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Department of Physics: 8.02
8.02 Final Exam Review Session Problems Solutions
Problem 1: Spherical Capacitor
A capacitor consists of two concentric spherical shells. The outer radius of the inner shell
is a = 0.1 m and the inner radius of the outer shell is b = 0.2 m .
a) What is the capacitance C of this capacitor?
b) Suppose the maximum possible electric field at the outer surface of the inner shell
before the air starts to ionize is Emax (a) = 3.0 × 106 V ⋅ m -1 . What is the maximum
possible charge on the inner capacitor?
c) What is the maximum amount of energy stored in this capacitor? Also show that
Q 2 / 2C = ∫ uE dV .
d) When E(a) = 3.0 × 106 V ⋅ m -1 what is the potential difference between the shells?
Solution:
The shells have spherical symmetry so we need to use spherical Gaussian surfaces.
Space is divided into three regions (I) outside r ≥ b , (II) in between a < r < b and
(III) inside r ≤ a . In each region the electric field is purely radial (that is E = Er̂ ).
Region I: Outside r ≥ b :
Region III: Inside r ≤ a :
These Gaussian surfaces contain a total charge of 0, so the electric fields in these regions
must be 0 as well.
Region II: In between a < r < b : Choose a Gaussian sphere of radius r . The electric flux
 
2
E
on the surface is 
∫∫ ⋅ d A = EA = E ⋅ 4π r
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The enclosed charge is Qenc = +Q , and the electric field is everywhere perpendicular to
the surface. Thus Gauss’s Law becomes
E ⋅ 4π r 2 =
Q
Q
⇒E=
ε0
4πε 0 r 2
That is, the electric field is exactly the same as that for a point charge.
Summarizing:
⎧
Q
 ⎪
r̂ for a < r < b
E = ⎨ 4πε 0 r 2
⎪ 
elsewhere
⎩ 0
We know the positively charged inner sheet is at a higher potential so we shall calculate
a
 
ΔV = V (a) − V (b) = − ∫ E ⋅ d s = − ∫
a
b
b
a
Q
Q
Q ⎛ 1 1⎞
dr =
=
−
>0
2
4πε 0 r b 4πε 0 ⎜⎝ a b ⎟⎠
4πε 0 r
which is positive as we expect.
We can now calculate the capacitance using the definition
C=
C=
4πε 0
4πε 0 ab
Q
Q
=
=
=
b− a
ΔV
Q ⎛ 1 1⎞ ⎛ 1 1⎞
−
−
4πε 0 ⎜⎝ a b ⎟⎠ ⎜⎝ a b ⎟⎠
4πε 0 ab
(0.1 m)(0.2 m)
=
= 2.2 × 10−11 F .
b− a
(9 × 109 N ⋅ m 2 ⋅ C2 )(0.1 m)
Note that the units of capacitance are ε 0 times an area ab divided by a length b − a ,
exactly the same units as the formula for a parallel-plate capacitor C = ε 0 A / d . Also note
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that if the radii b and a are very close together, the spherical capacitor begins to look very
much like two parallel plates separated by a distance d = b − a and area
⎛ a + b⎞
⎛ a + a⎞
≈ 4π
= 4πa 2 ≈ 4πab
⎝ 2 ⎠
⎝ 2 ⎠
2
A ≈ 4π
2
So, in this limit, the spherical formula is the same at the plate one
4πε 0 ab ε 0 4π a 2 ε 0 A
.
C = lim

=
b→ a b − a
d
d
b) Suppose the maximum possible electric field at the outer surface of the inner shell
before the air starts to ionize is E(a) = 3.0 × 106 V ⋅ m -1 . What is the maximum
possible charge on the inner capacitor?
Solution:
The electric field E(a) =
Q
. Therefore the maximum charge is
4πε 0 a 2
Qmax = 4πε 0 Emax (a)a 2 =
(3.0 × 106 V ⋅ m -1 )(0.1 m)2
= 3.3 × 10−6 C
(9 × 109 N ⋅ m 2 ⋅ C2 )
c) What is the maximum amount of energy stored in this capacitor?
Solution:
The energy stored is
U max
2
Qmax
(3.3 × 10−6 C)2
=
=
= 2.5 × 10−1 J .
−11
2C
(2)(2.2 × 10 F)
In the region between the two shells, the electric field energy density is given by
2
⎛ Q ⎞
Q2
.
uE = (1/ 2)ε 0 E = (1/ 2)ε 0 ⎜
=
2⎟
32π 2ε 0 r 4
⎝ 4πε 0 r ⎠
2
To determine the total stored energy, integrate the energy density over a spherical shell of
radius r , thickness dr , and volume 4π r 2 dr ,
r=b
r=b
Q2
Q2
dr
Q2 ⎛ 1 1 ⎞ Q2
2
U E = ∫ uE dV = ∫
4
π
r
dr
=
=
2
4
∫ r 2 8πε 0 ⎜⎝ a − b ⎟⎠ = 2C .
8πε 0 r=a
r=a 32π ε 0 r
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d) When E(a) = 3.0 × 106 V ⋅ m -1 what is the potential difference between the shells?
Solution:
We can find the potential difference two different ways.
Using the definition of capacitance we have that
ΔV =
2
Q 4πε 0 E(a)a (b − a) E(a)a(b − a)
=
=
C
4πε 0 ab
b
(3.0 × 106 V ⋅ m -1 )(0.1 m)(0.1 m)
ΔV =
= 1.5 × 105 V
(0.2 m)
.
We already calculated the potential difference in part a):
ΔV =
Q ⎛ 1 1⎞
.
−
4πε 0 ⎜⎝ a b ⎟⎠
Q
Q
or
= E(a)a 2 . Substitute this into our expression for
2
2
4πε 0 a
4πε 0
potential difference yielding
Recall that E(a) =
⎛ 1 1⎞
(b − a)
(b − a)
ΔV = E(a)a 2 ⎜ − ⎟ = E(a)a 2
= E(a)a
ab
b
⎝ a b⎠
in agreement with our result above.
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Problem 2 Faraday’s Law, Induced Electric Fields, and Lorentz Force Law.
An electron of the mass me , and charge q = −e is constrained to move in a circle of
radius r by a time changing non-uniform magnetic field. Assume that the magnet has
cylindrical symmetry about the central axis passing through the poles and that the plane
of the orbit is perpendicular to that symmetry axis.

Denote the magnetic field by B(r) = Bz (r)k̂ where k̂ points form the north pole to the
south pole. Assume that z -component of the magnetic field, Bz (r) , varies as a function
of the distance r from the symmetry axis, that it is symmetric about this axis, and that it
is perpendicular to the plane of the orbit of the electron. The magnitude of the average
magnetic field over the electron’s circular orbit is
Bave =
1
∫∫ B dA .
π r 2 disk
Find a condition relating the rate of change in time of the average magnetic field
dBave / dt , to the rate of change in time of the magnetic field, dB / dt , in order for the
electron to stay in a circular orbit.
Solution: Let’s apply Faraday’s Law to a circle of radius r between the poles of the
magnet and shown in the figure below.
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Then Faraday’s Law
 
 
d
E
⋅
d
s
=
−
B
∫
∫∫ ⋅ d a ,
dt disk
circle
implies that
E2π r = −π r 2
 ⎞
d⎛ 1
2 dBave
.
B
⋅
d
a
=
−
π
r
∫∫
⎟
dt ⎜⎝ π r 2 disk
dt
⎠
We can solve this equation for the electric field and find that

r dBave
E=−
θ̂ .
2 dt
The force on the electron that is in a circular orbit or radius r with a tangential
component of the velocity vθ , is then

  
⎛ r dBave
⎞
⎛ r dBave
⎞
Fe = −e(E + v e × B) = −e ⎜ −
θ̂ + vθ θ̂ × Bz (r)k̂ ⎟ = −e ⎜ −
θ̂ + vθ Bz (r)r̂ ⎟ ,
⎝ 2 dt
⎠
⎝ 2 dt
⎠
where Bz (r) is the z -component of the magnetic field at radius z -component of the


magnetic field r . We now apply Newton’s Second Law, Fe = mea , to the electron and
obtain tangential and radial equations of motion
dv
er dBave
θ̂ = me θ θ̂ ,
2 dt
dt
−evθ Bz (r)r̂ = −(mevθ 2 / r)r̂ ⇒ Bz (r) =
mevθ
.
er
We can integrate the tangential equation and find that
vθ =
er
B .
2me ave
Then the z -component of the magnetic field becomes
Bz (r) =
mevθ
m er
B
= e Bave = ave .
er
2meer
2
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Problem 3: RC Circuit
(a) With the assumed directions of the current as shown, what does Kirchhoff’s 2nd Law
give you if you move clockwise around the lower loop (e.g. through the battery and
R1 ). Be careful of your signs.
− I1 R1 + V = 0 .
(b) With the assumed directions of the current as shown, what does Kirchhoff’s 2nd Law
give you if you move clockwise around the outer loop (e.g. through the battery, the
capacitor, and R2 ). Be careful of your signs! What is the relation between I 2 and
Q(t) ?
−
Q
− I R +V = 0.
C 2 2
The relation between I 2 and Q(t) is dQ / dt = I 2 , i.e., the rate at which the capacitor
plate is charged is equal to the current across the resistor R2 .
(c) Just after the switch is closed what are the values of I1 , I 2 , and Q(t = 0) ?
After the switch is closed, the currents are I10 =
V
,
R1
I 20 =
V
, and Q(t = 0) = 0 .
R2
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(d) A long, long time after the switch is closed what are the values of I1 , I 2 , and Q(t)
?
A long, long time after the switch is closed, I1 =
V
,
R1
I 2 = 0 , and Q = CV .
(e) Find expressions for I 2 (t ) and Q(t) good for all t > 0 .
Note that since R1 is directly across the battery it is really just a distracter – it in no way
influences the current through the capacitor. You can also see this in the equation from
(b):
Q
Q
dQ
0 = − − I 2 R2 + V = − − R2
+V .
C
C
dt
So we can just write down the solution for the time dependence of something that
increases in time (the charge) and that decreases in time (the current) with time constant
t = R2C :
Q(t) = Q f (1− e−t/τ ) and I 2 (t) = I 0 e−t/τ .
Q f just depends on the final voltage, which is the same as the voltage across the battery:
Q f = CV .
The initial current is what you get when the capacitor is essentially a short:
I0 =
V
.
R2
(f) The switch is now opened again after a long time. How long after the switch is
opened does it take the charge on the capacitor to fall to 1/ e of its value just before
the switch was opened?
After the switch is reopened again, the upper loops acts as a discharging RC -circuit,
with Req = R1 + R2 (since the two resistors are connected in series.)
The time it takes for the charge to fall to 1/ e of its initial value is given by the time
constant
τ = Req C = (R1 + R2 )C .
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Problem 4 Consider a plane parallel capacitor of plate separation d and plate area A .
The capacitor is fully charged with charge Q on the positive plate. Ignore edge effects.
Express your answers in terms of the given quantities.
a) Derive an expression for the magnitude of the electric field between the capacitor
plates. Draw the direction of the electric field on the sketch above.
Solution: We use Gauss’s Law
  qenc
E

∫∫ ⋅ dA = ε 0 to find that
EAcap =
σ Acap
ε0
.

Choose a unit vector k that points from the positively charged plate to the negatively
charged plate. Then the electric field between the plates is

σ 
Q 
E=
k=
k
ε0
Aε 0
Note: we have assumed that there are no edge effects i.e. the electric field is zero outside
the region between the plates.
b) What is the electric potential difference between the plates? Which plate is at a
higher potential?
Solution: The potential difference between the plates is given by
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+
z =0
z =0
 

Q 
Q
Q
φ (+) − φ (−) = − ∫ E ⋅ dr = − ∫
k ⋅ dz k = − ∫
dz =
d.
Aε 0
Aε 0
Aε 0
−
z=d
z=d
The positive plate is at a higher potential.
c) What is the capacitance C ?
Solution: The capacitance is given by
C=
Aε
Q
Q
=
= 0
φ (+) − φ (−) (Q / Aε 0 )d
d
A current I flows counterclockwise as seen from above through each turn of a long air
core cylindrical shaped solenoid that has N turns, length h , and radius a . Ignore edge
effects.
d) Derive an expression for the magnitude of the magnetic field inside the solenoid.
Draw the direction of the magnetic field on the sketch above.
Solution: We use Ampere’s Law to calculate the magnetic field
∫
 
B ⋅ dr = µ0 I through .
loop
If we choose a rectangular loop of length l , then Ampere’s Law becomes
Bl = µ0
N
lI
h
So the magnetic field inside the solenoid is given by

N
B = µ0 I k̂
h
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where k̂ is a unit vector pointing upward.
e) What is the self-inductance L of the solenoid?
Solution: The self-inductance of the solenoid is given by
L=
Φtotal
mag
I
=N
Φturn
mag
I
=
N
I
  NBπ a 2 N 2 µ0π a 2
B
∫∫ ⋅ da = I = h
turn
The capacitor is now connected to the solenoid to form a resonant circuit without any
resistance. The capacitor is initially charged with charge Q0 on the positive plate, and
when the switch is closed the current in the circuit is found to undergo sinusoidal
oscillations.
f) What is the period T of oscillation of the charge on the capacitor plate?
Solution: The period of oscillation is given by
⎛ N 2 µ0π a 2 ⎞ ⎛ Aε 0 ⎞
⎛ π A⎞
2π
T=
= 2π LC = 2π ⎜
= 2π Na µ0ε 0 ⎜
⎟
⎜
⎟
ω
h
⎝ hd ⎟⎠
⎝
⎠⎝ d ⎠
g) A time interval Δt = T / 4 has passed since the switch was closed. What is energy
stored in the capacitor? What is energy stored in the inductor?
Solution: One-quarter cycle later the capacitor is uncharged and the current is maximum.
The energy in the capacitor is zero and the energy in the inductor is equal to the initial
energy in the circuit,
Q 2 Q 2d
U mag = E = 0 = 0
2C
ε0 A
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Problem 5: Coaxial Cable and Power Flow
A coaxial cable consists of two concentric long
hollow cylinders of zero resistance; the inner
has radius a , the outer has radius b , and the
length of both is l , with l >> b , as shown in
the figure. The cable transmits DC power from
a battery to a load. The battery provides an
electromotive force ε between the two
conductors at one end of the cable, and the load
is a resistance R connected between the two
conductors at the other end of the cable. A current I flows down the inner conductor
and back up the outer one. The battery charges the inner conductor to a charge −Q and
the outer conductor to a charge +Q .

(a) Find the direction and magnitude of the electric field E everywhere.
Answer:
Consider a Gaussian surface in the form of a cylinder with radius r and length l, coaxial
with the cylinders. Inside the inner cylinder (r<a) and outside the outer cylinder (r>b) no
charge is enclosed and hence the field is 0. In between the two cylinders (a<r<b) the
charge enclosed by the Gaussian surface is –Q, the total flux through the Gaussian
cylinder is
 
ΦE = 
∫∫ E ⋅ dA = E(2π rl)
Thus, Gauss’s law leads to E(2π rl) =
qenc
, or
ε0

q
Q
E = enc r̂ = −
r̂ (inward) for a < r < b, 0 elsewhere
2π rl
2πε 0 rl

(b) Find the direction and magnitude of the magnetic field B everywhere.
Answer:
Just as with the E field, the enclosed current Ienc in the Ampere’s loop with radius r is
zero inside the inner cylinder (r<a) and outside the outer cylinder (r>b) and hence the
field there is 0. In between the two cylinders (a<r<b) the current enclosed is –I.
Applying Ampere’s law,


∫ B ⋅ d s = B(2π r) = µ I
0 enc
, we obtain

µI
B = − 0 ϕ̂ (clockwise viewing from the left side) for a < r < b, 0 elsewhere
2π r
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
(c) Calculate the Poynting vector S in the cable.
Answer:
For a < r < b , the Poynting vector is
 1  
⎞ ⎛ µ I ⎞ ⎛ QI ⎞
1 ⎛
Q
S=
E×B =
−
r̂ ⎟ × ⎜ − 0 ϕ̂ ⎟ = ⎜ 2 2 ⎟ k̂ (from right to left)
⎜
µ0
µ0 ⎝ 2πε 0 rl ⎠ ⎝ 2π r ⎠ ⎝ 4π ε 0 r l ⎠

On the other hand, for r < a and r > b , we have S = 0 .

(d) By integrating S over appropriate surface, find the power that flows into the coaxial
cable.
Answer:

With dA = (2π r dr )kˆ , the power is
 
P=
∫∫ S ⋅ dA =
S
b 1
⎛ b⎞
QI
QI
(2π rdr) =
ln ⎜ ⎟
∫
2
2
2π ε 0 l ⎝ a ⎠
4π ε 0 l a r
(e) How does your result in (d) compare to the power dissipated in the resistor?
Answer:
Because
 
b
ε = ∫E⋅d s = ∫
a
Q
Q
⎛b⎞
dr =
ln ⎜ ⎟ = IR
2π rlε 0
2π lε 0 ⎝ a ⎠
the charge Q is related to the resistance R by Q =
2π ε 0 lIR
. The above expression for P
ln(b / a)
becomes
⎛ 2π ε 0lIR ⎞ I
⎛ b⎞
P=⎜
ln ⎜ ⎟ = I 2 R
⎟
⎝ ln(b / a) ⎠ 2π ε 0 l ⎝ a ⎠
which is equal to the rate of energy dissipation in a resistor with resistance R.
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Problem 6: Charging Capacitor
A parallel-plate capacitor consists of two circular plates, each with radius R , separated
by a distance d . A steady current I is directed towards the lower plate and away from
the upper plate, charging the plates.

a) What is the direction and magnitude of the electric field E between the plates?
You may neglect any fringing fields due to edge effects.
Solution: If we ignore fringing fields then we can calculate the electric field using
Gauss’s Law,

∫∫
closed
surface
  Q
E ⋅ d a = enc .
ε0
By superposition, the electric field is non-zero between the plates and zero everywhere
else. Choose a Gaussian cylinder passing through the lower plate with its end faces
parallel to the plates. Let Acap denote the area of the end face. The surface charge density
is given by σ = Q / π R 2 . Let k̂ denote the unit vector pointing from the lower plate to the
upper plate. Then Gauss’ Law becomes

σ Acap
.
E Acap =
ε0
which we can solve for the electric field
 σ
Q ˆ
E = kˆ =
k.
ε0
π R 2ε 0
b) What is the total energy stored in the electric field of the capacitor?
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Solution: The total energy stored in the electric field is given by
1
1
U elec = ε 0 ∫ E 2 dV = ε 0 E 2π R 2 d .
2 volume
2
Substitute the result for the electric field into the energy equation yields
2
U elec
1 ⎛ Q ⎞
1 Q2d
2
.
= ε0 ⎜
π
R
d
=
⎟
2 ⎝ π R 2ε 0 ⎠
2 π R 2ε 0
c) What is the time rate of change of the energy stored in the electric field?
Solution: The rate of change of the stored electric energy is found by taking the time
derivative of the energy equation
d
Qd dQ
.
U elec =
dt
π R 2ε 0 dt
The current flowing to the plate is equal to
I=
dQ
.
dt
Substitute the expression for the current into the expression for the rate of change of the
stored electric energy yields
d
QId
.
U elec =
dt
π R 2ε 0

d) What is the magnitude of the magnetic field B at point P located between the
plates at radius r < R (see figure above). As seen from above, is the direction of
the magnetic field clockwise or counterclockwise. Explain your answer.
Solution: We shall calculate the magnetic field by using the generalized Ampere’s Law,
∫
closed path
 
B ⋅ d s = µ0
 
 
d
J ⋅ da + µ0ε 0
E ⋅ da
∫∫
dt open surface
open surface
∫∫
We choose a circle of radius r < R passing through the point P as the Amperian loop
and the disk defined by the circle as the open surface with the circle as its boundary. We
choose to circulate around the loop in the counterclockwise direction as seen from above.
This means that flux in the positive k̂ -direction is positive.
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The left hand side (LHS) of the generalized Ampere’s Law becomes
LHS =
∫
  
B ⋅ d s = B 2π r .
circle
The conduction current is zero passing through the disk, since no charges are moving
between the plates. There is an electric flux passing through the disk. So the right hand
side (RHS) of the generalized Ampere’s Law becomes

dE 2
 
d
RHS = µ0ε 0 ∫∫ E ⋅ da = µ0ε 0
πr .
dt disk
dt
Take the time derivative of the expression for the electric field and the expression for the
current, and substitute it into the RHS of the generalized Ampere’s Law:
RHS = µ0ε 0

dE
dt
π r2 =
µ0 I π r 2
π R2
Equating the two sides of the generalized Ampere’s Law yields

µ Iπ r 2
B 2π r = 0 2
πR
Finally the magnetic field between the plates is then

µI
B = 0 2 r; 0<r <R.
2π R
The sign of the magnetic field is positive therefore the magnetic field points in the
counterclockwise direction (consistent with our sign convention for the integration
direction for the circle) as seen from above. Define the unit vector θ̂ such that is it
tangent to the circle pointing in the counterclockwise direction, then

µI
B = 0 2 r θ̂ ; 0 < r < R .
2π R
e) Make a sketch of the electric and magnetic field inside the capacitor.
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
f) What is the direction and magnitude of the Pointing vector S at a distance r = R
from the center of the capacitor?
Solution: The Poynting vector at a distance r = R is given by

1  
S(r = R ) =
E×B .
r=R
µ0
Substituting the electric field and the magnetic field (setting r = R ) into the above
equation, and noting that k̂ × θ̂ = −r̂ , yields

µI
1 Q
Q
I
S(r = R) =
k̂ × 0 θ̂ =
(−r̂) .
2
2
µ0 π R ε 0
2π R
π R ε 0 2π R
So the Poynting vector points inward with magnitude

S(r = R ) =
Q
I
.
2
π R ε 0 2π R

g) By integrating S over an appropriate surface, find the power that flows into the
capacitor.
Solution: The power flowing into the capacitor is the closed surface integral
P=

∫∫


S(r = R) ⋅ da .
closed surface
The Poynting vector points radially inward so the only contribution to this integral is
from the cylindrical body of the capacitor. The unit normal associated with the area
vector for a closed surface integral always points outward, so on the cylindrical body
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
da = da rˆ . Use this definition for the area element and the power is then
P=
∫∫
cylindrical
body


S ( r = R ) ⋅ da =
Q
I
(−rˆ ) ⋅ da rˆ
2
π R ε 0 2π R
cylindrical
∫∫
body
The Poynting vector is constant and the area of the cylindrical body is 2π Rd , so
P=
Q
I
Q
I
QId
(−r̂) ⋅ da r̂ = −
2π Rd = −
.
2
2
2
2
π
R
2
π
R
π
R
ε
π
R
ε
π
R
ε
cylindrical
0
0
0
∫∫
body
The minus sign correspond to power flowing into the region.
h) How does your answer in part g) compare to your answer in part c)?
Solution: The two expressions for power are equal so the power flowing in is equal to the
change of energy stored in the electric fields.
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics: 8.02
Problem 7: Superposition of Two Travelling Waves
Suppose the electric field of an electromagnetic wave is given by the superposition of two

waves E = E0 cos(kz − ω t) î + E0 cos(kz + ω t) î . You may find the following identities
useful cos(kz ± ω t) = cos(kz)cos(ω t)  sin(kz)sin(ω t) .

a) What is the associated magnetic field B(x, y, z,t) ?
Answer: We treat each contribution to the electric field separately. Then we have that
 E
E
B = 0 cos(kz − ω t) ĵ − 0 cos(kz + ω t) ĵ ,
c
c
where we used that fact that that the amplitude of the magnetic field is related to the
amplitude of the electric field by B0 = E0 / c , and the direction of the field satisfy
 

dir(E × B) = dir(propagation) . The contribution E1 = E0 cos(kz − ω t) î is propagating in

the positive k̂ -direction, and the contribution E2 = E0 cos(kz + ω t) î is propagating in the
negative k̂ -direction.
b) Use the identity cos(a)sin(a) = (1/ 2)sin(2a) to show that the energy per unit
area per unit time (the Poynting vector) transported by this wave is

S = (E0 2 / cµ0 )sin(2kz)sin(2ω t)k̂ .
Answer: We first use the identities cos(kz ± ω t) = cos(kz)cos(ω t)  sin(kz)sin(ω t) , to
rewrite the electric field as

E = E0 cos(kz − ω t) î + E0 cos(kz + ω t) î = 2E0 cos(kz)cos(ω t) î .
The corresponding expression for the magnetic field is
 E
E
2E0
B = 0 cos(kz − ω t) ĵ − 0 cos(kz + ω t) ĵ =
sin(kz)sin(ω t) ĵ .
c
c
c
The Poynting vector is then
 1  
⎡ 2E
⎤
1
S=
E × B = ⎡⎣ 2E0 cos(kz)cos(ω t) î ⎤⎦ × ⎢ 0 sin(kz)sin(ω t) ĵ⎥
µo
µo
⎣ c
⎦
 4E 2
S= 0 cos(kz)cos(ω t)sin(kz)sin(ω t)k̂
c µ0
.
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics: 8.02
1
Now use the identity cos(a)sin(a) = sin(2a) to rewrite the Poynting vector as
2
 E0 2
S=
sin(2kz)sin(2ω t)k̂ .
c µ0
c) What is the time-average of the Poynting vector? Briefly explain your answer.
Answer: Note the time average of the Poynting vector is given by
T

1 
S ≡ ∫ Sdt .
T 0
Therefore
T

1 E0 2
S = ∫
sin(2kz)sin(2ω t) dtk̂
T 0 µ0 c
T
E2 1
= 0 sin(2kz) ∫ sin(2ω t) dtk̂
µ0 c T
0
.
The integral is then
T
T
1
1
sin(2ω t) dt = −
cos(2ω t) 0 = 0 ,
∫
T 0
2ω T
where we used the fact that 2ω T = 4π and cos(2ω T ) = cos(4π ) = 1 . Thus


S = 0.
This result may seem surprising at first. If we rewrite the z -component of the Poynting
vector as
S z (t) = A(t)sin(2kz) ,
where
E0 2
A(t) =
sin(2ω t)
µ0 c
is the time dependent amplitude. Then the spatial dependence sin(2kz) remains fixed
with respect to time i.e. there is no propagation. Only the amplitude A(t) varies
sinusoidally with time.
E2
In the graph below, we plot S z (t) = 0 sin(2ω t)sin(2kz) vs. z for various values of
µ0 c
2ω t = 0, ± π / 12, ± π / 6, ± π / 4, ± π / 3, ± π / 2 .
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics: 8.02
We see that the nodal structure does not change. If we time average S z (t) we see that
the sinusoidal amplitude averages to zero over one cycle.
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics: 8.02
Problem 8 Measuring the Wavelength of Laser Light
Suppose you shine a red laser through a pair of narrow slits, each of width a = 40 µm ,
separated by a distance d = 250 µm and allow the resulting interference pattern to fall on
a screen a distance L = 40cm away. This set up is shown in the figure to the right.
a) Will the center of the pattern (directly between the two holes) be an interference
minimum or maximum?
Answer: The center of the pattern will be a maximum because the waves from both slits
travel the same distance to get to the center and hence are in phase.
(b) For the sizes given above, will these maxima be roughly equally spaced, or will
they spread out away from the central peak?
Answer: We get a maximum every time that the extra path length is an integral number
of wavelengths:
δ = d sin θ = mλ
The spacing is the distance between these locations, ym+1 − ym . We can get ym from θ :
sin θ m =
ym
L +y
2
2
m
=
mλ
.
d
Let α m ≡ ym / L2 + ym2 . Then
αm =
(
ym2
⇒ 2
= α m2
2
2
2
L + ym
L + ym
ym
)
ym2 1− α m2 = α m2 L2 ⇒ ym =
⎛ α2 ⎞
≈ α m L ⎜ 1− m ⎟ .
2 ⎠
⎝
1− α m2
αmL
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics: 8.02
We have made the approximation that α m << 1 , which is valid for the wavelengths and
slit separations of this lab (it is order 10−3 ). As long as this approximation is valid, we
can also ignore the term that goes like α m 2 , and hence we find the maxima are equally
spaced:
λL
ym+1 − ym ≈
.
d
(c)
Approximately how many interference maxima will you see on one side of the
pattern before their intensity is significantly reduced by diffraction due to the finite width
a of the slit?
Answer: The first single slit minimum appears at asin θ = λ . So when we approach:
m=
d sin θ d λ d
=
=
λ
λa a
we will lose signal due to the diffraction minimum.
(d) Derive an equation for calculating the wavelength λ of the laser light from your
measurement of the distance Δy between interference maxima.
Answer: Using what we derived for part b,
Δy = ym+1 − ym ≈
(e)
λL
⇒
d
λ=
dΔy
.
L
In order to most accurately measure the distance between maxima Δy , it helps to
have them as far apart as possible. (Why?) Assuming that the slit parameters and
light wavelength are fixed, what can we do in order to make Δy bigger? What
are some reasons that can we not do this ad infinitum?
Answer: We can increase the distance to the screen and measure the distance between
distant interference maxima (e.g. m = 1 and m = 4), which increases distances, making
them easier to measure, and then allows us to divide down any measurement errors.
Single Slit Interference
Now that you have measured the wavelength λ of the light you are using, you will want
to measure the width of some slits from their diffraction pattern. When measuring
diffraction patterns (as opposed to the interference patterns of problem 1) it is typically
easiest to measure between diffraction minima.
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics: 8.02
(f)
Derive an equation for calculating the width a of a slit from your measurement of
the distance Δy between diffraction minima.
Answer: Single slit minima obey the relationship a sin θ = mλ , which is the same
formula as two slit maxima. So we can calculate the slit width from what we derived in
part 1b (replacing the distance between the slits d with the width of the single slit a ):
a=
(g)
λL
.
Δy
What is the width of the central maximum (the distance on the screen between the
m = −1 and m = +1 minima)? How does this compare to the distance Δy
between other adjacent minima?
Answer: The central minimum is twice as wide as the distance between other minima. It
is:
λL
Δycentral = 2
.
a
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics: 8.02
Problem 8 Second-Order Bright Fringe
A monochromatic light is incident on a single slit of width 0.800 mm, and a diffraction
pattern is formed at a screen which is 0.800 m away from the slit. The second-order
bright fringe is at a distance 1.60 mm from the center of the central maximum. What is
the wavelength of the incident light?
Solution:
The general condition for destructive interference is
sin θ = m
λ y
≈
a L
(1)
where small-angle approximation has been made. Thus, the position of the m-th order
dark fringe measured from the central axis is
ym = m
λL
a
(2)
Let the second bright fringe be located halfway between the second and the third dark
fringes. That is,
1
1
λ L 5λ L
y2b = ( y2 + y3 ) = (2 + 3)
=
2
2
a
2a
(3)
The approximate wavelength of the incident light is then
λ≈
2a y2b 2(0.800 ×10− 3 m)(1.60 ×10− 3 m)
=
= 6.40 ×10− 7 m .
5L
5(0.800 m)
(4)