Download Chapter 5.3 Q1 The positive charge on the rod will attract electrons

Chapter 5.3
Q1 The positive charge on the rod will attract electrons to sphere A, making sphere C
positively charged. So when they are separated, sphere A will be negative, B will be
neutral and C positive.
Q2 (a) The distance of the center of the square from each of the vertices is
a = 0.050 2 + 0.050 2 = 0.0707 cm . So the potential at the center is
kQ
8.99 "109 " 5 "10!6
= 4"
= 2.5 "106 V . (b) The field at the center is clearly
a
0.0707
zero – a positive charge placed there will be repelled equally from the four charges at
the vertices resulting in a net force of zero. (c) The electric field is the gradient of the
potential. The fact that the field is zero at the center means that the potential near the
center is a stationary point.
V = 4"
Q3 (a) V = 2 !
kQ
4kQ
kQ
kQ
. (b) V =
=
!
=0.
d /2
d
d /2 d /2
Q4 A diagram is:
The potential at P is V =
8.99 "109 " 2.0 "10!6 8.99 "109 " 4.0 "10!6
!
= !1.5 "104 V .
0.4
0.6
Q5 The work done is
kQ kQ
8.99 "109 "10 8.99 "109 "10
!3
W = q#V = q (
!
) = 1.0 "10 " (
!
) = 3.6 "107 J .
r2
r2
2.0
10
Q6 The work done ion the electron is
kQ
8.99 "109 " (!10)
!19
W = q#V = q (
! 0) = (!1.6 "10 ) "
= +1.4 "10!7 J .
r
0.10
1
Q7 The work done ( W = q!V ) is equal to the change in kinetic energy ( mv 2 ).
2
Hence
1
! 9.1 ! 10 "31 ! v 2 = 1.6 ! 10 "19 ! (200 " 100)
2
#v=
2 ! 1.6 ! 10 "19 ! 100
= 5.9 ! 10 6 m s "1
"31
9.1 ! 10
Q8 (a) The forces are roughly as follows.
They have magnitudes:
8.99 "109 "1"10!6 " 2 "10!6
F1 =
= 7.19 N
0.052
8.99 "109 " 4 "10!6 " 2 "10!6
F2 =
= 14.4 N
0.052 + 0.052
8.99 "109 " 3 "10!6 " 2 "10!6
F3 =
= 21.6 N
0.052
We must find the components of F2 :
F2 x = F2 cos 45! = 10.2 N and F2 y = F2 sin 45! = 10.2 N . So the net force has
components:
Fx = 10.2 ! 7.2 = 3.0 N and Fy = 10.2 ! 21.6 = 11.4 N . The net force is then
F = (11.4)2 + (3.0)2 = 11.8 N . The direction of the net force is
" !11.4 %
arctan $
= !75! .
# 3.0 '&
(b) The distance of the center of the square from each of the vertices is
a = 0.025 2 + 0.025 2 = 0.0354 cm . So the potential at the center is
kQ kQ kQ kQ 8.99 "109
V= 1+ 2+ 3+ 4 =
" (!1"10!6 + 2 "10!6 ! 3 "10!6 + 4 "10!6 )
a
a
a
a
0.0354
5
V = 5.1"10 V
(c) The work done is W = q!V = q(V " 0) = 1.0 # 10 "9 # 5.1 # 10 5 = 5.1 # 10 "4 J .
Q9 (a) The dipole moment is the product of one of the charges in the dipole times
6.2 ! 10 "30
= 3.9 ! 10 "11 m . (b) Their equal and opposite
their separation and so a =
1.6 ! 10 "19
forces on each of the charges of the dipole and so the net force is zero. (c) The largest
toque is when the dipole is normal to the field and equals
a
a
! = F + F = Fa = qEa = 1.2 " 10 #23 N m .
2
2
Q10 (a) Charge will move until both spheres are at the same potential. Then
kq1 kq2
. By conservation of charge, q1 + q2 = Q where Q is the charge on the one
=
r1
r2
q
q
sphere originally. Thus 1 = 2 ! 3q1 = 2q2 and q1 + q2 = 2.0 . Hence
10 15
2
3
q1 = ! 2.0 = 0.80 µC and q2 = ! 2.0 = 1.2 µC . (b)
5
5
#6
0.80 " 10
1.2 " 10 #6
#6
#2
!1 =
= 6.4 " 10 C m and ! 2 =
= 4.2 " 10 #6 C m #2 . (c)
2
2
4$ " 0.10
4$ " 0.15
kq
E1 = 21 = 4! k" 1 = 4! $ 8.99 $109 $ 6.4 $10#6 = 7.2 $105 N C#1 and
r1
E2 = 4! k" 2 = 4! $ 8.99 $109 $ 4.2 $105 = 4.8 $105 N C#1 . (d) The electric field is
largest for the sphere with the larger charge density. The wire has to be long so that
the charge of one sphere will not affect the charge distribution on the other so that
both are uniformly charged.
Q11 You must draw lines that are normal to the equipotentials.
Q12 (a) The potential a distance x from the bottom plate is given by
250 ! (!250)
V = !250 +
x = (!250 + 3.33 " 10 3 x) V and so at x = 3.00 cm ,
0.15
V = (!250 + 3.33 " 10 3 " 0.0300) = !150 V . Therefore the electric potential energy of
the charge is EP = qV = (!2.00 " 10 !6 ) " (!150) = 0.300 mJ . (b) The potential at
x = 12.0 cm is V = (!250 + 3.33 " 10 3 " 0.120) = 150 V and hence
EP = qV = (!2.00 " 10 !6 ) " 150 = !0.300 mJ . (c) The work done must be
W = q!V = !EP = "0.300 " 0.300 = "0.600 J .
Q13 (a) The kinetic energy of the electron
1
1
EK = mv 2 = ! 9.1 ! 10 "31 ! (1.59 ! 10 6 )2 = 1.15 ! 10 "18 J gets converted to electric
2
2
potential energy eV at the point where the electron stops. Hence the potential at P is
1.15 ! 10 "18
V=
= 7.19 V . Since the charge Q is negative the potential just found
1.6 ! 10 "19
must be negative i.e. V = !7.19 V . (b)
kQ
Vr ("7.19) # 2.0 # 10 "10
V=
!Q =
=
= "1.6 # 10 "19 C .
9
r
k
9 # 10
Q14 (a) The field due to each of the charges has the direction shown. It is clear that
the net field will point in the negative y – direction.
kQ
kQ
. The y = 2
2
r
a + d2
a
kQa
and so the net field
= 2
2 3/2
a 2 + d 2 (a + d )
The magnitude of the field due to one of the charges is E =
component is Ey =
kQ
kQ
sin ! = 2
2
a +d
a + d2
2
2kQa
.
(a + d 2 )3/2
(b) For two negative charges:
is Enet =
2
The net field is clearly directed to the left. It has magnitude
kQ
kQ
d
kQd
.
Enet = 2Ex = 2
cos! = 2
= 2
2
2
a +d
a + d a 2 + d 2 (a + d 2 )3/2
(c) We have Ea =
Eb =
kQd
(a + d 2 )3/2
2
in units of
kQa
kQ
1
and
= 2
3/2
2 3/2
(a + d )
a !
d2 $
#" 1 + a 2 &%
kQ
d
kQ
d /a
The plots are (the vertical axis is
= 3
3/2 =
3/2
2
2
a !
a !
d $
d2 $
#" 1 + a 2 &%
#" 1 + a 2 &%
2
kQ
):
a2
Q15 Since the field is the same as that of the charge Q and an equal and opposite
image charge !Q inside the plane the field at P is just
kQ kQ 8kQ
8Q
E= 2 ! 2 =
(=
) and is directed upwards.
2
d
9d
9d
36"# 0 d 2
Q16 (a) Radially in towards the center. (b) The net force on the charge is the electric
kq 2 mv 2
kq 2
2
!v =
force and so 2 =
. (c) The total energy is the sum of the kinetic
r
r
mr
and the electric potential energies i.e.
1 2 k(!q)(q) 1 kq 2 kq 2
kq 2
ET = mv +
=
!
=!
. (d) At the new orbit the total energy is
2
r
2 r
r
2r
kq 2
kq 2 " kq 2 % kq 2
greater: ET! = "
. Hence the energy required is !
.
! !
=
4r
4r $# 2r '& 4r
Q17 The initial potential energy of the three protons is zero. When at the vertices of
k(e)(e) 3ke2
=
the triangle of side a the potential energy is EP = 3 !
since there are
a
a
three pairs of charges a distance a apart. This evaluates to
3 " 8.99 "109 " (1.6 "10!19 ) 2
EP =
= 1.4 "10!12 J # 8.6 MeV . This is the energy that
5.0 "10!16
must be supplied.