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3
••
Determine the Concept During this sequence of events, negative charges are attracted
from ground to the rectangular metal plate B. When S is opened, these charges are trapped
on B and remain there when the charged body is removed. Hence B is negatively charged
and (c) is correct.
6
•
Determine the Concept The forces acting
on +q are shown in the diagram. The force
acting on q due to Q is along the line
joining them and directed toward Q. The
force acting on q due to Q is along the
line joining them and directed away from
Q.
*8 •

Determine the Concept E is zero wherever the net force acting on a test charge is zero.
At the center of the square the two positive charges alone would produce a net electric
field of zero, and the two negative charges alone would also produce a net electric field
of zero. Thus, the net force acting on a test charge at the midpoint of the square will be
zero. (b) is correct.
30 ••
Picture the Problem The configuration of
the charges and the forces on the fourth
charge are shown in the figure … as is a
coordinate system. From the figure it is
evident that the net force on q4 is along the
diagonal of the square and directed away
from q3. We can apply Coulomb’s law to



express F1, 4 , F2, 4 and F3, 4 and then add
them to find the net force on q4.
Express the net force acting on q4:




F4  F1, 4  F2, 4  F3, 4
Express the force that q1 exerts on
q4:

kq q
F1, 4  12 4 ˆj
r1, 4

Substitute numerical values and evaluate F1, 4 :

 3 nC  ˆ
 j  3.24  105 N ˆj
F1, 4  8.99  109 N  m 2 /C 2 3 nC 
2 


0
.
05
m







kq q
F2, 4  22 4 iˆ
r2, 4
Express the force that q2 exerts on q4:

Substitute numerical values and evaluate F2, 4 :

 3 nC  ˆ
 i  3.24  105 N iˆ
F2, 4  8.99  109 N  m 2 /C 2 3 nC 
2 
 0.05 m 





kq q
F3, 4  32 4 r̂3, 4 , where r̂3, 4 is a unit vector
r3, 4
Express the force that q3 exerts on q4:
pointing from q3 to q4.





r3, 4  r3,1  r1, 4

Express r3, 4 in terms of r3,1 and r1, 4 :
 0.05 m  iˆ  0.05 m  ˆj


r
0.05 m iˆ  0.05 m ˆj
ˆr3, 4  3, 4 
r3, 4
0.05 m2  0.05 m2
Convert r3, 4 to r̂3, 4 :
 0.707 iˆ  0.707 ˆj

Substitute numerical values and evaluate F3, 4 :



3 nC
 0.707 iˆ  0.707 ˆj
F3, 4  8.99 109 N  m 2 /C 2  3 nC 
 0.05 2 m 2 


5
5
ˆ
ˆ
  1.14 10 N i  1.14 10 N j




 





Substitute and simplify to find F4 :

F4  3.24 10 5 N ˆj  3.24 10 5 N iˆ  1.14 10 5 N iˆ  1.14 10 5 N ˆj


 2.10 10 5
 
 
N iˆ  2.10 10 N  ˆj
5
 

35 ••
Picture the Problem By considering the symmetry of the array of charges we can see that
the y component of the force on q is zero. We can apply Coulomb’s law and the principle
of superposition of forces to find the net force acting on q.
Express the net force acting on q:



Fq  FQ on x axis, q  2FQ at 45, q
Express the force on q due to the
charge Q on the x axis:

kqQ
FQ on x axis, q  2 iˆ
R
Express the net force on q due to the
charges at 45:
Substitute to obtain:

kqQ
2 FQ at 45, q  2 2 cos 45iˆ
R
2 kqQ ˆ

i
2
2 R
 kqQ
2 kqQ ˆ
Fq  2 iˆ 
i
2
R
2 R
kqQ 
2 ˆ
1 
i
2 
R 
2 
kqQ
Note: I think the right answer of this problem should be 2 1  2 iˆ . Since the answer
R
on the textbook is also wrong, this problem will not be graded.



40 •
Picture the Problem We can compare the electric and gravitational forces acting on an
electron by expressing their ratio. We can equate these forces to find the charge that
would have to be placed on a penny in order to balance the earth’s gravitational force on
it.
(a) Express the magnitude of the
electric force acting on the electron:
Fe  eE
Express the magnitude of the
gravitational force acting on the
electron:
Fg  me g
Express the ratio of these forces to
obtain:
Fe eE

Fg mg
Substitute numerical values and
evaluate Fe/Fg:



Fe
1.6  10 19 C 150 N/C

Fg
9.11  10 31 kg 9.81 m/s 2


 2.69  1012
or


Fe  2.69  1012 Fg , i.e., the electric
force is greater by a factor of 2.691012.
(b) Equate the electric and
gravitational forces acting on the
penny and solve for q to obtain:
q
Substitute numerical values and
evaluate q:
q
mg
E
3  10
3

kg 9.81m/s 2
150 N/C

 1.96  10 4 C
44 ••
Picture the Problem The diagram shows the locations of the charges q1 and q2 and the

point on the x axis at which we are to find E . From symmetry considerations we can

conclude that the y component of E at any point on the x axis is zero. We can use
Coulomb’s law for the electric field due to point charges to find the field at any point on
the x axis. We can establish the results called for in parts (b) and (c) by factoring the
radicand and using the approximation 1    1 whenever  << 1.
(a) Express the x-component of the
electric field due to the charges at y
= a and y = a as a function of the
distance r from either charge to
point P:

kq
E x  2 2 cos  iˆ
r
Substitute for cos and r to obtain:

kq x
2kqx
2kqx ˆ
E x  2 2 iˆ  3 iˆ 
i
r r
r
x 2  a 2 3 2
2kqx

iˆ
2
2 3 2
x  a 
and
(b) For x  a, x2 + a2  a2, so:
For x  a, x2 + a2  x2, so:
2kqx
Ex 
x
Ex 
2kqx
Ex 
2
 a2

32
a 

2kqx
a3
2kqx

2kq
x2
2 32
x 
2 32
For x  a, the charges separated by a would appear to be a single charge
(c)
2kq
of magnitude 2q. Its field would be given by E x  2 .
x
Factor the radicand to obtain:
For a << x:
  a 2 
Ex  2kqx x 2 1  2 
x 
 
a2
1 2 1
x
3 2
and
 
Ex  2kqx x 2
3 2

2kq
x2
55 ••
Picture the Problem We can use constant-acceleration equations to express the x and y
coordinates of the electron in terms of the parameter t and Newton’s 2nd law to express
the constant acceleration in terms of the electric field. Eliminating the parameter will
yield an equation for y as a function of x, q, and m. We can decide whether the electron
will strike the upper plate by finding the maximum value of its y coordinate. Should we
find that it does not strike the upper plate, we can determine where it strikes the lower
plate by setting y(x) = 0.
Express the x and y coordinates of the
electron as functions of time:
x  v0 cos t
and
y  v0 sin  t  12 ayt 2
F
eE
a y  net, y  y
me
me
Apply Newton’s 2nd law to relate the
acceleration of the electron to the net
force acting on it:
eE y
Substitute in the y-coordinate equation
to obtain:
y  v0 sin  t 
Eliminate the parameter t between the
two equations to obtain:
yx   tan  x 
To find ymax, set dy/dx = 0 for
extrema:
eE y
dy
 tan  
x'
dx
mev02 cos 2 
2me
t2
eE y
2m v cos 
2
e 0
2
x 2 (1)
 0 for extrema
me v02 sin 2
(See remark below.)
2eE y
Solve for x to obtain:
x' 
Substitute x in y(x) and simplify to
obtain ymax:
ymax 
me v02 sin 2 
2eE y
Substitute numerical values and evaluate ymax:
9.1110 kg5 10 m/s  sin 45  1.02 cm

21.6  10 C3.5  10 N/C
31
ymax
2
6
19
2
3
and, because the plates are separated by 2 cm, the electron does not strike the upper
plate.
To determine where the electron will
strike the lower plate, set
y = 0 in equation (1) and solve for x to
obtain:
x
me v02 sin 2
eE y
Substitute numerical values and evaluate x:
9.1110 kg5 10 m/s  sin 90 
x
1.6 10 C3.5 10 N/C
31
2
6
19
3
4.07 cm
Remarks: x is an extremum, i.e., either a maximum or a minimum. To show that it is a
maximum we need to show that d2y/dx2, evaluated at x, is negative. A simple alternative is to
use your graphing calculator to show that the graph of y(x) is a maximum at x. Yet another
alternative is to recognize that, because equation (1) is quadratic and the coefficient of x2 is
negative, its graph is a parabola that opens downward.
72 ••
Picture the Problem Choose the
coordinate system shown in the diagram
and let Ug = 0 where y = 0. We’ll let our
system include the ball and the earth. Then
the work done on the ball by the electric
field will change the energy of the system.
The diagram summarizes what we know
about the motion of the ball. We can use
the work-energy theorem to our system to
relate the work done by the electric field to
the change in its energy.
Using the work-energy theorem,
relate the work done by the electric
field to the change in the energy of
the system:
Welectricfield  K  U g
 K 2  K1  U g,2  U g,1
or, because K1 = Ug,2 = 0,
Welectricfield  K 2  U g,1
Substitute for Welectric field, K2, and
Ug,0 and simplify:
qEh  12 mv12  mgh
Solve for m:
m
*78 ••

 12 m 2 gh
qE
g

2
 mgh  mgh
Picture the Problem Each sphere is in
static equilibrium under the influence of


the tension T , the gravitational force Fg ,

and the electric force FE . We can use
Coulomb’s law to relate the electric force
to the charge on each sphere and their
separation and the conditions for static
equilibrium to relate these forces to the
charge on each sphere.
(a) Apply the conditions for static
equilibrium to the charged sphere:
 Fx  FE  T sin  
kq2
 T sin   0
r2
and
F
y
 T cos   mg  0
Eliminate T between these equations
to obtain:
tan  
Solve for q:
qr
kq2
mgr 2
mg tan 
k
r  2L sin 
Referring to the figure, relate the
separation of the spheres r to the
length of the pendulum L:
Substitute to obtain:
q  2 L sin 
mg tan 
k
(b) Evaluate q for m = 10 g, L = 50 cm, and  = 10:
q  20.5 m sin 10
0.01kg 9.81m/s 2 tan 10
8.99 109 N  m 2 /C 2
80 ••
Picture the Problem Let the origin be at
the lower left-hand corner and designate
the charges as shown in the diagram. We
can apply Coulomb’s law for point charges
to find the forces exerted on q1 by q2, q3,
 0.241 C
and q4 and superimpose these forces to find
the net force exerted on q1. In part (b),
we’ll use Coulomb’s law for the electric
field due to a point charge and the
superposition of fields to find the electric
field at point P(0, L/2).
(a) Using the superposition of forces,
express the net force exerted on q1:

Apply Coulomb’s law to express F2 ,1 :




F1  F2,1  F3,1  F4,1

kq q
kq q 
F2,1  22 1 rˆ2,1  23 1 r2,1
r2,1
r2,1


Apply Coulomb’s law to express F4 ,1 :




kq q
kq q 
F4,1  42 1 rˆ4,1  43 1 r4,1
r4,1
r4,1

Apply Coulomb’s law to express F3,1 :
2
k  q q
ˆj  kq ˆj

L
L3
L2
2
k  q q
ˆ  kq iˆ

L
i
L3
L2



kq q
kq q 
F3,1  32 1 rˆ3,1  33 1 r3,1
r3,1
r3,1

kq2
 L iˆ  L ˆj
23 2 L3
kq2
  3 2 2 iˆ  ˆj
2 L


Substitute and simplify to obtain:


 kq2
kq2
F1  2 ˆj  3 2 2 iˆ 
L
2 L
2
kq ˆ ˆ
kq2
 2 i  j  32 2
L
2 L


(b) Using superposition of fields,
express the resultant field at point P:

kq2
L2



 iˆ  ˆj 
1 ˆ ˆ

1 
i j
 2 2





E P  E1  E2  E3  E4
2
ˆj  kq iˆ
L2

(1)

Use Coulomb’s law to express E1 :
 kq
kq  L
E1  2 1 rˆ1, P  3 
r1, P
r1, P  2


Use Coulomb’s law to express E 2 :


 kq  L

3 
L  2
 
2
ˆj 

ˆj   4kq ˆj
2
 L

kq
kq 
L
E3  2 3 rˆ3, P  3   Liˆ 
r3, P
r3, P 
2

Use Coulomb’s law to express E 4 :
ˆj   4kq ˆj
2
 L

kq
k  q   L
E 2  2 2 rˆ2, P  3 
r2, P
r2, P  2

Use Coulomb’s law to express E3 :
kq  L
3 
L  2
 
2
ˆj 

8kq  ˆ 1
 i 
53 2 L2 
2
ˆj 

ˆj 


kq
k  q  
L
E 4  2 4 rˆ3, P  3  Liˆ 
r4, P
r4, P 
2

8kq  ˆ 1
i 
53 2 L2 
2
ˆj 

ˆj 

Substitute in equation (1) and simplify to obtain:

4kg
4kg
8kg  1
E P  2 ˆj  2 ˆj  3 / 2 2  -iˆL
L
5 L  2
ˆj   8kg  iˆ- 1
3/ 2 2
 5 L  2
ˆj   8kg 1  5 ˆj 
2
25 
 L 
85 •••
Picture the Problem We can use Coulomb’s force law for point masses and the
condition for translational equilibrium to express the equilibrium position as a function
of k, q, Q, m, and g. In part (b) we’ll need to show that the displaced point charge
experiences a linear restoring force and, hence, will exhibit simple harmonic motion.
(a) Apply the condition for
translational equilibrium to the point
mass:
kqQ
 mg  0
y02
Solve for y0 to obtain:
(b) Express the restoring force that
acts on the point mass when it is
displaced a distance y from its
equilibrium position:
kqQ
mg
y0 
F

kqQ
kqQ
 2
2
 y0  y  y0
kqQ
kqQ
 2
y  2 y0 y
y0
2
0
because y << y0.
Simplify this expression further by
writing it with a common
denominator:
F 


2 y0 ykqQ
y04  2 y03 y
2 y0 ykqQ

y 
y04 1  2 
y0 

2ykqQ
y03
again, because y << y0.
From the 1st step of our solution:
kqQ
 mg
y 02
Substitute to obtain:
F 
Apply Newton’s 2nd law to the
displaced point charge to obtain:
m
2mg
y
y0
d 2 y
2mg

y
2
dt
y0
or
d 2 y 2 g

y  0
dt 2
y0
the differential equation of simple
harmonic motion with   2 g y0 .