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Physics 9 Fall 2011
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. A solid, nonconducting sphere of radius R has a nonuniform charge density, ρ(r) = Cr,
where C is a constant, such that the total charge on the sphere is Q.
(a) Determine C in terms of the total charge Q and radius R of the sphere.
(b) What is the electric field at a distance r outside the sphere? (You don’t have to
calculate this, just give the expression.)
(c) Find the electric field at a distance r inside the sphere.
(d) Show that the two fields agree on the boundary of the sphere, when r = R.
Hint: recall that the volume element of a sphere is dV = 4πr2 dr.
————————————————————————————————————
Solution
(a) The total charge may be found
by integrating the charge density over the total
R
volume of the sphere, Q = ρ dV . So,
Z
Z R
R4
= πR4 C,
Q = ρ dV = 4πC
r3 dr = 4πC
4
0
from which we find that C =
Q
.
πR4
(b) Outside the sphere the field is simply that of a point charge,
Eout =
Q
,
4π0 r2
which can be easily found using Gauss’s law.
(c) To determine the field inside the sphere we use Gauss’s
law,
I
~ · dA
~ = Qencl .
E
0
The first step is to choose a Gaussian surface over which
to integrate. Based on the symmetry of the charge distribution we choose a spherical surface of radius r ≤ R.
As we’ve seen many times, along this surface the electric field is constant and points along the direction of the
normal to the surface.
1
r
R
H
H
~ · dA
~ = EdA cos θ =
Thus,
the
left-hand
side
of
Gauss’s
law
becomes
E
H
E dA = EA = E (4πr2 ). To determine the enclosed charge we again integrate
the charge density, only this time out to the same radius r ≤ R. Then
Z
Z r
r4
r3 dr = πCr4 = Q 4 ,
Qencl = ρ dV = 4πC
R
0
after plugging in for C. Notice that, if r = 0 we get Qencl = 0, while if r = R we
have Qencl = Q, as expected. Thus, Gauss’s law gives
Q r4
Q
⇒ Ein =
r2 ,
E 4πr2 =
4
0 R
4π0 R4
which grows with distance.
(d) When r = R the electric field outside then
Eout =
while
Q
,
4π0 R2
Q
,
4π0 R2
and so the fields agree on the boundary, as we expected.
Ein =
2
2. An uncharged, infinitely long conducting cylinder of radius a is placed in an initially
~ = E0 ĵ, such that the cylinder’s axis lies along the z axis. The
uniform electric field E
resulting electrostatic potential is V (x, y, z) = V0 for points inside the cylinder, and
V (x, y, z) = V0 − E0 y +
E0 a2 y
x2 + y 2
for points outside the cylinder, where V0 is the (constant) electrostatic potential on the
~
conductor. Use this expression to determine the resulting electric field, E.
————————————————————————————————————
Solution
~ = −∇V .
The electric field can be found from the voltage by finding the gradient, E
So,
Ex = − ∂V
∂x h
i
a2 y
∂
= − ∂x
V0 − E0 y + xE20+y
2
2x
2
= 0 + 0 + E0 a y (x2 +y2 )2
=
2E0 a2 xy
.
(x2 +y 2 )2
Next,
Ey = − ∂V
∂y
i
h
a2 y
∂
= − ∂y
V0 − E0 y + xE20+y
2
h 2 2
i
x
+y
−y(2y)
2
= 0 + E0 − E0 a
2
2 2(x2+y2 )
−x
= E0 + E0 a2 (xy2 +y
.
2 )2
Finally, the z-component is simplest of all,
Ez = −
∂V
= 0,
∂z
since nothing depends on z. Thus, the electric field is
2
2
2
2E
a
xy
y
−
x
0
2
~ =
E
î + E0 + E0 a
ĵ
(x2 + y 2 )2
(x2 + y 2 )2
This electric field is made up of the original external electric field, E0 ĵ, and the induced
field, which is everything else
2
2
2
2E
a
xy
y
−
x
0
2
~ ind =
E
î + E0 a
ĵ
(x2 + y 2 )2
(x2 + y 2 )2
3
3. Calculate the current through and power
dissipated in each resistor in the circuit
seen to the right. Enter your results in the
table below with the given resistances.
Resistor
Current (A) Power (W)
R1 = 2 Ω
3.76
28.3
R2 = 0.5 Ω
1.88
1.77
R3 = 0.5 Ω
1.88
1.77
R4 = 2 Ω
3.76
28.3
‚…
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€ ‚…
„„
„„
„„
ƒ
ƒ
„ƒ
16 V
R
R
…
…
…
„„
„„
„„
R
‚ƒ
ƒ€ ‚ € ‚€
ƒ
1
2
4
————————————————————————————————————
Solution
In order to find the current through each
resistor we need to determine the current
coming from the battery. We begin by redrawing the circuit, combining the two resistors in parallel to an equivalent circuit.
Recalling that
1
1
1
1
1
=
+
=
+
,
Reqv
R2 R3
0.5 0.5
we see that the equivalent resistor is 1/4
= 0.25 Ω, as drawn in the diagram to the
right.
Now we simply have three resistors in series, for which the equivalent resistance can
be found simply by adding the individual
resistances,
Reqv = 2 + 2 + .25 = 4.25 Ω,
…€ ‚2.0 Ω € ‚…
„„
„„
ƒ
„ƒ
16 V
0.25 Ω
…
…
„„
„„
2.0 Ω
ƒ€ ‚ € ‚ƒ
…€ ‚4.25 Ω € ‚…
„ƒ
„„
16 V
„„
€ ‚ƒ
which gives us our equivalent circuit seen
at right.
Now, finding the P
current produced by the battery is trivial using Kirchhoff’s loop law
and Ohm’s law, i ∆Vi = E − IR = 0, such that I = E/R = 16/4.25 = 3.76 amps.
This is also the current through R1 and R4 in the original circuit since these resistors
are in series with the battery. From Kirchhoff’s junction rules, the 2 amp current is
split evenly between resistors 2 and 3 since they both have the same resistance. The
power in each resistor is just P = I 2 R, which gives P1 = P4 = (3.76)2 × 2 = 28.3 W,
and P2 = P3 = (1.88)2 × (0.5) = 1.77 W.
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3
4. The most reactive isotope of uranium, 235 U , which has 92 protons and 143 neutrons,
undergoes fission under the influence of an additional neutron. One possible (simplified ) decay reaction is given below
235
U + n0 → 236 U → 91 Kr + 141 Ba + 3n0 ,
with the energy released being carried away by the kinetic energy of the products.
(a) Suppose we wanted to break the uranium nucleus apart using a proton, instead.
How fast would we need to fire the proton at the nucleus in order to overcome
the electrostatic repulsion so that the proton hits the 235 U nucleus, which has a
radius of r235 U ≈ 7.4 × 10−15 m? (Note - you can neglect the motion of the much
heavier nucleus!)
(b) The neutron breaks the nucleus apart into a 92 Kr nucleus, with 36 protons, and
a 141 Ba nucleus, with 56 protons (and also three additional neutrons). If these
nuclei are still separated by the same distance, 7.4 × 10−15 m, then what is the
electrostatic potential energy between the nuclei?
————————————————————————————————————
Solution
(a) In order to get the proton close enough to hit the nucleus, we have to give it enough
energy to overcome the electrostatic repulsion that it feels due to the positively
charged nucleus. This means that the kinetic energy, 21 mv 2 , of the proton has to
be equal to the potential energy between the proton and the nucleus, kQq
. Setting
r
these equal and solving for the velocity gives
s
r
2kQq
2 · 8.99 × 109 · 92 (1.6 × 10−19 )2
=
= 5.85 × 107 m/s.
v=
mr
1.67 × 10−27 · 7.4 × 10−15
(b) The electrostatic potential energy is just the potential energy between the Kr
and Ba nuclei, P E = kQr1 Q2 , where Q1 = 36e, and Q2 = 56e. So,
kQ1 Q2
8.99 × 109 · 36 · 56 · e2
PE =
=
= 6.3 × 10−11 J.
−15
r
7.4 × 10
5
Extra Credit Question!!
The following is worth 10 extra credit points!
Embodied in Kirchhoff’s Laws are two conservation laws. Explain what they are.
————————————————————————————————————
Solution
Kirchhoff’s loop rule, which says that the net change in voltage along a closed circuit
is zero,
X
∆Vi = 0,
i
expresses the law of conservation of energy. The change in voltage that a charge q
passes through changes its potential energy by an amount ∆P E = q∆V . So, we could
write
X
X
X
q
∆Vi =
q∆Vi =
∆P Ei = 0.
i
i
i
Thus, the loop law says that the net change in potential energy around the loop is
zero. Since ∆E = 0 around the loop for the closed system, then ∆KE = −∆P E = 0,
and so energy is conserved.
The Kirchhoff’s junction rule, which says that the sum of the currents in to a junction
is equal to the sum of the currents out of the junction,
X
X
Iin =
Iout ,
i
i
expresses the conservation of charge. This is because the current is made up of moving
charges, and since the charges don’t get lost in the junction, whatever charges come
in to the junction have to go back out again. So, charge in equals charge out, which
means that current in also equals current out.
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Some Possibly Useful Information
Some Useful Constants.
Coulomb’s Law constant k ≡
1
4π0
= 8.99 × 109
N m2
.
C2
The magnetic permeability constant µ0 = 4π × 10−7
N
.
A2
Speed of Light c = 2.99 × 108 m/s.
Newton’s Gravitational Constant G = 6.672 × 10−11
N m2
.
kg 2
The charge on the proton e = 1.602 × 10−19 C
The mass of the electron, me = 9.11 × 10−31 kg.
The mass of the proton, mp = 1.673 × 10−27 kg.
Boltzmann’s constant, kB = 1.381 × 10−23 J/K.
1 eV = 1.602 × 10−19 Joules ⇒ 1 MeV = 106 eV .
1 Å = 10−10 meters.
Planck’s constant, h = 6.63 × 10−34 J s = 4.14 × 10−15 eV s.
The reduced Planck’s constant, ~ ≡
h
2π
= 1.05 × 10−34 J s = 6.58 × 10−16 eV s.
Some Useful Mathematical Ideas.
( n+1
x
R n
n 6= −1,
x dx = n+1
ln (x) n = −1.
√
R dx
2 + x2 .
√
=
ln
x
+
a
2
2
a +x
R
√ x dx
a2 +x2
=
√
a2 + x 2 .
Other Useful Stuff.
The force on an object moving in a circle is F =
mv 2
.
r
Kinetmatic equations x(t) = x0 + v0x t + 21 ax t2 , y(t) = y0 + v0y t + 12 ay t2 .
The binomial expansion, (1 + x)n ≈ 1 + nx, if x 1.
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