hw4_solutions - Egg 309m kw ficf gal/010 1 Wm...

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Unformatted text preview: Egg 309m kw ficf gal/010% 1;) Wm; UHF» [“23",]: [bi—‘3 m VIM: [W11 [53%;] PWGGW W M WTDrYfi ems: Dz”! 5 1 .1 [g] M Newf U9""CDFSW%T Am 17> WW AV?) A g Rabbi. SrMCCE “W2V1‘ —~—§ {W1:CVE[A‘] (NUWQMSLE Upi‘fé GCVE“: 1 [5,}: [hi] (JUNO/K mm 90 THE STATIC ELECTRIC FIELD 4-4. 4-5. 4-6. whereg = lgl. Substituting Q = 150 nC, m 22.5 g, and l = 12 cm,‘ gives 6 a: 0.522 rad = 299°. The equation can be solved numerically by simple iteration. Zero force. (a) By inspection, it is clearly seen that the charge Q3 must be located to the right of charge Q3, i.e., the :1: coordinate of Q3 must satisfy :53 > 0. The net force on charge Q2 due to charges Q1 and Q3 can be written as leQ2 szQ3 F F = -fi-——-———-— fi—-———-— = 12 + 32 ($2 - 361)2 + ($3 - 232)2 where Q1 = +40 11C, Q2 = ~20 11C, Q3 = +10 nC, 1:; = ~~2, x2 = 0, and x3 is the unknown location of charge Q3 on the 2: axis. Substituting these values yields 2:3 = 1. Therefore, Q3 = +10 11C must be located at (1,0,0). (b) By inspection, it is clear that the location of Q3 must satisfy 2:3 > 0. The net force on Q3 is k k F13+F23=fi Qlez _fi QzQ32 = ($3 - 551) ($3 - 22) Substituting the above values, we get x§~423-—4=0 -—> $3214.83 Therefore, the charge Q3 = +10 nC should be located at (~ 4.83, 0, 0). Three charges. The force on one of the charges due to the other two (identical each having a charge Q) at the corners of an equilateral triangle of side a is given by 2 2 F=2 (fl) cosBO° r- 13:2 a2 Three point charges. (a) The force on charge Q; = +150 nC located at (0,0,0) due to charges Q2 : +100 nC and Q3 = +200 11C located at points (1,0,0) and (0,1,0) can be calculated as F1=~R [“91ng »- 91:22:23 = ~£135 — 9270 uN from 'which the magnitude of the force is [F]! 2 302 uN, where we used [c 2 9 x 109 m-F". The force on Q2 due to Q1 and Q3 can be found as _ 3—57 szQ3 leQ2 F2~(——-) (J2): +2 12 v’i 23363.6 —— 363.6 +3135 5: 2199 - $63.6 uN from which {le :2: 209 MN. The force on charge Q3 due to charges Q1 and Q2 can be evaluated as _ “91% 43+? szQs N . F3 -9 12 +( ) (fif _. 5:63.6+9334 an (fif THE STATIC ELECTRIC FIELD Fig. 4.1. Figure for Problem 4-6. Pl(0,0,3a) —————— :Iflmw Fig. 4.2. Figure for Problem 4-7. The electric fields at points P1 and P2. from which |F3| 2 340 iLN. Therefore, charge Q3 = +200 nC experiences the largest force. (b) This problem is identical to part (a) except that charge Q2 = —100 nC. Following a similar approach, we have F1 = 2135 — 9270 iLN with |F1| 2 302 p.N, F2 2 —iil99 + y63.6 p.N with lF2| 2 209 uN, and F3 2 9270 + x636 — y63.6 = 263.6 + y206.4 uN with |F3| 2 216 p.N respectively. Therefore, in this case, charge Q1 experiences the largest force. ' . Two point charges. The total electric field produced at point P1(0,0,3a) due to two point charges Q and ——Q located at points (0,0,0) and (0,4a,0) is given by kQ _ W 1662 E14 2 (0.6)+y(5a)2 (3a)2 — (5a)2 2%[90032 + 20.0871] (0.8) Similarly, the total electric field at point P2(0,4a,3a) is _ W W _ kQ E2 _y(3a)2 (0.8) +2(5a)2 (0.6) 2(3a)2 2%[90032 — 20.0871] Both vectors E1 and E2 at points P1 and P2 are sketched as shown. THE STATIC ELECTRIC FIELD 9? Fig. 4.7. Figure for Problem 4-16. which can be integrated to find the total field at the origin due to the half—ring, i.e., 1T 77 I , _ _ 393 Z f- 2Q1/7r [Exlhalrmg- [0 diff -411'6032 / sin (R) all _47r€0 2 so that the total field at the origin due to both the halfiring and the two charges is 26m 02 fi_ Q2 ____H 1 39:39.2 47reoR2it 4W€0R2 47reo(2R)2 47T€oR2 E(0, 0)- _ 4 Setting E = 0 we find Q2 = 8621 / (571'). 4-17. Semicircular line charge. (a) We choose an elemental line charge element pgdi’ = adef: along the semicircle. The electric field at the origin is given by [4.14], namely 1 (r —- r’ p; (1")dl’ E(0 0 0): f0.— 47reo Ir - r’l3 wherer = 0, p10") = p; = const, dl’ = adqfi’, (r-—~r’) = ——(a cos (Vina sin d)’ 9), and Il'—-r’|3 = :23. Thus, ~. E (0 O 0) = p; ’r —(a cos (13’ i + a sin 43" wedge" ‘ ’ ’ 4’H'60 9 a3 =-pl? fisingéid¢l=— Pl 9 47mm 21mm 0 (b) With pz(r’) = p0 sin (15’, we have" 7r _ I . I _ ’ I E(0,0,0)=4_€L/ WW _ ...<._:A,;.W;r.—V,.Mg “51;,fflnx4’ ”:5" “AL; V N «.3»; ,VZA’le—g” 100 THE STATIC ELECTRIC FIELD zero potential. We can choose this point to be the middle point of the hole, i.e., let (DA = 0 at zA = 0, in which case we have 4-20. Spherical charge distribution. (a) The total charge Q in the spherical region 0 < r < a is given by 211 it a ’ 2 Q: p(r’)dv' =K / f f 6"” 1" sine?’dr’d3’d¢' V’ 0 0 0 2 2a .. 2 —ba ‘3 2 —K(27§)(2) [5-3- 8 (b + b2 + b3” where for the r" integral, we used 2 fuzewdu=em [1-2—:+-33] a a a which can easily be shown using integration by parts twice. (b) Since there is spherical symmetry, we consider a spherical Gaussian surface S with radius r and apply Gauss’s law for the cases when r < a and r > a with the result of part (a) as r2 "27‘ 2 [560E ‘ d5 = 60133-041772) = Qenc = (497K) [*3 " fl — £3] resultingin” I K 2 4,, r2 27' 2 -—-- ~—-—- -— —- - < E- éor2[b3 e (6+b2+b3)] r_a 2"" i Ewe"b“(gi+-2—E+~2- r>a and b3 b _b2 b3 (c) The electric potential can simple be found by integrating the electric field, namely 2 (Mr) = -— / Er(r)dr For 1' > a we simply find For 3* S a we have _K 2 K 4,, 2K '6'!" 2K '5'”? $09-25 (5-37) + (fobz) B 4‘ (ob L 77-d1'4-60b3L72—d7‘ THE STATIC ELECTRIC FLEELD 101 where the last two terms are the somcalled Exponential Integrals for which there is no closed form solution but which are well tabulated? (d) The result can be shown by simple substitution into the cylindrical coordinate version of Poisson’s equation; namely 1 a 3" 1 324’ 82¢ war) qu’: "M” (Tm) + *W + w = 'r 3?" 8r 1-2 6452 6z2 59 No variation in ‘5 No variation in 2 --+ 1 a (1-5931) — ”KB—b1. : $207 8? 60 Note that differentiation of the Exponential Integral terms simply yield the integrands. 4-21. The electron charge density in a hydrogen atom. (3) Since there is Spherical symmetry, we consider a sphericai Gaussian surface S with radius r and apply Gauss's law as / 60E -ds = 60mm?) = cm 3 where " q 21r 1r 1‘ I ’ 2 Q8“: =.—33- / f e“?! 1’0?! sinB’dr’dfl’dq’)’ ’M o 0 n Qe r 12 -2 ’/ I =(21r)(2) 3 f r e 7‘ 6dr 7T3 0 =39: 93+e—2r/a -Efi_9_2.’£__9.3. a3 4 2 2 4 where we used the integral which can easily be shown using integration by parts twice. As a result, the electric field due to the electron cloud only can be found as t 3 2 2 3 :~ E, 1 fle[9_se~2rxa(£+s:+s.)] t =47r60'r2 (:3 4 2 2 4 2 1‘ = 48 1 —- 6’2”“ 2(3) +2 (3:) +1 a 47r€or2 a a 3 Note that the constant a in the above expression is called the Bohr radius given by a c: 0.529 x 10'“ 10 m. The electric potential can be found by integrating the electric field. We find (11(7) 2 w] Er(r)drl= £3 ear/G [—1. + 1] ~ - qe -1- {- m 471'60 a r 47r€0 :r _....._. i See Chapter 5 of M. Abramowitz and I. A, Stegun, Handbook of Mathematical Functions With Formulas, 5 it 'aka, and mthematical Tables, National Bureau of Standards, Tenth Printing, 1972. 1 ...
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