Download Problem Set 3: Solutions

University of Alabama
Department of Physics and Astronomy
PH 102 / LeClair
Summer II 2012
Problem Set 3: Solutions
1. In Rutherford’s famous scattering experiments that led to the planetary model of the atom,
alpha particles (having charge +2e and masses of 6.64 × 10−27 kg) were fired toward a gold nucleus
with charge +79e. An alpha particle, initially very far from the gold nucleus, is fired at a speed of
vi = 2.00 × 107 m/s directly toward the nucleus, as shown below.
a) How close does the alpha particle get to the gold nucleus before turning around? Assume the
gold nucleus remains stationary, and that energy is conserved.
b) What will the acceleration of the alpha particle be at the moment it reverses direction?
+79e
+2e
+2e
vi
vf = 0
d
In essence, this is a conservation of energy problem. The two energies of interest are the kinetic
energy of the alpha particle, and the potential energy of the alpha particle-gold nucleus pair. Since
the two are both positively charged, they will repel each other, and their electrical potential energy
will be positive at any finite distance. If the alpha particle is initially very, very far away, we can
approximate their starting separation as infinite, meaning their initial electrical potential energy is
zero, and the total energy of the system is just the alpha particle’s kinetic energy.
How close does the alpha particle get? When it used all of its initial kinetic energy up as electrical
potential energy. The closer it gets, the larger the electrical potential energy, and the more kinetic
energy it must spend. At some point, it is all gone, and the particle instantaneously stops and then
turns around. At that point of closest approach, the alpha particle’s kinetic energy is zero. Comparing the energy in the initial and final cases will allow us to find the distance of closest approach d.
To go further, we must make one approximation: even though v/c ∼ 0.067, and we should in
principle use the relativistic equation for kinetic energy, we will ignore it for simplicity. The error
we will incur will be only of order v/c, so . 10% at best. With that out of the way, we can apply
conservation of energy. Let the alpha particle have charge qa and the gold nucleus charge qg , we
can plug in numbers later.
Ki + Ui = Kf + Uf
ke qa qg
1
mvi2 + 0 = 0 +
2
d
8.99 × 109 2 · 1.60 × 10−19 79 · 1.60 × 10−19
2ke qa qg
d=
≈
mvi2
(6.64 × 10−27 ) (2.00 × 107 )2
≈ 2.74 × 10−14 m = 27.4 fm
(1)
(2)
(3)
(4)
What is the acceleration at the moment it reverses? At this point the alpha particle is a distance
d from the gold nucleus. The force between them at that instant is just the electric force, and the
alpha particle’s acceleration will be its net force divided by its mass. Basically: acceleration comes
from force, and there is only one force present here.
a=
1
1 k e qa qg
F =
m
m d2
(5)
Using our previous expression for d,
a=
ke qa qg
ke qa qg m2 vi4
mvi4
=
=
≈ 7.3 × 1027 m/s2
md2
m 4ke2 qa2 qg2
4ke qa qg
(6)
2. An interstellar dust grain, roughly spherical with a radius of 3 × 10−7 m, has acquired a negative
charge such that its electric potential is −0.15 Volts.
(a) How many extra electrons has it picked up?
(b) What is the strength of the electric field at its surface?
If it is spherical, Gauss’ law tells us that we may treat it as a point charge (so long as we are outside
the dust grain, anyway). The excess charge must therefore be equivalent to a point charge which
at a distance d = 3 × 10−7 m creates a potential of −0.15 Volts. If there are n excess electrons on
the dust grain, the net charge is qnet = −ne. Thus,
−kne
kqnet
=
d
3 × 10−7 m
n ≈ 31 electrons
−0.15 V =
(7)
Here we rounded to the nearest integer for n. The same point charge would produce an electric
field at a distance of 3 × 10−7 m of
E=
−kne
(3 × 10−7 m)2
≈ 5 × 105 V/m
(8)
3. Calculate the total electric potential energy in the following two cases:
a) Four particles of equal charge q are located at the corner of a square of side a while a fifth
charge of value −4q is placed at the center (Figure (a) below).
b) Four equal and opposite charges (of magnitude 2q each) are located at the corner of a square
of side a (Figure (b) below)
q
q
(a)
2q
-2q
(b)
-4q
q
a
q
-2q
a
2q
Using the principle of superposition, we know that the potential energy of a system of charges is
just the sum of the potential energies for all the unique pairs of charges. The problem is then
reduced to figuring out how many different possible pairings of charges there are, and what the
energy of each pairing is. The potential energy for a single pair of charges, both of magnitude q,
separated by a distance d is just:
P Epair =
ke q 2
a
We need figure out how many pairs there are, and for each pair, how far apart the charges are.
Once we’ve done that, we need to figure out the two different arrangements of charges and run the
numbers.
Case (a): Label the upper left charge q1 and number the other corner charges q2 -q4 clockwise,
with the center charge as q5 . The possible pairings are then only
q1 q2 , q1 q3 , q1 q4 , q1 q5
q2 q3 , q2 q4 , q2 q5
q3 q4 , q3 q5
q4 q5
There are ten possible pairings, but only three different distances between paired charges, i.e., there
is a lot of redundancy in this list, many of the terms are the same. Pairs 1 − 2, 2 − 3, 3 − 4, and 4 − 1
are all separated by a distance a and involve two charges q. Pairs 1 − 3 and 2 − 4 are separated by
√
a distance a 2 and involve two charges q. All pairings with charge 5 (4 of them) involve a charge
√
q and −4q, and are separated by a distance a 2/2. Thus, there are only three different types of
potential energy terms to account for.
PE =
ke q1 q2 ke q1 q3 ke q1 q4 ke q1 q5 ke q2 q3 ke q2 q4 ke q2 q5 ke q3 q4 ke q3 q5 ke q4 q5
+ √ +
+ √
+
+ √ + √
+
+ √
+ √
(9)
a
a
a
a
a 2
a 2/2
a 2
a 2/2
a 2/2 a 2/2
Noting that q1 = q2 = q3 = q4 = q and q5 = −4q, and that every term involves ke q 2 /a:
PE = 4
ke q 2
ke q 2
ke 4q 2
ke q 2
+2 √ −4 √
=
a
a
a 2
a 2/2
2
32
4+ √ − √
2
2
=
√ ke q 2
ke q 2 4 − 15 2 ≈ −17.2
(10)
a
a
Case (b): We can do the exact same thing for the second crystal: number the corner charges 1 − 4.
We don’t have a fifth charge, so things are a bit simpler:
PE =
ke q1 q2 ke q1 q3 ke q1 q4 ke q2 q3 ke q2 q4 ke q3 q4
+ √ +
+
+ √ +
a
a
a
a
a 2
a 2
(11)
Now must take care that q1 = q3 = 2q and q2 = q4 = −2q.
4ke q 2 4ke q 2 4ke q 2 4ke q 2 4ke q 2 4ke q 2
+ √ −
−
+ √ −
a
a
a
a
a 2
a 2
2
2
√
ke q
4
ke q
ke q 2
=
−4 · 4 + 2 · √
=
−16 + 4 2 ≈ −10.3
a
a
a
2
PE = −
(12)
(13)
Importantly, both crystals have a negative potential energy, meaning they are favorable arrangements compared to having isolated charges. The first is a bit more negative, lower in energy, so it
is more stable than the second.
4. In the circuit below, C1 = 2.0 µF, C2 = 6.0 µF, C3 = 3.0 µF, and ∆V = 10.0 V. Initially all
capacitors are uncharged and the switches are open. (a) What is the charge on C2 when switch S1
is open and switch S2 is closed? (b) What is the charge on C1 when S1 is closed and switch S2 is
open?
∆V
C1
C2
S1
C3
S2
When S1 is open and S2 closed, the battery doesn’t make a complete circuit - only one pole is
connected. The circuit does nothing, and the charge on all capacitors is zero. When S1 is closed
and S2 open, now C3 isn’t connected on one end, so it does nothing. Capacitors C1 and C2 are in
series, giving an equivalent capacitance
1
1
1
=
+
Ceq
C1 C2
=⇒
Ceq =
C1 C2
= 1.5 µF
C1 + C2
(14)
This equivalent capacitance is connected to a 10.0 V battery, so it holds charge Q = Ceq ∆V = 15 µC.
The individual series capacitors C1 and C2 making up the equivalent must have the same charge
as the equivalent, so they each have 15 µC.
5. In the circuit diagram below, the resistors represent light bulbs. In these three circuits, all the
batteries are identical and have negligible internal resistance, and all the light bulbs are identical.
Rank all 5 light bulbs (A, B, C, D, E) in order of brightness from brightest to dimmest. Justify
your rankings briefly.
(a)
(b)
(c)
(d)
(e)
∆V
∆V
∆V
Let each bulb have resistance R. By symmetry, we can expect that b and c are the same, and d and
e must also be the same. The brightness of the bulb will be proportional to the power it receives,
so ranking by brightness means ranking by power. In the first case, we have a current in the circuit
∆V /R, and the voltage is ∆V , so the power delivered to bulb a is
(∆V )2
Pa = I∆V =
R
(15)
In the second case, the current in the circuit is ∆V /2R, since the total resistance is 2R. This is
the current in each bulb. Since bulbs a and b are in series, they split the total supply voltage, and
since they are the same, they each take half of it. Thus, for either b or c the power delivered is
Pb = Pc =
∆V ∆V
(∆V )2
1
=
= Pa
2R 2
2R
2
(16)
Thus, b and c get half the power of a, since in series they only get half the voltage. In the last
case, both bulbs d and e are connected directly to the battery and have the same voltage ∆V .
That means both d and e look just like the first situation - a bulb of resistance R connected to a
battery delivering ∆V , so the power in each is Pe = Pd = Pa . Overall, we now see the ranking must
be A = D = E > B = C.