29.1. model - wordpress.com...model: the mechanical energy of the proton is conserved. a parallel...

THE ELECTRIC POTENTIAL

29.1. Model: The mechanical energy of the proton is conserved. A parallel plate capacitor has a uniform electric field. Visualize: After Before

* . v = o . E‘ *

I , I x 0 1 .o 2.0

The figure shows the before-and-after pictorial representation. The proton has an initial speed vi = 0 m / s and a final speed vf after traveling a distance d = 2.0 mm. Solve: The proton loses potential energy and gains kinetic energy as it moves toward the negative plate. The potential energy U is defined as U = U, + qEx, where x is the distance from the negative plate and U, is the potential energy at the negative plate (at x = 0 m). Thus, the change in the potential energy of the proton is

The change in the kinetic energy of the proton is

The law of conservation of energy is AK + AUp = 0 J. This means

2(+1.60 x C)(50,000 N / C)(2.0 x m)

(1.67 x IO-*’ kg) =1.38x1OS m / s

Assess: As described in Section 29.1, the potential energy for a charge 4 in an electric field E is U = U, + qEx, where x is the distance measured from the negative plate. Having U = Uo at the negative plate (with x = 0 m) is completely arbitrary. We could have taken it to be zero. Note that only AU, and not U, has physical consequences.

29.2. Model: electric field. Visualize: Before After

The mechanical energy of the electron is conserved. A parallel plate capacitor has a uniform

+ - I=+ * E - + -

, I x (-) 0 0.5 1.0 mm

29-1

29-2 Chapter 29

The figure shows the before-and-after pictorial representation. The electron has an initial speed vi = 0 m / s and a final speed v, after traveling a distance d = 1 .O mm. Solve: The electron loses potential energy and gains kinetic energy as it moves toward the positive plate. The potential energy U is defined as U = U, +qEx, where x is the distance from the negative plate and U,, is the potential energy at the negative plate (at x = 0 m). Thus, the change in the potential energy of the electron is

The change in the kinetic energy of the electron is = Kf - K , = +mv: - 3mv' 1 - - 1 2 mv:

Now, the law of conservation of mechanical energy gives AK + AU = 0 J. This means

+mv: + qEd = 0 J

(-2)(-1.60 x C)(20,000 N / C)(l.O xIO-~ m) =2.65x106 m / s

9.11~10"' kg

Assess: Note that AU, = qEd is the change in the potential energy of the electron. It is negative because q = -e for the electron. Thus, the potential energy becomes more negative as d increases, that is, the potential energy of the electron decreases with an increase in d (or x).

29.3. Model: field. Visualize:

The mechanical energy of the proton is conserved. A parallel plate capacitor has a uniform electric

Charge on each plate is Q Charge on each plate is 2Q

After Before After Before

4 v ; = o v,=o * 2 2 4 4

! X X 0 d 0 d

The figure shows the before-and-after pictorial representation. Solve: The proton loses potential energy and gains kinetic energy as it moves toward the negative plate. The potential energy is defined as U = U, + qEx, where x is the distance from the negative plate and U, is the potential energy at the negative plate (at x = 0 m). Thus, the change in the potential energy of the proton is

AUp = U, -Vi = (U, + O J)-(U, ++Ed)= -qEd

The change in the kinetic energy of the proton is

AK = Kf - Ki = +mv: -+mv? = +mv:

Applying the law of conservation of energy AK + AU, = 0 J, we have

2qEd +mv: + (-qEd) = 0 J 3 v: = - m

When the amount of charge on each plate is doubled, then the final velocity of the proton is

2qE'd m

vi' =-

Dividing these equations,

For a parallel-plate capacitor E = q / E , = Q/AE,, . Therefore,

V ; = f $ v f = &(50,000 m / s) = 7.07 x 10' m / s

Assess: a higher electric field between the plates and hence to an increased force on the proton.

The proton's velocity is expected to increase because an increased charge on the capacitor plates leads to

The Electric Potential 29-3

29.4. Model: The mechanical energy of the charged particles is conserved. A parallel plate capacitor has a uniform electric field.

Visualize: Proton Helium ion After Before

*

I

d 0 The figure shows the before-and-after pictorial representation. Solve: The potential energy is defined as U = U, + q&, where the potential energy at the negative plate (at x = 0 m). Thus, the moves from the positive plate to the negative plate is

Before

. I X

d 0

x is the distance from the negative plate and U,, is change in the potential energy of the proton as it

AUp =Uf -U , = ( U , + O J ) - ( U , + e E d ) = - e E d

This decrease in potential energy appears as an increase in the proton’s kinetic energy:

AK = K , - K , = tmpv:p - +mpv:p = t m P v 2 f P

Applying the law of conservation of mechanical energy AK + AUp = 0 J, we have 2eEd

+mvip + (-eEd) = 0 J * v:, = - m_ F

When the proton is replaced with a helium ion and the same experiment is repeated,

Dividing the two equations,

Assess: Being a heavier particle, the helium ion’s velocity is expected to be smaller compared to the proton’s velocity.

29.5. Model: Visualize: Solve: For a system of point charges, the potential energy is the sum of the potential energies due to all distinct pairs of charges:

The charges are point charges. Please refer to Figure Ex29.5.

1 = ( 9 x lo9 N m‘ / c 2 ) ( 2 x 10-~ c ) ( 2 x 10-~

40.03 m)’ + (0.04 m)‘

= 1.20 X IO6 J + 0.90 X 10” J + 0.72 x lod J = 2.82 x IOd J Assess: Note that U,? = U?,, U,, = U,,, and U2, = U,?.

29.6. Model: Visualize:

The charges are point charges. Please refer to Figure Ex29.6.

29-4 Chapter 29

Solve: charges:

For a system of point charges, the potential energy is the sum of the potential energies due to all pairs of

1 nC)(-l.O n c ) + (2 nC)(-l.O nC) (-1 nC)(-1.0 nC) + 0.03 m 0.03 m 0.03 m

= (9.0 x lo9 N m' / C')

= -0.60 x 10" J - 0.60 x 10" J + 0.30 x 10" J = -0.90 x 10" J

Assess: Note that U,, = U2,, U,, = U,,, and U,, = U,2.

29.7. Model: The charges are point charges. Visualize: Solve:

Please refer to Figure Ex29.7. The electric potential energy of the electron is

'elecmn = '13 + ' 2 3

( 1 . 6 0 ~ 1 0 - ' ~ C)(-1.60~10- '~ C) ( 1 . 6 0 ~ 1 0 - ' ~ C)( -1 .60~10- '~ C )

(2.0 x m)' + (0.5 x m)' d(2.O x m)' + (0.5 x m)' + = (9.0 x lo9 N m2 / C')

=-1.12x1o-I9 ~ - 1 . 1 2 ~ 1 0 - ' ~ ~ = - 2 . 2 4 ~ 1 0 - ' ~ J


Solve: We are given that 11, = rZ3 = q3 = 1.0 x m . From the geometry of the figure,

The contributions to the total potentid energy are

(9.0 x lo9 N m2 / C*)(-1.60 x C)(-1.60 x

1.0 x 1 0 - ~ m u,, = u,, = U', =

( 9 . 0 ~ lo9 N m' / C')(-1.60 x C)(1.60 x C) U,, = u2, = u, = = -3.990 x J

0.5774 x m


29.9. Model: A water molecule is at the point of minimum potential energy when it is aligned with an electric field. However. an external force can rotate the water molecule causing its dipole moment to make an angle with the field. Solve: The potential energy of an electric dipole moment in a uniform electric field is given by Equation 29.23: U,,,e = -2. E = - pE cos 0 . This means

-

1 . 0 ~ 1 0 - ~ ' J 1 . 0 ~ 1 O - ~ ' J = 1 . 6 1 ~ 1 0 ~ N / C - 9 E = - P 6.2 x C m

Assess: Note that the units with E are J/C m. Because 1 JIC m = 1 N m/C m = 1 N/C, the units of E are N/C.

29.10. Model: An external electric field supplies energy to a dipole. Visualize: On an energy diagram, the oscillation occurs between the points where the potential-energy curve crosses the total energy line.

u (PJ)

I 2 -

1 -

-60" 60"

- 1 80" 180" I l l

Kinetic energy at 0 = 0"

Total energy

1 Turning point ; I ; I I I I

line

I '\ I Oscillation

Solve: 29.23: U = [email protected] = -pEcosB. This means

(a) The potential energy of an electric dipole moment in a uniform electric field is given by Equation -

The mechanical energy Emh = U + K . We know that at 8 = 60°, KB=6r = 0 J. So,

E-,, = U,=,, + KO=,-,, -1 /IJ + 0 J = -1 4 (b) Conservation of mechanical energy gives

U,=,. + KB=,o = U,=,, + -1 @ + 0 J = -2 ,uJ i- K,,, a KO@ = 1 /.LI

29.11. Model: Mechanical energy is conserved. The potential energy is determined by the electric potential. Visualize: Before After

"f L', = 0 d s

0 o---.-

The figure shows a before-and-after pictorial representation of an electron moving through a potential difference of 1000 V. A negative charge speeds up as it moves into a region of higher potential (U + K).

29-6 Chapter 29

Solve: potential V, is

The potential energy of charge q is U = qV. Conservation of energy, expressed in terms of the electric

Kf + qVf = K, +qV, *Kf = K, + q(V, - Vf) = K, - q(Vf- V,)

47 iZ(I.6 x C)(lOOO V) = 1.87 x io7 m/s * j m v : = O J - ( - e ) ( A V ) d v , = -= 9 . 1 1 ~ 1 0 - ~ ' kg

Assess: Note that the electric potential difference of 1000 V already existed in space due to other charges or . sources. The electron of OUT problem has nothing to do with creating the potential.

29.12. Model: visualize: Before After

Energy is conserved. The potential energy is determined by the electric potential.

vi = 0 m/s

0 Vf

t I I X

vi=ov vf= -1Ooov

-AV= -1OooV- The figure shows a before-and-after pictorial representation of a proton moving through a potential difference of -1000 V. A positive charge speeds up as it moves into a region of lower potential (U + K ) . Solve: The potential energy of charge q is U = qV. Conservation of energy, expressed in terms of the electric potential V, is

Kf + qVf = K, + qV,, Kf = K, + q(v - Vf) = 0 J -qAV

2 ( 1 . 6 0 ~ C)(lOOO V)

1.67 x kg = 4.3s x io5 dS 3 +mv: = -q(-IOOO V) * vf =

Assess: Note that the proton of our problem has nothing to do with creating the potential difference of -1000 V.

29.13. Visualize: Before After

Model: Energy is conserved. The potential energy is determined by the electric potential.

v, = 0 d s v f = 1.0 x 106ds

o--, "f

@ v, I--------AV=V,- V,-

The figure shows a before-and-after pictorial representation of a He' ion moving through a potential difference. The ion's initial speed is zero and its final speed is 1.0 x lo6 d s . A positive charge speeds up as it moves into a region of lower potential (U + IC). Solve: The potential energy of charge q is U = qV. Conservation of energy, expressed in terms of the electric potential V , is

Kf + qVf = K, + qV, 3 q(V, - V,) = K, - K f

K, - K , - 0 J -+mv: - 4(1 .67~10-~ ' kg)(l.0x106 mls) ' *AV=-- _ - = - z . 0 9 ~ 1 0 ~ v

4 4 2(1.60 x C)

Assess: This result implies that the helium ion moves from a higher potential toward a lower potential.

29.14. Model: Visualize: Before After


vi = 0 d s vf= 1.0 x 1 0 6 d s

Vf

A V = Vf- Vi ____((

The figure shows a before-and-after pictorial representation of an electron moving through a potential difference. Because the negative electron gains speed as it travels, it moves into a region of higher potential (U K ) .


Solve: The potential energy of charge q is U = qV. Using the conservation of energy equation,

1 1 Kf i- qV, = K, + qV, 3 V, - = AV = - ( Ki - K, ) = -(O J - 5 mvi )

4 (-4 ( 9 . 1 1 ~ 1 0 - ~ ~ kg)(l.0x106 mis)’

= 2.85 V mvf - a A V = - - 2e 2(1.60 x 10-l~ c)

Assess: higher potential.

A positive value of AV shows that the electron moved from a region of lower potential to a region of



vi = 500,000 m / s Vf = 0 m / s - 0 “i “f

A V = Vf- V, ____/

The figure shows a before-and-after pictorial representation of an electron moving through a potential difference. Solve: region of higher potential to a region of lower potential. (b) Using the conservation of energy equation,

(a) Because the electron is a negative charge and it slows down as it travels, it must be moving from a

K, i- Uf = K, i- U, 3 Ki + qV, = K, i- qV,

1 1 3 V, - = - (K , - K ) = -(+mv,? - 0 J)

4 (-4 mv,* 2e 2(1.60 x C)

(9.1 1 x 10”’ kg)(5.0 x IO5 m / s)2 = -0.712 V j AV= -- =-

Assess: potential region.


The negative sign with AV verifies that the electron moves from a higher potential region to a lower


v, = 800,Ooo m / s Vf = 0 m/s - 0 VI Vf +-----AV=V,-

The figure shows a before-and-after pictorial representation of a proton moving through a potential difference. Solve: of lower potential to a region of higher potential. (b) Using the conservation of energy equation,

(a) Because the proton is a positive charge and it slows down as it travels, it must be moving from a region

K, + U, = K, + U, * Kf + gV, = K, i- qV,

3 V, - = -( 1 K, - Ki ) = -(+mv? 1 - 0 J) 9 ( e )

mv; (1.67 x IO-” kg)(8.0 x lo5 m / s)’ *AV=-= = 3340 V

2e 2( I .60 x 10-l~ c) Assess: A positive AVconfms that the proton moves into a higher potential region.

29.17. Solve: By definition 1 V = 1 J/C and 1 J = 1 N m. Consequently,

V J N m N 1 - = I - = I - = I -

m C m C m C

29-8 Chapter 29

29.18. Model: The electric potential difference between the plates is determined by the uniform electric field in the parallel-plate capacitor. Solve: (a) The potential of an ordinary AA or AAA battery is 1.5 V. Actually, this is the potential difference between the two terminals of the battery. If the electric potential of the negative terminal is taken to be zero, then the positive terminal is at a potential of 1.5 V. (b) If a battery with a potential difference of 1.5 V is connected to a parallel-plate capacitor, the potential difference between the two capacitor plates is also 1.5 V. Thus,

AVc= 1.5 V = V+- V - = E d

where d is the separation between the two plates. The electric field inside a parallel-plate capacitor is

(1.5 V)(AE,) (1.5.V)r(2.0 x m)2(8.85x C2 / Nm2) = 8 . 3 4 ~ C . - * Q = -

d 2.0x10-~ m Thus, the battery moves 8.34 x lo-'' C of electron charge from the positive to the negative plate of the capacitor.

29.19. Model: The electric potential difference between the plates is determined by the u n i f m electric field in the parallel-plate capacitor. Solve: (a) The potential difference AV, across a capacitor of spacing d is related to the electric field inside by

E = - AVC * AVc = Ed = (1.0 x IO5 V / m)(0.002 m) = 200 V d

(b) The electric field of a capacitor is related to the surface charge density by

E=-=- 77 Q/A €0 Eo

* Q = &,AE = (8.85 x IO-'* C2 / N m2)(4.0 x IO4 m2)(1.0 x lo5 V / m) = 3.54 x lo-'' C

29.20. Model: The electric potential between the plates of a parallel plate capacitor is determined by the uniform electric field between the plates. Solve: (a) The potential difference across the plates of a capacitor is

(0.708 x C)(1 .O x m) AV , - - E d = ( z ) - d=-d=-= (QIA) Qd =2wv

E, (4.0 x IO4 m2)(8.85 x C2 / N m2)

(b) Ford = 2.0 111111, AV, = 400 V. Assess: Note that the units in part (a) are N m/C. But Exercise 29.17 showed that 1 N/C = 1 Vlm, so 1 N m/C = 1 V. We also see that the potential difference across a parallel-plate capacitor is directly proportional to the plate separation.

29.21. Model: The electric field inside a parallel-place capacitor is determined by the potential difference between the plates. The proton's potential energy inside the capacitor is also determined by the capacitor's potential difference. Visualize: Solve: right and has a potential of 300 V. (b) The electric field strength inside the capacitor is

Please refer to Figure Ex292 1. (a) Because the right plate is at a higher potential compared with the left plate, the positive plate is on the

AV, -3OOV-OV E = - - - - =1.0x105 V / m

d 3.0 x w3 m (c) The potential energy of a charge q is U = qAV. A proton on the left plate will have zero potential energy. A proton at the midpoint of the capacitor is at a potential of 150 V. Thus, its potential energy is

U = (1.6 x C)(150 V) = 2.40 x lo-'' J

29.22. Model: The charge is a point charge. Solve: (a) The electric potential of a charge q is

2 5 ~ 1 0 - ~ C - 2 2 5 N m 2 / C V=-- ' q,r=--=(9.0x109 1 q N m 2 / C 2 ) - 4mo r 4X&, v V V


For V = 1000 V,

225Nm’IC 1000 v = 0.225 m ‘loo0 =

; for V = 4000 V, = 0.056 m .

29.23. Model: The charge is a point charge. Visualize: Solve:

Please refer to Figure Ex29.23. (a) The electric potential of the point charge q is

V = - - = ( 9 . O x l O 9 1 q N m 2 / C z ) (2.0 x r i C) = 18.0 N m2 I C 47re0 r r

For points A and B, r = 0.01 rn. Thus,

C 1 8 . 0 N m 2 / C

0.01 m v = v = \ B

For point C, r = 0.02 m and V, = 900 V . (b) The potential energy of a charge q’ at a point where the electric potential is V is U = q’ V . The expression for the potential in part (a) assumes that we have chosen V = 0 V to be the potential at r = 00. So, we are obtaining potential/potentiaI energy relative to a zero of potential/potential energy at infinity. Thus,

U , = U , = (q’)V = ( -e ) (V) = -(1.60~ C)(1800 V) = -2.88 x J

u, = (-1.60 X C)(900 V) = - 1 . 4 4 ~ J

(c) The potential differences are

AV,, = V, - V, = 1800 V -1800 V = 0 V A V ~ ~ = V, - v, = 900 v - 1800 v = -900 v 29.24. Solve: at the center. That is,

Outside a charged sphere of radius R, the electric potential is identical to that of a point charge Q

The potential of the ball bearing is the potential right on the surface of the ball bearing. Thus,

(9.0 x IO9 N m’ / C2)(2.0 x 109)(-1 .60 x C) V = = -5760 V 0.5 x m

29-10 Chapter 29

29.25. Model: Visualize: Glass bead

Outside a charged sphere the electric potential is identical to that of a point charge at the center. 0 v z p m v j m m +++++ It2mm-?

-4mm------rl 1IIlIll

Solve: For r 1 R, V = Q / 4 n g 0 r . The potentials at the two points are

v’m = (1mm+2mm) 3.0 x m (9.0 x lo9 N m2 / C2)Q (9.0 x lo9 N m2 / C ’ ) Q (9.0 x lo9 N m’ / C 2 ) Q

5 . 0 ~ 1 0 - ~ m v,m = - -

- 1 aV,,,-V,,,=+500V = ( 9 . 0 x 1 0 9 N m 2 / C 2 ) ( 3 . 0 ~ 1 0 ” m 5 . 0 ~ 1 0 - ~ m

* Q =

Assess:

29.26. Model: Solve:

Do not forget to include the radius of the glass bead in r.

Outside a charged sphere the electric potential is identical to that of a point charge at the center. (a) For a proton, assumed to be a point charge, the electric potential is

1.60 x 10-1~ c = 27.2 V = -- 1 (+e) = (9.0 x lo9 N m’ / C’) 4m0 r 0 . 0 5 3 ~ 1 0 - ~ m

(b) The potential energy of a charge q at a point where the potential is V is U = qV. The potential energy of the electron in the proton’s potential is

U = (-1.60 x C ) (27.2 V) = -4.35 x J

29.27. Viualiie: Solve:

Model:

The potential at the dot is

The net potential is the sum of the potentials due to each charge. Please refer to Figure Ex29.27.

V=-- q’+--+-- 1 % 1 q 3 4 q , 5 4m0 r, ~ZE,, r,

2.0 x 1 0 - ~ c + 2.0 x io4 c + 2.0 x io+ c = + 1410

0.040 m 0.050 m 0.030 m 1 = (9.0 x lo9 N m’ / C ’ )

Assess:

29.28. Visualize: Solve:

Potential is a scalar quantity, so we found the net potential by adding three scalar quantities.

Model:

From the geometry in the figure,


The potential at the dot is

2.0 x 10-~ c 1.0 x 1 0 - ~ c - 1.0 x io-9 c = 1 - 0.01732 m 0.01732 m 0.01732 m

= (9.0 x lo9 N m’ / C’)

Assess:


Potential is a scalar quantity, so we found the net potential by adding three scalar quantities.


The Electric Potential 29-1 1

, /

q, = -3.0 nC /

-8, I

0 - 16

\

Solve: (a) Let q1 = +5 nC, q2 = -5 nC, and q3 = 10 nC. Also, r, = 2 cm, r, = 4 cm, and r3 =

4- = 4.47 cm. Potential is a scalar, not a vector, so the net potential is simply the sum of the potentials of each of the charges. Each individual potential is simply that of a point charge, so

,, Zero potential point / \ , ‘

‘ x q7 = +4.0nC . - I @- x (cm) 8 ,’ 16

4 4 2 43 v, = + v, + v, = A + -

+ 0.02 m 0.04 m 0.0447 m

= (9.0 x lo9 N m’ / C‘)

(b) The potential energy of a proton at point A is

Assess: Note that the units in part (a) were N m/C. But Problem 29.17 showed that 1 N/C = 1 V/m, so 1 N m/C = 1 V. Up,,, = qprotanVA = eV, = (1.6 x C)(3140 V) = 5.02 x 10-l6 J

2930. Visualize: Solve: (a) Let ql = -5 nC, q, = 10 nC, and q3 = 10 nC. Also, rl = 2 cm, r2 = 4 cm, and r3 =

Model: The net potential is the sum of the potentials due to each charge. Please refer to Figure Ex29.30.

d ( 2 cm)’ + (4 cm)2 = 4.47 cm. Each individual potential is simply that of a point charge, so

q2 43 1 VB = v, + v, + v, = 41 +-

-5 x 1 0 - 9 c + 10 x 10-9 c + 10 x 10-9 c 0.02 m 0.04 m 0.0447 m

= (9.0 x lo9 N m2 / C’)

(b) The potential energy of an electron at point B is

Assess: Note that the units in part (a) are N m/C. But Problem 29.17 showed that 1 N/C = 1 V/m, so 1 N m/C = 1 V. U,,,, = q&-vB =-eV, = (-1.6 X C)(201O V) = -3.22 X J

29.31. Model: Visualize: q1 = +3.0 nC q2= -1.0nC

The net potential is the sum of the scalar potentials due to each charge.

Solve: point, V, + V2 = 0 V. This means

Let the point on the x-axis where the electric potential is zero be at a distance x from the origin. At this

L{ 3.0 x r 4 c + -1.0 x 1 0 - ~ c = O V 3 -x+31x-4.O cmJ = O cm

4n&Eo Ix - 4.0 cml

Either -x + 3(x - 4.0 cm) = 0 cm, or-x + 3(4.0 cm - x) = 0 cm. In the first case, x = 6.0 cm and, in the second case, x = 3 cm. That is, we have two points on the x-axis where the potential is zero.

29.32. Model: The net potential is the sum of the scalar potentials due to each charge. Visualize: Y

Zero potentid point

-y

29-12 Chapter 29

Solve: point, V, + V, = 0 V. This means

Let the point on the y-axis where the electric potential is zero be at a distance y from the origin. At this

1 - 3 . 0 ~ 1 0 ~ C 4 . 0 ~ 1 0 ~ C

a 34- = 44- 9(256 cm2 + y’) = 16(81 cm’ + y’)

* 7y2 = 1008 cm2 3 y = +12 cm.

29.33. vector sum of the fields due to each charge. Visualize: Please refer to Figure Ex29.33. Solve: (a) By the symmetry of the drawing about the middle, it appears that the magnitudes of the charges are the same. E, points left for x > b because E, is negative. So, the charge at x = b is negative. For x < a , E, points left, so the charge at x = a is positive. For a < x < b, E, is positive which is consistent with the charge choices.

Model: The potential is the sum of the scalar potentials due to each charge. But, the electric field is the

(b) V

The graph of the electric potential shown in the figure is consistent with the electric field as well as the charges.

29.34. vector sum of the electric fields due to each charge. Visualize: Solve: (a) By the symmetry of the drawing about the middle, we infer that the magnitudes of the charges are the same. As V is always positive, both charges must be positive. The positive signs for the two equal charges at x = a and at x = b are also consistent with the behavior of the potential in the range a < x < b.

Model: While the potential is the sum of the scalar potentials due to each charge, the electric field is the

Please refer to Figure Ex29.34.

OD) Ex I I I I I I I I I I I

I I

The graph of Ex, the x-component of the electric field, as a function of x is shown in the figure.



Consider the rod to be a line of charge. Divide the rod into small segments, each with charge h i q .

Symmetrically placed segments

I+ + +Kj+ + + + + I - - - - - w - - -1 + Aq -a4

Solve: Consider two segments, one positive and one negative, equally distant from the center of the rod. These segments are the same distance r from the dot. Thus the contribution of this pair of segments to the potential at the dot is

v, + v- = --+-- 1 +Aq 1 -Aq = ov 41tE0 r 4m0 r

Since we can divide the entire rod into pairs of symmetrically placed segments, the net result of adding the potentials due to each pair is V = 0 V. Assess: This conclusion depends on the dot being directly outward from the midpoint of the rod. The potential is not zero at other points.

29.36. Solve: Let the unknown charges be Q, and Q2. Then Q, + Q2 = 30 x C and

-(180 x lo6 J)(2.0 x lo-* m)

9.0 x lo9 N mz / C2 = -4.0 x 10-l6 c2 * QiQz =

Solving the first equation for Q2 and substituting into the second equation, C - Q,) = -4.0 x Q,(30 x C2 * Q: - (30 x C)Q, - (4.0 x C') = 0

(30 x C) k ,/(-30 x C)' + 4(4.0 x C')

2 3 Q, = +40 nC and -10 nC * Qi =

That is, the two charges are -10 nC and 40 nC. Assess: potential energy equation yield U = -180 x lod J.

29.37. Solve: 5.0 x lo-' m . Their electric potential energy is

As they must, the two charges when added yield a total charge of 30 nC, and when substituted into the

Let the two unknown, positive charges be Q, and Q2. They are separated by a distance q2 =

(72 x 10" J)(5.0 x lo-' m) = 4.0 x C' U = 1 -- QiQZ - - 72 x 10" J 9 Q,Q2 =

47% 5 2 9.0 x IO9 N m' / C2 The electric force between these two charges is

4.0 x C'

(5.0 x IO-' m)' = 1.44 x N 1 QiQ, - F = -- - ( 9 . 0 ~ 1 0 ~ N m' /C ' )

4E&, 5; Assess: As far as the magnitudes are concerned,

= I.MxIO-~ N U 72x10" J Y 5.0x10-' m

F = - =

29-14 Chapter 29

29.38. Model: The charged beads are point charges. Visualize: Before

qA= -5nC qB= -1OnC n

0 4 8 12 mA= 15g mB=25g

I 1 x (cm) - -

viA=omls Vie = 0 d s

After

q =-10nC q = -5nC B 7 Infinitely separated ______)(

“€4 VtB

Solve: the force between the beads is given by Coulomb’s law:

Once the beads are released they will accelerate according to the force acting on them. The magnitude of

(5.0 x 10” c)(io.o x c) F’ =-T=(9.0x109 b A k ! B / N m 2 / C 2 ) = 3.125 x lo-’ N 4m0 rm (0.12 m)’

The same force acts on bead A and bead B. Since F =ma,

where a, is directed toward the left and a,, is directed toward the right. To find the maximum speeds vfA and vm it is easier to use the conservation of momentum and conservation of energy equations than kinematics. The momentum conservation equation along x-direction paiter = pMm = 0 kg d s means

-mAvfA + m,vm = 0 kg m/s *-(0.015 kg)vfA + (0.025 kg)vm = 0 kg m/s 3 vm = $v,

The conservation of energy conservation is U, + Kf = U, + K,. Noting that U, + 0 J as the masses are infinitely separated and

(9.0 x lo9 N m’ / C2)(5.0 x C)(lO.O x C) u, =-- 1 Q A Q , = = 3.75 x lo4 J 4ne0 rm 0.12 m

a 0 J + (+mAv:, + +m,v;) = 3.75 x 10“ J + 0 J

i(0.015 kg)viA + +(0.025 kg)(+vfA)2 = 3.75 x 10” J 3 0.012~:~ = 3.75 x 10“ J

Solving these equations yields vfA = 1.77 cm/s and vm = f(1.77 cm/s) = 1.06 cm / s . These are the maximum speeds of the two beads and occur when they are infinitely separated from each other.


29.39. Model: While the net potential field is the vector sum of the electric fields. Visualize:

is the sum of the scalar potentials due to each charge, the net electric

E

Q, = -2.0 nC Q2 = +2.0 nC

1 I P x(cm) - - 1.0 0 1 .o

k ( x - l . O c m ) - - +

- x V k- (x + 1 .O cm)--(

V

Solve: (a) Let the two charges be labeled as Q, = -2.0 nC and Q2 = +2.0 nC. As pictorially shown in the figure, the net electric field will not become zero anywhere except at infinity. We will now show this result mathematically. Let the point of zem electric field be at a distance x from the origin. Then,

- - - I 2 . 0 ~ 1 0 - ~ c 1 2 . 0 ~ 1 0 - ~ c Enet = E, - E , = 0 N 1 C * E, = E2 * -

4m0 (x - 1 .O cm)' -

9 -- 4m0 (x + 1 .O cm)*

a (x + 1 .O cm)' = (x - I .O cm)'

Obviously, this equation can only be satisfied if x = 00. That is, the electric field is zero at +oo and at -. (b) Because potential is a scalar quantity, the potential of charge Q, is always negative and the potential of charge Q2 is always positive. These two scalar numbers will add to give zero potential at x = 0 m and at +-. Mathematically this can be shown as follows. When -1.0 cm 5 x I 1.0 cm, let the point of zero potential be at x. Then, the condition V, + V, = 0 V/m is

( - 2 x 1 0 ~ c) 1 ( 2 ~ 1 0 - ~ c) +- = 0 a -(1.0 cm + x) + (1.0 cm -a) = 0 3 x = 0 cm. 4m,, 1.0 cm + x 4m0 (1.0 cm - x)

Whenx > 1.0 cm,

1 ( - 2 ~ 1 0 - ~ c) 1 ( 2 ~ 1 0 - ~ c) +- = O * (x+l.Ocm)=(x-l .Ocm) 4m0 (x + 1 .O cm) 47rc0 (x - 1 .O cm)

This equation can only be satisfied at x = +a. (c) The graphs are shown in the figure above.

29.40. Model: the vector sum of the electric fields. Visualize:

While the net potential is the sum of the potentials due to each charge, the net electric field is

Q , = 20.0 nC 2, Q2 = - 10.0 nC - - E , 0 - 0 z2 - - -

E2 -5 -- gq

I I - x Icrn) 0 P k--- 15.0cm-x- 15.0crn---+

p x -

29-16 Chapter 29

The charge Q, = 20.0 nC is at the origin. The charge Q2 = -10.0 nC is 15 cm to the right of the charge Q, on the x-axis. Solve: (a) As the pictorial representation shows, the point P on the x-axis where the electric field is zero can only be on the right side of the charge Q2, that is, at x 5 15 cm. At this point E, = E 2 , so we have

I 20.0 x c 1 10.0 x c, axz = 2(x - 15.0 cm)2 =- 4n€, X2 4 a 0 ( x - 15.0 cm)

(60.0 cm) f 43600 cm2 - 1800 cm2 a2 - (60.0 cm)x + 450 cm2 = 0 * x =

-x=51.2cmand8.8cm. The root x = 8.8 cm is not possible physically. So, the electric fields cancel out at x = 51.2 cm. The electric potential at this point is

= +lo3 V 1 2 0 . 0 ~ 1 0 - ~ c - 1 0 . 0 ~ 1 0 - ~ c + 0.512m (0.512m-0.150m)

V = --+Le, 1 Qi = ( 9 . 0 ~ lo9 N m2 / C') 4 ~ 5 , 5 4nsO r2

(b) The point on the x-axis where the potential is zero can be obtained from the condition VI + V2 = 0 V, which is

= O 1 Qi 1 Q2 - 20.0 x c + -10.0 x 10-~ c -- +- - -OVd 4m0 5 4m0 r, X (15 cm - x)

*2(15 cm - x) - x = 0 a x = 10 cm The electric field 10 cm away from charge Q, is

29.41. Visualize: r,, = 5 cm

Model: Mechanical energy is conserved. Metal spheres are point particles and they have point charges.

A B q A = ' P C 4g=2cLC

m , = 2 g % = 4 g

Solve: system is

(a) The system could have both kinetic and potential energy, although here K = 0 J. The energy of the

(9.0 x lo9 N m2 / C2)(2.0 x lo4 C)(2.0 x lo4 C) Eo = KO + U, = 0 J + * = = 0.720 J

4%rO 0.05 m

(b) In static equilibrium, the net force on sphere A is zero. Thus

lqA1lqB1 4m0r02 (0.05 m)'

(9.0 x lo9 N m2 / C2)(2.0 x 10" C)(2.0 x 10" C) T=F,...=-- - = 14.4 N

(c) The spheres move apart due to the repulsive electric force between them. The surface is frictionless, so they continue to slide without stopping. When they are very far apart (r, + -), their potential energy U, -+ 0 J. Energy is conserved, so we have

E, = K , + U, = 3mAv:, + +m,v;, + 0 J = E,

Momentum is also conserved: Pa,, = mAvAl + m,v,, = Pbefore = 0 kg d s . Note that these are velocities and that v,, is a negative number. From the momentum equation,


Substituting this into the energy equation,

Using this result, we can then find vAl = -mBvB, / m A = -21.9 m / s. These are the velocities, so the final speeds are 21.9 m/s for the 2 g sphere and 10.95 m / s for the 4 g sphere.


Apply the principles of conservation of energy and conservation of momentum.

mp, vp = 0.01c ma, va = 0.01c

e CFS _ _ _ _ _ _ _ _ _ _ _ - - _ _ - _ _ Far apart

Proton ‘ d

0-t Alpha particle

The figure shows a before-and-after pictorial representation of the charged particles moving toward each other. The proton’s physical quantities are denoted by the subscript p and that of the alpha particle by the subscript a. Solve: The conservation of energy equation is Kf + U, = K, + U,. Initially when the charges are far away, U, = 0 J and K, = +m,vi + + m a v i . At the distance of closest approach,

The particles are not at rest at closest approach. That would violate conservation of momentum. Instead, the particles are instantaneously moving with the same velocity v, = vrp = v, so that the distance between them is (instantaneously) not changing. The conservation of energy equation simplifies to

The conservation of momentum equation PdE, = Pbefore is

mavf + mpvf = m,(O.Olc) - m,(O.Olc)

(O.Olc)(m, - m p ) - (0.01)(3 x 10’ m / s)(4 u - 1 u) 3 Vf = - =1.8x106 m / s

ma +mp 4 u + l u

Substituting into the energy-conservation equation,

+(4 u + 1 u)(1.8 x lo6 m / s)‘ +(9.0 x lo9 N m* / C ’ ) 2(1.60 x 10-l9 c)’

= +(4 u + 1 u)(3.0 x lo6 m / 5)’ d

4.608 x lo-’’ N m’ d

* = 1.44 x 10” u m’ / sz

4.608 x lo-’* N m’ (1.44)(1.661~10-” kg)x10L3 (m’ I s * )

* d = = 1.93 x m

29.43. Model: Mechanical energy is conserved. Visualize: and so on.

Please refer to Figure P29.43. Label the 5.0 nC charge with subscript 1, the 3.0 nC with subscript 2 ,

29-1 8 Chapter 29

Solve: The conservation of energy equation Kf + U, = K, + U, is

K,+ 0 J = O J + Vi * K, = U, = U,, + U,, + U,, + U,, + U, + U3,

1 qlq? (9.0 x lo9 N m2 / C2)(5.0 x lo4 C)(3.0 x C) = 3.857 x 10” J u,, = -- =

4 ~ & , 112 0.035 m

(9.0 x lo9 N m2 / C2)(5.0 x lo4 C)(4.0 x C) = 4.727 x 10“ J qIq3 - u,, = -- - 4% 5 3 J(0.035 m)’ + (0.015 m)‘

(9.0 x lo9 N m2 / C2)(5.0 x C)(2.0 x C ) = 6.000 x 10“ J qlq4 - u,, = -- - 4 ~ & , 5, 0.015 m

Likewise, U,, = 7.200 x 10“ J , U, = 1.418 x 10“ J , and U , = 2.057 x 10” J . The sum of all the potential energies is 25.3 x lod J, which is the final kinetic energy 4.

29.44. Model: Energy is conserved. The potential energy is determined by the electric potential. Visualize: Solve: The proton at point A is at a potential of 30 V and its speed is 50,000 d s . At point B, the proton is at a potential of -10 V and we are asked to find its speed. Clearly, the proton moves into a lower potential region, so its speed will increase. The conservation of energy equation Kf + U, = K, + U, is

+mv: + (+e)(-IO V) = +rn(50,000 m / s)’ + (+e)(30 V)

Please refer to Figure P29.44.

2(1.60 x C)(40 V) 3 vf = (50,000 m/s)’ + =1.01x1O5 m / s d 1.67 x lo-’’ kg

Assess: The speed of the proton is higher, as expected.

29.45. capacitor’s electric potential. Solve:

Model:

(a) The voltage across the capacitor is

Energy is conserved. The electron’s potential energy inside the capacitor can be found from the

AVc = Ed = (5.0 x lo5 V / m)(2.0 x IO-, m) = 1000 V

(b) Because E = q/&, for a parallel-plate capacitor, with q = Q / A , the charge on each plate is

Q = nR2E&, = rr(l.O x lo-’ m)’(5.0 x l o5 V / m)(8.85 x C 2 / N m2) = 1.39 x C

(c) The electron has charge q = -e, and its potential energy at a point where the capacitor’s potential is V is U = -eV. Since the electron is launched from the negative (lower potential) plate toward the positive (higher potential) plate, its potential energy becomes more negative (because of the negative sign of the electron charge). That is, the potential energy decreases, which must lead to an increase in the kinetic energy. Conversely, the electron’s speed as it is launched is smaller than 2.0 x lo7 d s . The conservation of energy equation is

K,+ qV,= K,+ qV,* 3m.2 = +mv: +q(V, - y ) 2 m

v: = v: + -(-e)(lW V)

2( I .bo x 1 0-19 c)( 1000 v) 3 v, = ( 2 . 0 ~ 1 0 ~ m / s ) - 9.11~10-~’ kg =7.Ox1O6 m / s d

29.46. Model: Energy is conserved. Solve: (a) The conservation of energy equation Kf -+ U, = K, + U, is

Kf + qV, = 0 J + qVt * Kf = q(V, - Vi)

The above equation can be rewritten separately for the proton and the electron as

K , ~ = +m,v i = ( + ~ ) ( A v ~ ) K, = +me.: = ( - c ) ( - A ~ )


Note that we have written V, - V, as AVp for the proton, but speed up a proton, and V, > V, to speed up an electron. With vrp = v,, the above two equations give

AVP - mP - 1.67 x AY me 9.11~10-~ ’ kg

(b) The first two equations of part (a) can be divided to give

- V, = -AV, for the electron. T h i s is because Vi > V, to

kg = 1833 - _ - _

K , - AVP Kre AK -_ -

For the same final kinetic energy of the proton and the electron AVp/AK = 1

29.47. Model: Energy is conserved. Solve: (a) v (V)

I

-10 -8 -6 -4 -2 0 2 4 6 8 10 (b) The potential energy of the positive charge as a function of x is U(x) = 5000 qx’. This is analogous to the potential energy of a mass on a spring. Thus the motion is simple harmonic motion. (c) Turning points at k8.0 cm mean that x,, = f8.0 cm. At the turning points, the energy is all potential energy and the kinetic energy is zero. The mechanical energy of the charged particle is

E = U + K = U,, + 0 J = 5000 q(x,,)* = 50OO(10.0 x C)(0.08 m)’ = 3.20 x IO-’ J

(d) The conservation of energy equation K , + U , = K2 + U, is

2(3.20 x J)

(1.0 x kg) = 2.53 cm I s K,, + O J = 0 J + Urn, 3 f m v l =3.20x10-’ J * v,,

29.48. capacitor’s potential difference. Visualize: Solve: (a) The electric potential at the midpoint of the capacitor is 250 V. This is because the potential inside a parallel-plate capacitor is V = Es where s is the distance from the negative electron. The proton has charge q = e and its potential energy at a point where the capacitor’s potential is V is U = eV. The proton will gain potential energy

if it moves all the way to the positive plate. This increase in potential energy comes at the expense of kinetic energy which is

Model: Energy is conserved. The proton’s potential energy inside the capacitor can be found from the

Please refer to Figure P29.48.

AU = eAV = e(250 V) = 1.60 x C (250 V) = 4.0 x lo-’’ J

This available kinetic energy is not enough to provide for the increase in potential energy if the proton is to reach the positive plate. Thus the proton does not reach the plate because K < AU. (b) The energy-conservation equation K , + U, = K, + U, is

Z(1.60 x C)(250 V -0 V) =2.96x105 m / s

1.67 x IO-” kg

29.49. Model: its final potential energy to be zero. Visualize: reach r, = - when v, = 0 d s .

Energy is conserved. The proton ends up so far away from the two charges that we can consider

Please refer to Figure P29.49. The minimum speed to escape is the speed that allows the proton to

29-20 Chapter 29

Solve: energy once and multiply by two. The conservation of energy equation K, + U, = Ki + U, is

Both charges contribute equally to the proton’s initial potential energy, so we can calculate the potential

(1.60 x C)2(9.0 x lo9 N m2 f C2)(2.0x lo4 C) * t(1.67 x kg)v’ = 5.0 x 10-~ m

*vi =1.17x106 m f s

29.50. Model: Energy is conserved. The electron ends up so far away from the plastic sphere that we can consider its potential energy to be zero. Visualize: Before After

The minimum speed to escape is the speed that allows the electron to reach r, = - when v, = 0 d s . Solve: The conservation of energy equation K, + U, = K, + Vi is

0 J + 0 J = +mv,? + q y 3 0 J = + mv? + ( -e ) -- ( 4jE0 :) (9.0 x 10’ N m2 / C’) lox = 7.95 x 10’ m / s 0 0.5 x 10” m

C)

29.51. Solve:

Model: The conservation of energy Kf + U, = K, + U, is


Kf+ qV, = K, + qV, 3 Kf = Ki + q(V, - Vf)

In this equation, K , = tmv’ = 0 J , the final proton potential is zero, and the initial proton potential is V. We have K , = eV. That is, the maximum kinetic energy or maximum speed is directly proportional to the potential of the surface the proton is launched from. From Equation 29.36, Example 29.1 1, and Example 29.12, when z + 0 for the ring and disk:

Because the potential on the disk is a factor of 2 larger than the potential on the sphere or the ring, the proton’s kinetic energy or the speed will be maximum in the case of the disk.

The dipole initially is at the higher potential energy. As it moves to its minimum energy state, the decrease in the potential energy appears as rotational kinetic energy.


Solve: The potential energy of a dipole is UaPle = -j. and the electric field. The energy conservation equation K, + U, = K, + Vi is

= -pEcos8, where 8 is the angle between the dipole

4 IO2 + U , = 0 J + Ui a +ZW' = Vi - U, = -pEcos90° - (-pEcosO")

The moment of inertia of the dipole is I,,,, = 2m(+s)'. Substituting into the above equation,

= 0.283 rad f s 1.0 x kg)(O.lO m)'

29.53. Model: assume that the nucleus does not move (no recoil) and that the proton is essentially a point particle with no diameter. Visualize: Before After

Energy is conserved. Because the iron nucleus is very large compared to the proton, we will

* . t-,

'i

The proton is fired from a distance much greater than the nuclear diameter, so r, = - and U, = 0 J. Because the nucleus is so small. a proton that is even a few atoms away is, for all practical purposes, at infinity. As the proton approaches the nucleus, i t is slowed by the repulsive electric force. At the end point, the proton has just reached the surface of the nucleus (r( = nuclear diameter) with vf = 0 m/s. (The proton won't remain at this point but will be pushed back out again, but the subsequent motion is not part of this problem.) Solve: Initially, the proton has kinetic energy but no potential energy. At the point of closest approach, where v, = 0 m/s, the proton has potential energy but no kinetic energy. Energy is conserved, so Kf + U, = K, + U,. This equation is

where r, is half the nuclear diameter. The initial speed of the proton is

2(1.6 x C)(26 x 1.6 x

(1.67 x lo-'' kg)(4.5 x lo-'' m) C)(9.0 x lo9 N m2 / C')

= 3.99 x 10' m / s - - 4% 5 mproton

29.54. Model: assume that the nucleus does not move (no recoil) and that the proton is essentially a point particle with no diameter. Visualize: Before After

Energy is conserved. Because the mercury nucleus is very large compared to the proton, we will

. . - d

The proton is fired from a distance much greater than the nuclear diameter, so r, = - and U, = 0 J. Because the nucleus is so small, a proton that is even a few atoms away is, for all practical purposes, at infinity. As the proton approaches the nucleus, it is slowed by the repulsive electric force. At the end point of the problem, the proton reaches the distance of closest approach (d) with vf = 0 d s . (The proton won't remain at this point but will be pushed back out again, but the subsequent motion is not part of the problem.) Solve: Initially, the proton has kinetic energy but no potential energy. At the point of closest approach, where v, = 0 d s , the proton has potential energy but no kinetic energy. Energy is conserved, so K, + U, = K, + V, . This equation is

e(80e) 4n&,d

0 J+-==' mprolm V: + o J

1 6 0 ~ ' - (9.0 x IO9 N m' f C2)(160)(1.6 x C)' * d = - = 1 . 3 8 ~ m = 13.8 fm 4 ~ E 0 m ~ ~ , ~ ~ v 2 (1.67 x kg)(4.0 x IO' m f s)?

The radius of the nucleus is 7.0 fm, so the proton's closest approach to the surface is 6.8 fm.

29-22 Chapter 29

I

29.55. Model: Energy is conserved. Vial ize:

Before After

Thorium Thorium q = 90e

The alpha particle is initially at rest (vi dw = 0 m/s) at the surface of the thorium nucleus. The potential energy of the alpha particle is Ui alpha. After the decay, the alpha particle is far away from the thorium nucleus, U, alpha = 0 J, and moving with speed v, alpha.

Solve: Initially, the alpha particle has potential energy and no kinetic energy. As the alpha particle is detected in the laboratory, the alpha particle has kinetic energy but no potential energy. Energy is conserved, so Kf alpha + U, alpha = K, alpha + Vi This equation is

1 (2e)(90e) imv:,,, + 0 J = 0 J + - 41r&0 r,

1 (360e2) (9.0 x lo9 N m2 / C2)360(1.60 x C) - = 4.07 x 107 m / s 3 VfdPha = -- - 4(1.67 x kg)(7.5 x m)

29.56. Model: Energy is conserved. Before M e r visualize: Undergoes dpcav ”-

Vf-+ 0

rf -+ m 0 q = f 2 e

Hydrogen Helium Solve: (a) The neutron has zero charge. In the decay n i, p’ + e- +v, where the neutrino v is chargeless, the net charge of the final state is e + (-e) + 0 = 0. So charge is conserved. (b) As shown in the figure, the tritium nucleus has one proton and two neutrons. It is the single proton that makes this an isotope of hydrogen, the Z = 1 element in the periodic table. One of the neutrons undergoes the decay n i, p + e +v. The electron and neutrino are ejected from the nucleus as beta radiation, but the new proton remains in the nucleus. The nucleus still has three nucleons, but now two are protons and only one is a neutron. An atom whose nucleus contains two protons is the Z = 2 element of the periodic table, namely helium. (c) As shown in the figure, the electron is “launched” from the surface of the 3He nucleus with speed vi. The initial energy is

If the electron just barely escapes, it slows to vf + 0 m / s as rf + -. Its final energy is Ef = Kf + U, = 0 J + 0 J = 0 J. Energy is conserved, E, = E,, so

(-e)@) 41r&0r,

+mev,2 + - = 0 J

2 (-e)(&) (2)(9.Ox1O9 N m’ /C2)(1)(1.60X C)* =8.21x108 m / s (9.1 1 x kg)( 1 S x lo-’’ m)

Assess: Our “classical” analysis gives an answer that exceeds the speed of light (3 x 10* d s ) , which is forbidden by Einstein’s theory of relativity. We would need to perform a relativistic analysis, which is beyond the scope of this chapter, to get the correct minimum escape speed.


29.57. Model: The electric field inside a capacitor is uniform. Solve: (a) Because the parallel-plate capacitor was connected to the terminals of a 15 V battery for a long time, the potential difference across the capacitor right after the battery is disconnected is AV= 15 V. The electric field strength inside the capacitor is

E=%= lSv = 3 0 0 0 V / m d 0 .5~10-* m

Because E = q / ~ , for a parallel-plate capacitor and 77 = Q / A , the total charge on each plate is

Q = - E , = (3000 V / m)n(0.05 m)2(8.85 x C 2 / Nm') = 2.09 x lo-'' C

(a) After the electrodes are pulled away to a separation of d'= 1.0 cm, the charges on the plates are unchanged. That is, Q'= Q. Because A' = A , the electric field inside the capacitor is also unchanged. So, E' = E. The potential difference across the capacitor is AV: = E'd' . Because d increases from 0.5 cm to 1 cm (d = 1 cm), the potential difference AV; increases from 15 V to 30 V. (c) When the electrodes are expanded, the new area is A' = z(r')* = Ir(2r)' = 4A. The charge Q' on the capacitor plates, however, stays the same as before (Q' = Q). The electric field is

The potential difference across the capacitor plates AV; = E'd' = (750 V)(O.OS m) = 3.75 V .

29.58. Model: The electric field inside a capacitor is uniform. Solve: (a) While the capacitor is attached to the battery, the plates are at the same potentials as the terminals of the battery. Thus, the potential difference across the capacitor is AVc = 15 V. The electric field strength inside the capacitor is

E = - - - - A", - = ~ O O O V / ~ d 0.5 x IO-* m

Because E = q / ~ ~ = Q / A E , , the charge on each plate is

Q = EAE, = (3000 V/m) k (0.05 m)2 (8.85 X lo-" C2/N m2) = 2.09 x lo-'' C

(b) After the electrodes are pulled away to a new separation of d' = 1.0 cm, the potential difference across the capacitor stays the same as before. That is, AV: = AV, = 15 V. The electric field strength inside the capacitor is

The charge on each plate is

Q' = E'AE,= (1500 V)n(O.OS m)'(8.85 x lo-'' C2/N m') = 1.04 x lo-'' C

(c) The potential difference AV: = AV, = 15 V is unchanged when the electrodes are expanded to area A' = 4 A . The electric field between the plates is

The new charge is Q' = E'A'E, = ~ E A E , = 4Q = 8.34 x IO-'' C

29.59. at the center. For r 2 R,

Solve: (a) Outside a uniformly charged sphere, the electric field is identical to that of a point charge Q

Evaluating this at r = R and using Equation 29.36,

(b) The field strength is

29-24 Chapter 29

29.60. Visualize: QA = 0.1 nC QB = 0.1 nC (V0)*= 300 v (Voh = 300 v

9rJ;;\ Solve: A sphere of radius R and charge Q has a surface potential V, = Q/4moR. When the two drops A and B merge to form drop C, two quantities are conserved: the total charge and the total volume of mercury. The conservation of charge yields

Qc=Qm,, = QA + QB =0.1 nC +0.1 nC=0.2 nC

Since the quantity of mercury does not change, the volumes of drops A, B, and C are related by

[Vol, = $I&] = [Vol, + Vol, = $ z R ~ + $&:I 3 R, = (Ri + R;),,’

We’re not given the radii of drops A and B, but we know their surface potentials (Vo), and (Vo)B. Thus,

(0.1 x C)(9.0 x lo9 N m2 / C’) = 3.00 x m = RB 3 RA = QA = (“A = 4 ~ ~ 0 ( ~ 0 )A 300 v

Combining these radii gives R, = 3.78 x m, and

(0.2 x lo4 C)(9.0 x lo9 N m2 / C’)

3.78 x lo-’ m = 476 V

29.61. Solve: (a) Because the excess charge resides on the outer surface of a conductor, the charge placed on the inside surface of a hollow metal sphere will move rapidly to the outside surface. (b) Because a spherical shell of charge has the same electric potential as a point charge Q at the center, the potential on the surface is

0.15 m

(c) The electric field inside a conductor is zero, so the electric field just inside the sphere is zero. From Gauss’s Law. the electric field outside a charged conductor is

= 3.33 x lo6 V / m v Q 8.33 x 10“ C

E, A&, 41~(0.15 n1)~(8.85 x lo-’‘ C 2 / Nm’) E=-=-=

29.62. Solve: A charged particle placed inside a uniformly charged spherical shell experiences no electric force. That is, E = 0 V/m inside the shell. We know from Section 29.5 that a difference in potential between two points or two plates is the source of an electric field. Since the potential on the surface of the shell is V = Q/4m0 R , the potential inside must be the same. This ensures that the potential difference is zero and hence the electric field is zero inside the shell. The potential at the center of the spherical shell is thus the same as at the surface. That is, V,m, = Y d x e = Ql4x~oR.

29.63. uniformly charged sphere, the electric potential is identical to that of a point charge Q at the center. Visualize: Solve:

Model: The potential at any point is the superposition of the potentials due to all charges. Outside a

Please refer to Figure P29.63. Sphere A is the sphere on the left and sphere B is the one on the right. The potential at point a is the sum of the potentials due to the spheres A and B:

1 Q A 1 QB y = v,,, + V,,,, = -- + -- ~YE, R, 4m0 0.70m

25 x lo-’ C = ( 9 . 0 ~ l O ~ N m ’ / C ’ ) ~ ~ ~ ~ ~ - ~ 0.30 m c + ( 9 . 0 ~ 1 0 9 Nm’/C’) 0.70 m = 3000 V + 321 V = 3321 V


Similarly, the potential at point b is the sum of the potentials due to the spheres A and B:

v, = v,,, +VAatb=--+-- 1 QB 1 QA

4ZE0 RB 4?E0 0.95m

2 5 ~ 1 0 ~ C + lOOxlO-’ C 0.05 m 0.95 m

= (9.0 x lo9 N m’ / C’)

= 4500 v + 947 v= 5447 v Thus, the potential at point b is higher than the potential at a. The difference in potential is V, - V, = 5447 V - 3321 V = 2126 V. Assess: V, I a = 3000 V and the sphere B has a potential of 225 V at point a. The spherical symmetry dictates that the potential on a sphere’s surface be the same everywhere. So, in calculating the potential at point a due to the sphere B we used the center-to-center separation of 1 .O m rather than a separation of 100 cm - 30 cm = 70 cm from the center of sphere B to the point a. The former choice leads to the same potential everywhere on the surface whereas the latter choice will lead to a distribution of potentials depending upon the location of the point a. Similar reasoning also applies to the potential at point b.

29.64. Model: The charges making the dipole are point charges. The potential is the sum of potentials from each charge. visualize: Y

I

I Point P is far away compared to the separation s of the dipole charges, that is, y >> S.

Solve: (a) The potentials at P due to the positive and negative charges of the dipole are

1 (-4) v- =-- 4m0 y+s f2

1 q v, =-- 4m0 y - 4 2

p =‘P y’ - S 2 / 4 - 4RE0 y 2

(b) The potential due to the dipole moment of water is 6.2 x C m (1 .O x IO-’ m)‘

Vbale = (9.0 x 10’ N m2 / C’) = 0.056 V

29-26 Chapter 29

+-si2

@

:: st2 --I xi 8

X o \

Proton

Point P is such that 1x1 << s. Solve: (a) The potentials at P due to the two positive charges are

4

v"ght = L(L) 4Z&, s/2-x

f l I 1

I \ I \

yen = L( L) 4Z&, s/2+x

S

We can use the binomial approximation (1 + u)" = 1 + nu if x << 1 to write

vp=(&)( [ S 4:]

214) 1--

(b) The first term is a constant that does not affect the motion. Because of the quadratic dependence on x in the second t e n , the motion is simple harmonic motion.

0

Point P is located on the x-axis and the two positive charges are located on the y-axis. The separation between the two charges is s.


Solve: (a) The potentials at P due to the two positive charges are

. (b) The V-versus-x graph is shown in the figure. The potential due to a charge 2q located at the origin is V2, = 2 q / 4 m 0 x , and is also plotted. This expression differs from the expression obtained in part (a) by the factor

(1 + s2 /4x2) -+ . For x >> s , this factor becomes 1 and the two potentials become the same.

29.67. Model: Visualize: Please refer to Figure P29.67. Point P at which we want the net potential due to the linear electric quadrupole is far away compared to the separation s, that is, y >> s. Solve:

The net potential is the sum of potentials from all the charges.

The net potential at P is

v,, = y + v, + v, = -A+L%+-- 1 q 3 1

4ZE0 y - s 4 m 0 y 4Z€ , y + s

=-- 1 [(+q) +- (-2q) +- (+q)]- -- q [ y ~ + y s - 2 y ~ + 2 s ~ + y ~ - y s - q 2s2

4 m 0 y - s y y + s 4m0 Y ( Y 2 - s2) ]-Ky(y2 - 2 ) At distances y >> s ,

where Q = 2qs2 is called the electric quadrupole moment of the charge distribution. Assess: completely cancel.

This charge distribution is in fact a combination of two dipoles. As seen above their effects do not

29.68. Model: Visualize: Please refer to Figure 29.32. Solve:

The disk has uniform surface charge density 17 = Q f A .

The on-axis potential of a disk of radius R and charge Q was found in Example 29.12 to be

We’re interested in the case where z >> R so R l z << 1. To make use of the binomial approximation, which relies on very small numbers, the expression in brackets can be written as

The binomial approximation is (1 + x ) ” ~ = 1 + 4 x when x << 1. The expression becomes

As a result, the potential of the disk becomes

T h i s is the potential of a point charge Q at distance r = z on the z-axis. Assess: The disk appears as a point charge from far away. This is understandable and reasonable.

.- .. “-I._... .

29-28 Chapter 29

29.69. Visualize: Please refer to Figure P29.69. The point P is a distance d from the origin. Divide the charged rod into N small segments, each of length Ax and with charge Aq. Segment i, located at position x,, contributes a small amount of potential V, at point P. Solve:

Model: Assume the thin rod is a line of charge with uniform linear charge density.

The contribution of the ith segment is

y = A = 4 - QhIL 4 7 ~ ~ ~ 4 n ~ , ( d - x , ) - 4 m E , ( d - x , )

Where Aq = ;ths and the linear charge density is I , = Q/ L. We are placing the point P at a distance d rather than x from the origin to avoid confusion with xi. The V, are now summed and the sum is converted to an integral giving

Replacing d with x, the potential due to a line charge of length L at a distance x along the axis is

v = --In( Q -) x + L / 2 4n&0L x - L / 2

29.70. Visualize: Please refer to Figure F'29.69. The point P is located a distance z from the origin. Divide the charged rod into N small segments, each of length Ax and with charge AQ. Segment i , located at position xi, contributes a small amount of potential V, at point P. Solve:

Model: Assume the thin rod is a line of charge with uniform linear charge density.

The contribution of the ith segment is

Charge AQ is related to length Ax through the linear charge density: AQ = ;ths = (Q/L)Ax . Thus

Qd. y = 4 m o L , / 7 7

Potential is a scalar, so we can find the net potential at P by summing all the V,: Q Ax

If we now let Ax + du, the sum becomes an integral over the length of the rod, from x,, = -L/2 to x,, = +W2:

Ll? Q dr v = - Jyy2 4Z&,L - &qp

We can also get a simpler expression by using symmetry to integrate from 0 to L/2 and multiplying by 2. This gives

This expression can be shown to be equivalent to the one obtained earlier.

29.71. Visualize: Please refer to Figure P29.71. The bent rod lies in the xy-plane with point P as the center of the semicircle. Divide the semicircle into N small segments of length As and of charge AQ = (Q/nRR)&, each of which can be modeled as a point charge. The potential Vat P is the sum of the potentials due to each segment of charge. Solve: The total potential is

Model: Because the rod is thin, assume the charge lies along the semicircle of radius R.


All of the terms come to the front of the summation because these quantities did not change as far as the summation is concerned. The summation does not have to convert to an integral because the sum of all the A€J around the semicircle is x. Hence, the potential at the center of a charged semicircle is

29.72. Visualize: into rings of equal width Ar. Ring i has radius ri and charge AQ,. Solve:

Model: The disk has a uniform surface charge density q = Q / A = Q/.(Rtn - R i ) . Please refer to Figure 29.31. Orient the disk in the xy-plane, with point P at distance z. Divide the disk

Using the result of Example 29.1 1 , we write the potential at distance z of ring i as

Noting that AQ, = @A, = ~2m,Ar ,

In the limit R,, + 0 m,

This is the same result obtained for a disk of charge in Example 29.12.

29.73. Solve: energy of the charge arrangement is 90 pJ. What are the two charges? (b) We have qlq2 = 3.0 x from the second equation,

(a) A charge of 40 nC is separated into two charges which are placed 3.0 cm apart. The potential

C2 and q, + q2 = 40 x lo4 C. Substituting into the first equation an expression for q,

(40 x C - qr)q2 = 3.0 x C’ * qi -(40 x lo4 C)q2 + 3.0 x Cz = 0

The solutions to this equation are q2 = 30 nC and 10 nC. This means the two charges are 10 nC and 30 nC.

29.74. proton’s speed when it is 1 .O mm from the charge? (b) Solving the equation yields y = 1.67 x IO6 m/s.

Solve: (a) A proton is shot toward a 2.0 nC point charge with a speed of 2.5 x lo6 m/s. What is the

29.75. Solve: (a) Charges +Q are placed on the plates of a parallel-plate capacitor. Each plate is 2.0 cm x 2.0 cm and they are 1.0 mm apart. Find the magnitude of the charge on each plate if the voltage across the capacitor is 100 v. (b) Solving the equation yields Q = 0.35 nC.

29.76. Solve: opposite the other charge, the potential is 1200 V. Find the separation d between the charges. (b) The given equation simplifies to

(a) Two 3.0 nC charges are distance d apart. At a point 3.0 cm from one charge, on the side

2 7 N m 2 / C 2 7 N m ’ / C = 1200 v 3 0.03 m + d 0.03 m + d

900 N m / C + = 300 V d = 0.060 m = 6.0 cm

29-30 Chapter 29

- - - - _

I I

/ I

/ (-0.5cm.O)

‘.,P (x, Y ) \ 1 1

3 3 4 m = 4- 3 9 2 +9y2 = 16 +2 +$ - 8x

a 82 + 8x - 16 + 89 = 0 S(x +$)’ - 2 - 16 + 8y2 = 0 3 ( x + $)* + y2 = 9 = $

This is the equation of a circle, x” + y” = R2, whose center is at ( x = -3 cm, y = 0 cm) and whose radius is

R = & = 3 = 1.5 cm. All values of (x, y) that satisfy this equation are the points with zero potential. This zero potential contour map is shown as a dotted line in the figure. Assess: It is clear from the contour map that two zero-potential points lying on the x-axis are at x = 1.0 cm and x = -2.0 cm. Each position is nearer to the smaller charge and away from the bigger charge, as it must be.

29.78. Model: Energy is conserved. The charged spheres are point charges. Visualize: in the upper left-hand comer being sphere 1. Solve:

Please refer to Figure CP29.78. Label the spheres 1,2, 3, and 4 in a clockwise manner, with the sphere

The total potential energy of the four spheres before they are allowed to move is

u, = W,? + u23 + u34 + U41) + w , 3 + u,> All the charges are identical, r,? = r23 = r34 = r4] = 1 cm, and q3 = rZ4 = & cm. The initial potential energy is

u , = 4 - 1 ( i o ~ i o ~ C ) ( I O X I O - ~ c) ] + 2 [ k ( i o ~ 1 0 - 9 c ) ( i o ~ i o - ~ c) = 48.73 x 10” J [ 4nEo 0.01 m 0.01414 m

Since all charges are at rest, K, = 0 J. As the spheres are allowed to move away from one another and they are far apart, Ut = 0 J. The final kinetic energy is

K , = 4(+mv:) = 2(1.0 x kg)$

From the energy conservation equation Ut + Kf = U, + K,,

2(1.0 x 10-j kg)vj = 48.73 x IO” J a v f = 0.494 m/s

29.79. Model: Energy and momentum are conserved. Visualize: when they are very far apart.

Please refer to Figure CP29.79. Let the two spheres have masses mA and mB, and speeds vA and vB


Solve: The energy conservation equation Kf + U, = K, + Ui is

( 2 . 0 ~ 1 0 - ~ C)(I.OXIO-~ c) + 9(0.002 kg)vi + +(0.001 kg)vi = (9.0 x IO9 N m’ / C’) 2 . 0 ~ 1 0 - ~ m

* v i + 0.5 vi = 0.0090 mz 1 s2

To solve for vA and v,, we need another equation relating vA and vB. From the momentum conservation equation PdW = pbefare we get

Substituting this expression into the energy conservation equation,

mAVA + m B v B = 0 kg m / S (2.0 g)VA + (1 .o g)VB = 0 kg m / S 3 VB = -2VA

v i + 0.5(2vA)’ = 0.009 m2 / s‘ * 3vi = 0.009 m’ / s2

Solving these equations, we get v, = -0.0548 m / s and vB = d.110 d s . Thus the speeds are 0.0548 m / s for A and 0.1 10 m/s for B.

29.80. Model: Energy and momentum are conserved. Visualize: Please refer to Figure CP29.80. Let the two spheres have masses rn, and rn,, and speeds v, and v, when the two spheres collide. Solve: The energy Conservation equation Kf + U, = Ki + U, is

The momentum conservation equation P&,, = P,, is

m,v, + mDv, = 0 kg m/s 3 vc = - - vD = -2v, (;:: ,9) Substituting this expression in the energy equation,

+(O.OOl kg)4vi + +(0.002 kg)vi = ( 9 . 0 ~ lo9 N m‘ / C2)(2.0 x lo4 C)(-1.0 x

a (0.003 kg)vi = 7.2 x lo4 N m Solving these equations gives vD = 0.049 d s = 4.9 cm/s and hence v, = -9.8 c d s . The speed of C is +9.8 c d s .

29.81. Model: The charges making the dipole are point charges. The potential is the sum of potentials from each charge. Visualize: Y

I Point P is located at a distance r from the center of the dipole and it makes an angle 8 with the dipole axis.

29-32 Chapter 29

Solve: The potential at P due to the dipole is

Because x, y >> s , (d2)’ can be ignored compared with 2, y’ and ys. The potential is

Using the binomial expansion,

where p = Qs is the dipole moment of the dipole.


The potential energy is determined by the electric potential. (a) The electric field in the parallel-plate capacitor when the plates have charges kq is

The increase in the potential energy dU due to moving charge dq through potential difference AV, is

(b) Integrating the expression for dU from q = 0 (uncharged) to q = Q (fully charged),

where we have used the expression for AV, from part (a).


The potential energy is determined by the electric potential. (a) A sphere of radius R that is uniformly charged with a charge q is at a potential V, (see Equation 29.36):

1 9 v, =-- 4m0 R

The increase in the potential energy dU when an infinitesimal amount of charge dq is moved from infinity (where the potential is zero) to the surface of the sphere (where the potential is V,) is

dU=dqV, =-- ‘ d q 4m0 R

(b) Integrating the expression for U from q = 0 (uncharged) to q = Q (fully charged),

(c) The self-energy of a proton is

(1.60 x 10-l9 c)’ = 2.30 x io-l3 J 1 Q2

4m0 2R u = -- = (9.0 x lo9 N m2 / C’)

(2)(0.5 x lo-’’ m)


29.84. Model: Assume the cylinder is uniformly charged. Visualize: Divide the cylinder into narrow rings of width dz, radius R, and charge dq.

Ring with charge dq

I' L Solve: of a ring of charge was found in Example 29.1 1 to be

The potential at the center is the sum of the potential of all the rings. The potential at distance z on the axis

The surface area of the ring is dA = (2ldi)dz, so the amount of charge is dq = q a!A = (2nRq)dz. The surface area of the cylinder is A = (2ldi)L, so the surface charge density is 77 = Q12zRL. Thus dq = (Q/L)dz and the potential of the ring is

The potential at the center of the cylinder is found by summing (i.e., integrating) the potential due to all rings. That is,

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