Midterm Exam
Name:Phys221 Fall 2007 Dr. P. Hanlet
Show your work!!! If I can read it, I will give you partialcredit!!! Correct answers without work will NOT get full credit.
Concept (5 points)
1. For the following travelling wave, label the following plot for:
• Wavelength λ
• Amplitude A
• Velocity ~v
• axes (both)
2 4 6
-5
-4
-3
-2
-1
1
2
3
4
5
position x
disp
lace
men
t y
A
λ v
2. Describe the following equations and discuss their differences:
• y(x, t) = 2A sin (kx) cos (ωt)This equation, as does the next one, comes from the superpositionprinciple, and describes two waves which interfere. Because the oscil-lating factor has no kx, it means that the source waves are travellingin opposite directions, and their intereference creates a standing wave(not a travelling wave).
• y(x, t) = 2A cos
(
φ
2
)
sin
(
kx − ωt +φ
2
)
(Hint is in the phase offset)
Because the oscillating term keeps the kx part, it means that thesource waves are in the same direction, and remains a travelling wave.The phase offset also provides a clue since it is the average value ofthe two source waves.
1
3. Consider two identical conducting spheres of radius d and separated by a distance D (D >> d).One sphere has a different magnitude positive net charge and the other a negative net charge. Aconducting wire (handled by an insulated device) is placed so that it simultaneously touches bothspheres, and then is removed.
• Describe the resultant charge on each sphere and discuss why there is this charge distribution.When the conductor touches both spheres simultaneously, the forcesbetween the spheres cause a rearrangement of the charges. Now thatthere is a path on which the charges may travel, then the only way tominimize the force is to make the charges on the two spheres identical(not necessarily neutral), which is exactly what occurs.
• What is the sign of the initial force on one sphere due to the other sphere?
Since the sign of the charges is opposite, then the force will be nega-tive. This comes from Coulomb’s Law:
~F = kq1q2
r2
12
r12
If q1 = q and q2 = −q then the product gives q1q2 = −q2, and the forceis negative, i.e. attractive.
• What is the sign of the final force on one sphere due to the other sphere?Since, by the first part, the charges on the two spheres are identical,then—again by Coulomb’s Law—the force is positive, i.e. repulsive.
2
4. In the figure below, the dashed lines are equipotential surfaces and the solid arrows are electric field
lines. Points A, B, and C lie on the equipotential surfaces and have potentials VA, VB, and VC ,
respectively. Consider a positive test charge, q. In terms of the variables given, what is the change
in potential energy in moving the q from, and for each will ∆U be positive or negative?
The potential energy is related to the potential by the equation:
∆V = Vf − Vi =∆U
q=⇒ ∆U = q(Vf − Vi)
FieldLines
SurfacesEquipotential
C
BA
• A to B
∆U = q(VB − VA)
where ∆U <0 since q>0 and VA >VB
• B to A
∆U = q(VA − VB)
where ∆U >0 since q>0 and VA >VB
• B to C
∆U = q(VC − VB) = 0
where ∆U =0 since B & C are same potential
• A to C
∆U = q(VC − VA) = q(VB − VA)
where ∆U <0 since q>0 and VA >VB
5. In your own words, state Gauss’ Law and for what is it used.
Gauss’ Law is
q = ε0Φ = ε0
∮
~E · d~s
This means that Gauss’ Law gives a relationship between the total chargeenclosed in a Gaussian surface and the electric field produced by thecharge. It can therefore be used to calculate the field is the charge isknown, or to calculate the charge if the field is known. All of this requiresa judicious choice of the Gaussian surface to simplify the calculations.
3
6. What property must an electric field posses which has Φ 6=0 through a Gaussian surface?In order that the flux not equal zero, the field must vary. If it did notvary, then the same flux would be entering and leaving the Gaussiansurface such that the net flux is zero. Note that in order for the field tovary, there must be a net charge inside the Gaussian suface.
7. Why is the net electric field of a dipole not zero, since the net charge of a dipole is zero?Though the net charge of a dipole is zero, there will always exist anelectric field associated with the dipole, the “dipole field”. This is becausethe charges are separated, and therefore the electric field at any pointdue to the nearer charge will always be stronger than the further charge.This means that there will always be a net field.
4
Problem (10 points)
8. Given: two blocks (m=1kg and M =10kg) and a spring (k=200 N/C). The blocks are arranged as
in the figure below and the surface on which the larger block moves is frictionless. The coefficient of
static friction between the two blocks is µ = 0.4. At what amplitude of simple harmonic motion of
the spring-block system puts the smaller block on the verge of slipping?
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m
M
µk
The only way for the small block to slip offof the larger block is if there is a sufficientforce to overcome the force of friction be-tween the two blocks. This force comesfrom the acceleration of the two block sys-tem due to the spring; as the blocks oscil-late, they accelerate, which (by N2L)
means that there must be a force. The reason that the amplitude playsa roll is that the amount of acceleration depends on the maximum am-
plitude; this is because the frequency depends only on the mass and thespring constant, k; i.e.:
ω =
√
k
mTherefore, the two block with spring system will always oscillate at thesame frequency. Since the frequency is fixed for the system, then the
acceleration must depend on how large is the amplitude of oscillation;the large the amplitude, the larger the acceleration so that a completecycle may occur at that fixed frequency.
Setting the frictional force equal to the force yielding the acceleration of
the smaller block, and noting that the acceleration is maximum at theturning points, one can solve for the maximum amplitude, A:
Ff = −µN = −µmg
= ma = m(
−ω2A cos (ωt + φ))
−µmg = −mω2A = −m
(
k
(m+M)
)
A
µmg =
(
m k
(m+M)
)
A
Solving for A:
A =µ(m+M)g
k=
0.4 · (1kg + 10kg)(9.8m/s2)
200N/m
such that:
A = 0.22m
5
9. An electron is accelerated eastward at 1.8× 109m/s2 by an electric field. Given the electron mass asme = 9.11 × 10−31kg, and the electron charge as qe = 1.6 × 10−19C, determine:
• the magnitude of the electric field.
Coulomb’s Law gives:
~F = kqe Q
r2r = qe
~E
for an electric field due to charge Q.
Equating this to Newton’s second law:
~F = m~a = qe~E
Considering only the magnitudes, and solving for | ~E|:
E =qe
ma=
9.11 × 10−31kg · 1.8 × 109m/s2
−1.6 × 10−19C
or
E = 1.02 × 10−2 N/C
• the directions of the electric field. Why?
Since the electron is travelling is eastward, and since the electron will
be attracted to the positive charge and simultaneously repelled bythe negative charge; this means that the charge distribution which
creates the field has positive charge on the the east side and negativeon the west. Hence, the direction of the electric field is westward, or
~E = −1.02 × 10−2 N/C i
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10. A sinusoidal wave is traveling on a string, of linear density µ = 4g/cm, with speed 40cm/s. Thedisplacement of the particles of the string at x = 10cm is found to vary with time according to theequation:
y = 5cm · sin[
1 −(
4s−1)
t]
If the wave is of the formy(x, t) = A sin (kx ± ωt),
what are the
(a) frequency of the wave?
Solving these problems means comparing the two equations. Here,one can see that ωt =
(
4s−1)
t, or ω = 4rad/s. Since ω = 2πf , then:
f =ω
2π= 0.637Hz
(b) wavelength of the wave?
v = λf =⇒ λ =v
f=
40cm/s
0.637/s= 62.83cm/s = 0.63m/s
(c) amplitude?
Comparing the two equations: A = 5cm
(d) k?
Comparing the two equations:
kx = 1 =⇒ k =1
x=
1
10cm= 0.1/cm
(e) ω?
ω = 4rad/s , see part (a)
(f) correct choice of sign for ω? Why?Comparing the two equations: The direction of the wave isdetermined by the sign of the time component of the phase.−ωt = −
(
4s−1)
t, which means that the wave is travelling in the posi-tive x-directions. Hence here the sign must be negative, since bothtime and frequency must have positive values.
(g) the tension in the string?
Since:
v =
√
τ
µ=⇒ τ = µv2 = 4g/cm · (40cm/s)2 = 6, 400g · cm/s2
orτ = 6, 400dyn = 0.064N
7
11. A particle of charge q = 3.0 × 10−6 C is d = 12cm distance from a second particle of charge Q =
−1.5× 10−6 C. Calculate the magnitude of the electrostatic force between the two charges. Will the
force be attractive or repulsive? Why?
(ε0 =8.85 × 10−12 C2/N/m2 and k=8.99 × 109 N · m2/C2)
Coulomb’s Law yields:
~F = kqQ
r2r = 8.99 × 109 N ·m2/C ·
(
3 × 10−6 C) (
−1.5 × 10−6 C)
(0.12m)2r
such that:~F = −2.8 r N
The magnitude, therefore is 2.8 N , and the force is attractive since thesign is negative; i.e. opposite charges attract.
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12. Consider four charges: q1 = q, q2 = −2q, q3 = 2q, q4 = −q, which originate infinitely far apart and
are brought into the configuration below. How much work is required to set up this arrangement?y
x
q 2 q 3
q 4q 1 d
d
Bringing each charge in to its final position frominfinity will require work to be done on the par-
ticle. The amount of work depends on the fieldthat must be overcome to bring in the charge:
W = qV
For the first charge, there is no field to overcome,
hence no work is required.For the second charge, it will have to overcome the field due to the first
charge; hence:
W2 = q2V1 = kq2 q1
d= −k
2q2
dFor the third charge, note that potentials sum:
V =
n∑
i=1
Vi
such that:
W3 = q3V1 + q3V2 = kq3 q1√
2d+ k
q3 q1
d= k
2q2
√2d
− k4q2
d=
kq2
d
(
2√2− 4
)
Then, for the fourth charge:
W4 = q4V1+q4V2+q4V3 = kq4 q1
d+k
q4 q2√2d
+kq4 q3
d= −k
q2
d+k
2q2
√2d
−k2q2
d=
kq2
d
(
2√2− 3
)
Finally, the total work is:
W = W1 + W2 + W3 + W4 =kq2
d
[
(−2) +
(
2√2− 4
)
+
(
2√2− 3
)]
such that:
W =kq2
d
(
4√2− 9
)
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13. A uniformly charged conducting sphere of diameter d = 1m has a surface charge density of σ =8.1 µC/m2.(ε0 =8.85 × 10−12 C2/N/m2 and k=8.99 × 109 N · m2/C2)
(a) Determine the net charge on the sphere.
The net charge on the surface of the sphere can be found from:
σ = Q/A =⇒ Q = A · σ
Here:
Q = 4π ·(
d
2
)2
· σ = 4π · (0.5m)2 · 8.1 × 10−6C/m2
such thatQ = 2.54 × 10−5C
(b) What is the total electric flux leaving the surface of the sphere?
From Gauss’ Law
q = ε0Φ =⇒ Φ =q
ε0
Here:
Φ =q
ε0
=2.54× 10−5 C
8.85× 10−12 C2/N/m2
such that
Φ = 2.87 × 106 N · m2/C
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14. The electric potential at points in an xy-plane is given by V = (2.0 V/m2)x2 − (3.0 V/m2)y2. In
unit-vector notation, what is the electric field at the point (3m, 2m)?
The s-component of the electric field can be determined from the poten-tial by:
Es = −∂V
∂sSo:
Ex = −∂V
∂x= − ∂
∂x
(
(2.0 V/m2)x2 − (3.0 V/m2)y2)
= (−4.0 V/m2)x
and
Ey = −∂V
∂y= − ∂
∂y
(
(2.0 V/m2)x2 − (3.0 V/m2)y2)
= (6.0 V/m2)y
such that:~E = −i(4.0 V/m2)x + j(6.0 V/m2)y
Finally, evaluating at (3m, 2m):
~E = −i(12.0 N/C) + j(12.0 N/C)
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