29.3. Model: The mechanical energy of the proton is conserved. A parallel plate capacitor has a uniform electric
field.
Visualize:
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 = U0 + qEx, where x is the distance from the negative plate and U0 is the potential
energy at the negative plate (at x = 0 m). Thus, the change in the potential energy of the proton is
∆Up = Uf − Ui = (U0 + 0 J ) − (U0 + qEd ) = −qEd
The change in the kinetic energy of the proton is
∆K = K f − K i = 12 mvf2 − 12 mvi2 = 12 mvf2
Applying the law of conservation of energy ∆K + ∆Up = 0 J, we have
1
2
mvf2 + ( −qEd ) = 0 J ⇒ vf2 =
2qEd
m
When the amount of charge on each plate is doubled, then the final velocity of the proton is
vf′2 =
2qE ′d
m
Dividing these equations,
vf′2 E ′
=
⇒ vf′ =
E
vf2
E′
vf
E
For a parallel-plate capacitor E = η ε 0 = Q Aε 0 . Therefore,
vf′ =
Q′
vf = 2 ( 50,000 m/s) = 7.07 × 10 4 m/s
Q
Assess: The proton’s velocity is expected to increase because an increased charge on the capacitor plates leads to
a higher electric field between the plates and hence to an increased force on the proton.
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29.5. Model: The charges are point charges.
Visualize: Please refer to Figure Ex29.5.
Solve: For a system of point charges, the potential energy is the sum of the potential energies due to all distinct
pairs of charges:
Uelec = ∑
i,j
Kqi q j
rij
= U12 + U13 + U23

1
1
= 9 × 109 N m 2 /C2 2 × 10 −9 C 2 × 10 −9 C 
+
+
 0.03 m 0.04 m

(
)(
)(
)
= 1.20 × 10 −6 J + 0.90 × 10 −6 J + 0.72 × 10 −6 J = 2.82 × 10 −6 J
Assess: Note that U12 = U21, U13 = U31, and U23 = U32.
1
( 0.03 m ) + ( 0.04 m )
2
2




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:
r r
Udipole = − p ⋅ E = − pE cosθ . This means
Udipole parallel = − pE cos0° = − pE
Udipole perpendicular = − pE cos90° = 0 J
⇒ Udipole perpendicular – Udipole parallel = 1.0 × 10−21 J = 0 J – (–pE)
⇒E=
1.0 × 10 −21 J
1.0 × 10 −21 J
=
= 1.61 × 108 N/C
p
6.2 × 10 −30 C m
Assess: Note that the units with E are J/C m. Because 1 J/C m = 1 N m/C m = 1 N/C, the units of E are N/C.
29.23. Model: The charge is a point charge.
Visualize: Please refer to Figure Ex29.23.
Solve: (a) The electric potential of the point charge q is
V=
 2.0 × 10 −9 C  18.0 N m2 /C
1 q
= 9.0 × 109 N m2 /C2 
=
4πε 0 r
r
r


(
)
For points A and B, r = 0.01 m. Thus,
VA = VB =
18.0 N m2 /C
Nm
V
= 1800
= 1800   m = 1800 V
0.01 m
C
m
For point C, r = 0.02 m and VC = 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 = ∞. So, we are obtaining
potential/potential energy relative to a zero of potential/potential energy at infinity. Thus,
(
)
UA = UB = ( q′ )V = ( −e )(V ) = − 1.60 × 10 −19 C (1800 V ) = −2.88 × 10 −16 J
(
)
UC = −1.60 × 10 −19 C ( 900 V ) = −1.44 × 10 −16 J
(c) The potential differences are
∆VAB = VB − VA = 1800 V −1800 V = 0 V
∆VBC = VC − VB = 900 V − 1800 V = −900 V
29.31. Model: The net potential is the sum of the scalar potentials due to each charge.
Visualize:
Solve: Let the point on the x-axis where the electric potential is zero be at a distance x from the origin. At this
point, V1 + V2 = 0 V. This means
1  3.0 × 10 −9 C −1.0 × 10 −9 C 
+

 = 0 V ⇒ − x + 3 x − 4.0 cm = 0 cm
4πε 0 
x
x − 4.0 cm 
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.
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29.53. Model: Energy is conserved. Because the iron nucleus is very large compared to the proton, we will
assume that the nucleus does not move (no recoil) and that the proton is essentially a point particle with no diameter.
Visualize:
The proton is fired from a distance much greater than the nuclear diameter, so ri ≈ ∞ and Ui ≈ 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, the proton has just reached the
surface of the nucleus (rf = 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 vf =
0 m/s, the proton has potential energy but no kinetic energy. Energy is conserved, so Kf + Uf = Ki + Ui. This
equation is
0 J+
( e)( 26e) = 1 m
4πε 0 rf
2
v +0J
2
proton i
where rf is half the nuclear diameter. The initial speed of the proton is
vi =
2 ( e )( 26e )
4πε 0 rf mproton
=
(
)(
)(
2 1.6 × 10 −19 C 26 × 1.6 × 10 −19 C 9.0 × 10 9 N m 2 / C2
(1.67 × 10
−27
)(
kg 4.5 × 10
−15
m
)
) = 3.99 × 10
7
m/s