TECHNICAL PHYSICS II TEST NO. 2 SPRING 2008
1. Two very thin spherical shells of charge with different radii are
arranged concentrically (centered on the same point) and insulated
from each other. The inner shell has a charge of +20 nC and a radius
of 20 cm. The outer shell has a charge of -20 nC and a radius of
40 cm. Use the principle of superposition to find the electric
potential as a function of the radius r for r < 20 cm, 20 cm < r
< 40 cm, and r > 40 cm. What is the electric field as a function of
r between the shells? Express your answers in SI units.
The region where r is greater than 40 cm is the easiest in this case,
because the shells look like point charges as far as the electric
potential is concerned. Since they have equal and opposite charges,
the net point charge is zero, hence the electric potential must be
zero for r > 40 cm, taking the zero level at infinity.
Inside the outer shell but outside the inner shell, you see a constant
potential due to the outer shell and a point-charge potential due to
the inner shell. The potential due to the outer shell must be the
same as that due to the shell at its surface (calculated at a very
small distance, practically equal to 40 cm, above the surface such that
the shell still appears to be a point charge). Therefore the
potential inside the outer shell due to that shell must be
Vout = k(-q) / r = (9 × 109 V·m/C)(-20 × 10-9 C) / (0.4 m) = -450 V, for r < 40 cm.
The potential outside the inner shell is that of a point charge of
+ 20 nC. Hence that potential is
Vout = kq / r = (180 V·m) / r for r > 0.20 m.
To this potential you have to add that due to the outer shell,
therefore the potential for 20 cm < r < 40 cm is
V = (180 V·m) / r + -450 V.
At the edge of the inner shell (an "infinitesimal" distance above
the surface), the potential due to that shell is
V = (180 V·m) / (0.20 m) = 900 V.
The potential inside a spherically symmetric charge distribution (with
no other charges in the vicinity) has to be constant, and potential
has to be continuous. Therefore the potential for r < 0.20 m due to
the inner shell charge is 900 V and that due to the outer shell is
-450 V. Hence, for r < 20 cm.
V = 450 V.
The electric field is zero in regions of constant potential, so it is
only nonzero between the shells. There, it can be calculated from
E = -dV/dr = 180 V·m / r2 in the positive radial direction.
2. One form of pair production occurs when a gamma-ray photon passes
near an atomic nucleus and is converted into an electron-positron
pair. An electron and positron have the same mass (9.11E-31 kg) and
equal and opposite charges (e = 1.60E-19 C). Say a pair is created
this way, each having a speed of 2.18E+6 m/s and moving in opposite
directions when separated by a distance of 5.29E-11 m. Assuming no
other charges are in the vicinity (and they are now distant from the
nucleus), how far apart will they get before coming to (a momentary)
rest? Use conservation of energy.
If v is the common speed of the particles, and r1 is their
initial separation, the energy of the system is
K1 + U1 = 2(1/2)mv2 + k(e)(-e) / r1,
where q is the charge on the sphere and e is the magnitude of the
electron charge. The energy computes to
K1 + U1 = (9.11 × 10-31 kg)(2.18 × 106 m/s)2 + (9 × 109 N·m2/C2)(1.6 × 10-19 C)(-1.6 × 10-19 C) / (5.29 × 10-11 m) = -2.60× 10-20 J.
Since the electrostatic force is conservative, this must also be its
energy when it is at maximum separation. At this point, its energy is
given algebraically by
K2 + U2 = 0 + k(e)(-e) / r2,
where K2 = 0 and U2 = -(2.30 × 10-28 J·m) / r2.
Since the initial and final energies must be the same, you equate them
and solve for r2.
r2 = 8.89 × 10-9 m.
3. A parallel-plate air capacitor is made using two plates, each with an
area of 0.0150 square meter spaced 2.50 mm apart. It is connected
to a 300-V power supply and fully charged, and then disconnected from
the power supply. What are the capacitance, charge on the plates,
and energy stored in the capacitor? Now, with the power supply still
disconnected, the plates are moved together to a separation of 1.00
mm. What are the capactance, voltage between the plates, and stored
energy now? What happened to the missing energy?
The capacitance of a parallel-plate air capacitor is given by
C = εoA / d.
Plug and chug the numbers above and C = 53.1 pF.
The charge on each plate is
Q = VC = (300 V)(5.31 × 10-11 F) = 15.9 nC.
The energy stored can be gotten from
U = (1/2)QV = (0.5)(1.59 × 10-8 C)(300 V) = 2.39 μJ.
From the equation for capacitance you note the capacitance scales as
the inverse of the separation of the plates. Therefore the new
capacitance is
C = (53.1 pF)(2.50 mm / 1.00 mm) = 133 pF.
The charge on each plate is still 15.9 nC, because it is trapped.
Since the capacitance has increased by a factor of 2.5, the relation-
ship V = Q/C tells you the voltage has decreased by that factor. The
new voltage is therefore
V = (300 V) / (2.5) = 120 V.
Finally, if the charge is the same and the voltage has decreased by a
factor of 2.5, the energy has also decreased by a factor of 2.5. The
new energy stored is
U = (2.39 μJ) / (2.5) = 0.956 μJ.
In order to move the plates together you had to do negative work due
to the attraction of the oppositely charged plates. Doing negative
work on the capacitor reduced the stored energy.
4. 4.00 m of wire consists of 2.50 m of steel (resistivity 2.00E-7 ohm-m)
followed by 1.50 m of aluminum (resistivity 2.75E-8 ohm-m) of equal
diameter (both 5.00 mm). A potential difference of 20.0 mV is applied
across the composite wire. (a) What is the total resistance of the
wire? (b) How much current flows through the wire and what are the
voltages across the steel part and the aluminum part?
(a)Use the formula relating resistance to resistivity, R = ρ l / A,
where R is the resistance, ρ is the resistivity, l is the length of
the wire, and A the cross-sectional area. Since there are two wires,
with different lengths but the same diameter,
R = ρsteellsteel/A + ρAllAl/A,
where
A = π(0.0025 m)2 = 1.96 × 10-5 m2
Then
R = [(2.00 × 10-7 Ω·m)(2.5 m) + (2.75 × 10-8 Ω·m)(1.5 m)] / (1.96 × 10-5 m2)
= 2.76 × 10-2 Ω = 27.6 mΩ
(b) I = 20 mV / (027.6 mΩ) = 0.726 A
is the current through the wire. From the first calculation in part
(a), you can find the resistance across the steel is 25.5 mΩ.
The voltage across the steel is then
Vsteel = (0.726 A)(25.5 mΩ) = 18.5 mV.
Since the total voltage is 20 mV, the voltage across the Al has to be
VAl = 20.0 mV - 18.5 mV = 1.5 mV.
5. Determine the unknown emf E and the unknown resistance R in the
circuit below. What is the change in voltage going from point a
to point b?
<- 0.50 A 8 ohms E |
----------/\/\/\/\/---*------| |----------
| a | |
| |
| <- 1.00 A 8 ohms 20 V | |
|----------/\/\/\/\/---*------| |----------|
| b | |
| R |
----------/\/\/\/\/-----------------------
Top loop, starting at upper right and going counterclockwise:
E - (0.5 A)(8 Ω) + (1 A)(8 Ω) - 20 V = 0
E - 4V + 8 V - 20 V = 0
E = 16 V
Outer loop, starting at upper right and going counterclockwise:
16 V - (0.5 A)(8 Ω) - (1.5 A)R = 0
R = 8 Ω
Va→b = -4 V + 8 V = 4 V