 2! Ch 30 - Souces of Magnetic Field 1.) Example 1 Detemine the magnitude and diection of the magnetic field at the point O in the diagam. (Cuent flows fom top to bottom, adius of cuvatue.) Fo staight segments,

More information

Between any two masses, there exists a mutual attractive force.

Between any two masses, there exists a mutual attractive force. YEAR 12 PHYSICS: GRAVITATION PAST EXAM QUESTIONS Name: QUESTION 1 (1995 EXAM) (a) State Newton s Univesal Law of Gavitation in wods Between any two masses, thee exists a mutual attactive foce. This foce

More information

PHYS 2135 Exam I February 13, 2018

PHYS 2135 Exam I February 13, 2018 Exam Total /200 PHYS 2135 Exam I Febuay 13, 2018 Name: Recitation Section: Five multiple choice questions, 8 points each Choose the best o most nealy coect answe Fo questions 6-9, solutions must begin

More information

EELE 3331 Electromagnetic I Chapter 4. Electrostatic fields. Islamic University of Gaza Electrical Engineering Department Dr.

EELE 3331 Electromagnetic I Chapter 4. Electrostatic fields. Islamic University of Gaza Electrical Engineering Department Dr. EELE 3331 Electomagnetic I Chapte 4 Electostatic fields Islamic Univesity of Gaza Electical Engineeing Depatment D. Talal Skaik 212 1 Electic Potential The Gavitational Analogy Moving an object upwad against

More information

PY208 Matter & Interactions Final Exam S2005

PY208 Matter & Interactions Final Exam S2005 PY Matte & Inteactions Final Exam S2005 Name (pint) Please cicle you lectue section below: 003 (Ramakishnan 11:20 AM) 004 (Clake 1:30 PM) 005 (Chabay 2:35 PM) When you tun in the test, including the fomula

More information

Electrostatics. 1. Show does the force between two point charges change if the dielectric constant of the medium in which they are kept increase?

Electrostatics. 1. Show does the force between two point charges change if the dielectric constant of the medium in which they are kept increase? Electostatics 1. Show does the foce between two point chages change if the dielectic constant of the medium in which they ae kept incease? 2. A chaged od P attacts od R whee as P epels anothe chaged od

More information

Chapter Sixteen: Electric Charge and Electric Fields

Chapter Sixteen: Electric Charge and Electric Fields Chapte Sixteen: Electic Chage and Electic Fields Key Tems Chage Conducto The fundamental electical popety to which the mutual attactions o epulsions between electons and potons ae attibuted. Any mateial

More information

Gauss s Law Simulation Activities

Gauss s Law Simulation Activities Gauss s Law Simulation Activities Name: Backgound: The electic field aound a point chage is found by: = kq/ 2 If thee ae multiple chages, the net field at any point is the vecto sum of the fields. Fo a

More information

Welcome to Physics 272

Welcome to Physics 272 Welcome to Physics 7 Bob Mose mose@phys.hawaii.edu http://www.phys.hawaii.edu/~mose/physics7.html To do: Sign into Masteing Physics phys-7 webpage Registe i-clickes (you i-clicke ID to you name on class-list)

More information

Hopefully Helpful Hints for Gauss s Law

Hopefully Helpful Hints for Gauss s Law Hopefully Helpful Hints fo Gauss s Law As befoe, thee ae things you need to know about Gauss s Law. In no paticula ode, they ae: a.) In the context of Gauss s Law, at a diffeential level, the electic flux

More information

Today in Physics 122: getting V from E

Today in Physics 122: getting V from E Today in Physics 1: getting V fom E When it s best to get V fom E, athe than vice vesa V within continuous chage distibutions Potential enegy of continuous chage distibutions Capacitance Potential enegy

More information

Chapter 7-8 Rotational Motion

Chapter 7-8 Rotational Motion Chapte 7-8 Rotational Motion What is a Rigid Body? Rotational Kinematics Angula Velocity ω and Acceleation α Unifom Rotational Motion: Kinematics Unifom Cicula Motion: Kinematics and Dynamics The Toque,

More information

UNIT 3:Electrostatics

UNIT 3:Electrostatics The study of electic chages at est, the foces between them and the electic fields associated with them. UNIT 3:lectostatics S7 3. lectic Chages and Consevation of chages The electic chage has the following

More information

AST 121S: The origin and evolution of the Universe. Introduction to Mathematical Handout 1

AST 121S: The origin and evolution of the Universe. Introduction to Mathematical Handout 1 Please ead this fist... AST S: The oigin and evolution of the Univese Intoduction to Mathematical Handout This is an unusually long hand-out and one which uses in places mathematics that you may not be

More information

Chapter 21: Gauss s Law

Chapter 21: Gauss s Law Chapte : Gauss s Law Gauss s law : intoduction The total electic flux though a closed suface is equal to the total (net) electic chage inside the suface divided by ε Gauss s law is equivalent to Coulomb

More information

Magnetic Field. Conference 6. Physics 102 General Physics II

Magnetic Field. Conference 6. Physics 102 General Physics II Physics 102 Confeence 6 Magnetic Field Confeence 6 Physics 102 Geneal Physics II Monday, Mach 3d, 2014 6.1 Quiz Poblem 6.1 Think about the magnetic field associated with an infinite, cuent caying wie.

More information

PHY2061 Enriched Physics 2 Lecture Notes. Gauss Law

PHY2061 Enriched Physics 2 Lecture Notes. Gauss Law PHY61 Eniched Physics Lectue Notes Law Disclaime: These lectue notes ae not meant to eplace the couse textbook. The content may be incomplete. ome topics may be unclea. These notes ae only meant to be

More information

? this lecture. ? next lecture. What we have learned so far. a Q E F = q E a. F = q v B a. a Q in motion B. db/dt E. de/dt B.

? this lecture. ? next lecture. What we have learned so far. a Q E F = q E a. F = q v B a. a Q in motion B. db/dt E. de/dt B. PHY 249 Lectue Notes Chapte 32: Page 1 of 12 What we have leaned so fa a a F q a a in motion F q v a a d/ Ae thee othe "static" chages that can make -field? this lectue d/? next lectue da dl Cuve Cuve

More information

ev dm e evd 2 m e 1 2 ev2 B) e 2 0 dm e D) m e

ev dm e evd 2 m e 1 2 ev2 B) e 2 0 dm e D) m e . A paallel-plate capacito has sepaation d. The potential diffeence between the plates is V. If an electon with chage e and mass m e is eleased fom est fom the negative plate, its speed when it eaches

More information

Physics 122, Fall October 2012

Physics 122, Fall October 2012 Today in Physics 1: electostatics eview David Blaine takes the pactical potion of his electostatics midtem (Gawke). 11 Octobe 01 Physics 1, Fall 01 1 Electostatics As you have pobably noticed, electostatics

More information

Your Comments. Do we still get the 80% back on homework? It doesn't seem to be showing that. Also, this is really starting to make sense to me!

Your Comments. Do we still get the 80% back on homework? It doesn't seem to be showing that. Also, this is really starting to make sense to me! You Comments Do we still get the 8% back on homewok? It doesn't seem to be showing that. Also, this is eally stating to make sense to me! I am a little confused about the diffeences in solid conductos,

More information

10. Force is inversely proportional to distance between the centers squared. R 4 = F 16 E 11.

10. Force is inversely proportional to distance between the centers squared. R 4 = F 16 E 11. NSWRS - P Physics Multiple hoice Pactice Gavitation Solution nswe 1. m mv Obital speed is found fom setting which gives v whee M is the object being obited. Notice that satellite mass does not affect obital

More information

anubhavclasses.wordpress.com CBSE Solved Test Papers PHYSICS Class XII Chapter : Electrostatics

anubhavclasses.wordpress.com CBSE Solved Test Papers PHYSICS Class XII Chapter : Electrostatics CBS Solved Test Papes PHYSICS Class XII Chapte : lectostatics CBS TST PAPR-01 CLASS - XII PHYSICS (Unit lectostatics) 1. Show does the foce between two point chages change if the dielectic constant of

More information

16.1 Permanent magnets

16.1 Permanent magnets Unit 16 Magnetism 161 Pemanent magnets 16 The magnetic foce on moving chage 163 The motion of chaged paticles in a magnetic field 164 The magnetic foce exeted on a cuent-caying wie 165 Cuent loops and

More information

The Millikan Experiment: Determining the Elementary Charge

The Millikan Experiment: Determining the Elementary Charge LAB EXERCISE 7.5.1 7.5 The Elementay Chage (p. 374) Can you think of a method that could be used to suggest that an elementay chage exists? Figue 1 Robet Millikan (1868 1953) m + q V b The Millikan Expeiment:

More information

Electric Field. y s +q. Point charge: Uniformly charged sphere: Dipole: for r>>s :! ! E = 1. q 1 r 2 ˆr. E sphere. at <0,r,0> at <0,0,r>

Electric Field. y s +q. Point charge: Uniformly charged sphere: Dipole: for r>>s :! ! E = 1. q 1 r 2 ˆr. E sphere. at <0,r,0> at <0,0,r> Electic Field Point chage: E " ˆ Unifomly chaged sphee: E sphee E sphee " Q ˆ fo >R (outside) fo >s : E " s 3,, at z y s + x Dipole moment: p s E E s "#,, 3 s "#,, 3 at

More information

(r) = 1. Example: Electric Potential Energy. Summary. Potential due to a Group of Point Charges 9/10/12 1 R V(r) + + V(r) kq. Chapter 23.

(r) = 1. Example: Electric Potential Energy. Summary. Potential due to a Group of Point Charges 9/10/12 1 R V(r) + + V(r) kq. Chapter 23. Eample: Electic Potential Enegy What is the change in electical potential enegy of a eleased electon in the atmosphee when the electostatic foce fom the nea Eath s electic field (diected downwad) causes

More information

Electricity Revision ELECTRICITY REVISION KEY CONCEPTS TERMINOLOGY & DEFINITION. Physical Sciences X-Sheets

Electricity Revision ELECTRICITY REVISION KEY CONCEPTS TERMINOLOGY & DEFINITION. Physical Sciences X-Sheets Electicity Revision KEY CONCEPTS In this session we will focus on the following: Stating and apply Coulomb s Law. Defining electical field stength and applying the deived equations. Dawing electical field

More information

Continuous Charge Distributions: Electric Field and Electric Flux

Continuous Charge Distributions: Electric Field and Electric Flux 8/30/16 Quiz 2 8/25/16 A positive test chage qo is eleased fom est at a distance away fom a chage of Q and a distance 2 away fom a chage of 2Q. How will the test chage move immediately afte being eleased?

More information

Math 2263 Solutions for Spring 2003 Final Exam

Math 2263 Solutions for Spring 2003 Final Exam Math 6 Solutions fo Sping Final Exam ) A staightfowad appoach to finding the tangent plane to a suface at a point ( x, y, z ) would be to expess the cuve as an explicit function z = f ( x, y ), calculate

More information

B. Spherical Wave Propagation

B. Spherical Wave Propagation 11/8/007 Spheical Wave Popagation notes 1/1 B. Spheical Wave Popagation Evey antenna launches a spheical wave, thus its powe density educes as a function of 1, whee is the distance fom the antenna. We

More information

Charge is that property of material due to which it shows electric and magnetic phenomena.

Charge is that property of material due to which it shows electric and magnetic phenomena. Electostatics Electostatics deals with the study of foces, fields and potentials aise fom static chages. In othe wods it is study of chages at est. Chage Chage is that cetain something possessed by mateial

More information

Module 05: Gauss s s Law a

Module 05: Gauss s s Law a Module 05: Gauss s s Law a 1 Gauss s Law The fist Maxwell Equation! And a vey useful computational technique to find the electic field E when the souce has enough symmety. 2 Gauss s Law The Idea The total

More information

DEVIL PHYSICS THE BADDEST CLASS ON CAMPUS IB PHYSICS

DEVIL PHYSICS THE BADDEST CLASS ON CAMPUS IB PHYSICS DEVIL PHYSICS THE BADDEST CLASS ON CAMPUS IB PHYSICS LSN 10-: MOTION IN A GRAVITATIONAL FIELD Questions Fom Reading Activity? Gavity Waves? Essential Idea: Simila appoaches can be taken in analyzing electical

More information

ELECTROMAGNETISM (CP2)

ELECTROMAGNETISM (CP2) Revision Lectue on ELECTROMAGNETISM (CP) Electostatics Magnetostatics Induction EM Waves based on pevious yeas Pelims questions State Coulomb s Law. Show how E field may be defined. What is meant by E

More information

Electrostatic Potential

Electrostatic Potential Chapte 23 Electostatic Potential PowePoint Lectues fo Univesity Physics, Twelfth Edition Hugh D. Young and Roge A. Feedman Lectues by James Pazun Copyight 2008 Peason Education Inc., publishing as Peason

More information

13. The electric field can be calculated by Eq. 21-4a, and that can be solved for the magnitude of the charge N C m 8.

13. The electric field can be calculated by Eq. 21-4a, and that can be solved for the magnitude of the charge N C m 8. CHAPTR : Gauss s Law Solutions to Assigned Poblems Use -b fo the electic flux of a unifom field Note that the suface aea vecto points adially outwad, and the electic field vecto points adially inwad Thus

More information

Math Notes on Kepler s first law 1. r(t) kp(t)

Math Notes on Kepler s first law 1. r(t) kp(t) Math 7 - Notes on Keple s fist law Planetay motion and Keple s Laws We conside the motion of a single planet about the sun; fo simplicity, we assign coodinates in R 3 so that the position of the sun is

More information

q r 1 4πε Review: Two ways to find V at any point in space: Integrate E dl: Sum or Integrate over charges: q 1 r 1 q 2 r 2 r 3 q 3

q r 1 4πε Review: Two ways to find V at any point in space: Integrate E dl: Sum or Integrate over charges: q 1 r 1 q 2 r 2 r 3 q 3 Review: Lectue : Consevation of negy and Potential Gadient Two ways to find V at any point in space: Integate dl: Sum o Integate ove chages: q q 3 P V = i 4πε q i i dq q 3 P V = 4πε dq ample of integating

More information

AH Mechanics Checklist (Unit 2) AH Mechanics Checklist (Unit 2) Circular Motion

AH Mechanics Checklist (Unit 2) AH Mechanics Checklist (Unit 2) Circular Motion AH Mechanics Checklist (Unit ) AH Mechanics Checklist (Unit ) Cicula Motion No. kill Done 1 Know that cicula motion efes to motion in a cicle of constant adius Know that cicula motion is conveniently descibed

More information

( ) Make-up Tests. From Last Time. Electric Field Flux. o The Electric Field Flux through a bit of area is

( ) Make-up Tests. From Last Time. Electric Field Flux. o The Electric Field Flux through a bit of area is Mon., 3/23 Wed., 3/25 Thus., 3/26 Fi., 3/27 Mon., 3/30 Tues., 3/31 21.4-6 Using Gauss s & nto to Ampee s 21.7-9 Maxwell s, Gauss s, and Ampee s Quiz Ch 21, Lab 9 Ampee s Law (wite up) 22.1-2,10 nto to

More information

AP Physics Electric Potential Energy

AP Physics Electric Potential Energy AP Physics lectic Potential negy Review of some vital peviously coveed mateial. The impotance of the ealie concepts will be made clea as we poceed. Wok takes place when a foce acts ove a distance. W F

More information

Review for Midterm-1

Review for Midterm-1 Review fo Midtem-1 Midtem-1! Wednesday Sept. 24th at 6pm Section 1 (the 4:10pm class) exam in BCC N130 (Business College) Section 2 (the 6:00pm class) exam in NR 158 (Natual Resouces) Allowed one sheet

More information

Review of Potential Energy. The Electric Potential. Plotting Fields and Potentials. Electric Potential of a Point Charge

Review of Potential Energy. The Electric Potential. Plotting Fields and Potentials. Electric Potential of a Point Charge eview of Potential negy Potential enegy U() can be used to descibe a consevative foce. efeence point (U) can be chosen fo convenience. Wok done by F : W F d s F d (1D) Change in P.. : U U f U i W Foce

More information

Charges, Coulomb s Law, and Electric Fields

Charges, Coulomb s Law, and Electric Fields Q&E -1 Chages, Coulomb s Law, and Electic ields Some expeimental facts: Expeimental fact 1: Electic chage comes in two types, which we call (+) and (). An atom consists of a heavy (+) chaged nucleus suounded

More information

Chapter 2: Basic Physics and Math Supplements

Chapter 2: Basic Physics and Math Supplements Chapte 2: Basic Physics and Math Supplements Decembe 1, 215 1 Supplement 2.1: Centipetal Acceleation This supplement expands on a topic addessed on page 19 of the textbook. Ou task hee is to calculate

More information

Introduction: Vectors and Integrals

Introduction: Vectors and Integrals Intoduction: Vectos and Integals Vectos a Vectos ae chaacteized by two paametes: length (magnitude) diection a These vectos ae the same Sum of the vectos: a b a a b b a b a b a Vectos Sum of the vectos:

More information

Review. Electrostatic. Dr. Ray Kwok SJSU

Review. Electrostatic. Dr. Ray Kwok SJSU Review Electostatic D. Ray Kwok SJSU Paty Balloons Coulomb s Law F e q q k 1 Coulomb foce o electical foce. (vecto) Be caeful on detemining the sign & diection. k 9 10 9 (N m / C ) k 1 4πε o k is the Coulomb

More information

working pages for Paul Richards class notes; do not copy or circulate without permission from PGR 2004/11/3 10:50

working pages for Paul Richards class notes; do not copy or circulate without permission from PGR 2004/11/3 10:50 woking pages fo Paul Richads class notes; do not copy o ciculate without pemission fom PGR 2004/11/3 10:50 CHAPTER7 Solid angle, 3D integals, Gauss s Theoem, and a Delta Function We define the solid angle,

More information

3.8.1 Electric Potential Due to a System of Two Charges. Figure Electric dipole

3.8.1 Electric Potential Due to a System of Two Charges. Figure Electric dipole 3.8 Solved Poblems 3.8.1 Electic Potential Due to a System o Two Chages Conside a system o two chages shown in Figue 3.8.1. Figue 3.8.1 Electic dipole Find the electic potential at an abitay point on the

More information