Unformatted text preview: Electric Forces and Electric Fields 7 Problem Solutions
15.1 Since the charges have opposite signs, the force is one of attraction Its magnitude is I I. 15.2 The electricalforce would need to have the same magnitude as the current gravitational force, or giving This yields ,...,.,,...,.,,...,.~ q= (6.67xiO ll N.m /kg )(5.98Xi024 kg)(7.36Xi022 / 8.99xi09 N .m2 C2 2 2 kg) = I 5.71xiO C I
13  15.3 = ( 8.99xio . 9 N.m2)(158)(1.60XiO19
2 ct 2 C (2.0xiO14 m) 19iN ~ (repulsion) I 8 CHAPTER 15 15.4 The attractive forces exerted on the positive charge by the negative charges are shown in the sketch and have magnitudes F1 q . I I I I I
I a ~q
, ,'1 I I I I r =F. =!L a2
2 k 2 , , , , aJi: ' , " I and a
I I I
I q LFx = F2 + F3 cos 45 = kef + ke : (0.707) a 2a +q = 1.35(ke~2)
. a "q I I I so F R = 1.91( )
( along the diagonal toward the negative charge k (2e)2 15.5
(a) F= e r2 = 8.99xl0 9 .N ':n C 2 ) [4(1.60X10.19)2]
2 (5.00XI0.15 m) =136.8NI  . (b) The mass of an alpha particle is m = 4.002 6 u where u = 1.66X1027 kg is the
I unified mass unit. The acceleration of either alpha particle is then a 36.8N  m  4.0026(1.66x10.27 kg) I .554xl0 27 m/s2 1 I
b Electric Forcesand Electric Fields 9 15.6 The attractive force between the charged ends tends to compress the molecule. Its magnitude is F = k()2 Ie e
,2 = 8.99xl09 ( N.m. C2 2)(160XI0 C)2 (2.17xl0...{; mt 19 = 4.89xl017 N. The compression of the spring" is
II x =(0.010 0), = (O.OlO 0)(2.17 XlO...{; = 2.17xl08 m) m, so thespring constant is k = F = 4.89XI0: N 12.25XlO9 N/m x 2.17xl0 m ~.. I 15.7 1.00 g of hydrogen contains Avogadro's number of atoms, each containing one proton and one electron. Thus, each charge has magnitude = N A e. The distance separating Iql these charges is , = 2RE, where RE is Earth's radius. Thus, 15.8 The magnitude of the repulsive force between electrons must equal the weight of an electron, Thus, kee2 = meg /,2 or 15.9 (a) The spherically symmetric charge distributions behave as if all charge was located at the centers of the spheres. Therefore, the magnitude of the attractive force is I I F= keql1q21 ( 2 = 8.99xl0 , 9 N .m2) (12xl09 C)(18XI09 C) 2 2 . =2.2xl0 C (0.30 m)  I 5 N I I, 10 CHAPTER 15 (b) When the spheres are connected by a conducting wire, the net charge qnet = q} + q2 = 6.0 X109 C will divide equally between the two identical spheres. Thus, the force is now F= ke(qnet/2)2
, 2 = 8.99xl0 ( 9 N.m2)(6.0X109 2 2 C 4(0.30 m) ct or F =!9.0XlO7 N (repulsion) I
.+6.0? JLC.
I _ 15.10 The forces are as shown in the sketch at the right. ~:.~~_~~ __
:
I I I ?__u a
Ii Ii 3 3 ~;~O_O_~C F1 :
I I I F2 em :
I I I :'3.00
I
I :. 2.00 em+:
I
I I
I Ii ~ keq~q2 =(8.99XI09 N '12 C. ':n2)(6.00X106 C)(1.50X;06 C) =89.9 N
(3.00X10,2 m) The net force on the 6 pC charge is F6 = F}  F2 = 146.7 N (to the left) I
I The net force on the 1.5pC charge is Ii.s = Ii + F3 = 1157 N (to the right) The net force on the 2 pC charge is F_2 = F2 + F3 = 1111 N (to the left) I Electric Forcesand Electric Fields 11 15.11 In the sketch at the right, FR is the resultant of the forces F6 and Fs that are exerted on the charge at the origin by the 6.00 nC and the 3.00 nC charges respectively.
9 _ 5.00nC
 .~ _ F6~FR O~300~ 6.00nC
I F 0.100m ~r '3.00nC N .m2)( C
2 6.00x109 C)(5.00x109 C) (0.300 m)
2 F6= ( 8.99x10 = 3.00x106 N
9 Fs = 8.99x10
( N.m2)(3.00X109
2 C)(5.00X109 C)
2. C (0.100m) = 1.35x10 5 N or FR = 11.38X105 N at 77.5 below  x axis I
I 15.12 Consider the arrangement of charges shown in the sketch at the right. The distance r is ~3.00nC
I " I , ~ I 1:1, 01 lO' " , O' ,
" r, ' o. I
I "
' ~_ .. _:'
The forces exerted on the 6.00 nC charge are
9 : 0.500m,'
I I " 1:1 0 ' 01 c:: F2 = ( 8.99x10 N .m2) (6.00X109 C)(2.00X109 C) .2 2 C (0.707 m) . ~: , . ,,
" I' ", ,
r,' ' 6.00nC ~2.00nC = 2.16x107 N 9 N.m2)(6.00X109C)(3.00X109C)_7
2 2 and Fs = ( 8.99xlO C . (0.707m) =3.24xlO N 12 CHAPTER 15 The resultant force on the 6.00 nC charge is then or FR = 13.89X107 Nat 11.3 below +x axis I
y 15.13 The forces on the 7.00 j:. charge are shown at the right. 1:; = ( 8.99x109 =0.503 N N.m2)(7.00X106 C)(2.00X106 C) C2 (0.500 m)2  Fz
60.0 + 2.00 JLC 0.500 m 4.00 JLC &~ \ =1.01 N and ~y =(1:; F2)sin60.00=0.436 N The resultant force on the 7.00 j:. charge is or FR = I 0.872 N at 30.0 below the +x axis I Electric Forces and Electric Fields 13 15.14 Assume that the third bead has charge Q and is located at 0 < x < d. Then the forces exerted on it by the +3q charge and by the +lq charge have magnitudes These forces are in opposite directions, so charge Q is in equilibrium if F3 = Fl' This gives 3( d  X)2 = x2 , and solving for x, the equilibrium position is seen to be This is a position of stable equilibrium if Q > 0 I I. In that case, a small displacement from the equilibrium position produces a net force directed so as to move Q back toward the equilibrium position. 15.15 Consider the freebody diagram of one of the spheres given .at the right. Here, T is the tension in the string and F" is the repulsive electrical force exerted by the other sphere. r.F
y ,. 5.0,, ,
I +Y:, T =0 ~ T cos 5.0 = mg, or T = mg cos5.0 +x ~r =0 ~ F" = Tsin5.0 = mgtan5.0 mg
At equilibrium, the distance separating the two spheres is r = 2Lsin5.0.
r Thus, F"=mgtan5.0 becomes k (2Lsin5.00)2 eq 2 mgtan5.0 and vields q=(2Lsin5.00) mgtan5.0 ke [ (  2 O.300m sm5.0 )' oJ 3 (0.20xlO kg)(9.80 m/s2)tan5.0 8.99xl09 N.m2/C2 =~7_.2_n_C_. .1 I 14
15.16 CHAP'I'ER 15 The required position is shoWn in the sketch at the right. Note that this places q closer to the smaller charge, which will allow the two forces to cancel. Requiring that 6.00 nC
., 3.00 nC q
~ ., 0.600m .  x ,I ' 6  ke(6.00nC)q (x + 0.600 m)2 ="'2 ''' , or ke(3.00nC)q 2 2=( x x 0600 )2 x +. m Solving for x gives the equilibrium position as x = 0.600 m = 1.45 m beyond the  3.00 nC charge .Ji 1. I . I 15.17 For the object to "float" it is necessary that the electrical force support the weight, or qE=mg or in = qE = (24x106 C)( 610 N/C) g 9.S m/s2 1 1.5X103 kg I . 15.18 (a) Taking to the right as positive, the resultant electric field at point P is . given by 2 =(S.99X109 N .m )[6.00X106 C + 2.00x106 C _1.50XlO6C] 2 C (0.0200mt (0.0300m)2 (0.0100mt This gives ER = +2.00x107 N/C (b) F=qER = (2.00X106 C)(2.00X107 N/C) =40.0 N or F = 140.0 N to the left I Electric Forces and Electric Fields 15 15.19 We shall treat the concentrations as point charges. Then, the resultant field consists of two contributions, one due to each concentration. The contribution due to the positive charge at 3 000 m altitude is
2 E =k M=(8.99X109
+ e ,2 N.m ) C2 (40.OC) =3.60x105 (1 000 m t NfC (downward) The contribution due to the negative charge at 1 000 m altitude is
2 E =k M=(8.99X109
e ,2 N.m ) C2 (40.OC) =3.60x105 NJC (downward) (1 000 m )2 The resultant field is then 15.20 (a) The magnitude of the force on the electron is F =
19 IqlE = eE, and the acceleration
I is F eE (1.60X10 C)(300 NJC) 13 21 a===~~=.5.27x10 ms 31 me me 9.11xlO kg "'' (b)
V I =vo +at =0+(5.27x1013 m/s2)(l.00X10"'<! s) =I5.27X105 mfs I 15.21 If the electric force counterbalances the weight of the ball; then qE=mgor E== q mg (5.Ox10 kg)(9.8 m/S2) I . 4 1 6 = .1.2x10 NJC 4.Ox10 C ~~~ 3 15.22 The force an electric field exerts on a positive change is in the direction of the field. Since this force must serve as a retarding force and bring the proton to rest, the force and hence the field must be in the direction opposite to the proton's velocity The workenergy theorem,
Wild I I. = KEf  KEj, gives the magnitude of the field as E = KEj = 3.25 X10 J =11.63X104 NJC (1.60X10.19 C)(1.25 m) 15 ( qE)b.x = 0  KEj or q(b.x) I 16 CHAPTER 15 15.23 (a) a= F_qE m mp (1.60X1019C)(640N/C)_1
1.673x 10. kg
6
27. 6.12x10 .10 m s . /21 (b) t=!:J.v 1.20x10 m/s = 1.96X105 s=119.6/lsl a 6.12x1010 m/s2 '. _.
VI Vo 1.20x10 m s 0 !:J.x=== 10 . 2a 2(6.12x10 m/s2) 2 2 ( 6 1)2 (c) I11.8 m. I
' 15.24 The altitude of the triangle is ql =3.00nG9
, , , , ,'1\ \ I
I I I I I
I h=(0.500 m)sin60.0 = 0.433 m
J:'!
,."." I \ \ \ \ \.
\ "" ....,..,,' and the magnitudes of the fields due to each of the charges are ~' t::;:,.,
, , h
I " \ I I 0 I  \ \ " " .: q2 = 8.00 nG o~ ,'J600 E = keq1 = (8.99x109 N .m2/C2)(3.00XlO9 C)
1 j.0.250 m.. _ " E 0 E q2 = 5.00 nG 3 2 E1 h 2 (0.433 m)2 =144 N/C I
! and Electric Forces and Electric Fields 17 Thus, ~Ex =E2 +E3 =1.87x103 N/C and ~Ey =~E1 =144 N/C giving and (J = tan 1 (~Ey Hence /~Ex ) = tan
3 1 ( 0.0769) = 4.400 I ER= 1.88x10 N/C at 4.400 below the +x axis I
", 15.25 From the symmetry of the charge distribution, students should recognize that the resultant electric field at the center is .... .....  .. "" .... q,,., ,, , E2 30 , 1 I I : , , , If one does not recognize this intuitively, consider: ', ,
" I E3
.....  .........
"" q and 18
" ,CHAPTER 15 ,I ,II.'1. "
iii' ! 15.26 If the resultant field is to be zero, the contributions of the two charges must be equal in magnitude and must have opposite directions. This is only possible at a point on the line between the two negative charges. Assume the point of interest is located on the yaxis at 4.0 m < y < 6.0 m . Then, for equal magnitudes, 8.0 pC rr i
r I q1 =9.0 pC I 6.0m  FT
, ,   r~~t, +x ,
,, ,, ,, (y+4.0 m)2 tq2=8.0 j.LC Solving for y gives y + 4.0 m = J! (6.0 m  y) , or y = I + 0.85 m I 15.27 If the resultant field is zero, the :+y , , contributions from the two charges must :"r1 = d.:. 1.0 m .: be in opposite directions and also have " , +x ~ Q    .                      .   equal magnitudes. Choose the line E2 : E1 :q1 = 2.5 pC q2 = 6.0 pC connecting the charges as the xaxis, with , :. , the origin at the 2.5 j.. charge. Then, the r2= 1.0m+d , two contributions will have opposite directions only in the regions x < 0 and x > 1.0 m . For the magnitudes to be equal, the point must be nearer the smaller charge. Thus, the point of zero resultant field is on the xaxis at x < 0 .
I , I .'
I Requiring equal magnitudes gives ke Iq11 = ke Iq21 or 2.5 pC _ 6.0 pC r12 r22 d2 (1.0 m + d)2 Thus, (1.0 m+d)~2.5
=d 6.0 Solving for d yields
d=1.8 m, or 11.8 m to the left of the  2.5 pC charge I 15.28 Themagnitude of q2 is three times the magnitude of q1 because 3 times as many lines Iq21 = 31% emerge from q2 as enter ql . (a) Then, I Electric Forces and Electric Fields 19 (b) I q2 > 0 I because lines emerge from it, and % < 0 because lines terminate on it. I I 15.29 Note in the sketches at the right that electric field lines originate on positive charges and terminate on negative charges. The density of lines is twice as great for the 2q charge in (b)as it is for the lq charge in (a).
(a) (b) 15.30 Rough sketches for these charge configurations are shown below. (a) (b) (c) 15.31 (a) The sketch for (a) is shown at the right. Note that four times as many lines should leave ql as emerge from q2 although, for clarity, this is not shown in this sketch. (b) The field pattern looks the same here as that shown for (a) with the exception that the arrows
are reversed on the field lines. II!
p 20 . CHAPTER 15 I' I! 15.32 (a) In the sketch for (a) at the right, note that there are no lines inside the sphere. On the outside of the sphere, the field lines are uniformly spaced and radially outward. (b) In the sketch for (b) above, note that the lines are (a) (b) perpendicular to the surface at the points where they emerge. They should also be SYmmetrical about the sYmmetry axes of the cube. The field is zero inside the cube. 15.33 (a) I Zero I net charge on each surface of the sphere. (b) The negative charge lowered into the sphere repels and leaves + 5 pC on the inside 15 pC to the outside I surface, I I surface of the sphere. (c) The negative charge lowered inside the sphere neutralizes the inner surface, leaving I zero charge on the inside I. This leaves 1 5pC on the outside I surface of the I zero charge on the inside I. sphere. (d) When the object is removed, the sphere is left with surface and 1 5.00 pC on the outside I 15.34 (a) The dome is a closed conducting surface. Therefore, the electric field is zero everywhere inside it. At the surface and outside of this spherically sYmmetric charge distribution, the field is as if all the charge were concentrated at the center of the sphere. (b) At the surface, kq E=_e2
R =~=
(1.0m)2 (8.99XI0 9 N .m /C )(2.0XI04 2 2 C) I 1.8xlO
 6 NjC I Electric Forces and Electric Fields 21 (c) Outside the spherical dome, E = kif. Thus, at r = 4.0 m, r E = (8.99xI0 9 2 C2 N .m / 1(2.oxlo4 C) =ll.IXI05 (4.0 m) N/C I 15.35 For a uniformly charged sphere, the field is strongest at the surface. Th us, Emax = keqmax R
2' or q max R.2E max ke  6 (2.0m)2(3.0XI0 N/C) 9 N .m2jC2 8.99xl0 = .1 1.3xlO 3 CI
_ 15.36 H the weight of the drop is balanced by the electric force, then mass of the drop must be mg = Iql E = eE or the m = g eE = 9.8'mj's2 19 (1.6XIO C)(3XI04  N/C) == 5x 10 16 kg But, m = pV = p(: m' ) and the radius of the drop is r = [ :;;;, r r= 3(5xIO16 kg) [ 41Z'( 858 kg/m3) .
1f 3 = 5.2xlO 7 m or 15.37 (a) F = qE = (1.60XlO19 C)(3.0XI04 N/C) = 14.8XlO15N I 15.38 The flux through an area is <1> = EAcosB, where Bis the angle between the direction of the field E and the line perpendicular to the area A. (b) In this case, B = 90 and <1> [email protected]] 22 CHAPTER 15 2 15.39 The area of the rectangular plane is A = (0.350 m) (0.700 m)= 0.245 m . (a) When the plane is parallel to the yz plane, 0 = 0, and the flux is (b) When the plane is parallel to the xaxis, 0 = 90 and cP = @] 15.40 In this problem, we consider part (b) first. (b) Since the field is radial everywhere, the charge distribution generating it must be spherically sYmmetric Also, since the field is radially inward, the net charge I I. inside the sphere is negative charge I I.
IQI/R
r
2 , (a) Outside a spherically sYmmetric charge distribution, the field is E = ke~ . Thus, just outside the surface where r = R , the magnitude of the field is E = ke R2 E (0.750 m)2 (890 N/C)
.8 so /QI=k= 89 9x 109 N m 2/C2 e 5.57xlO C=55.7nC Since we have determined that Q < 0, we now have Q = 155.7 nC I 15.41 cP = EA cos 0 and cP = cPt:, maxwhen 0 = 0 cP cP Thus, E=E,max = \max =. A 1td 4(5.2xl05 /4 N .m2/C) 2 =14.1X106 N/CI 1t(OAOm) ~. 15.42 cP = EAcosO = (~;) 41tR2 )cosoo = 41tkeq Electric Forces and Electric Fields 23 15.43 We choose a spherical gaussian surface, concentric with the charged spherical shell and of radius r. Then, ~EAcos(} = E( 4nr2 )cosO = 4nr2E . (a) For r> a (that is, outside the shell), the total charge enclosed by the gaussian surface is Q = +q  q = 0 . Thus, Gauss's law gives 4nr2 E = 0, or E = 0 . .(b) Inside the shell, r < a, and the enclosed charge is Q = +q . Therefore, from Gauss' s law, 4nr2 E = .!L, or E Eo q
4nEo r2 = keq r2 The field for r < a is E = ke~
r directed radially outward 15.44 Construct a gaussian surface just barely inside the surface of the conductor, where E = 0 . Since E = 0 inside, Gauss' law says .9..= 0
Eo inside. Thus, any excess charge residing on the conductor must be outside our gaussian surface (that is, on the surface of the conductor). E = 0 at all points inside the conductor, and cos () = cos 900 = 0 on the cylindrical surface. Thus, the only flux through the gaussian surface is on the outside end cap and Gauss's law reduces to ~EAcos(} = EArop =.9.. .
Eo 15.45 The charge enclosed by the gaussian surface is Q = u A, where A is the crosssectional area of the cylinder and also the area of the end cap, so Gauss's law becomes uA EA=,or
Eo E=~
Eo 15.46 Choose a very small cylindrical gaussian surface with one end inside the conductor. Position the other end parallel to and just outside the surface of the conductor. Since, in static conditions, E = 0 at all points inside a conductor, there is no flux through the inside end cap of the gaussian surface. At all points outside, but very close to, a conductor the electric field is perpendicular to the conducting surface. Thus, it is parallel to the cylindrical side of the gaussian surface and no flux passes through this cylindrical side. thE! total flux through the gaussian surface is then ct> = EA , where A is the crosssectional area of the cylinder as well as the area of the end cap. 24 CHAPTER15 The total charge enclosed by the cylindrical gaussian surface is Q = (f A ,where charge density on the conducting surface. Hence, Gauss's law gives
(fA ~ EA= or E=Eo Eo (f is the v=t' F me = (0.53xW" m)(8.2xlO<I N) 9.11xl031 kg  2.2xlO _I 6 m/s I 15.49 The three contributions to the resultant electric field at the point of interest are shown in the sketch at the right. The magnitude of the resultant field is E =(S.99XI09
R N .m C2 . 2 )[ 4.0xlO9 C 5.0xl09 C 3.0xl09 2 + (2.0 m) 2 + (1.2 m) 2 (2.5 m) c] ER =+24 N/C, or ER =124 N/C in the +x direction I Electric Forces and Electric Fields 25 .15.50 Consider the freebody diagram shown at the right. 1:F =0 => TcosO=mg
y or T= mg cosO x 1:Fr=0 => ~ =TsinO=mgtanO Since Fe = qE, we have mg tan 0 qE=mgtanO, or q==E3 (2.00XIO kg)(9.80 m/s2)tan15.0 1.00xlO
3 q= N/C  5.25 x 10 6 C = 5.25,llC I I  15.51 (a) At a point on the xaxis, the contributions by the two charges to the resultant field have equal magnitudes . keq gIven bEE. 1 = 2 = 2 . Y
r :y
I
I H" a:
a' The components of the resultant field are l....",' ....~.... ~JC~'::~ : b . () x
,'r
" " '''r' <{I
E
~ I I I " qt' , . cosO blr b Smce 2 = 2 = '3 = b r r r (a2 +b2) . . 3/2, the resultant field IS 26 CHAPTER15 (b) Note that the result of part (a) may be written as ER = ke (Q)~/2 where Q = 2q is (a2 + b2) the total charge in the charge distribution generating the field. In the case of a uniformly charged circular ring, consider the ring to consist of a very large number of pairs of charges uniformly spaced around the ring. Bach pair consists of two identical charges located diametrically opposite each other on the ring. The total charge of pair number Us Qi. At a point on the axis of the ring, this pair of charges generates an electric field contribution that is parallel to the axis and . h as magmtu d e Ei = kebQi (a 3/2 2 + b2) The resultant electric field of the ring is the summation of the contributions by all pairs of charges, or where Q = :Qi is the total charge on the ring. 15.52 (a) _ v~V~y ..;.21.0 m.jst 0 ( ay  2(~y) ~ 2(5.00 m) 2 44.1 mls (downward) I Since ay > g , the electrical force must be directed downward, aiding the gravitational force in accelerating the bead. Because the bead is positively charged, the electrical force acting on it is in the direction of the electric field. Thus, the field is directed ! Idownward I. (b) Taking downward as positive, :Fy = qE + mg = may. Therefore, _ (1.00XIO3 kg)[(44.19.80) 1.00xl04 NjC m/s2 J 3.43xl06 C=13.43 pC I I I Electric Forces and Electric Fields 27 15.53 Because of the spherical symmetry of the charge distribution, any electric field present will be radial in direction. If a field does exist at distance R from the center, it is the same as if the net charge located within r S R were concentrated as a point charge at the center of the inner sphere. Charge located at r > R does not contribute to the field at r=R. (a) At r = 1.00 em, IE = 0 I since static electric fields cannot exist within conducting materials. (b) The net charge located at r S 3.00 em is Q = +8.00 pC. Thus, at r = 3.00 em, =
(c) At r (8.99x10 9 ( N .m jC )(8.00XI06 3.00xlO2 m)
2 2 2 C) = I7.99xlO 7 ~'' N/C (outward) I = 4.50 em, IE = 0 I since this is located within conducting materials. (d) The net charge located at r S 7.00 em is Q = + 4.00 pC. Thus, at r = 7.00 em, _ (8.99XI09 N .m2jC2)( 4.00xl06 C) ~17.34X106 2 (7.00XIO mr ~. N/C (outward) 'J. I 28 CHAPTER 15 15.54 The charges on the spheres will be equal in magnitude and opposite in sign. From F = keq2 / r2 , this charge must be The number of electrons transferred is n = 9.. 3 1.05X10 C = 6.59x1015 19C 1.60X10 e The total number of electrons in 100g of silver is N = (47 electrons). (6.02X1023 atoms)( 1 mole )(100 g) = 2.62x1025 atom . mole 107.87 g Thus, the fraction transferred is .!!. = 6.59X10~ = 12:51XlO10 (that is, 2.51 out of every 10 billion).
N 2.62x10 . I 15.55 <1> =EAcos8 =(2.00X104 NfC)[(6.00 m)(3.00 m)]cos10.0o=13.55X105 N .m2/C I 15.56 (a) The downward electrical force acting on the ball is ~ =qE = (2.00 X106 C)(1.00X105 NfC) =0.200 N
The total downward force acting on the ball is then F = ~ +mg =0.200 N+(1.00x103 kg)(9.80 m/s2) =0.210 N Thus, the ball will behave as if it was in a modified gravitational field where the effective freefall acceleration is "g" == 0.210 N =210 m/S2 1.00x 103 kg .m Electric Forces and Electric Fields 29 The period of the pendulum will be T =2tr~ L =2tr 0.500 m =10.307 s "g" 210 S2 ~,~, m/ I
of the total downward force (b) IYes I. The force of gravity is a significant portion
/I acting on the ball. Without gravity, the effective acceleration would be "= Fe =
m g
.. 0.200 N = 200 m/s2 1.00xl03 kg 0.500 m 0 314 m/ 2 =. s 200 s gJ.vmg T = 2 1r a 2.28% difference from the correct value with gravity included. 15.57 The sketch at the right gives a freebody diagram of the positively charged sphere. Here, F1 = ke Iq12/,2 is the attractive force exerted by the negatively charged sphere and F2= qE is exerted by the electric field.
~r ~y x 0 ==> T cos 10 =mg or T = mg
cos 10 :U =0 => F2= F1+T sin 10 or qE= x kel;12 , +mgtanl0 Thus, At equilibrium, the distance between the two spheres is r=2(Lsin100). E= kelql
4(Lsinl00)2
9 +mgtanl0 q C) + (2.0XI03 kg)(9.80 m/s2)tanl0 (5.0xl08 C) E=14.4X105 NfC = (8.99xl0 N.m2/C2)(5.0Xl08 4[(0.100 m)sinl00J or the needed electric field strength is I 30 CHAPTER 15 As shown in the sketch, the. electric field at any point on the xaxis consists of two parts, one due to each of the charges in the dipole.
   . 15.58 :y
I
I /E
q a ! :;
: a
         . I I
I It , E+
I       I"       . :E
I Thus, if x a . 2 2 , this gIves E:::: keq " [4ax] = f4k:qiil "7""" m
qEx
sin 37.0 x 15.59 (a) Consider the freebody diagram for the ball given in the sketch. or T= and 3 2 mg Thus q= (1.00xlO kg)(9.BO m/s ) =_ , Ey + Ex cot 37.0  [5.00 + (3.00) cot 37.0] x 105 N/C = 1.09x 108 C = 110.9 nC I
. (b) From 1:Fx = 0, we found that T qE
x sin 37.0 Hence, T= (1.09xl08 C)(3.00X105 N/C) 3  5.44xlO N sin 37.0 ~__ I I 15.60 (a) At any point on the xaxis in the range 0 < x < 1.00 m, the contributions made to the resultant electric field by the two charges are both in the positive x direction. Thus, it is not possible for these contributions to cancel each other and yield a zero field. Electric Forces and Electric Fields 31 (b) Any point on the xaxis in the range x < 0 is located closer to the larger magnitude charge (q = 5.00 pC) than the smaller magnitude charge (Iql = 4.00 pC) . Thus, the contribution to the resultant electric field by the larger charge will always have a greater magnitude than the contribution made by the smaller charge. It is not possible for these contributions to cancel to give a zero resultant field. (c) H a point is on the xaxis in the region x> 1.00 m, the contributions made by the two charges are in opposite directions. Also, a point in this region is closer to the smaller magnitude charge than it is to the larger charge. Thus, there is a location in this region where the contributions of these charges to the total field will have equal magnitudes and cancel each other. (d) When the contributions by the two charges cancel each other, their magnitudes must be equal. That is,
k (5.00 pC)
e x2. k (4.00 pC) e(x1.00m)2 or x 1.00 m = + {45 x V5 Thus, the resultant field is zero at x= 1.00m 1+9.47m
1~4/5 l . 15.61 We assume that the two spheres have equal charges, so the repulsive force that one exerts on the other has magnitude F. = ke q2 / r2 From Figure P15.6I in the textbook, observe that the distance separating the two spheres is r=3.0 cm+2[(5.0 cm)sinlooJ=4.7 cm=0.047 m From the freebody diagram of one sphere given above, observe that Uy =0 ~ Tcos10o=mg
or T=mg/cos10 and r.F =0 ~ F =TSinIOO=(
r e mg )Sin100=mgtan100 cosl00 .. , 32 CHAPTER 15 or q = mgr2tan10 = (O.015kg)(9.8 m/s2)(0.047m)2tan100'
ke 8.99x109 N .m2/C2 giving q = 8.0 X108 C or q _107 C I I
T=mg/cos8 15.62 Consider the freebody diagram of the rightmost charge given below. LFy =0 => Tcos8=mg or and LFx =0 => F.,=Tsin8=(mg/cos8)sin8=mgtanO But, Thus, If 8=45, m=0.10 kg, and L=0.300 m then or q = 2.0x106 C =12.0,uC I 15.63 (a) When an electron (negative charge) moves distance /1x in the direction of an electric field, the work done on it is W = Fe (/1x)cosO= eE(/1x) cos 180 = eE(/1x) From the workenergy theorem eE(tix)=KE. (Wnet = KEf  KEj) with KEf
17 = 0, we have
N/cl
f l' orE= KE; = 1.60x10 J 11.00X103 e(tix) (1.60XlO19 C)(O.100 m)  f Electric Forcesand ElectricFields 33 (b) The magnitude of the retarding force acting on the electron is Fe =eE, and Newton's second law gives the acceleration as a =  Fe /m = eE/m . Thus, the time required to bring the electron. to rest is t==~=~a eE/m eE or
12(9.11XlO31 kg)(1.60XlO17 J) t=~"=3.37xI08 (1.60XIO19C)(l.OOXI03 N/C) VVa O~2(KEj)/m ~2m(KEj) s= 33.7 ns .  I I (c) After bringing the electron to rest, the electric force continues to act on it causing the electron'to accelerate in the direction opposite to th~ field at ii rate of I I a II == eE (1.60XlO C)(l.OOX10 N/C) . '31 m9.11xlOkg 19 3  ~~ I 1.76xlO 14 m s . I 21 15.64 (a) The acceleration of the protons is downward (in the direction of the field) and I F eE a I =!..=Y m m (1.60xlO C)(720 N/C) 6.90xlO 1.67xlO27 kg 19 10 ms I 2 The time of flight for the proton is twice the time required to reach the peak of the are, or _~ _. R ~i The horizontal distance traveled in this time is ,,:~~ww'**"~~:$,
101( ~I L 34 CHAPTER 15 Thus, if R = 1.27 X 103 m, we must have giving 28 = 73.9 or 28 = 18073.9 = 106.1 . Hence, 8 = 137.0 or 53.00 [
\ (b) The time of flight for each possible angle of projection is: For 8 = 37.0:  2vosin8 _ 2(9550 mjs)sin37.0 ay 6.90x10 10 m /2 s t II I1.66x10
~ 7 s I For 8 = 53.0 : t _2VOSin8_2(9550mjs)Sin53.00_j.. ay 6.90x10 10 m/ s 2 II  7 2.21x10  s I L, ..... ...
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 Winter '07
 BRIANKEATING
 Charge, Current, Electric Fields, Force, Electric charge

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