Download Phys 202A Homework 7 Solutions 7. Since point P lies directly

Phys 202A
Homework 7 Solutions
7. Since point P lies directly between the two identical +5.0q charges, the field due to one of
those charges is equal in magnitude and opposite in direction to the field due to the other charge.
Their sum is zero. Note that the +3.0q charge and the –12q charge produce electric fields at P in
opposite directions. The –12q charge has a magnitude four times that of the +3.0q charge but is
twice as far away. Since E q r 2 , the forces due to these two charges also sum to zero. So the
net field at the point P is E net ( P) 0N / C.
9. We choose the coordinate axes as shown on the diagram that follows. At the center of the
square, the electric fields produced by the charges at the lower left and upper right corners are
both along the x axis and each points away from the center and toward the charge that produces
2a / 2 a / 2 away from the center, the net field due to
it. Since each charge is a distance d
these two charges is
2q
q
q
k 2
a2 / 2 a2 / 2
a /2
8.99 109 N m 2 / C2 (1.0 10
Exnet
k
8
2
(0.050 m) / 2
C)
7.2 104 N/C.
At the center of the square, the field produced by the charges at the upper left and lower right
corners are both along the y axis and each points away from the charge that produces it. The net
field produced at the center by these charges is
E ynet
The magnitude of the field is
k
2q
a2 / 2
q
a2 / 2
k
q
a2 / 2
7.2 104 N/C.
E net
Exnet
2
E ynet
2
2 7.2 104 N/C
2
1.0 105 N/C
and the angle it makes with the x axis is
tan
1
Ey
Ex
tan 1 (1)
45 .
It is upward in the diagram, from the center of the square toward the center of the upper side.
20. We set a coordinate system with the origin at the center of the dipole, the x axis through the
point P and the y axis through the charge +q. The fields due to the individual charges make an
angle
with the x-axis. The x-components of the individual fields sum to zero, while the ycomponents are equal and point in the ˆj direction. The net electric field is given by
E ynet
d 2
since sin
d 2
For r
d
2
q
2k
d 2
sin
2
qd
k
r2
d 2
2
3/ 2
r2
.
r2
, we write [(d/2)2 + r2]3/2
r3 so the expression above reduces to
E ynet
k
qd
.
r3
E net
k
p
.
r3
Since p (qd )ˆj ,
39. (a) The magnitude of the force on the particle is given by F elec qE, where q is the magnitude
of the charge carried by the particle and E is the magnitude of the electric field at the location of
the particle. Thus,
E
F elec
q
3.0 10 6 N
2.0 10 9 C
1.5 103 N C .
The force points downward and the charge is negative, so the field points upward.
(b) The magnitude of the electrostatic force on a proton is
F elec
eE
1.60 10
19
C 1.5 103 N C
2.4 10
16
N.
A proton is positively charged, so the force is in the same direction as the field, upward.
(c) The magnitude of the gravitational force on the proton is
F grav
mg
1.67 10
27
kg 9.8 m s
2
1.64 10
26
N.
The force is downward.
(d) The ratio of the forces is
2.4 10 16 N
1.64 10 26 N
F elec
F grav
1.5 1010 .
41. (a) The magnitude of the force acting on the proton is F elec eE , where E is the magnitude of
the electric field. According to Newton’s second law, the acceleration of the proton is
a F elec mp eE mp , where mp is the mass of the proton. Thus,
19
1.60 10
a
C 2.00 104 N C
1.67 10
27
kg
2
1.92 1012 m s .
(b) We assume the proton starts from rest v1 0 and that the force has only an x-component.
We can use the kinematic equation v22 x v12x 2ax x to show that
v2 x
2 1.92 1012 m s
2ax x
2
0.0100 m
1.96 105 m s .
47. (a) We put the origin of a coordinate system at the initial position of the electron. We take the
x axis to be horizontal and positive to the right; take the y axis to be vertical and positive toward
the top of the page. The electric field is upward in the diagram and the charge is negative, so the
force of the field on it is downward. The magnitude of the acceleration on the electron is
ay eEy me . Its numerical value is
ay
1.60 10
19
C 2.00 103 N C
9.11 10
31
kg
2
3.51 1014 m s .
The kinematic equations are
x
v1 x t ,
y
v1 y t
1
a y ( t )2 ,
2
and
v2 y
v1 y
a y t.
First, we find the greatest y coordinate attained by the electron. If it is less than d, the electron
does not hit the upper plate. If it is greater than d, it will hit the upper plate if x L at the point
where y d . The greatest y coordinate occurs when v2 y 0. So at the greatest y coordinate,
Therefore, using the kinematic equation
v2 y v1 y ay t v1 sin
ay t 0 or t v1 sin ay .
y
v1 y t
1
2
a y ( t )2 ,
v12 sin 2
ay
y max
v 2 sin 2
1
ay 1 2
2
ay
6.00 106 m s
2
1 v12 sin 2
2
ay
sin 2 45
14
2 3.51 10 m s
2.56 10 2 m.
2
Since this is greater than d = 2.00cm, the electron will hit the upper plate if x L at the point
where y d 2.00 cm.
(b) To demonstrate that the electron will in fact hit the upper plate, we need to find the x
coordinate of the electron when y d . To begin, what is the elapsed time interval when y d?
y
v1 y t
6.00 106 m s
t
1
a y ( t )2
2
1
2
d
6.00 106 m s
v12y
v1 y
t
1
2
4
ay
d
ay
2
1
2
3.51 1014 m s
2 3.51 1014 m s
2
.0200m
2
From this, the time at which the electron strikes the upper plate is t 6.43 10 9 s, where
the negative root was used to obtain the solution because we want the earliest time for which y
= d. The x coordinate is
x
v1 x t
6.00 106 m s cos 45
6.43 10 9 s
2.72 10 2 m.
This is less than L 10.0 cm, so the electron hits the upper plate at x = 2.72 cm.
1. (a) An Ampere is a Coulomb per second, so
84 A h
84 C h
s
3600 s
h
(b) The change in potential energy is U = q V = (3.0
3.0 10 5 C.
105 C)(12 V) = 3.6
106 J.
15. First, we observe that V (x) cannot be equal to zero for x > d. In fact V (x) is always negative
for x > d. Now we consider the two remaining regions on the x axis: x < 0 and
0 < x < d. For x < 0 the separation between q1 and a point on the x axis whose coordinate is x is
given by d1 = –x; while the corresponding separation for q2 is d2 = d – x. We set
V(x)
q1
k d
1
q2
d2
kq
1
x
3
d
x
0 V
to obtain x = –d/2. Similarly, for 0 < x < d we have d1 = x and d2 = d – x. Let
q1
k d
1
V(x)
q2
d2
1
kq x
3
d
0V
x
and solve: x = d/4.
21. A charge –5q is a distance 2d from P, a charge –5q is a distance d from P, and two charges
+5q are each a distance d from P, so the electric potential at P is
V
kq
5
2d
5
d
5
d
5
d
5kq
.
2d
The zero of the electric potential was taken to be at infinity.
23. We choose the zero of electric potential to be at infinity. The initial electric potential energy
U 1 of the system before the particles are brought together is therefore zero. After the system is set
up the final potential energy is found from Eq. 25-26, where we write the distance between
charges qi and q j as ri j :
U2
k
qi q j
kq 2
all pairs
ri
kq 2
a
2 4
j
1 1
a a
2.58q 2
.
a
1
2a
1
a
1
a
1
2a
Thus the amount of work required to set up the system is given by
W ext
25. Let
U
U 2 U1
kq 2
4
a
2.
0.15m be the length of the rectangle and w = 0.050 m be its width.
(a) Charge q1 is a distance
at A is
VA
from point A and charge q2 is a distance w, so the electric potential
k
q1
q2
w
(8.99 10 9 N m2 / C2 )
5.0 10 6C
0.15m
2.0 10 6 C
0.050 m
6.0 104 V.
(b) Charge q1 is a distance w from point B and charge q2 is a distance , so the electric potential
at B is
VB
q1
k w
q2
(8.99 10 9 N m2 / C2 )
7.8 105 V.
5.0 10 6C
0.050 m
2.0 10 6 C
0.15m
(c) Since the kinetic energy is zero at the beginning and end of the trip, the work done by an
external agent equals the change in the potential energy of the system. The potential energy is the
product of the charge q3 and the electric potential. If UA is the potential energy when q3 is at A
and UB is the potential energy when q3 is at B, then the work done in moving the charge from B
to A is W ext = UA – UB = q3(VA – VB) = (3.0 10–6 C)(6.0 104 V + 7.8 105 V) = 2.5 J.
(d) The work done by the external agent is positive, so the energy of the three-charge system
increases.
(e) and (f) The electrostatic force is conservative, so the work is the same no matter which path is
used.