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Physics 9 Fall 2010
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 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 constant 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 g.
————————————————————————————————————
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.)
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2. A nonconducting solid sphere of radius 10.0 cm has a uniform volume charge density.
The magnitude of the electric field 20.0 cm from the sphere’s center is 1.88 × 103 N/C.
(a) What is the sphere’s volume charge density, ρ?
(b) Find the magnitude of the electric field at a distance of 5.00 cm from the sphere’s
center. (Hint: use Gauss’s law to figure out the field at this position.)
————————————————————————————————————
Solution
(a) As we’ve discussed many times, the field outside the sphere is just that of a point
charge, E = 4πQ0 r2 . We can determine the charge on the sphere by solving for
Q = 4π0 r2 E. Plugging in the numbers gives
Q = 4π0 r2 E =
1
(.200)2 (1.88 × 103 ) = 9.36 × 10−9 C.
9 × 109
For a constant charge density ρ = Q/V , and for a sphere of radius R, V = 4π
R3 ,
3
so
3 × 9.36 × 10−9
3Q
=
= 2 × 10−6 C/m3 .
ρ=
3
3
4πR
4π(.1 )
H
~ ·dA
~ = Qencl /0 .
(b) Now we want the field inside the sphere. We need Gauss’s law, E
Taking our Gaussian surface as a sphere of radius r the spherical symmetry of the
charge distribution
tells us that the field is constant on the Gaussian surface, and
H
~
~ = EA, where A = 4πr2 is the surface area. Now, because
so, as always, E · dA
the charge density is uniform, the total charge enclosed in the Gaussian surface
3
3Q
4πr3
= Qr
, and so Gauss’s law gives
is ρVencl = 4πR
3 ×
3
R3
E(4πr2 ) =
Q
Qr3
⇒
E
=
r.
0 R3
4π0 R3
Plugging in the numbers gives
E=
Q
9 × 109 (9.36 × 10−9 )
r
=
(.05) = 4212 N/C.
4π0 R3
.13
2
3. A radioactive 210 Po nucleus emits an α particle that has a charge +2e. When the
α particle is a large distance from the nucleus, it has a kinetic energy of 5.30 MeV.
Assume that the α particle had negligible kinetic energy as it left the surface of the
nucleus. The “daughter” (or residual) nucleus 206 Pb has a charge of +82e. Determine
the radius of the 206 Pb nucleus. (Neglect the radius of the α particle and assume the
206
Pb nucleus remains at rest.) Express your answer in terms of fermis (1 fm = 10−15
meters).
————————————————————————————————————
Solution
The α particle starts off at the edge of the 206 Pb nucleus, say at a distance R. If the
α particle starts with no kinetic energy, then the initial energy is purely potential, so
. The α particle is repelled from the 206 Pb nucleus, and picks up
Ei = P E = 82e×2e
4π0 R
speed. When it’s far away all the energy is kinetic, since the potential goes to zero at
large distances. The final energy is KE = 5.30 MeV, and so equating the two energies
164e2
gives 4π
= 5.30 MeV. Solving for R gives
0R
R=
164e2
4π0 × (5.30 MeV)
Now, 1 MeV = 106 ×1.602×10−19 = 1.602×10−13 joules, and so 5.30 MeV = 8.5×10−13
joules. Thus,
R=
164 × (1.602 × 10−19 )2
164e2
= 9 × 109 ×
= 4.45 × 10−14 m,
4π0 × (5.30 MeV)
8.5 × 10−13
or about 45 fm.
3
4. Consider the circuit in the figure to the
right.
(a) What is the equivalent resistance between points a and b?
(b) How would adding a fifth resistor
that has resistance R between point
c and d affect the equivalent resistance between point a and b? (Hint:
what is the potential difference between points c and d?)
————————————————————————————————————
Solution
…€ ‚
(a) This circuit is equivalent to a par„
allel circuit with two resistors in series on each side, as we see in the a „
„
figure to the right. The equivalent ðƒ„
resistance of each side of the paral„
„
lel circuit is R + R = 2R.
„€
ƒ
‚
Now we have an equivalent parallel
circuit with just two resistors, each
of resistance 2R, as seen in the next
figure. Since this is a parallel circuit, the equivalent resistance is
1
Requiv
=
„
€ ‚€ ‚
€ ‚
2R
R
R
€ …‚
„
„
b
„ƒ…ò
„
„
€ „ƒ‚
€ ‚…
„
„
„
a
„
b
„
ƒ…ò
„
„
„
and so the equivalent resistance of
the entire circuit is Requiv = R.
R
…€ ‚
ðƒ„
1
1
1
+
= ,
2R 2R
R
R
„€
ƒ
‚
2R
„
€ ‚„ƒ
(b) Because all of the resistors are identical, the points c and d are at the same
potential. This means that if we connected a wire between points c and d, then
no current would flow along it. Thus, if we added a resistor between these two
points, it would do nothing, and so the equivalent resistance of the entire circuit
would still be R.
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Extra Credit Question!!
The following is worth 10 extra credit points!
A charge q sits at the back corner of a cube, as
~ through
shown in the figure. What is the flux of E
the shaded side? If the charge was in the center,
the flux would be ΦE = 6Q0 (the total flux divided
evenly into six cube faces), but the charge is in
the corner. (Hint: Think of the symmetry of the
problem - you don’t have to do any complicated
integrals!)
————————————————————————————————————
Solution
We could try to figure out the components of the
electric field, and integrate it over the surface of
the cube, but there’s a much easier and more
clever way. Suppose that we imagine stacking
other boxes around the cube, keeping the charge
at the center, as seen in the figure to the right.
The net flux of the electric field through a volume doesn’t depend on the shape of the volume,
~ through the whole set of
and so the net flux of E
boxes is just q/0 . We only want to know the flux
through the shaded side. Since the charge is at
the center, each face of the cube has 1/6 of the
flux passing through it. The shaded side is one
quarter of the face of the cube, and so the net
flux passing through there is 1/4 of 1/6, or 1/24
of the total flux. So,
passing through
R the net flux
~ · dA
~= q .
the shaded face is E
240
5
q
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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