Physics 9 Fall 2009
Midterm 1 Solutions
For the midterm, you may use one sheet of notes with whatever you want to put on it, front
and back. Please sit every other seat, and please don’t cheat! If something isn’t clear, please
ask. You may use calculators. All problems are weighted equally. PLEASE BOX YOUR
FINAL ANSWERS! You have the full length of the class. If you attach any additional
scratch work, then make sure that your name is on every sheet of your work. Good luck!
1. Recalling that the electric field at a distance r inside a uniformly charged sphere, of
radius R, carrying charge Q, is
~ (~r) =
E
Q
rr̂,
4π0 R3
determine the electric potential V (r) at a point r inside the sphere, relative to infinity.
————————————————————————————————————
Solution
The potential is given in terms of the electric field as
Z r
~ · d~`.
E
V (r) = −
∞
Inside the sphere, the electric field is given as above. Outside the sphere, the electric
field is just that of a point charge,
~ (~r) =
E
1 Q
r̂.
4π0 r2
So, in order to find the potential, we have to integrate the electric field over both
regions, breaking up the integral from ∞ to R, the radius of the sphere. Then we
integrate the internal field from R to r. Since we’re integrating over a radial direction,
we have r̂ · d~` = dr. So, integrating gives
Z R
Z
1
1 dr
1 Q r
rdr .
V (r) = −
Q
+
2
4π0
4π0 R3 R
∞ 4π0 r
Performing the integrations gives
h
i
R R 1 dr
Rr
1
1 Q
V (r) = − 4π
Q
+
rdr
2
3
4π0 R
∞ 4π0 r
0
R
Q 1 R
Q
1
2 r
=
−
3r
=
4π0 r ∞
2 4π0 R
1 Q
1 1 Q
(r2
−
4π0 R
2 4π0 R3
R
− R2 ) .
Combining the first and last terms gives the potential inside
Q
r2
V (r) =
3− 2 .
8π0 R
R
1
2. Show that the electric field due to an infinitely long, uniformly charged thin cylindrical
= η is given by the following
shell of radius a having a surface charge density Q
A
expressions:
0 for 0 ≤ r < a
E=
ηa
for
r ≥ a.
0 r
————————————————————————————————————
Solution
We can solve this problem using Gauss’s
law, which says that
I
~ · dA
~ = Qencl .
E
0
The situation is as seen in the figure to the
right. Because of the cylindrical symmetry
we choose a cylinder of radius r and length
L as our Gaussian surface. As we’ve seen
many times, the electric field is constant
on the Gaussian surface and points along
the direction of the normal to the surface,
and so
I
~ · dA
~ = EA,
E
η
r
a
r
where A = 2πrL is the surface area through which the electric field is fluxing (note that
the ends of the cylinder don’t contribute to the flux since the surface is perpendicular
to the electric field there). So, the net flux is 2πrLE. Now we just need to work out the
enclosed charge. For a Gaussian surface inside the cylinder, the enclosed charge is zero,
= 0,
since all the charge lies on the surface. So, inside the cylinder, 2πrLE = Qencl
0
and so inside the cylinder, E = 0.
Outside the cylinder, a Gaussian surface of length L encloses a certain amount of
charge Qencl . This can be given in terms of the surface charge density, η, writing
Qencl = ηAencl , where Aencl is the area enclosed by the Gaussian surface. The area of
the charge is just the area of the shell, Aencl = 2πaL, since all the charge is located on
the surface at r = a. So, Qencl = η (2πaL). Putting everything together gives
(2πrL) E =
η (2πaL)
ηa
⇒E=
.
0
0 r
So, we find
E=
for 0 ≤ r < a
for
r ≥ a,
0
ηa
0 r
which is exactly right.
2
3. Suppose we have two identical metal spheres, each of
mass m, and carrying a positive charge Q. These two
spheres are placed inside a plastic cylindrical can of raQ
dius equal to that of the spheres so that they just fit and
can move freely up and down inside the can. The can
and charges are sitting upright in Earth’s gravity, and
so gravity pulls them down. However, since they both
carry positive charges they repel each other. The charge
h
Q
is such that the top charge floats a height h above the
bottom charge, which sits on the floor of the can. Find
the height, h, of the top sphere above the center of the
bottom sphere in terms of the charges on the spheres and
the acceleration due to gravity.
————————————————————————————————————
Solution
Because the spheres are charged and are separated by a distance, h, at equilibrium,
they have an electrostatic force
Felec =
1 Q2
.
4π 0 h2
Now, in a gravitational field the top sphere experiences a gravitational force
Fgrav = mg.
If the system is in equilibrium then the two forces are equal. So, comparing the two
2
forces gives Felec = 4π10 Qh2 = mg = Fgrav . Solving for the height gives
s
h=
3
Q2
.
4π 0 mg
4. Consider the RC circuit shown in the diagram. In your homework, you showed that
if the switch is closed at time t = 0, then
the charge on the capacitor at time t was
given by the expression
Q = Qmax 1 − e−t/τ ,
where the time constant, τ ≡ RC.
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C
(a) In terms of τ , how long does the capacitor take to charge to half of its maximum
value?
(b) What is the potential across the capacitor, ∆VC , at a time t?
(c) What happens to the potential across the capacitor, ∆VC at very late times?
(d) Show that Qmax = CE.
(e) What is the current, I(t), in the capacitor at a time t in terms of R, C, and E?
————————————————————————————————————
Solution
=
(a) At a time t0 , the
charge is half of the maximum charge, so Q(t0 ) = Qmax
2
Qmax 1 − e−t0 /τ . So, canceling the common factor of Qmax and rearranging gives
e−t0 /τ = 12 . Taking natural logarithms gives t0 = ln (2) τ ≈ 0.69τ .
(b) Since Q = CV , the potential is just
V =
Qmax
Q
=
1 − e−t/τ .
C
C
(c) At very late times the capacitor must saturate to the full potential given by the
battery, E.
(d) Using our result from part (c), as t → ∞, the exponential term vanishes, and so
V → E = Qmax
, and so Qmax = CE.
C
(e) The current in the capacitor is just I = dQ
. Taking the derivative gives, I(t) =
dt
Qmax −t/τ
d
−t/τ
Qmax 1 − e
= τ e
. But, since Qmax = CE, and τ = RC, we find
dt
I(t) =
4
E −t/RC
e
.
R
Extra Credit Question!!
The following is worth 10 extra credit points!
where and σ are constants,
and r is the distance between the molecules. The
potential energy is plotted in
the figure to the right. The
vertical axis is in units of ,
while the horizontal axis is
in units of σ.
Energy
The potential energy between a pair of neutral
atoms or molecules is very
well-approximated by the
Lennard-Jones
Potential,
given by the expression
σ 12 σ 6
,
P E(r) = 4
−
r
r
Molcular Bond Energy
8
7
6
5
4
3
2
1
0
0.75
-1
1
1.25
1.5
1.75
2
2.25
2.5
2.75
3
-2
-3
-4
Distance
(a) Why does the potential energy approach zero as the distance gets bigger?
(b) At what separation distance, in terms of σ and , is the potential energy zero?
(c) At approximately what distance is the system in equilibrium? What is the potential energy at that distance? (Express your answers in terms of σ and .)
(d) How much energy would you need to add to the system at equilibrium in order
to break the molecular bonds holding it together? Why?
(e) How much energy is released in the breaking of those molecular bonds? Why?
Note - no calculation is needed to answer these problems!
————————————————————————————————————
Solution
(a) As the two molecules get further apart, the attractive force between them gets
weaker and weaker. When the are very far apart, they hardly interact at all they are basically free molecules. The potential energy of a free particle is zero,
since potential energy depends on the interaction between multiple particles.
(b) We can just read the value off from the graph. We see that the potential energy
crosses the x axis when x = 1, which means that r = σ. We can see this from the
equation, too: setting r = σ gives P E(σ) = 0.
5
3.25
(c) The system is in equilibrium when the net force on it is zero. Since the force is
the slope of the potential energy graph, this happens when the slope is zero. The
potential energy graph has zero slope when it’s at it’s minimum point. Checking
the graph, we see that this happens right around x ≈ 1.15, or r ≈ 1.15σ. We could
d
(P E(r)) = 0, which gives r = 21/6 σ ≈ 1.12σ,
check the exact answer by finding dr
and so we were close on our guess. The energy at this distance can just be read
off the graph, giving y = −3, or P E = −3.
(d) In order to break the molecular bonds apart, we’d need to raise the energy to
zero. At equilibrium the energy is P E = −3, and so we’d need to add +3 units
of energy.
(e) There is no energy released in breaking these molecular bonds - we had to add
the energy to break these bonds. Energy is never released in the breaking of
bonds! One can obtain energy by breaking a less stable bond, then forming a
more stable bond. The more stable bond has a more negative potential energy
(a deeper potential “well”). The difference in energy between the initial and final
states is released to the environment. This is where the energy comes from in the
ATP reactions, and not by releasing energy from the breaking of bonds!
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