8.02
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Problem Set 5 Solutions
Spring 2013
Problem 1 Concept Questions
Part A.
Three pairs of charged conducting spheres are connected with wires and switches. The spheres are all very far apart. The large spheres have twice the radius of the small spheres. Each sphere on the left has a charge of +20 nC and each sphere on the right has a charge of +70 nC before the switches are closed.
All of the switches are closed and then, after a long time, opened again. Which of the below rankings of the electric potentials on four of these spheres (A, C, D, E) is correct?
Explain your reasoning.
1.
V
A
< V
C
= V
D
< V
E
2.
V
E
< V
C
= V
D
< V
A
3.
V
A
= V
D
< V
C
= V
E
4.
V
C
= V
E
< V
A
= V
D
5.
V
A
< V
D
< V
C
< V
E
6.
V
E
< V
C
< V
D
< V
A
7.
The electric potential is the same for all four spheres (A, C, D & E)
8.
The ranking of the electric potential cannot be determined or is none of the above
Answer 1.
Again, when connected the spheres will come to the same potential. For AB and EF this means the same charge (45 nC) will be on each sphere. For CD less charge will be on C than on D (you can calculate it will be half as much, 30 nC on C, 60 nC on
D, but that isn’t necessary to answer the question).

Part B. Consider two hollow concentric conducting shells. They carry no net charge.
We place a negative charged point-like object at the center of the shells. Let V
O 1
be the potential at O1 and V be the potential at O2.
O 2
Which of the following statements is true? Explain your reasoning: a) The surface charge at O2 is positive and V
O 1
>
V
O 2
. b) The surface charge at O2 is positive and V
O 1
<
V
O 2
. c) The surface charge at O2 is positive and V
O 1
=
V
O 2
. d) The surface charge at O2 is negative and V
O 1
>
V
O 2
. e) The surface charge at O2 is negative and V
O 1
<
V
O 2
. f) The surface charge at O2 is negative and V
O 1
=
V
O 2
Answer e): When we place a negative charge appears on the inner surface I1. No matter where we placed the negative charge, a negative charge
-
Q appears symmetrically distributed on the outer surface O1, a positive charge
+
Q
-
Q in the center, a positive charge
appears symmetrically distributed on the inner surface I2, and a negative charge
+
Q
-
Q appears symmetrically distributed on the outer surface O2. The electric field is non-zero between the two conductors and points radially inward. We know that the electric field point sin the direction of maximally decreasing electric potential, therefore V
O 1
<
V
O 2
.
Recall that every point on the outside conductor is at the same potential V
O 2
, in particular

V
O 2
=
V
I 2
, and every point on the inner conductor is at the same potential could use Gauss’s Law to find that
E
= kQ r
ˆ
and then integrate and find that r
2
V
O 1
. We also
V
I 2
-
V
O 1
= -
I 2
O 1
×
E
×
ds r
I 2
= - r
O 1
× kQ r
ˆ × dr r
ˆ r
2
= r
I 2 × kQ r
2 dr
= kQ r
O 1
×
1
×× r
I 2
-
1 r
O 1
×
××
>
0.

Part C. A parallel plate capacitor, initially with vacuum between the plates, is connected to a battery and charged to
+
Q on the upper plate and
-
Q on the lower plate respectively.
The battery is then disconnected. A slab of material with dielectric constant k
is then inserted between the plates.
Which of the following statements is true? Explain your reasoning. a) The charge stored in the capacitor increases and the energy stored increases. b) The charge stored in the capacitor remains the same and the energy stored increases. c) The charge stored in the capacitor decreases and the energy stored increases. d) The charge stored in the capacitor increases and the energy stored remains the same. e) The charge stored in the capacitor remains the same and the energy stored remains the same. f) The charge stored in the capacitor decreases and the energy stored remains the same. g) The charge stored in the capacitor increases and the energy stored decreases. h) The charge stored in the capacitor remains the same and the energy stored decreases. i) The charge stored in the capacitor decreases and the energy stored decreases.
Answer h): The capacitor without the dielectric is initially charged and then the battery is disconnected. The free charge on the two plates must remain on those plates (there is no conducting path that would draw the charge off the plates.) Once the dielectric is introduced the induced charge produces a field in the opposite direction from the field produced by the free charge and so the electric field between the plates decreases by a factor of of
1/ k k
(because
. The potential across the plates has decreased because the electric field has decreased. Because the free charge does not change, the capacitance increases by a factor

Q
=
C k
D
V k
=
C k
æ
ææ
E
0 k d
æ
ææ
=
C k k
E
0 d
=
C k k
D
V
0
=
C
0
D
V
0
æ
.
C k
= k
C
0
Therefore the stored energy decreases because it is inversely proportional to the capacitance
U k
=
Q
2
C k
=
1 k
Q
2
C
0
=
1 k
U
0
.

Problem 2: Electrostatic Shielding
We have a visualization of a hollow conducting shell with free charges to help you better understand shielding. Open it up: http://web.mit.edu/viz/EM/visualizations/electrostatics/ChargingByInduction/shielding/shielding.htm
and play with it for a while. You can move the free charge around the outside of the shell, in the hollow region inside the shell, or even in the shell using the parameters
“radius pc” and “angle pc.” You can change which field you are looking at: the free charge + induced charges, the free charge only, or the induced charge only. You can display the electric field with grass seeds by clicking “Electric Fields: Grass Seeds” or display the equipotential streaks by clicking “Electric Potential.”
Part A. Place a free charge outside the conductor (the grey shell). Look at the electric fields of (i) the free charge + induced charges, (ii) the free charge only, (iii) the induced charge only. a) The induced charge distribution (the orange and blue small balls) is located on the outer surface of the spherically symmetric conducting shell. Explain why the electric field associated with this induced charge distribution is non-zero inside the shell.

Answer: The free charge by itself has a non-zero field inside the shell and therefore the induced charge must cancel this field in the metal of the shell and in the hollow region inside as well. b) If you placed a charged particle anywhere in the hollow region inside the conducting shell, will that charge experience a force? In the context of this case, explain what shielding means.
Answer: The charged particle will induce a charge distribution on the inner surface of the conducting shell and the outer surface of the conducting shell. There will be a force on that charge due to the electric field produced by the induced charges from the inner shell. The free charge will be attracted to the surface of the shell. Shielding now means that the induced charge distribution on the inner surface of the conducting shell will exactly cancel the electric field of the free charge placed in the hollow region everywhere in the conductor and outside. Therefore the induced charge distribution on the outer surface of the conducting shell is completely screened from any effects of the free charge in the hollow region. c) Does the electric potential have the same value at every point in the hollow region of the conductor as at every point in the conducting shell? Click “Show total field of free charge + induced charges”, and then click on “Electric Potential”. Explain how your answer is represented in the “Electric Potential” display.
Answer: yes. Recall that
E
= -Ñ
V
. Because the electric field is zero inside the hollow region, the electric potential must be constant and hence equal everywhere in the interior to the value of the potential on the conductor. There are no equipotential contours in the shell or hollow region indicating that all those points have the same value as the equipotential contour just on the outside surface of the shell but not outside the shell.
Part B. Now place a free charge inside the hollow region close to the inner surface of the conductor.

a) (i) Which surfaces of the conducting shell are the induced charges now located on? (ii) Is the inner surface charge distribution uniform or non-uniform? Explain.
(iii) How is the total charge on the inner surface related to the free charge placed in the hollow region? (iv) Is the outer surface charge distribution uniform or nonuniform? Explain. (v) What is the sum of the induced charges on both surfaces of the conducting shell?
Answer: (i) The induced charges are located on both the inner and outer surface of the conductor. (ii) The charge distribution on the inner surface is non-uniform because the free charge is not located exactly at the center of the conductor. (iii) The total charge on the inner surface is equal in magnitude but opposite in sign to the free charge located in the hollow region. (iv) The charge distribution on the outer surface is uniform because the electric field of the charge distribution on the inner surface exactly canceled the electric field of the free charge. Therefore the charge distribution on the outer surface
“sees” no effect from the interior and hence distributes itself uniformly on the outer surface. (v) Because the conducting shell is neutral the sum of the induced charges on the inner and outer surface is zero. b) (i) Click on “Show Field of Free Charge ONLY” and then click on “Electric
Field: Grass Seeds”. Describe the electric field pattern outside the conducting shell. (ii) Click on “Show field of Induced Charge ONLY” and then click on
“Electric Field: Grass Seeds”. Describe the electric field pattern in the conducting shell of the induced charges. (iii) Now click on “Show total field of free charge
+induced charges”, and then click on “Electric Field: Grass Seeds”.

Answer: (i) The electric field pattern, outside the conducting shell due to the free charge, radiates outward from the location of the free charge. It is not radially symmetric outside the shell. (ii) The electric field from the induced charges on the inner surface of the conductor produce a field in the conducting shell (and outside the conducting shell) that exactly cancels the electric field from the free charge in the hollow region. (Note: the electric field of the induced charges on the inner surface of the conductor is identical to the electric field of a point charge with the same magnitude but opposite sign of the free charge and is located at the identical point as the free charge. This is an example of the
“method of images”. The induced charge on the inner surface of the shell can be replaced by an “image point like charge”.) c) For the three regions (i) outside the conducting shell, (ii) in the conducting shell,
(iii) in the hollow region inside the conducting shell, explain in words what type of field pattern the sum of the electric fields of the free charge and induced charges produce.
Answer: (i) The electric field outside the conducting shell is identical to an electric field generated by a charged object placed at the origin with charge equal to the free charge inside the hollow shell. (Outside the conducting shell, the induced charges produced a complicated field, which when added to the non-radially symmetric field of the free charge, produce a radially symmetric field.) (ii) In the conducting shell the electric field is zero. (iii) In the hollow region inside the conducting shell, the electric field is a complicated pattern where the field lines start from the free charge and end on the induced opposite charges on the inner surface. Because the inner induced charge distribution is non-symmetric with more opposite charges nearer to the free charge, the field lines have more curvature in the region of the shell closest to the free charge. d) Is the hollow region of the conductor at the same potential as at every point in the conducting shell? Click on “Show total field of free charge + induced charges”, and click on “Electric Potential”. Explain how your answer is represented in the
“Electric Potential” display.
Answer: no. The potential is varying in the hollow region because the electric field is non-zero. We expect many equipotential contours in the hollow region.

Problem 3: Conductors
A spherical neutral conducting shell, with inner radius a and outer radius
3 a
, is wrapped around a uniformly positively charged sphere (insulator) of radius a
and positive volume charge density r as pictured below. A conducting wire is then attached to the outside surface of the conductor. The other end of the wire is attached to the outer surface of a neutral conducting spherical shell of radius
2 a
, which is located a very large distance away. a) After the wire is connected, are any electrons transferred through this wire? If yes, describe which direction the electrons will move and why? If no, why not? b) What are the final charges on the following three surfaces: (i) surface of shell of radius
2 a , (ii) outer surface of the conducting shell (outer radius 3 a ), and (iii) inner surface of conducting shell (inner radius a )? c) What is the potential (choose
V
(
¥
)
=
0
) on the two surfaces attached to the wire?
Solution: The insulating sphere has positive charge density r
Q
=
. Therefore an induced negative charge equal to surface of the conductor . Hence a charge r
(4 / 3)
-
Q p a
3 due to the charge
will appear on the inner
+
Q
appears on the outer surface of the conducting shell. Therefore the potential difference between the outer surface of the conductor ( r
=
3 a ) and infinity is positive (the outer surface is at the higher potential).
Therefore if a second neutral shell is placed very far away, the potential on that sphere is zero hence lower than the potential on the outer surface of the first conductor. If the two surfaces are then connected by a wire, electrons will move to the higher potential surface so they will move inwards along the wire until deposited on the outer surface of the conductor of radius
2 a r
=
3 a
. They will stop moving when the potential on the outer surface of the conducting shell of outer radius radius
3 a is equal to the potential on the shell of k e
Q
3 a
3 a
= k e
Q
2 a
2 a
Þ
Q
3 a
=
3
2
Q
2 a
.

Because charge is conserved, the charge on two shells must equal +
Q
, therefore
Q
3 a
+
Q
2 a
=
Q
Þ
Q
2 a
=
2 Q
5 and Q
3 a
=
3 Q
5
The potential on the shell of radius 2 a is then
V
2 a
= k e
Q
2 a
2 a
= k e
Q
5 a
.
As a check, the potential on the outer surface of the conducting shell of outer radius should be the same
3 a
V
3 a
= k e
Q
3 a
3 a
= k e
Q
,
5 a which it is.

Problem 4: Capacitors
Part A. With a roll of aluminum foil and a roll of plastic wrap what is the capacitance of the largest capacitance capacitor that one could make that would comfortably fit in a pocket?
Answer: To make this we will essentially build a parallel plate capacitor (foil, plastic, foil) and then fold or roll it up to fill the volume of our pocket. So let’s start by thinking about the capacitance of this parallel plate capacitor –
C
= e
A d
– where  
0
. The area A is going to depend on how much volume we think we can stuff in our pocket, while the dielectric constant and thickness d depends on the plastic wrap. I’ll approximate the thickness to be a little less than a piece of paper, or d


3. These are both simply guesses. I know the am pretty sure that plastic wrap is thinner, but probably only by a factor of 2 or 3, not an
1-6, so I just chose a value in that range.
Now, rather than trying to estimate A
, I’m instead going to think about volume. The volume of this sandwich is V



2 d foil


3 Ad , assuming that the thickness of the foil is about the same as the plastic wrap (given the stiffness of aluminum, it can’t be as thick as paper). With this all I am left estimating is the volume of a pocket. This obviously has a lot of variation, but I’ll approximate it as 10 cm x 10 cm x 2 cm or
V

200 cm 3 . Then:
C
 
A d
 
0
V 3 d
2 

 
12
F m

 
6 3
 
 
6

2 
2 μF
Part B.
The English Channel tunnel consists of a series of tunnels, each with a set of railroad tracks. As an order of magnitude estimate, calculate the capacitance of an idealized tunnel, cylindrical in shape with a monorail which is also cylindrical and is coaxial to the tunnel. Both the outer and inner cylinders are conducting. The length of tunnel is l
=
5.0
´
10
1 km , the diameter of monorail is d
=
5.0
´
10
1 of tunnel is a
=
1.0
´
10
1 m .
(a) Find the capacitance of this idealized tunnel. cm , and the diameter
(b) Does the capacitance increase, decrease, or stay the same when a train enters the tunnel? Explain your reasoning.
Solution From Gauss’s Law, the electric field as a function of distance of a coaxial cylinder that has a charge per unit length l is given by r from the center

E
=
æ
æ
æ
2
0 l pe
0 r
1 r
ˆ
, d / 2
< r
< a / 2
, r
< d / 2, r
> a / 2
æ
æ
æ
The voltage difference between the cylinders is the integral of the electric field
V ( a / 2)
-
V ( d / 2)
= a / 2
ò d / 2
E
× dr
= a / 2
ò d / 2 l
2 pe
0 r
1 r
ˆ × dr
= a / 2
ò d / 2
So the absolute value of the voltage difference is given by l
2 pe
0
1 r dr
= l
2 pe
0 ln a / 2 d / 2
D
V
= l
2 pe
0 ln a d
The capacitance is
C
=
Q
D
V
= l
D
V l
=
2 pe
0 l ln( a / d )
=
5
æ
10
4 m
(1.8
æ
10
10
N
æ m
2 æ
C
-
2
)(ln(20))
=
9.3
æ
10
-
7
F b) Does the capacitance increase, decrease, or stay the same when a train enters the tunnel? {Briefly} explain your reasoning.
Answer.
The capacitance increases because the ratio a/d decreases, hence 1/ln(a/d) increases.
Essentially the two cylinders are closer together.

Problem 5: Two capacitors with the same capacitance
C
and charge
Q
are placed next to each other as shown in the figure.
A wire then connects the two positive plates. Will charge flow in the wire? Consider two possible reasonings:
(A) Before the plates are connected, the potential differences of the two capacitors are the same (because
Q
and C are the same). So the potentials of the two positive plates are equal. Therefore, no charge will flow in the wire when the plates are connected.
(B) Number the plates 1 through 4, from left to right. Before the plates are connected, there is zero electric field in the regions between the capacitors, so plate 3 must be at the same potential as plate 2. But plate 2 is at a lower potential than plate 1.
Therefore, plate 3 is at a lower potential than plate 1, so charge will flow in the wire when the plates are connected.
Which reasoning is correct, and what is wrong with the wrong reasoning.
Answer.
The second reasoning is correct. Plate 3 is indeed at a lower potential than plate
1, so charge will flow. The error in the first reasoning is encompassed in the word, “So.”
Although it is true that the potential differences of the two capacitors are the same, this does not imply that the potentials of the two positive plates are equal. If we arbitrarily assign zero potential to plate 1, and if the common potential difference is V , then the potentials of the four plates are, from left to right, 0,
-
V ,
-
V , and
-
2 V . No matter where we define the zero of potential, the potential of the leftmost plate is V larger than the potential of the third plate, and 2 V larger than the potential of the rightmost plate.

Problem 6: Energy Stored in Capacitors
A parallel-plate capacitor has fixed charges + Q and – Q . The separation of the plates is then doubled.
(a) By what factor does the energy stored in the electric field change?
(b) How much work must be done if the separation of the plates is doubled from d to
2 d ? The area of each plate is A .
Consider now a cylindrical capacitor with inner and outer radii a and b , respectively.
(c) Suppose the outer radius b of a cylindrical capacitor is doubled, but the charge is kept constant. By what factor would the stored energy change? Where would the energy come from?
(d) Repeat (c), assuming the voltage remains constant.
(a) By what factor does the energy stored in the electric field change?
Since the capacitor has fixed charges + Q and – Q ,
U
=
1
2
Q
2
C
Then the ratio of the energy stored is
U
U after before
=
æ
ææ
1
2
Q
C
2 after
æ
ææ
æ
ææ
1
2 C
Q
2 before
æ
ææ
=
C
C before after
=
æ
ææ d e
0
A before
æ
ææ
æ
ææ e d
0
A after
æ
ææ
= d d after before
=
2
(b) How much work must be done if the separation of the plates is doubled from d to 2 d ?
The area of each plate is A .
The electric field on the top plate due to the bottom plate is given by
E bottom
The force acting on the top plate is
= s
2 e
0
ˆ = -
2 e
Q
0
A
ˆ
F top
= q top
E bottom
=
Q
-
Q
2 e
0
A
ˆ = -
2 e
Q
2
0
A
ˆ

There fore the work done by an external agent to separate the plate from d to 2 d is calculated as
W
= -
F z
=
ò
2 d d top
×
d s
2 d
= - d
ò
2 e
Q
0
2
A dz
=
2
Q e
2 d
A
0 but this work is just the amount of additional energy that appeared in the electric field when we moved the plate. Therefore this extra energy in the capacitor comes from the fact that we had to do work exactly equal to that amount of extra energy to separate the plates.
Consider now a cylindrical capacitor with inner and outer radii a and b , respectively.
(c) Suppose the outer radius b of a cylindrical capacitor is doubled, but the charge is kept constant. By what factor would the stored energy change? Where would the energy come from?
From part (a)
U after
=
C before
U before
C after
Thus,
U
U after before
=
æ
2 pe
0 l
ææ ln( b / a )
æ
ææ
æ
2 pe l
0
ææ ln(2 b / a )
æ
ææ
= ln(2 ln( b b
/
/ a a
)
)
The energy stored increases when the outer radius b is doubled. It is because the force acting between the two cylindrical shells is always attractive and positive work is required to separate them.
(d) Repeat (c), assuming the voltage remains constant.
Now the voltage remains constant instead of the charge stored. The energy stored is
U
=
1
2
CV 2
Then the ratio of the energy stored is
U
U after before
=
æ
ææ
1
2
C after
V
2
æ
ææ
æ
ææ
1
2
C before
V
2
æ
ææ
=
C
C after before
=
æ
2 pe l
0
ææ ln(2 b / a )
æ
ææ
æ
2 pe l
0
ææ ln( b / a )
æ
ææ
= ln( ln(2 b b
/
/ a a
)
)
The energy stored goes down in this case when the outer radius b is doubled. This is because a part of the charge initially stored on the capacitor flows out of the capacitor

through the battery opposite the direction charge usually flows through a battery to keep the voltage constant. This charge flow in the “wrong” direction actually charges up the battery, e.g. increases its stored energy. The battery gains both the energy stored in the capacitor and the work we do to separate the shells, which is always positive.

Problem 7 Coaxial Cable with Dielectric
A certain coaxial cable consists of a copper wire, radius a , surrounded by a concentric copper tube of inner radius c .
The space between is partially filled
(from
b
out to c dielectric constant
) with material of k
. Find the capacitance per unit length of this cable.
You may neglect edge effects.
Solution: Assume that the copper wire has uniform positive charge per unit length copper tube has uniform negative charge per unit length on it’s inner surface field is zero inside the copper wire and outside the coaxial cable,
E
l l
and the
. The electric
=
0 , r
< a , and r
> c .
Let’s apply Gauss’ Law òò
closed surface k
E
×
da
For
a
= q free , enc
< r
< b : The dielectric constant is e
0
for each region. k =
1 so we have our usual Gauss’s Law
òò
closed surface
E
×
da
= q free , enc e
0
. If we choose a Gaussian cylinder of radius
òò k
òò
r such that a
< r
<
. If we choose a Gaussian cylinder of radius b , and length
For that
b
b
l
<
< r r
, we have that
<
< c c
closed surface
, and length l
E
×
da
E 2 p rl
=
=
2 p rl and q l e
: The dielectric constant is l
0
, we have that
Þ
E
=
2 pe k free , enc l
0
E r
×
= l l and so Gauss’ law becomes
, a
< r
< b .
da
= k
2 p rl and q free , enc
= l l
r such
because the
closed surface induced charge distributions in the dielectric are not free charges and so Gauss’ law becomes k
E 2 p rl
= l l e
0
Þ
E
= l k
2 pe
0 r
, b
< r
< c .
Thus the electric field is

E ( r )
=
æ
0 ,
æ
æ
ææ
2 l pe
0
æ
æ l k
2 pe
0 r
1
æ 0 , r
< a r
ˆ
, a
< r
< b r
1 r
ˆ
, b
< r
< c
æ
æ r
> c
æ
æ
æ
æ
æ
The electric potential difference between the wire and the tube is then
V ( c )
-
V ( a )
= a b
æ l
2 pe
0 r
1 dr
+ b c
æ l k
2 pe
0 r
1 dr
= l
2 pe
0
æ ln b a
+
1 k ln c b
æ
æ
.
The absolute value of the voltage difference is given by
D
V
= l
2 pe
0
æ ln b a
+
1 k ln c b
æ
æ
The capacitance per unit length is
C
D l
=
Q
D l
D
V
=
2 l D pe l
0
æ ln l D l b a
+
1 k ln c b
æ
æ
=
æ ln
2 pe
0 b a
+
1 k ln c b
æ
æ

Problem 8 Two Dielectrics
Two dielectric rectangular slabs both have length L , width dielectric constant

and the second has thickness
1 w , but the first has thickness d and dielectric constant
2 d and
1

. The two slabs
2 are inserted into a parallel-plate capacitor consisting of two conducting plates of width w , length
L , and thickness d
  d
2
. The capacitor is charged up such that the total free charge on the entire top and bottom plates is
+
Q and
-
Q respectively. The charging battery is then removed from the circuit. Neglect all edge effects. a) Find an expression for the magnitude of the electric field E in dielectric slab 1 in terms
1 of terms of Q ,

1
,

2
, d ,
1 d ,
2 w , L , e
0
, and/or d , as needed. b) Find an expression for the magnitude of the electric field E
2 of terms of Q ,

,
1

,
2 d
1
, d ,
2 w , L , e
0
, and/or d
in dielectric slab 2 in terms
, as needed. c) Using your expressions for the electric fields from (a) and (b), calculate the potential drop

V between the top and bottom plate of the capacitor. d) What is the capacitance of this system? Express your answer in terms of

,
1 w , L , e
0
, and/or d , as needed.

,
2 d ,
1 d ,
2 e) What is the bound surface charge density on the upper surface of dielectric slab 1?

Solutions: a) Find an expression for the magnitude of the electric field E in dielectric slab 1 in terms
1 of terms of Q ,

,
1

,
2 d
1
, d
2
, w , L , e
0
, and/or d , as needed.
Use Gauss’s Law for dielectrics with Gaussian surface shown in the figure that has an end-cap in dielectric material 1 :
òò k
1
E
1
× ˆ = q free , enc e o
. Computing both sides yields
The free surface charge density is given by field in dielectric 1 s free k
A cap
= s free
A cap
1
E
1 e o
=
Q / wL . We can now solve for the electric
.
E
1
= s k free
1 e o
ˆ j
=
Q ˆ j k
1 e o wL b) Find an expression for the magnitude of the electric field E
2 of terms of Q ,

,
1

,
2 d
1
, d
2
, w , L , e
0
, and/or d
in dielectric slab 2 in terms
, as needed.
We use an identical argument with a Gaussian surface that now ahs an end-cap in dielectric 2 yielding
E
2
= s k
2 free e o
ˆ j
= k
Q
2 e o wL
ˆ j c) Using your expressions for the electric fields from (a) and (b), calculate the total potential drop

V between the top and bottom plate of the capacitor.
D
V
=
E
1 d
1
+
E
2 d
2
= k
1 e
Q o wL d
1
+ k
2 e
Q o wL d
2 d) What is the capacitance of this system? Express your answer in terms of

,
1 w , L , e
0
, and/or d , as needed.

,
2 d ,
1 d ,
2

C
=
Q
D
V
=
Q d
2
= e o wL k
1 e
Q o wL d
1
+ k
2 e
Q o wL d
1 k
1
+ d
2 k
2 e) What is the bound surface charge density on the upper surface of dielectric slab 1?
In dielectric 1, the electric field is the superposition of the electric field due to the free charges and the electric field due to the induced sheet of charges.
E
1
=
E free
+
E ind ,1
We can use Gauss’s Law to calculate these two electric fields and find that
E free
E
= s ind ,1 e free o
= -
ˆ j
= e s e ind ,1 o o
Q wL
ˆ j .
ˆ j ,
Therefore
E
1
=
Q k
1 e o wL
ˆ j
=
Q e o wL
ˆ j
s e ind ,1 o
ˆ j .
We can now solve this last equation for the induced surface charge density in dielectric 1 s ind ,1
=
Q wL
æ
ææ
1
k
1
1
æ
ææ
.