solucion fisica un_22

7/27/2019 Solucion Fisica Un_22

1/27

22.1: a) C.Nm75.160cos)m(0.250N/C)14( 22 === AE

b) As long as the sheet is flat, its shape does not matter.

ci) The maximum flux occurs at an angle = 0 between the normal and field.cii) The minimum flux occurs at an angle = 90 between the normal and field.

In part i), the paper is oriented to capture the most field lines whereas in ii) thearea is oriented so that it captures no field lines.

22.2: a) nAAE wherecos AEA ===

CmN329.36cosm)(0.1)CN104((back)

CmN329.36cosm)(0.1)CN104(front)(

090cosm)(0.1)CN104(bottom)(

CmN24)9.36(90cosm)(0.1)CN104(right)(

090cosm)(0.1)CN104(top)(

CmN24)93690(cosm)(0.1)CN104(left)(

223

223

23

223

23S

223

66

55

44

33

22

11

===

=+=+=

===

+=+=+=

==+=

===

SS

SS

SS

SS

S

SS .

in

in

kn

jn

kn

jn

b) The total flux through the cube must be zero; any flux entering the cube must alsoleave it.

22.3: a) Given that lengthedge,,DCB AEkjiE

=+= L, and

.BL

.BL

.DL

.CL

.DL

.CL

26

25

24

23

22

21

66

55

44

33

22

11

+===

==+=

+===

+==+=

==+=

===

SS

SS

SS

SS

SS

SS

A

A

A

A

A

A

nEin

nEin

nEkn

nEjn

nEkn

nEjn

b) Total flux = ==6

1 i0

i

22.4: C.Nm16.670cos)m(0.240)CN0.75( 22 === AE


2/27

22.5: a) C.Nm1071.2)2( 25m)(0.400C/m)1000.6(2 06

00=====

lr

rlAE

b) We would get the same flux as in (a) if the cylinders radius was made largerthefield lines must still pass through the surface.

c) If the length was increased to m,800.0=l the flux would increase by a factor of

two: C.Nm105.42

25

=

22.6: a) C.Nm452C)1000.4( 209

011=== qS

b) C.Nm881C)1080.7( 209

022=== qS

c) C.Nm429C)10)80.700.4(()( 209

0213==+= qqS

d) C.Nm723C)10)40.200.4(()( 209

0214=+=+= qqS

e) C.Nm158C)10)40.280.700.4(()( 209

03215=+=++= qqqS

f) All that matters for Gausss law is the total amount of charge enclosed by the

surface, not its distribution within the surface.

22.7: a) C.Nm1007.4C)1060.3( 2506

0 === q

b) C.106.90)CNm780( 92000==== qq

c) No. All that matters is the total charge enclosed by the cube, not the details of

where the charge is located.

22.8: a) No charge enclosed so 0=

b) C.Nm678NmC108.85C1000.6 2

2212

9

0

2 =

==

q

c) C.Nm226NmC108.85

C10)00.600.4( 22212

9

0

21 =

=

+=

qq

22.9: a) Since E

is uniform, the flux through a closed surface must be zero. That is:

===== .0000 1 dVdVd qAE

But because we can choose any volume we

want, must be zero if the integral equals zero.

b) If there is no charge in a region of space, that does NOT mean that the electric fieldis uniform. Consider a closed volume close to, but not including, a point charge. The fielddiverges there, but there is no charge in that region.


3/27

22.10: a) If 0> and uniform, then q inside any closed surface is greater than zero.

>> 00 AE

d and so the electric field cannot be uniform, i.e., since an

arbitrary surface of our choice encloses a non-zero amount of charge, E must depend onposition.

b) However, inside a small bubble of zero density within the material with density ,

the field CAN be uniform. All that is important is that there be zero flux through thesurface of the bubble (since it encloses no charge). (See Exercise 22.61.)

22.11: C.Nm1008.1C)1060.9( 2606

0sides6 === q But the box is

symmetrical, so for one side, the flux is: .CNm1080.1 25side1 =

b) No change. Charge enclosed is the same.

22.12: Since the cube is empty, there is no net charge enclosed in it. The net flux,according to Gausss law, must be zero.

22.13: 0encl QE =

The flux through the sphere depends only on the charge within the sphere.

nC3.19)CmN360( 200encl === Q E

22.14: a) .CN44.7m)(0.550

C)1050.2(

4

1

4

1m)0.1m450.0(

2

10

02

0

=

==+=

r

q

rE

b) 0=E

inside of a conductor or else free charges would move under the influence offorces, violating our electrostatic assumptions (i.e., that charges arent moving).

22.15: a) m.62.1CN614

C)10180.0(

4

1

4

1||

4

1||

6

002

0

=

===

E

q

r

r

q

E

b) As long as we are outside the sphere, the charge enclosed is constant and the sphereacts like a point charge.

22.16:a) C.1056.7)m(0.0610)CN1040.1(/

825

000

===== EAqqEA b) Double the surface area: C.1051.1)m(0.122)CN1040.1( 7250

== q


4/27

22.17: C.1027.3m)(0.160)CN1150(44 9202

041

20

==== ErqEr

q

So the

number of electrons is: .1004.2 10C1060.1

C1027.3

e 19

9

==

n

22.18: Draw a cylindrical Gaussian surface with the line of charge as its axis. Thecylinder has radius 0.400 m and is 0.0200 m long. The electric field is then 840 N/C atevery point on the cylindrical surface and directed perpendicular to the surface. Thus

== )2)(())(( cylinder rLEAEdsE

/CmN42.2m)(0.0200m)(0.400)(2N/C)840( 2==

The field is parallel to the end caps of the cylinder, so for them 0= sE

d . From

Gausss law:

C1074.3

)C

mN2.42()

mN

C10854.8(

10

2

2

212

0

=

== Eq

22.19:

rE

2

1

0=


5/27

22.20: a) For points outside a uniform spherical charge distribution, all the charge canbe considered to be concentrated at the center of the sphere. The field outside the sphereis thus inversely proportional to the square of the distance from the center. In this case:

CN53cm0.600

cm0.200)CN480(

2

=

=E

b) For points outside a long cylindrically symmetrical charge distribution, the field isidentical to that of a long line of charge:

,2

0rE=

that is, inversely proportional to the distance from the axis of the cylinder. In this case

CN160cm0.600

cm0.200)CN480( =

=E

c) The field of an infinite sheet of charge is ;2/ 0E= i.e., it is independent of the

distance from the sheet. Thus in this case .CN480=E

22.21: Outside each sphere the electric field is the same as if all the charge of the sphere

were at its center, and the point where we are to calculate E

is outside both spheres.

21 andEE

are both toward the sphere with negative charge.

sphere.chargednegativelythetoward,CN1006.8

CN10471.5m)(0.250

C1080.3||

CN10591.2m)(0.250

C1080.1||

521

5

2

6

2

2

22

5

2

6

2

1

11

=+=

=

==

=

==

EEE

kr

qkE

kr

qkE


6/27

22.22: For points outside the sphere, the field is identical to that of a point charge of thesame total magnitude located at the center of the sphere. The total charge is given bycharge density volume:

C1060.1m)150.0)(3

4)(mCn50.7( 1033 == q

a) The field just outside the sphere is

CN4.42m)(0.150

C)10(1.06)/CmN109(

4 2

10229

20

=

==

r

qE

b) At a distance of 0.300 m from the center (double the spheres radius) the field willbe 1/4 as strong: 10.6 CN

c) Inside the sphere, only the charge inside the radius in question affects the field. Inthis case, since the radius is half the spheres radius, 1/8 of the total charge contributes tothe field:

CN2.21m)(0.075

C)1006.1()8/1()C/mN109(2

10229

==

E

22.23: The point is inside the sphere, so 3/ RkQrE= (Example 22.9)

nC2.10)m100.0(

m)(0.220)CN950( 33===

kkr

ERQ

22.24: a) Positive charge is attracted to the inner surface of the conductor by the chargein the cavity. Its magnitude is the same as the cavity charge: nC,00.6inner +=q since

0=E inside a conductor.b) On the outer surface the charge is a combination of the net charge on the conductor

and the charge left behind when the nC00.6+ moved to the inner surface:

nC.1.00nC6.00nC00.5innertotouterouterinnertot ===+= qqqqqq

22.25: 32 SandS enclose no charge, so the flux is zero, and electric field outside the

plates is zero. For between the plates,1

S shows that: .000

EAqEA ===


7/27

22.26: a) At a distance of 0.1 mm from the center, the sheet appears infinite, so:

=

====

.CN662m)800.0(2

C1050.7

22

20

9

00 A

qE

qAEdAE

b) At a distance of 100 m from the center, the sheet looks like a point, so:

.CN1075.6m)(100

C)1050.7(

4

1

4

132

9

02

0

=

= r

q

E

c) There would be no difference if the sheet was a conductor. The charge wouldautomatically spread out evenly over both faces, giving it half the charge density on anyas the insulator

00 2::).(

cE == near one face. Unlike a conductor, the insulatoris the

charge density in some sense. Thus one shouldnt think of the charge as spreading overeach face for an insulator. Far away, they both look like points with the same charge.

22.27: a) .2

2

RL

Q

RL

Q

A

Q ====

b) ==== .2

)2(000 r

RE

RL

QrLEdAE

c) But from (a), ,so,202

rER == same as an infinite line of charge.


8/27

22.28: All the s' are absolute values.

(a) at0

1

0

4

0

3

0

2

2222:

AEA ++=

left.thetoCN1082.2

)mC6mC4mC2mC5(21

)(2

1

5

2222

0

1432

0

=

++=

++=

EA

(b)

left.thetoCN1095.3

)mC5mC4mC2mC6(2

1

)(2

1

2222

5

2222

0

2431

00

2

0

4

0

3

0

1

=

++=

++=++=

EB

(c)

leftthetoCN1069.1

)mC6mC4mC2mC5(2

1

)(2

1

2222

5

2222

0

1432

00

1

0

4

0

3

0

2

=

+=

+=+=

EC

22.29: a) Gausss law says +Q on inner surface, so 0=E inside metal.b) The outside surface of the sphere is grounded, so no excess charge.c) Consider a Gaussian sphere with the Q charge at its center and radius less than

the inner radius of the metal. This sphere encloses net charge Q so there is an electricfield flux through it; there is electric field in the cavity.

d) In an electrostatic situation 0=E inside a conductor. A Gaussian sphere withthe Q charge at its center and radius greater than the outer radius of the metal encloses

zero net charge (the Q charge and the Q+ on the inner surface of the metal) so there isno flux through it and 0=E outside the metal.

e) No, 0=E there. Yes, the charge has been shielded by the groundedconductor. There is nothing like positive and negative mass (the gravity force is alwaysattractive), so this cannot be done for gravity.


9/27

22.30: Given ,)m)CN(00.3()m)CN(00.5( kiE zx +=

edge length

,m300.0=L ,m300.0=L and .011 1

=== ASs nEjn

==+= )CN(00.3(21 2A

SS nEkn

== zz )m)CN(27.0()m300.0)(m 2

.m)CN(081.0)m300.0)(m)CN(27.0( 2=

.033 3

==+= ASS nEjn

).0(0)m)CN(27.0(44 4

===== zzASS nEkn

xxASS

)m)N/C(45.0()m300.0)(m)CN(00.5( 25 55 ===+= nEin

).m)CN(135.0()m300.0)(m)N/C(45.0( 2==

).0(0)m)CN(45.0(66 6

==+=== xxASS nEin

b) Total flux:

CNm054.0m)CN()135.0081.0( 2252 ==+=

C1078.4 130== q

22.31: a)

b) Imagine a charge q at the center of a cube of edge length 2L. Then: ./ 0q=

Here the square is one 24th of the surface area of the imaginary cube, so it intercepts 1/24of the flux. That is, .24 0q=

22.32: a) .CmN750)m0.6)(CN125( 22 === EA

b) Since the field is parallel to the surface, .0= c) Choose the Gaussian surface to equal the volumes surface. Then: 750

EA= ,CN577)750C1040.2( 08

m0.6

10 2 =+=

Eq in the positive x -direction.

Since 0


10/27

22.33: To find the charge enclosed, we need the flux through the parallelepiped:

CmN5.3760cos)CN1050.2)(m0600.0)(m0500.0(60cos 2411 === AE

CmN10560cos)CN1000.7)(m0600.0)(m0500.0(120cos 2422 === AE

So the total flux is ,CmN5.67CmN)1055.37( 2221 ==+= and

.C1097.5)CmN5.67(10

0

2

0

=== q b) There must be a net charge (negative) in the parallelepiped since there is a netflux flowing into the surface. Also, there must be an external field or all lines would pointtoward the slab.

22.34:

The particle feels no force where the net electric field is zero. The fields cancancel only in regions A and B.

sheetline EE =

00 22

r=

cm16m16.0)C/m100(

C/m50

2====

r

The fields cancel 16 cm from the line in regions A andB.


11/27

22.35:

The electric field 1E

of the sheet of charge is toward the sheet, so the electric

field 2E

of the sphere must be away from the sheet. This is true above the center of the

sphere. Let rbe the distance above the center of the sphere for the point where theelectric field is zero.

32

00

121

4

1

2

so

R

rQ

EE ==

m097.0C10900.0

)m120.0)(C/m1000.8(229

329

2

31 =

==

Q

Rr


12/27


13/27

22.38: a) ,, 204

1

r

Q

Ear =< since the charge enclosed is Q.

,0, = , since the total enclosed charge is 2Q.

b) The surface charge density on inner surface: 24aQ = .

c) The surface charge density on the outer surface: .242

b

Q =

d)

e)


14/27

22.39: a)(i) ,0, =< Ear since 0=Q

(ii) ,0, =


15/27

22.41: a)(i) ,0, =< Ear since charge enclosed is zero.

(ii) ,0, =


16/27

22.43: a) The sphere acts as a point charge on an external charge, so:

,204

1

r

qQ

qEF == radially inward.

(b) If the point charge was inside the sphere (where there is no electric field) itwould feel zero force.

22.44: a)

=

=

+=

333

343

34inner

1

4

3

ab

q

ab

q

VV

q

ab

=

=

=

333

343

34outer

1

4

3

cd

q

cd

q

VV

q

cd

b) (i) ==< .00 Edar AE

(ii) inner33

0

2inner

0

)(

3

44

1 ar

rEdV

dbra ==


17/27

22.45: a) ,2

4

1,

0 rEbra

= radially outward, since again the charge enclosed is the

same as in part (a).

c)

d) The inner and outer surfaces of the outer cylinder must have the same amountof charge on them: .and, outerinnerinner === ll

22.46: a) (i) .2

)2(,000 r

E

l

qrlEar ===<

(ii) ,bra

b) (i) Inner charge per unit length is . (ii) Outer charge per length is .2+


18/27

22.47: a) (i) ,ErlEarr

l

q

000 2)2(, ===< radially outward.

(ii) ,bra there is no net charge enclosed, so the electric field is zero.

b) (i) Inner charge per unit length is . (ii) Outer charge per length is ZERO.

22.48: a) ,)2(,00

2

0 2

r

lr

qErlERr ===< radially outward.

b) .)2(,and, 22

22

00

2

0

2

0 rk

rr

R

lR

q ErlERRr ======>

c) .Rr= the electric field for BOTH regions is ,02

RE= so they are consistent.

d)

22.49: a) The conductor has the surface charge density on BOTH sides, so it has twicethe enclosed charge and twice the electric field.

b) We have a conductor with surface charge density on both sides. Thus theelectric field outside the plate is .)2()2( 00 EAAE === To find the field

inside the conductor use a Gaussian surface that has one face inside the conductor, andone outside.Then:

.00but)( inin0out0inout ====+= EAEEAAEAE


19/27

22.50: a) If the nucleus is a uniform positively charged sphere, it is only at its verycenter where forces on a charge would balance or cancel

b)3

03

3

0

2

0 44

R

erE

R

r

erE

qd =

=== AE

.4

13

2

0 R

re

qEF ==

So from the simple harmonic motion equation:

.4

1

2

1

4

1

4

13

2

03

2

03

2

0

2

mR

e

f

mR

e

R

re

rmF ====

c) If3

2

0

14

4

1

2

1Hz1057.4

mR

e

f

==

m.1013.3Hz)1057.4)(kg1011.9(4

C)1060.1(

4

1 103

214312

219

0

=

=

R

1Thompsonactual

rr

d) If Rr> then the electron would still oscillate but not undergo simple

harmonic motion, because for ,1, 2rFRr > and is not linear.

22.51: The electrons are separated by a distance ,2d and the amount of the positive

nucleuss charge that is within radius d is all that exerts a force on the electron. So:

.2/8/2)2(

332nucleus2

2

3 RdRdkeFd

keF

R

de =====


20/27

22.52: ===

raxar

dxexa

QQdddrre

a

QQdVQrQ

0

/22

30

2/2

30

004

sin)(a)

].1)/(2)/(2[)222(4

)( 02

0/222

33

0

0 ++==

ararQerrea

QeQrQ

arrr

0.)(,ifNote rQr

b) The electric field is radially outward, and has magnitude:

).1)(2)(2( 02

02

/2 0

++=

ararr

kQeE

ar

22.53: a) At N.94,2 215219

02Fee

0 m)101.7(4

C)106.1)(82(4

144

1e =====

R

qq

EqFRr

So: .m/s101.0kg1011.9N94 23231 === mFa

b) .m/s101.44,At 232(a) === aaRr

c) At8

1

8

1 ()82(,2 eQRr == because the charge enclosed goes like 3r ) so with the

radius decreasing by 2, the acceleration from the change in radius goes up by ,4)2( 2 =

but the charge decreased by 8, so .m/s101.2 232)b(84 == aa

d) At .0so,0,0 === FQr


21/27

22.54: a) The electric field of the slab must be zero by symmetry. There is no preferreddirection in the y -z plane, so the electric field can only point in the x -direction. But at

the origin in the x -direction, neither the positive nor negative directions should besingled out as special, and so the field must be zero.

b) Use a Gaussian surface that has one face of area A on in the y -z plane at

,0=x and the other face at a general value .x Then:

,:000

xE

Ax

QEAdx encl ====

with direction given by .||i

x

x

Note that E is zero at .0=x Now outside the slab, the enclosed charge is constant with :x

,:000

encl

dE

Ad

QEAdx ====

again with direction given by .||i

x

x

22.55: a) Again, E is zero at 0=x , by symmetry arguments.

b) i||

in,33

'':2

0

30

20

30

0

2

20

0

0

encl

x

x

d

xE

d

Axdxx

d

A

QEAdx

x

===== direction.

=====d

o

x

xin

dE

Addxx

d

A

QEAdx

0 00

02

20

0

0

encl ||

,33

'': i direction.

22.56: a) We could place two charges Q+ on either side of the charge :q+

b) In order for the charge to be stable, the electric field in a neighborhood around itmust always point back to the equilibrium position.

c) Ifq is moved to infinity and we require there to be an electric field always

pointing in to the region where q had been, we could draw a small Gaussian surface

there. We would find that we need a negative flux into the surface. That is, there has to be

a negative charge in that region. However, there is none, and so we cannot get such astable equilibrium.

d) For a negative charge to be in stable equilibrium, we need the electric field toalways point away from the charge position. The argument in (c) carries through again,this time inferring that a positive charge must be in the space where the negative chargewas if stable equilibrium is to be attained.


22/27

22.57: a) The total charge: ]/[4)/1(40

3

0 0

220 RdrrdrrdrrRrq

RR R

==

.3

12

4

12

4]43[4

3

30

333

0 Q

R

QRRRRq ====

b) ,Rr all the charge Q is enclosed, and: ,/)4( 204

10

2

r

Q

EQrE ===

the same as a point charge.

c) ).then, 33 Rrq(Q(r)Rr =

( ).4)1(4)(Also, 4302043

RrrdrrRrrQ ==

=

=

=

R

r

R

rkQ

R

r

R

r

r

kQ

R

r

R

r

r

kQrE 34

34

4

1

3

112)(

34

4

3

3

24

4

3

3

2

d)

e)22maxmax43 3

4)24(

3

2So.

3

20

64)(0

R

kQ

R

kQERrr

R

kQ

R

kQRr

r

E=====


23/27

22.58: a)

=

==

RR

R

drrR

drrdrrR

rdrrrQ

00

32

0

02

02

03

44

3

414)(4

= 043

43

443

0 =

QR

RR

b) 00,0

encl === Eq

dRr AE

c)

==r r r

rdrR

rdr

rErdrr

dRr

0 0 0

32

0

022

0 3

444)(

4, AE

=

=

R

rr

R

rr

r

E 1

333

1

0

043

2

0

0

d)

e)2

03

2

30 max

0

max0

0

0 RrR

r

r

E===

0

0

0

0

122

11

232

RR

RrE =

=

22.59: a) === .4sin

2

2

Gmr

dddrrGmdg

Ag

b) For any closed surface, mass OUTSIDE the surface contributes zero to the fluxpassing through the surface. Thus the formula above holds for any situation where m isthe mass enclosed by the Gaussian surface.

That is: == .encl4GMdg Ag


24/27

22.60: a) .masspointaforassametheiswhich,442

2

r

GMgGMrgg ===

b) Inside a hollow shell, the .0so,0encl == gM

c) Inside a uniform spherical mass:

,444 33

3

encl2R

GMrgR

rMGGMrgg =

===

which is linear in .r

22.61: a) For a sphere NOT at the coordinate origin:

,33

44

0

3

00

encl2

rE

r

QEr

=

==== brr

in the direction.-r

.3

)(

0

brE

=

b) The electric field inside a hole in a charged insulating sphere is:

.33

)(

3000

(a)spherehole

bbrrEEE

=

==

Note that E

is uniform.

22.62: Using the technique of22.61, we first find the field of a cylinder off-axis, then theelectric field in a hole in a cylinder is the difference between two electric fieldsthat of asolid cylinder on-axis, and one off-axis.

.2

)(

22

00

2

00

encl

rErl

QElr

brEbrr

=

=====

uniform.isthatNote.

22

)(

2 000abovecylinderhole E

bbrrEEE

=

==


25/27

22.63: a) :0=x no field contribution from the sphere centered at the origin, and theother sphere produces a point-like field:

iiE 44

1)2(4

1)0(

20

20 R

Q

R

Q

x ===

.

b) :2Rx = the sphere at the origin provides the field of a point charge of charge

8Qq = since only one-eighth of the charges volume is included. So:

.184

1)9/42/1(4

1)23()2(

)8(

4

1)2(

20

20

220

iiiER

Q

R

Q

R

Q

R

Q

Rx ==

==

c) :Rx = the two electric fields cancel, so .0=E

d) :3Rx = now both spheres contribute fields pointing to the right:

.9

10

4

1)3(4

1)3(

20

220

iiER

Q

R

Q

R

Q

Rx =

+==

22.64: (See Problem 22.63 with QQ for terms associated with right sphere)

a) iE 44

1)0(

20 R

Q

x +==

b) iiiE 18

17

4

19

4

24

1)23()2(

)8(

4

1

2 2022

022

0 R

Q

R

Q

R

Q

R

Q

R

Q

Rx =

+=

+=

=

c) iiE 2

4

1)(

20

220 R

Q

R

Q

R

Q

Rx =

+==

d) iiiE 9

8

4

194

1)3(4

1)3(

2

0

22

0

22

0

R

Q

R

Q

R

Q

R

Q

R

Q

Rx

=

=

==


26/27

22.65: a) The charge enclosed:

.5

8

24

1524

11

4

)16(

3

)8(8

)/()2(4and,63

)2(4where,

3

3

34433

2/

320

33

0

R

Q

RQ

R

R

RRRR

drRrrQRR

QQQQR

Rii

==

=

=

===+=

b) .15

8

33

44:2

3000

32

R

Qr

rE

rrERr ====

enclosed.ischargeallsince,4

:

).1)(48)(64(15

)1)(48)(64()4(24

4

)16(

3

)8(8

14:2

20

43

2

43

20

3

4433

00

2

r

QERr

RrRrr

kQRrRr

r

RE

R

RrRr

QrERrR i

=

==

+==

c) .267.015

4)15/4(===

Q

Q

Q

Qi

d) ,:2/ 3015

8reEFRr

R

eQ== so the restoring force depends upon displacement to

the first power, and we have simple harmonic motion.

e) .8

152

2,

15

8,

15

8,

30

30

30 eQ

mR

T

mR

eQ

m

k

R

eQkkrF e

ee

======

f) If the amplitude of oscillation is greater than ,2/R the force is no longer linear in,r and is thus no longer simple harmonic.


27/27

22.66: a) Charge enclosed:

.233

480

480

233

120

47

32

3,Therefore

.120

47

160

31

24

74))(1(4and

.32

3

164

16

2

34where

3

33

2/

33220

343

2/

00

R

QRRQ

RRdrrRrQ

RR

R

dr

R

rQQQQ

R

R

R

ii

==

+=

=

==

===+=

b) ===

==r

R

Qr

R

rE

R

rrd

R

r

rERr

0 40

2

0

2

0

43

0

2 .233

180

16

6

2

3

2

344:2

enclosed.ischargeallsince,4

:

.1920

23

5

1

3

1

233

480

480

17

5

1

3

14

4

128

3

1605243

4

))/(1(4

4:2

20

53

20

53

0

3

0

33

2

533

00

2/

22

00

2

r

QERr

R

r

R

r

r

QE

R

r

R

r

R

RR

R

rRr

Q

rdrRr

QrERrR

i

r

R

i

=

=

+

=

++=

+==

c) The fraction ofQ between :2 RrR

.807.0233

480

120

47==

Q

Qo

d) ,)2/( 204233

180

R

QRrE == using either of the electric field expressions above,

evaluated at .2/Rr= e) The force an electron would feel never is proportional to r which is necessary for

simple harmonic oscillations. It is oscillatory since the force is always attractive, but ithas the wrong power of r to be simple harmonic.

solucion fisica un_22

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