Download Physics 213 — Problem Set 4 – Solutions Spring 1998

Physics 213 — Problem Set 4 – Solutions
Spring 1998
1. Reading Assignment
Serway Sections 25.1–6. Also Sections 27.1–3,6 and 28.1–2 are useful preparation for the first lab.
2. The voltage drop across a peculiar kind of resistor, which we shall call an “xistor”, is given by
V = I2 X
(2.1)
when I is positive, where X is a constant. (The xistor doesn’t allow negative I’s).
a) An xistor with X = 1.0 V /A2 is connected in series with a 3 Ω resistor and a 4 V battery. Assuming
that the battery is hooked up in the correct way to allow current through the xistor, what is the total
current that flows through the battery?
b) Repeat the previous calculation for the case where the xistor, resistor and battery are in parallel.
SOLUTION:
a)The applied voltage V is equal to the sum of the voltage drops due to the resistor and the “xistor” :
V = XI 2 + RI.
The quadratic equation can be solved to give :
I=
−R ±
√
R2 + 4XV
.
2X
We assume I positive, so we take the solution with the + sign. Substituting R = 3 Ω, V = 4 V, X = 1.0
V/A2 gives I = 1 A.
b)When the “xistor”, resistor, and battery are in parallel, then the current through the battery divides into
IX and IR , going through the xistor and the resistor respectively. However the voltages are now equal :
2
V = IX
X = IR R
giving
r
V
V
+ .
X
R
Substituting the given values for R, V , and X gives I = 3.3 A.
I = IX + IR =
3. Serway 27.46+
A particular automobile battery is characterized as “360 ampere-hours, 12V.” What total energy can this
battery deliver? Esitmate how far (vertically) up a hill one could lift a car with this amount of energy.
SOLUTION: Power is given by P = IV , so energy is E = P t = (It)V . The car is characterized by It =
360 A-hr = 1296 × 106 A-s and V =12 V, therefore the energy delivered by the battery is :
E = 1296 × 106 × 12 = 1.55 × 107 J.
This energy may be converted to potential energy (ignoring losses) by driving uphill until the battery dies.
Thus :
E
E = mgh → h =
.
mg
Using g = 9.8 m/s2 and estimating m to be around 103 kg we find h ≈ 1.5 km.
4. Serway 27.64
A stereo system has speakers, each with resistance 4 Ω. The system is rated at 60 W in each speaker, and
each speaker has a 4 A fuse. Are the fuses adequate protection against electrical overload? Explain.
1
SOLUTION: The fuse is placed in series with the corresponding speaker, and breaks the circuit when the
current through the speaker exceeds 4 A. If we want to pump up the volume to its maximum, then the
speaker will be used at full power P=60 W. The power is given by P = I 2 R, so
Imax =
p
P/R ≈ 3.87 A < 4 A.
This means that the the fuses are pretty useless since they would allow a current greater than 3.87 A which
would overload the speakers and result in a funny smell.
5. Serway 25.6
What is the potential difference needed to stop an electron having an initial speed of 4.2 × 105 m/s?
SOLUTION: The condition for the electron to stop is NOT that the force on it is zero, rather that all the
kinetic energy, mv 2 /2, has been tranformed to potential energy, eV . So we have :
1
mv 2
mv 2 = eV → V =
.
2
2e
Substituting v = 4.2 × 105 m/s, m = 9.1 × 10−31 kg, and e = 1.6 × 10−19 C, gives V = 0.5 V.
6. Serway 25.12
Two parallel plates are separated by 0.30 mm. If a 20 V potential difference is maintained between those
plates, calculate the electric field strength in the region between them.
SOLUTION: The electric field between parallel plates is constant. Hence
Z
V =−
0
d
V
~ · d~
E
x=E·d→E =
= 6.6 × 104 N/m,
d
with the electric field pointing from higher to lower potential.
7. Serway 25.20A
A particle having charge q and mass m (see Figure P25.20 in text) is connected to a string of length L
which is tied to pivot point P . The particle, string, and pivot point all lie on a horizontal table. The
particle is released from rest when the string makes an angle θ with a uniform electric field of magnitude E.
Determine the speed of the particle when the string is parallel to the electric field (that is when the particle
is at point a in the figure).
SOLUTION: Though we could set up and solve the equations of motion by resolving all the forces on the
particle into components, it is much easier in this case to use conservation of energy. Since the electric field
is uniform in the +x-direction, the electric potential is
V = −Ex.
Let’s take the origin of the x-axis at the pivot point, so that the particle starts at x = L cos θ (and some
value of y which doesn’t matter for our purposes). Thus the initial energy of the particle is all potential
energy:
Einitial = qV = −qEL cos θ.
When the particle reaches the point a, where x = L, say its velocity is v. Then
Ef inal =
1
mv 2 − qEL.
2
Equating initial and final energies gives
1
mv 2 − qEL = −qEL cos θ,
2
or,
r
v=
2qEL
(1 − cos θ).
m
2