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THEELECTRICPOTENTIAL
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.
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
29-2 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 x I O - ~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
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. 6 0 ~ 1 0 - ' ~ 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
29.8. Model:
Visualize:
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
1.0x 1 0 - ~m
u,, = u,, = U', =
( 9. 0 ~lo9 N m' / C')(-1.60 x C)(1.60 x C)
U,, = u2, = u, = (^) 0.5774 x m =^ -3.990^ x^ J
The Electric Potential 29-
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
9 E = - - = 1. 6 1 ~ 1 0 ~N / C
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 ( P J )
I
-60" 60"
Kinetic energy at 0 = 0" Total energy 1 Turning point ; I ; I I I I
line
I '\ I Oscillation
Solve:
(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
L', = 0 d s "f
0 o---.-
The figure shows a before-and-afterpictorial 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).
The Electric Potential 29-
Solve: The potential energy of charge q is U = qV. Using the conservation of energy equation,
Kf i- qV, = K, + qV, 3 V, - =^ AV^ =^ - ( Ki -^ K, )^ =^ -(O^ J^ -^^5 mvi^ ) 4 (-
( 9. 1 1 ~ 1 0 - ~ ~kg)(l.0x106 mis)’ a A V = - - mvf^ - = 2.85 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
29.15. Model:
Visualize: Before After
Energy is conserved. The potential energy is determined by the electric potential.
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,
3 V, - =^ - ( K ,^ -^ K^ )^ =^ -(+mv,?^ -^^0 J ) 4 (-
mv,*
2e 2(1.60 x C)
(9.1 1 x 10”’ kg)(5.0 x IO5 m / s)
j AV= -- = - = -0.712 V
Assess:
potential region.
29.16. Model:
Visualize: Before After
The negative sign with AV verifies that the electron moves from a higher potential region to a lower
Energy is conserved. The potential energy is determined by the electric potential.
v, = 800,Ooo m / s Vf = 0 m/s
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 A V c o n f m s 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)
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)(8.85x 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
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 = ( 1 m m + 2 m m ) 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
v,m = 5. 0 ~ 1 0 - ~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
V = -- 1 (+e) = (9.0 x lo9 N m’ / C’) 1.60^ x^ 10-1~c^ =^ 27.
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
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:
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 net potential is the sum of the potentials due to each charge.
Please refer to Figure Ex29.28.
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:
29.29. Model:
Visualize:
Potential is a scalar quantity, so we found the net potential by adding three scalar quantities.
The net potential is the sum of the potentials due to each charge.
Please refer to Figure Ex29.29.
The Electric Potential 29-1 1
/
q, = -3.0 nC /
-8,
I
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)
v, = + v, + v, = A^4 + - 4 2^43
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-l6J
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
q 2 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 q 2 = -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
The Electric Potential 29-
29.35. Model:
Visualize:
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 - - -
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 = o v
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 / C
- QiQz = =^ -4.0^ x^ 10-l6 c
Solving the first equation for Q2 and substituting into the second equation,
Q,(30 x C^ -^ Q,)^ =^ -4.0^ 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
- Qi =^3 Q,^ =^ +40^ nC and^ -10^ nC
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)
U = -- 1 QiQZ - - 72 x 10" J 9 Q,Q2 = = 4.0 x C'
47% 5 2 9.0 x IO9 N m' / C
The electric force between these two charges is
4.0 x C'
(5.0 x IO-' m)'
F = --^1 QiQ,^ - - ( 9. 0 ~ 1 0 ~N m' /C') = 1.44 x N
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 m B = 2 5 g
viA=omls Vie = 0 d s
After
q = -5nC^ q B^ =-10nC 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.ox ’ F =-T=(9.0x109 b A k! B / N m 2 / C 2 ) c)= 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 vfAand 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)(5.0x 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 c m / s ) = 1.06 cm / s. These are the maximum
speeds of the two beads and occur when they are infinitely separated from each other.
29-16 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) 4n€, X2 4 a 0 ( x - 15.0 cm)
(60.0 cm) f 43600 cm2 - 1800 cm 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 r
(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 1 0 - ~c -- + - - - O V d 4 m 0 5 4 m 0 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
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 m 2 / C2)(2.0x lo4 C)(2.0 x lo4 C)
Eo = K O + U, = 0 J + *= =^ 0.720^ J
4%rO 0.05 m
(b) In static equilibrium, the net force on sphere A is zero. Thus
lqA1lqB
4m0r02 (0.05 m)'
(9.0 x lo9 N m2 / C2)(2.0x 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,
The Electric Potential 29-
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.
29.42. Model:
Visualize: Before After
Apply the principles of conservation of energy and conservation of momentum.
ma, va = 0.01c mp, vp = 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)(3x 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 ,
The Electric Potential 29-
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. 1 1 ~ 1 0 - ~ ’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
(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)
K,, + O J = 0 J + Urn, 3 f m v l =3.20x10-’ J * v,, = 2.53 cm I s
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)
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 ) -- ( 4 j E 0 :)
(9.0 x 10’ N m2 / C’) lox = 7.95 x 10’ m / s
0 0.5 x 10” m
C)
Solve:
Model:
The conservation of energy Kf + U, = K, + U, is
Energy is conserved. The potential energy is determined by the electric potential.
Kf+qV, = K, + qV, 3 Kf = Ki + q ( V , - V f )
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.
