Download (r)` However, the problem tells us that they are the same. So we can

Solutions and Tests for Exploring Creation With Physics
SOLUTIONS TO THE PRACTICE PROBLEMS T'ORMODULE #13
1. This is a simple application of Equation (13.1). We have the two charges
and the distance between
them, so we just plug those values into the equation to calculate the force:
r-:, kq,q,
1
t"
(9.0 x 10e
Newton .m2
e'
) . (6.7
e)
. (3.. 1
€)
- 1.3 x 101r Newtons
(r.2 n)'z
Since one charge is negative and the other positive, this force is attractive. Thus, the charges exert
1.3 x 101l-Newton force towards each other.
a
and the force and we need to determine the distance between
the charges. Equation (13.1) relates these variables. The only thing we have to watch is units. The
charges are in ffiC, but "k" uses C. Thus, we must convert mC to C.
2.Inthis problem, we know the charges
kq,q,
)
rNewton .m2
(9.0 x 10e
2.5 x105 Newtons
(s.0, ro,
C2
-
.
) (3
.2x 10-a C) . (5.5 x 10-' C)
(r)'
ryd)
.
(:.2, ro-oe)
.(s.s * ro-o e)
= 0.080 m
2.5 x 105 l$erntons
The distance between the objects, then, is 0.080 m.
are given everything but the charge of the two objects. You might thi* at fus
that there's no way to solve the problem, because there are two things you don't know: both charges.
However, the problem tells us that they are the same. So we can call them each "q." That will give ur
3. In this problem, we
only one variable:
kqrq,
r2
(9.0 x 10e
2.2x105 Newtons =
Newton .m2
C2
(0.45 m)'z
)'q'q
Ptactice Problem Solutions
q-
2.2 x1
6s
N"*ens
.
(0.45 m)2
- 2.2x 10-3 C
9.0 x 10e
charge has a magnitude of 2.2 mC.
taneous electrostatic force on the -3 .4-mC
e is zero because the forces exerted
of the other ch
cel out. The other charges are equal in magnitude and the same distance
from
the
charge
interest.
of
They each exert an attractive force, pulling the charge of interest in
1ite directions. Thus, equal and opposite forces work on the -3.4-mC charge, making the net force
At the same time, however, this is not a static system. Even though the net force on the -3.4-mC
is zero, the net forces acting on the other two
are not. The -3 .4-mC charge attracts each
&e other two charges. The other two charges repel each other, but they are farther away from each
than the -3 .4-mC charge is from them. As a result, the net force acting on each of the positive
ges is NOT zero. Thus, the other two charges will move towards the -3 .4-mC charge.
e* In this problem, we are only worried about the +6.4-mC charge. As a result, we only consider the
which act on thatparticular charge. The +1 .2-mC charge exerts a repulsive force whose
itude is:
o
kq,q,
- r'
1
(9.0 x l0e
F-
Newtons .m2
).(6.4x
e2
10-3
e).(t .2x10-'e)
(1.5 m)2
:
3.
1
x 104 Newtons
The other force acting on the +6.4-mC charge is the repulsive force exertedby the
o
-
kq,q,
)
r-
(9.0 x 10e
F_
+9.l-mC charge.
Newtons
.m'
).(6.4x
e2
10-3
q
.
(9.1x 10-3 e)
-
(2.0 m)2
1.3 x 105 Newtons
The fact that both forces are repulsive will give us the directions ofthe force vectors, making our force
diagram look like this:
1.3
x
105
N
3.1
x
104
N
Since the force vectors both point in the same dimension, we car treat them as one-dimensional
rectors. This means we can take care of direction with positives and negatives and then simply add the
91
92
Solutions and Tests for Explodng Creation $7ith Physics
magnitudes together. Using the convention that vectors pointing to the left are negative, the total force
is:
Ftotr -1.3x 10sNewtons + 3.1x 1O4Newtons = -1.0x 105Newtons
:
A negative force means that tle final vector points to the left. Thus, the final instantaneous
electrostatic force is 1.0 x 10) Newtons to the left.
go through all of the trouble of adding the electric fields from each of the remaining
charges together, or you could re alize that since you already calculated the force experienced by a
charge placed at that point, you can use that information to calculate the electric field using Equation
b. You could
(13.2):
E- F
-1.0x10s Newtons
eo
6.4x10-3 C
- -1.6x112
Newtons
C
Notice what I did here. Equation (13.2) tells you that the electric field is the force experienced by a
charged particle divided by that particle's charge. Thus, I took the force experienced by the 6-4-mC
charge as calculated previously, and I divided by 6.4 mC (converted to C)' The result, then, is the
electric field at that point. Since both "E" and "F" are vectors in the equation, I kep the negative sien
for "F," since it denotes that the force is pointed to the left. Thus, the electric field is l'6 x 10'
Newtons/Coulomb to the left.
Since we are only interested in the - 1 .2-mC charge, we only need concem ourselves with forces
which act on that charge. The +1.2-mC charge exerts an attractive force on it:
6.
rr kq,q,
l-)
r'
(9.0 x 10e
Newtons .m2
e2
F_
).(t.2x 10-3 e).(1 .2x10-',€)
- 8. 1x104 Newtons
(0.40 m)2
The +2.4-mC charge also exerts an attractive force on it:
rr
r=
kq,q,
,
r-
(9.0 x 10e
F_
Newtons .m2 .(1.2x
10-3 q-Q.4x 10-3 e)
)
e2
(0.20 m)'z
Our force diagram, then, looks like this:
6.5
x
10s
N
8.1
x
104
N
-
6.5x10s Newtons
Practice Problem
lotal fo
Solutions
rttre vectors do not point in the same dimension, we will
have to add them with trigonometry.
utl*'ever, let's define the angles properly. Vector angles are always defined counterclockwise
&e pnsitive x-axis. This means that the first angle is 341o, and the angle for the vector on the left
f' \ou' we can add these vectors:
lmg
)ya
uation
A*:
(8.1x 10a Newtons).cos(341o)
:
:
(8.1 x lOa Newtons) . sin(341o)
: - 2.6x 104 Newtons
B* = (6.5 x 10s Newtons).cos(255")
- - 1.7 x10s Newtons
A,
B,
bya
-mC
:
(6.5 x 105 Newtons) .sin(255o ) =
7.7 x10a Newtons
6.3x105 Newtons
-
C*: A^* Br:
7.7
Cy: A, + By:
-2.6 x lOaNewtons + -6.3 x 105 Newtons
x
10a
Newtons + -I.7 x
105
Newtons: -9 x
1e
'e sign
sf
Newtons
-6.6 x 105 Newtons
::.at's left to do now is convert these x- and y-components into vector magnitude and direction:
il
ES
:
104
naagnitude
0=
-
C*'+C u' = Jf-g
x 104 Newtons)2
+(-6.6x
105
Newtons)'
-
6.7 x 105 Newtons
,/C.),/-e.e"totxe.r*e*s),-^
tan-'l -l l= tan 'l --------------l= 80"
(-9xl0.Nev*ons/
l.C, l
-
S::ce both the x and y components ofthe vector are negative, we know that the vector is in quadrant
This means that we need to add 180o to the angle above to properly defrne the vector angle. The
electrostatic force on the -1.2 mC charge, then, is 6.7 x 10s Newtons at an anele of 260".
I.
=-<lantaneous
-. \\hen drawing electric field lines, the lines go out ofpositive charges and into negative charges.
ijs case, the negative charge is twice as large as the positive charge, so it has twice as many lines
_:!rmg into it as the positive charge has going out of it. I will draw 6 lines going out of the positive
:harge and into the negative charge. However, if 6 lines go out of the positive charge,
go into the negative charge, because it has twice the charge.
=ust
xt additional
In
6
93
94
Solutions and Tests for Exploring creation
with physics
. That,s where the field
lines are the most dense.
8' To calculate the electric field, I will have to sum up the
fields_created by each stationary charge.
are asked to calculate the field at the midpoint
between the charges, which is 1.25 m from each
charge' Let',s start with the positive charge.
,,81.,,
we will
we
call its electric field
E
Lr -kQ
--;-=
(9.0 x 10e
Newton .m2
C2
f'
) .(1.0
e)
- 5.g x l oe
(1.25 m)2
Newtons
C
Now rememb er, electric field,lines alwayspoint
directly awayfrom positive charges. Thus, this
field is pointed directly to your tigt t. Now let's
calculate the electric fielJ from the negative
:lt#
(9.0 x 10'g
E2
Igyton'm2 )'(2.0 e)
Ct
F
=*=
t-
- l.2x 1010
Newtons
Since electric field lines always point towards
negative charges, this electric field points
directly to
your right as well. Thus, the two electric
fields look like this:
5.9
x lOe N/C
o-+
1.2
>
x loto N/C
since the two vectors arcpatallel, we don't
need to use trigonometry to add them.
we can just add
them directly:
E,r,r,
Newtons
=E, *Ez =5.8 x 1g' IYtons + l.2x 10,, Newtons 1.8 x
10'o
=
C:
The electric field, then, is
. Note that this answer
with the qualitative diagram of the electriCn.t,o
trrut *. d..*ln thf previous problem.
The
diagram has the electric field i, b.t*.en
the two charges pointed to the right, as
does this result.
agrees
To calculate the force experienced
by aparticle, we could use Equation (13.1), but that
won :
be necessary' we already know the
electric-field at thepoint directly inietween
the
two
charges.
Thus, we can quickly calculate the force
using
Equation (r3.2):
E-I
9o
F = eoE - (-6.6x
10-a
e)
.
I .g x l0ro
Newtons
- -l
.2x10' Newtons
Practice Problem
Solutions
negative sign means that the force is pointed opposite the electric field. Thus, since the electric
i is pointed to the right, the charge experiences a 1.2 x l07-Newton force that is pointed to the left.
q
Srnce all of the stationary charges have equal magnitudes, they all have the same number of lines
umcln,e out of them. These lines bend away from each other because they cannot cross.
,\I the very center ofthe triangle, the magnitude ofthe electric field is zero, because the three particle's
ields cancel out at that point.
The ion in this case has two protons, so its nucleus has a positive charg e of 3.2 x l0-te C ltwice the
:harge of a single proton). Since there is only one electron in the ion, however, we can solve this
loblem by setting the centripetal force equal to the electrostatic force.
- r-t.
mv2 _kqrq2
r12
kq,qrr kq,q,
-.2
\/v-1
r-m
v = 3.lxl06
(9.0 x 10e
rtn
m
sec
This electron has a speed of 3.1 x 106 m/sec.
Newton.m2 .(3
.2x10-" e) . (1.6 x l0-le e)
)
e2
(5 .29 x 10-1'*) . (9. I x 10-3' kg;
95