Download Physics 9 Spring 2011 Homework 1

Physics 9 Spring 2011
Homework 1 - Solutions
Wednesday January 19, 2011
Make sure your name is on your homework, and please box your final answer. Because
we will be giving partial credit, be sure to attempt all the problems, even if you don’t
finish them. The homework is due at the beginning of class on Wednesday, January
26th. Because the solutions will be posted immediately after class, no late homeworks can
be accepted! You are welcome to ask questions during the discussion session or during office
hours.
1. A −2.0 µC point charge and a 4.0 µC point charge are a distance L apart. Where
should a third point charge be placed so that the third charge has no acceleration?
————————————————————————————————————
Solution
The third charge should be placed at a point where the electric force from the other
two charges cancels out. Suppose that the third charge is positive. Then it will be
attracted to the −2.0 µC charge, and repelled by the 4.0 µC charge. However, since
the positive charge is bigger we should place the third charge to the left of the negative
charge since the repulsion will be smaller than the attraction. Suppose that the third
charge is negative. In this case the forces will be opposite, but we still see that we
should place the charge to the left of the −2.0 µC charge.
So, suppose that we call the distance from the negative charge to the third charge x.
Then, the attractive force on the third charge (denoted q) from the negative charge
will be
1 (2 × 10−6 ) q
.
F− =
4π0
x2
The repulsive force from the other charge will be
F+ =
1 (4 × 10−6 ) q
.
4π0 (x + L)2
If the third charge is to be static then the two forces must cancel, F− = F+ . The third
charge cancels off (meaning it doesn’t matter whether it’s positive or negative), and
we have
2 × 10−6
4 × 10−6
=
⇒ (x + L)2 = 2x2 .
x2
(x + L)2
Squaring and rearranging gives
√
√
2L ± 4L2 + 4L2
2
2
x − 2xL − L = 0 ⇒ s =
= L ± 2L.
2
We want the third charge to be to the left of the negative charge, so we take the
negative root. So, we need the third charge on the left of the −2 µC charge, a distance
√ x = 1 − 2 L ≈ 0.41L.
1
2. A point particle that has charge +q and unknown mass m is released from rest in a
~ that is directed vertically downward. The
region that has a uniform electric field E
√
particle hits the ground at a speed v = 2 gh, where h is the initial height of the
particle. Find m in terms of E, q, and h.
————————————————————————————————————
Solution
When the particle starts out at height h, it has two forces acting on it, both pointing
downward: gravity, Fg = mg, and the electric force FE = qE. So, Fnet = qE + mg,
. For a constant
and is constant. So, the constant acceleration is a = Fnet /m = g + qE
m
acceleration, the particle travels a distance h in time t, such that
r
1 2
2h
h = at ⇒ t =
.
2
a
q
√
So, after this time, the particle is moving at velocity v = at = a 2h
=
2ha. But,
a
√
√
√
we’re told that it hits the ground with speed v = 2 gh, and so 2 gh = 2ha, or
a = 2g. Since a = g + qE/m, we see that
qE
qE
=g⇒m=
.
m
g
(You could figure out the form of this expression just by looking at the units.)
2
3. An electron (charge −e, mass m) and a positron (charge +e, mass m) revolve around
their common center of mass under the influence of their attractive Coulomb force.
Find the speed v of each particle in terms of e, m, k, and their separation distance L.
————————————————————————————————————
Solution
Since the two charges have the same mass, they will have the same speed. If they are
separated by a distance L then the electrical force between them is just the Coulomb
force,
ke2
F = 2.
L
However, the particles are moving around in a circle and so the net force on them is
the centripetal force, with radius r = L/2,
F =
2mv 2
ke2
mv 2
=
= 2.
L/2
L
L
Solving for the speed gives
r
v=
3
ke2
.
2mL
Applying Coulomb’s law for electric fields and the superposition of fields in
interval IV yields:
−
3.00 μC
(x − 0)
2
+
4.00 μC
(x − 0.200 m )
−
2
4. Two charged spheres, each of mass m and charge
The
roots
of thisfrom
equation
are point
at x =by0.170
q, are
suspended
a common
stringsm
of lengthIV.
L. Each string makes an angle θ with
interval
the vertical as shown in the figure to the right.
(a) Show thatthe electric field is zero
Summarizing,
p II:
at two locations in interval
q = 2L sin θ
97.2 μC
(x − 0.320 m )
2
=0
and x = 0.220. Neither of these are in
x = 0.0508 m and x = 0.169 m
(mg/k) tan θ,
where k is the Coulomb constant.
74 ••
Two point particles, each of mass m and charge q, are suspended from
(b) Find point
q if m by
= threads
10.0 g, Lof=length
50.0 cm,
and thread makes an angle θ with the
a common
L. Each
◦
θ = 10.0 .
vertical as shown in Figure 21-44. (a) Show that q = 2 L sin θ (mg k ) tan θ
————————————————————————————————————
where
k is the Coulomb constant. (b) Find q if m = 10.0 g, L = 50.0 cm, and
θ = 10.0º.
Solution
(a) The two spheres are in static equilibrium, such
that the
of the forcesEach
on each point
is zero. There
Picture
thesumProblem
are three forces acting on them, as seen in the figparticle is in static equilibrium under
ure to the right. There is the gravitational
force,
r
the influence
of
the
tension
T
,
the
pointing down, the tension pointing up along the
r
string, and
the F
electrical
force, pointing to the
gravitational
force
g , and the electric
right
r (for the right-hand sphere). Working out
the
We can use Coulomb’s law to
force FE .components,
P
relate the electric
force to the charge on
P Fx = −T sin θ + Fe = 0
F
θ − mg =and
0.
y =
each particle and
theirT cos
separation
the conditions
static equilibrium
to
So, solving for
for tension
in the first equation,
in intoto
the
second
tellson
us each
that
relate plugging
these forces
the
charge
particle.
Fe = mg tan θ.
y
r
T
θ
r
Fe
x
r
r
Fg = m g
and
2
, where r is the sepaNow, the electrical force is just the Coulomb law, Fe = kq
r2
ration distance between the charges. The separation distance can be written in
terms of the length of the string and angle as
r = 2L sin θ,
such that the forces equal to
kq 2
= mg tan θ.
4L2 sin2 θ
Solving for the charge gives
r
q = 2L sin θ
4
mg tan θ
,
k
as claimed.
(b) Here we just need to plug in the given numbers,
q
q = 2L sin θ mg ktan θ
q
= 2 × 0.5 × sin (10◦ ) 0.01×9.8
tan (10◦ )
9×109
= 0.241 µC.
5
5. A point particle of mass m and charge q is constrained
to move vertically inside a narrow, frictionless cylinder,
as seen in the figure to the right. At the bottom of the
cylinder is a point charge Q having the same sign as q.
(a) Show that the top particle will be in equilibrium at
a height
p
y0 = kqQ/mg.
(b) Show that if the particle is displaced from its equilibrium postion by a small amount and released, it
will exhibit simple harmonic motion with angular
frequency
p
ω0 = 2g/y0 .
Hint: use the binomial expansion
(1 + )α ≈ 1 + α,
for small .
————————————————————————————————————
Solution
(a) The equilibrium position, which we’ll call y0 , occurs when the electrical force
balances the gravitational force, FE = Fg , or
s
kqQ
kqQ
= mg ⇒ y0 =
.
2
y0
mg
(b) Now, suppose that we shift move the top charge by a small distance y away from
the equilibrium position. Then it’s no longer in equilibrium and so it experiences
a net force,
kqQ
− mg = ma,
(y0 + y)2
where a is the acceleration of the charge. Now, let’s rewrite the first term using
the binomial expansion, (1 + )α ≈ 1 + α, for small ,
kqQ
(y0 +y)2
=
kqQ
1
y02 (1+y/y0 )2
=
kqQ
y02
kqQ
y02
1+
y
y0
−2
≈
1 − 2 yy0
= mg − 2 mg
y,
y0
where we have used our result from part (a) to rewrite y0 . Thus, the total force
is
2mg
2mg
ma = mg −
y − mg = −
y.
y0
y0
6
Finally, the acceleration may be written as a =
d2 y
,
dt2
such that
d2 y 2g
+ y = 0.
dt2
y0
This is precisely the differential equation of a harmonic oscillator,
if we identify
p
ω0 = 2g/y0 .
7
d2 y
dt2
+ ω02 y = 0,