SM chapter 25 - Solucionario capitulo 25 Serway 7ma edición PDF

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Solucionario capitulo 25 Serway 7ma edición ...

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25 Electric Potential ANSWERS TO QUESTIONS

CHAPTER OUTLINE 25.1 25.2 25.3 25.4

25.5 25.6 25.7 25.8

Electric Potential and Potential Difference Potential Difference in a Uniform Electric Field Electric Potential and Potential Energy Due to Point Charges Obtaining the Value of the Electric Field from the Electric Potential Electric Potential Due to Continuous Charge Distributions Electric Potential Due to a Charged Conductor The Millikan Oil Drop Experiment Application of Electrostatics

Q25.1

When one object B with electric charge is immersed in the electric field of another charge or charges A, the system possesses electric potential energy. The energy can be measured by seeing how much work the field does on the charge B as it moves to a reference location. We choose not to visualize A’s effect on B as an action-at-adistance, but as the result of a two-step process: Charge A creates electric potential throughout the surrounding space. Then the potential acts on B to inject the system with energy.

*Q25.2 (i) The particle feels an electric force in the negative x direction. An outside agent pushes it uphill against this force, increasing the potential energy. Answer (a). (ii) The potential decreases in the direction of the electric field. Answer (c). *Q25.3

The potential is decreasing toward the bottom of the page, so the electric field is downward. Answer (f ).

*Q25.4

(i) At points off the x axis the electric field has a nonzero y component. At points on the negative x axis the field is to the right and positive. At points to the right of x = 500 mm the field is to the left and nonzero. The field is zero at one point between x = 250 mm and x = 500 mm. Answer (b). (ii) The electric potential is negative at this and at all points. Answer (c). (iii) Answer (d). (iv) Answer (d).

Q25.5

To move like charges together from an infinite separation, at which the potential energy of the system of two charges is zero, requires work to be done on the system by an outside agent. Hence energy is stored, and potential energy is positive. As charges with opposite signs move together from an infinite separation, energy is released, and the potential energy of the set of charges becomes negative.

Q25.6

(a) The equipotential surfaces are nesting coaxial cylinders around an infinite line of charge. (b) The equipotential surfaces are nesting concentric spheres around a uniformly charged sphere.

*Q25.7

Answer (b). The potential could have any value.

*Q25.8

The same charges at the same distance away create the same contribution to the total potential. Answer (b).

49

50

Chapter 25

*Q25.9 The change in kinetic energy is the negative of the change in electric potential energy, so we work out −q∆V = −q(Vf − Vi) in each case. (a) −(−e)(60 V − 40 V) = +20 eV (b) −(−e)(20 V − 40 V) = −20 eV (c) −(e)(20 V − 40 V) = +20 eV (d) −(e)(10 V − 40 V) = +30 eV (e) −(−2e)(50 V − 40 V) = +20 eV (f) −(−2e)(60 V − 40 V) = +40 eV With also (g) 0 and (h) +10 eV, the ranking is f > d > c = e = a > h > g > b. Q25.10 The main factor is the radius of the dome. One often overlooked aspect is also the humidity of the air—drier air has a larger dielectric breakdown strength, resulting in a higher attainable electric potential. If other grounded objects are nearby, the maximum potential might be reduced. *Q25.11 (i) The two spheres come to the same potential, so q /R is the same for both. With twice the radius, B has twice the charge. Answer (d). (ii) All the charge runs away from itself to the outer surface of B. Answer (a). Q25.12 The grounding wire can be touched equally well to any point on the sphere. Electrons will drain away into the ground and the sphere will be left positively charged. The ground, wire, and sphere are all conducting. They together form an equipotential volume at zero volts during the contact. However close the grounding wire is to the negative charge, electrons have no difficulty in moving within the metal through the grounding wire to ground. The ground can act as an infinite source or sink of electrons. In this case, it is an electron sink.

SOLUTIONS TO PROBLEMS Section 25.1 P25.1

(a)

Electric Potential and Potential Difference Energy of the proton-field system is conserved as the proton moves from high to low potential, which can be defined for this problem as moving from 120 V down to 0 V. K i + Ui + ∆E mech = K f + U f

0 + qV + 0 =

1 2 mv p + 0 2

(1. 60 × 10

1J ⎞ 1 27 2 = C ) (120 V ) ⎛ . × − kg ) v p ⎝ 1 V ⋅ C ⎠ 2 (1 67 10

−19

v p = 1.52 × 105 m s (b)

The electron will gain speed in moving the other way, from Vi = 0 to Vf = 120 V:

K i + Ui + ∆E mech = K f + U f 0+ 0+ 0= 0=

1 2 m ve + qV 2

1 ( 9.11 × 10 −31 kg ) ve2 + ( −1.60 × 10 −19 C) (120 J C ) 2

6 ve = 6.49 × 10 m s

P25.2

∆V = −14.0 V ∆V =

W , Q

and

Q = − N A e = −( 6. 02 × 1023 ) (1. 60 × 10 −19 ) = − 9. 63 × 10 4 C

so

W = Q ∆V = (− 9.63× 10 4 C) ( −14.0 J C) = 1.35 MJ

Electric Potential

Section 25.2

51

Potential Difference in a Uniform Electric Field 3 ∆V 25 .0 × 10 J C = 1. 67 × 10 6 N C = 1.67 MN C = d 1. 50 × 10−2 m

P25.3

E=

P25.4

      VB − VA = − ∫ E ⋅ d s = − ∫ E ⋅ d s − ∫ E ⋅ d s

B

C

A

A

B

C

0.500

VB − VA = ( − E cos180° )

∫

0.400

dy − ( E cos 90 .0° )

−0.300

∫

dx

−0.2 00

VB − VA = ( 325) ( 0.800) = +260 V

FIG. P25.4

P25.5

)

(

)

1 1 m v 2f − vi2 = − ( 9. 11× 10−31 kg ⎡( 1. 40 × 10 5 m s ⎣ 2 2 −18 = 6.23 × 10 J

∆U = −

) − ( 3.70 × 10 2

6

)

2 ms ⎤ ⎦

+6.23 ×10− 18 = ( −1.60 ×10 − 19 ) ∆V

∆U = q ∆V :

∆V = −38.9 V. The origin is at highest potentiial. P25.6

Assume the opposite. Then at some point A on some equipotential surface the electric field has a nonzero component E p in the plane of the surface. Let a test charge start from point A and move B   some distance on the surface in the direction of the field component. Then ∆V = − ∫ E ⋅d s is A

nonzero. The electric potential charges across the surface and it is not an equipotential surface. The contradiction shows that our assumption is false, thatE p = 0, and that the field is perpendicular to the equipotential surface. P25.7

(a)

Arbitrarily choose V = 0 at 0. Then at other points V = − Ex

and

U e = QV = −QEx

Between the endpoints of the motion,

( K + Us + Ue )i = ( K + Us + Ue ) f 0 +0 +0 = 0+ (b)

1 2 kx − QExmax 2 max

FIG. P25.7

so

At equilibrium,

∑F

x

= −Fs + Fe = 0

or

kx = QE

So the equilibrium position is at x = (c)

x max

2QE = k

QE . k

The block’s equation of motion is

∑F

QE , k

QE , k

Let

x′ = x −

or

x = x′ +

x

= − kx + QE = m

d2 x . dt 2

continued on next page

52

Chapter 25

so the equation of motion becomes: d 2( x + QE k ) QE ⎞ ⎛ + QE = m −k x ′ + , ⎝ dt 2 k ⎠

or

2 d x′ ⎛ k⎞ =− 2 ⎝m ⎠ x′ dt

This is the equation for simple harmonic motion ax′ = − ω2 x′

(d)

with

ω=

The period of the motion is then

T=

2π m = 2π k ω

( K + Us + Ue )i + ∆ Emech = ( K + Us + Ue ) f 0 + 0 + 0 − µk mgx max = 0 +

1 2 kx − QEx max 2 max

2( QE − µ k mg ) k

x max = P25.8

k m

Arbitrarily take V = 0 at point P. Then the potential at the original position of the charge is   − E ⋅ s = − EL cos θ. At the final point a, V = −EL. Because the table is frictionless we have ( K + U )i = ( K + U ) f 1 0 − qEL cos θ = m v2 − qEL 2 v=

P25.9

2qEL (1 − cosθ ) = m

2( 2 .00 × 10 −6 C )( 300 N C )(1.50 m )( 1 − cos 60. 0°) = 0.300 m s 0. 010 0 kg

Arbitrarily take V = 0 at the initial point. Then at distance d downfield, where L is the rod length, V = −Ed and U e = − λLEd . (a)

( K + U) i = ( K + U ) f 1 0 + 0 = µL v2 − λ LEd 2 v=

(b)

Section 25.3 P25.10

2λ Ed = µ

2( 40. 0 × 10−6 C m ) (100 N C ) ( 2. 00 m )

( 0.100

kg m )

= 0.400 m s

The same. Each bit of the rod feels a force of the same size as before.

Electric Potential and Potential Energy Due to Point Charges

(a)

Since the charges are equal and placed symmetrically, F = 0 .

(b)

Since F = qE = 0, E = 0 .

(c)

V = 2 ke

−6 q C⎞ 2 2 ⎛ 2 .00 × 10 9 = 2( 8.99 × 10 N ⋅m C ) ⎜ ⎝ 0.800 m ⎟⎠ r

4 V = 4. 50 × 10 V = 45.0 kV

FIG. P25.10

Electric Potential

P25.11

(a)

53

The potential at 1.00 cm is

−19 9 2 2 q (8.99 × 10 N ⋅ m C )(1.60 × 10 C ) = = 1 .44 × 10 − 7 V − r 1.00 × 10 2 m (b) The potential at 2.00 cm is

V1 = k e

V2 = ke

−19 2 2 9 q (8.99 × 10 N ⋅ m C ) (1.60 ×10 C ) = = 0. 719 ×10− 7 V − r 2.00 × 10 2 m

Thus, the difference in potential between the two points is ∆V = V2 − V1 = − 7.19 × 10− 8 V . (c)

The approach is the same as above except the charge is − 1.60 × 10 −19 C. This changes the sign of each answer, with its magnitude remaining the same. That is, the potential at 1.00 cm is −1.44 × 10 −7 V . The potential at 2.00 cm is − 0.719 × 10− 7 V, so

P25.12

(a)

Ex =

k eq 2 ke q1 + 2 = 0 2 − ( x x 2. 00 )

⎛ +q −2 q ⎞ E x =k e ⎜ 2 + =0 ⎝x ( x − 2. 00) 2 ⎟⎠

becomes 2qx 2 = q ( x − 2. 00)

Dividing by ke ,

8 ∆V = V2 − V1 = 7.19 × 10− V

2

2 x + 4.00x − 4.00 = 0

− 4.00 ± 16.0 + 16.0 = − 4.83 m 2 (Note that the positive root does not correspond to a physically valid situation.) Therefore E = 0

(b)

V=

when

kq ke q1 + e 2 =0 x 2. 00 − x

or

P25.13

(a)

(b)

when

x = 0.667 m

For x < 0

x = −2.00 m

and

− 2q q = x 2 −x

E=

Q 4π ∈0 r 2

V=

Q 4 π ∈0 r

r=

3 000 V V = = 6.00 m E 500 V m

V = − 3 000 V = Q=

⎛ +q − 2q ⎞ = V = ke ⎝ 0 x 2 .00 − x ⎠ 2qx = q ( 2 .00 − x )

Again solving for x, For 0 ≤ x ≤ 2.00 V = 0

x=

Q 4π ∈0 ( 6. 00 m )

− 3 000 V ( 6. 00 m ) = −2. 00 µ C ( 8.99 × 109 V ⋅ m C)

54

P25.14

Chapter 25

U=

(a)

qQ ( 5.00 × 10 = 4 π ∈0 r

−9

C )( − 3.00 × 10

−9

C)( 8.99 × 109 V ⋅ m C)

(0.350 m)

= −3.86 ×10 −7 J

−7 The minus sign means it takes 3.86 × 10 J to pull the two charges apart from 35 cm to a much larger separation.

V=

(b)

=

Q2 Q1 + π π 4 ∈0 r1 4 ∈0 r 2

(5 .00 ×10

−9

C) (8 .99 × 10 9 V ⋅m C ) 0.175 m

+

(−3 .00 × 10

−9

C )(8 .99 ×10 9 V ⋅m C ) 0. 175 m

V = 103 V P25.15

V = ∑k i

qi ri

1 ⎤ 1 ⎡ −1 + V = (8 .99 × 109 ) (7 .00 × 10 −6 ) ⎢ − 010 0 0 038 7 ⎥⎦ 0 010 0 0 . . . ⎣ V = − 1.10 × 10 V = − 11.0 MV 7

FIG. P25.15

P25.16

(a)

V=

ke q1 k eq 2 ⎛ k q⎞ + =2 e ⎠ ⎝ r r1 r2

⎛ ( 8. 99 × 10 9 N ⋅ m 2 C 2 ) ( 2. 00 × 10−6 C ) ⎞ V = 2⎜ ⎟ (1.00 m)2 + (0 .500 m)2 ⎝ ⎠ V = 3. 22 × 104 V = 32. 2 kV (b)

P25.17

FIG. P25.16

6 4 2 U = qV = ( −3.00 ×10− C ) (3.22 × 10 J C ) = − 9.65 × 10− J

⎛ 1 ⎞ ⎛ q1 q2 q3 ⎞ U e = q 4V1 + q 4V2 + q 4V 3 = q 4 ⎜ ⎜ + + r ⎟ ⎝ 4π ∈0 ⎟⎠ ⎝ r1 r2 3 ⎠ ⎛ ⎞ 2 1 1 1 + U e = ( 10. 0 × 10−6 C ) ( 8. 99 × 109 N ⋅ m2 C2 ) ⎜ + ⎟ 2 2 ( 0 .600 m ) + ( 0 .150 m ) ⎠ ⎝ 0 .600 m 0 .150 m U e = 8. 95 J

Electric Potential

*P25.18 (a)

The first expression, with distances squared, describes an electric field. The second expression describes an electric potential. Then a positive 7 nC charge is 7 cm from the origin. To create f

ield that is to the left and downward, it must be in the first quadrant, with position

vector 7 cm at 70° .A negative 8 nC charge

P25.19

55

3 cm from the origin creates an upward

(b)

electric field at the origin, so it must be at 3 cm at 90° . We evaluate the given expressions:  E = −4 .39 kN C ˆi +67 .8 kN C ˆj V = −1 .50 kV   9 3 F = q E = −16 ×10 − C −4.39ˆi + 67.8ˆj 10 N C = 7. 03ˆi − 109ˆj × 10 −5 N

(c)

U e = qV = −16 × 10 −9 C( −1 .50 × 10 3 J C) = +2 .40 × 10 −5 J

(

)

(

)

U = U1 + U 2 + U 3 + U 4 U = 0 + U12 + (U13 + U 23 ) + (U 14 + U 24 + U 34 ) U = 0+ U=

2 k eQ 2 ke Q 2 ⎛ 1 ⎞ + k eQ ⎛ 1 + 1 + 1⎞ 1 + + ⎟ ⎜ ⎜ 2 ⎠⎟ s s ⎝ 2 ⎠ s ⎝

ke Q 2 ⎛ k eQ 2 2⎞ ⎜⎝ 4 + 2 ⎟⎠ = 5.41 s s

FIG. P25.19

2 ⎛ We can visualize the term ⎜ 4 + ⎟⎞ as arising directly from the 4 side pairs and 2 face diagonal ⎝ 2⎠ pairs. P25.20

Each charge creates equal potential at the center. The total potential is: ⎡ k ( −q) ⎤ 5k eq V = 5⎢ e ⎥= − R ⎣ R ⎦

P25.21

(a)

Each charge separately creates positive potential everywhere. The total potential produced by the three charges together is then the sum of three positive terms. There is no point , at a finite distance from the charges, at which this total potential is zero.

(b)

V=

2 k eq ke q k eq + = a a a

56

P25.22

Chapter 25

(a)

V ( x) =

k (+Q) k e ( +Q) k eQ1 k eQ 2 + = e2 + 2 2 r1 r2 x +a x + (− a) 2

V ( x) =

2 ke Q x 2 +a2

V (x ) = ( k eQ a)

(b)

V ( y) =

=

⎛ ke Q ⎜ a ⎜⎝

2

( x a )2

⎞ ⎟ + 1 ⎟⎠

2

(x a )2 + 1

ke Q1 keQ 2 ke ( +Q) ke (− Q) + + = r1 r2 y− a y+ a FIG. P25.22(a)

1 ⎞ k Q⎛ 1 − V ( y) = e ⎜ a ⎝ y a − 1 y a +1 ⎟⎠ V (y )

( k eQ a )

⎛ 1 1 ⎞ = ⎜ − ⎝ y a − 1 y a + 1 ⎟⎠

FIG. P25.22(b)

P25.23

Consider the two spheres as a system. (a)

( )

v2 =

Conservation of momentum:

0 = m 1v 1ˆi +m 2v 2 −ˆi

By conservation of energy,

0=

and

1 m 12 v21 k eq1q 2 k eq1 q2 1 − = m1 v12 + 2 m2 r1 + r2 d 2

or

m1v1 m2

k ( − q1) q2 k e ( −q1 ) q2 1 1 = m1 v12 + m 2 v 22 + e r1 + r2 2 2 d

v1 =

2m 2k eq1q 2 ⎛ 1 1⎞ − ⎟ ⎜ + m1 (m1 + m2 ) ⎝ r1 r2 d ⎠

v1 =

− − 2 (0 .7 00 kg )(8.99 × 10 9 N ⋅ m 2 C 2) (2 × 10 6 C )( 3 × 10 6 C ) ⎛ 1 1 ⎞ − −3 ⎝ 8 × 10 m 1 .00 m ⎠ ( 0.100 kg ) (0.800 kg)

= 10. 8 m s v2 = (b)

m1v1 0. 100 kg( 10.8 m s) = 1.55 m s = m2 0.700 kg

If the spheres are metal, electrons will move around on them with negligible energy loss to place the centers of excess charge on the insides of the spheres. Then just before they touch, the effective distance between charges will be less than r1 + r2 and the spheres will really be moving faster than calculated in (a) .

Electric Potential

P25.24

Consider the two spheres as a system. (a)

( )

Conservation of momentum:

0 = m 1v 1ˆi + m 2v 2 − ˆi

or

v2 =

By conservation of energy,

0=

and

k eq1q 2 ke q1 q2 1 1 m12v21 . − = m1 v12 + 2 m2 r1 + r2 d 2

m1 v1 . m2

k e ( −q1 )q 2

v1 =

d

=

(b)

k e (− q1 ) q2 1 1 2 m1 v1 + m2 v22 + r1 + r2 2 2

2m 2ke q1q 2 ⎛ 1 1⎞ − m1 ( m1 + m 2 ) ⎜⎝ r1 + r2 d ⎟⎠

⎛m⎞ v2 = ⎜ 1 ⎟ v1 = ⎝ m2 ⎠

P25.25

57

2 m1 ke q1 q2 ⎛ 1 1⎞ − m2 ( m1 + m2 )⎜⎝ r1 + r2 d ⎟⎠

If the spheres are metal, electrons will move around on them with negligible energy loss to place the centers of excess charge on the insides of the spheres. Then just before they touch, the effective distance between charges will be less than r1 + r2 and the spheres will really be moving faster than calculated in (a) .

The original electrical potential energy is U e = qV = q

k eq d

In the final configuration we have mechanical equilibrium. The spring and electrostatic forces kq k q2 on each charge are − k ( 2d ) + q e 2 = 0. Then k = e 3 . In the final configuration the total (3d ) 18 d ke q 4 ke q2 1 2 1 ke q2 2 potential energy is kx + qV = = . The missing energy must have ( 2d ) +q 2 2 18d 3 3d 2 9 d 2 k q 4 k eq become internal energy, as the system is isolated: e = + ∆ Eint . d 9d 5 ke q 2 ∆Eint = 9 d P25.26

Using conservation of energy for the alpha particle-nucleus system, we have

K f +U f = K i +U i

But

Ui =

ke qα qgold

Also

and ri ≈ ∞ Thus, ri K f = 0 (v f = 0 at turning point),

so

U f = Ki

or

k eq α qgold rmin

Ui = 0

1 = m α v2α 2

2 ke qα q gold 2 (8 .99 ×10 9 N ⋅m 2 C 2) (2 ) (79 ) ( 1. 60 × 10 C ) = 2.74 × 10−14 m = 2 −27 7 mα vα2 ( 6.64 × 10 kg) ( 2.00 × 10 m s) −19

rmin =

= 27.4 fm

2

58

P25.27

Chapter 25

Each charge moves off on its diagonal line. All charges have equal speeds.

∑ (K + U ) = ∑ ( K + U ) i

f

4 k q 2 2 k q2 4k q 2 2 k q 2 ⎛1 0 + e + e = 4 mv 2⎞ + e + e ⎝2 ⎠ 2L 2L L 2 2L 2 k q 1 ⎞ e ⎛ = 2 m v2 ⎜⎝ 2 + ⎟ 2⎠ L 1 ⎞ k eq 2 ⎛ ⎜⎝1 + ⎟ 8 ⎠ mL

v=

P25.28

A cube has 12 edges and 6 faces. Consequently, there are 12 edge pairs separated by s, 2 × 6 = 12 face diagonal pairs separated by 2s, and 4 interior diagonal pairs separated by 3s. U=

ke q2 s

Section 25.4 P25.29

P25.31

Obtaining the Value of the Electric Field from the Electric Potential

V = a + bx = 10.0 V + ( −7.00 V m ) x (a)

P25.30

2 k eq ⎡ + 12 + 4 ⎤ = 12 22 . 8 ⎢⎣ 2 3 ⎥⎦ s

At x = 0,

V = 10.0 V

At x = 3.00 m,

V = −11.0 V

At x = 6.00 m,

V = −32.0 V

dV = − b = − ( −7.00 V m) = 7.00 N C in the + x direection dx

(b)

E=−