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# Barnet HW2 - 16.49 Choose a coordinate system with pointing...

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Unformatted text preview: 16.49 Choose a coordinate system with pointing up. 'Thc only forces on the cork are its weight, —mgj, and the electric force Felec = qE, where m = 2.00 X 10—3 kg is the mass of the cork and q = 3.00 X 10‘6 C is the charge on the cork. The cork is not accelerating, so the total force on it is zero. Also q > (J C. Hence —mgj+qf}= 0N 2:- mg: quE: 9E. In Example 16.17 on page 738 of the text, we saw that the magnitude E of the electric ﬁeld of a uniformly charged (inﬁnite) sheet is Thus, substituting this expression for E and solving for a, 0 mg 4am mg = E: -— : —— : ——‘— mg q q2eu'l’d‘0 260(9) (2ﬁ)(q) _ l ((2.00 X 10”3 kg)(9.81 III/82) ﬁ’ ‘7 _ (gﬁxgm x 109 N-mQ/CE) 3.00 >< 10-6 o ) = 1.15 X 10—7 0/1112. 16.58 I a} The force on the particle of mass m and charge 9 is the electric force, whose magnitude is Felectric : To ﬁnd the particle’s acceleration= apply Newton’s second law using vector magnitudes, quE FelectrKCZma : IQIEZma Z} a: m The acceleration of a charged particle in a uniform E ﬁeld is constant, so we can use the kinematic equations for motion with a constant acceleration= letting i be in the direction of motion and choosing the origin at the initial position of the proton: It? :2 82 20m +(0 m/s)t+ a: :}Z_ 2.1:_ 2am Vaz Ile' For a. proton to travel 1.00 In, the time is 2mm m )(1.67 x 10-27 kg) _6 : ——..—.' : _ ‘. ﬁrm“ V (1.502 X 10—19 o )(300 we) 8 34 X 5 = \$0 + Uzot + b) For an electron, the time is _ 2(1.00 m)(9.11 x 10-31 kg) te nearer; —— ‘_ I E _7 ‘ 1 f (1.602x10*190)(300N/c) 190x 10 b. c) The ratio of the two times is: tproton _ X 10—5 S - ——'——— = 42. . telectron X 10—7 S 8 Now use the equation for t at the end of part a) to evaluate the above ratio without any numbers: f] 2 3 mproton itprCItOl'l _ 9E ____ .' mproton 42 8 teleciron \/2 Emelectron melectron u— . I gE So the ratio of times depends only upon the ratio of the masses long as both particles have the same charge). __ _ ‘"""—u—-——_T.——.-_——._—__ 16.66 Referring to Figure P. 66 on page 761, choose a coordinate system with i pointing right, pointing up, and origin at the launch point of the electron. With these choices, we make the important observation that, since we are projecting an electron into the given uniform electric ﬁeld, the motion of this electron is equivalent to the motion of a massive projectile in a gravitational ﬁeld. In other words, we can treat this problem exactly like we treated projectile motion problems in Chapter 4, with the only diﬁerence being that cry 2 —eE/m rather than —g. The magnitude of the electrical force 0n the electron is F = |q|E 2 EB. Apply Newton’s second law to the electron to ﬁnd the magnitude of its acceleration. Using vectOr magni- tudes, we have Ftotal __ 8E Fwtal = mo. 2:;- a = —. m m In the chosen coordinate system the acceleration is down, and yo = 0 m. 6 :30 = 0 In U30 : “9 SE} 'Umo = '00 cos 6 6 [1y = _— em = 0 m/s2. m Therefore, the equations for the velocity and position components are _ eE alga) = 110 51:16 '— 15 I gig) = '00 C086 BE E 36) -_-_ vo(cos tilt y(t) = [1,10 31119)t —— 2 u I _ ___ Impact occurs where y = 0 m, so 813 t2 0 in : (cosindlt — Use the quadratic formula to solve this for t, and thus, to find the time the electron is in the air. The two solutions are t = 0 s, which corresponds to the launch time (Le, the time at which y = U m), and 21) 27er t : —-—-2 sinQ z 0 ' 6. 0y BE s1n Finally, use this time in the equation for :r to ﬁnd the distance the electron travels: I = (no cos Ejt 2mm sin 6 _ (to a) (T) _ 2mv§ '— 6E 2(9.11 X 10—31 kg)(5_00 X 106 III/5);: I ’ a W 601]" mm (1.602 x 10-19 C)(250 N/G) am (:05 . .. _ . ____ __ _ _ _ i . .. ____. 16.72 Gauss’s law for the electric ﬁeld states that the ﬂux of the electric ﬁeld through a closed surface is equal to the net (i.e., total) charge enclosed by that surface divided by Q). The closed surface shown in Figure 19.72 on page 762 of the text surrounds twr: of the four charges. Therefore, the ﬂux of the electric ﬁeld through this closed surface is _ —1.00 x 10-'3 C — 3.00 x 10—6 o _ —4.00 'x 10-6 C (I) — . EU 60 sinél cosﬂ One way of remembering the numerical value of so is to recall that 1 _ s _2 2 1 _ —12 2 _ 2 limo—QDUXlO Nm/C _:> Eo—4ﬁ(9-OUXIOQN‘m2/C2)~8.84X10 C /(N m ). Hence, 41.00 x 10—6 o ¢=+———e—-——-—-=—4.52 15N- 2C. 8.84x10—1202/(N-m2) X 0 m/ _ ——_ . __.— 17.11 a) The electric potential at P is the sum of the electric potentials caused by each of the two point charges. Since both charges are the same distance 1" from P, we have 13+ 1 (Pelz0v. V=Vl+V2= (taco T 41760 i" b) The potential energy of a proton placed at P is Psqu = (1.602 x 10—19 01(0 V) =0J. c) By deﬁnition, the work done by the total electrical force is Welectric = nAPE = "(Pﬂsnu '— Pﬂinitiaﬂ = —(quna1 — 0 J Each charge produces an electrical potential equal to zero at inﬁnity, so the total electrical potential there, Vﬁnal, also is zero. Hence Welectric = X 10—19 C = 0 J _ Notice that you get the same result if you evaluate the work explicitly as co m A Welectric : f - Felectric . dry P since the total electric force is perpendicular to d? at every point on the path. 17.13 a) Mom Example 16.17 on page 738 of the text, the magnitude E of the electric ﬁeld between tivo parallel plates is related to the magnitude of the Surface charge density or on each by 0' Ez— => E0 _ _ i 1 __ 1 l 4 _ _8 2 0'~—€[JE—- (471) 1 E_ (4%) (9-00X109Nm2/02) (1.00X10 N/Cj—8.84x10 C/m . 471'Eo Since electric ﬁeld lines go from positive to negative charge, the top plate has the negative charge. Thus, atop plate = X 18—8 0/1112 . b) The electric field lines go from positive to negative charge, so the bottom plate has positive charge. Thus, Ubottom plate : X 10—8 0/1112 . c) The absolute value of the potential difference between the parallel plates is re ated to the magnitude of the ﬁeld and the plate Separation by ' |AV| = Ed: (1.00 X 104 N/C)(0.100 m) = 1.00 x 103 V. Electric ﬁeld lines go from regions of high potential to regions of low potential. If the lower plate is “grounded,” then by deﬁnition Vbottom = 0 V. The upper plate has a lower potential, so Wop plate = —1.00 x 103 V. d) Choose a coordinate system with pointing up. Let in = 1.00 X 10—5 kg denote the mass of the particle. Then the two forces on the particle are: 1. its weight ﬁr = 40193; and 2. the electrical force Pele: = qu, Since the particle is not accelerating, the sum of these forces is zero, so my _ (1.00 X 10"5 kg)(9.81 111/52) E 1.00 X 10‘II N/C —mgi+qu=0N =>q= #9.81><10_QC. 1?.14 a) The electric ﬁeld at P is the vector sum of the ﬁelds produced at P by the individual charges. The magnitude of the electric ﬁeld of a point charge is 1 lQl 4050 72 ‘ The electric ﬁeld created by a positive charge points away from that charge, while that created by a negative charge points toward it. The ﬁeld at P created by the 4.00uC has magnitude (9.00 x 109 N-m2/02 )(4.00 x 10—‘5 C) v E1 2 (4.00 x 10“2 In)2 2 2‘25 X 107 E/C' and is directed away from the positive charge. Therefore, E; = (—2.25 x 107 N/C )i. From the Pythagorean theorem, the —2_.DU,uC charge is located 5.00 cm from P. The magnitude of the ﬁeld produced at P by this charge is _ (9.00 x 109 N-mQ/Cﬂll 4 2.00 x 10-6 0| — ' 7 (5.00 X 10—2 m)2 — 0.720 X 10 N/C. E2 This ﬁeld vector is directed t0ward the negative charge. Referring to Figure P.14 on page 797 of the text, and calling 5' the acute angle at P that a line between P and the —2.0,u.C charge makes with i, we have E2 = 0.720 x 107 N/C (cos 6i+ sin 793) = 0.720 x 107 N/C (0.600i + 06003) = (0.576 x 107 N/C)i+ (0.432 x 107 N/ojj. The total electric ﬁeld at P is E = E31 "l‘ f112 e (—2.25 x 107 N/C )i+ (0.576 x 10’“ N/C )i+ (0.432 x 107 N/C)j = (—1.67 >< 107 N/C)i+ (0.432 x 107 N/C )3. b) The magnitude of the electric ﬁeld at P is E = Jag + E3 = N/(~1.67 x 107 19/0)2 + (0.432 x 107 N/C) = 1.72 >< ioT N/C. c) The angle 6“) that the total ﬁeld makes with} (not i) is 11311 _ 1.67 x 10? N/C lEyl _ 0.432 x 107 N70 tan a = =e- a 2 755°. Note that d: is measured counterclockwise from d) The total electric potential at P is the sum of the potentials produced at P by the individual charges, V = 1/1 + V2 1 Q1 1 Q2 _ + _ Tree 7'1 47110 72 # (9.00 x 109 N-mZ/C2 )(400 x 10—6 C) + (9.00 x 109 N-m2/02 )(w2.00 x 10—6 C) _ 4.00 x 10-2 m 5.00 x 10-2 m = 9.00 x 105 V — 3.60 x 105 V = 5.40 x 1105 v. e) From the electric ﬁeld found in part a), we have that the force on the charge q placed at P is 13 = qE' = —3.00 x 10-6 C[(—1.67 x 107 N/C)§+ (0432 X 107 Mom = (50.1 N)f— (13 mi: (-50N)i— (13.0 mi. The potential energy of the charge is PE : qV = (#300 x 10—6 c )(5.40 x 105 V) 2 -1.02 J. 17.33 ——I —- — — a a) The electric potential on the surface of a uniformly distributed spherical charge is 1 V : ﬁg- 471'60 R where Q is the total charge and R is the radius of the sphere. Hence, for the 2.00 cm sphere, 3.00 x 10—9 C _ Q —‘ 2 '12 W 1V2.00 cm — (9-00 X 10 N m K} )(200 x 10—2 In )=1.35><103V. For the surface of the 4.00 cm sphere, 3.00 x 10—9 C Igloo Cm = X 109 >=675V. b) When the two are connected by a conducting wire their potentials equalize, so 1 £22.00 cm 1 Q4110 cm — _—\, Cm = 2' 0 cm ' area (2.00 x 10'2 In > 4050 (4.00 X 10—2 In > Q4330 0 Q2110 Since charge is conserved, the total charge remains the same, hence €22.00 cm + @403 cm i 3-00 X 10—9 C +3.00 X 10—9 C : 6.00 X 10—9 C. :> 362200 Cm i 6.00 X IUTQ C => Q2100 a... = 2.00 x 10—9 C and Q4100 m =‘ 4.00 x 10—9 c. The potential on the surface of the 2.00 cm sphere is now —9 V200 cm 2 1 (m) = (9.00 ><10g Nina/CZ] (200 X 10 C 2.00 x 10“2 m 41m) 2.00 x104 m j = 900 V. The potential on the surface of the 4.00 cm is, of course. 110w the same at 1 Q V. m = w 2 g, ., 4.00x10‘90 4 00 4m (4.00 x 10—2 m) (9‘00 X 10 N'mE/C') <————4_00 X 1W ) = 000 v. The original charge 011 each was 3 00 x 10“} C a . , 11d th ' ch * . ‘9 “ ‘ so 1.00 x 10—9 C was exchanged in the process. e new ages are 2.00 X 10 C amd 4-00 x 10 9 C) ...
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