MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Exam 2 Practice Problems Part 1 Solutions
Problem 1 Electric Field and Charge Distributions from Electric Potential
An electric potential V ( z ) is described by the function
)(!2V " m -1 )z + 4V ; z > 2.0 m
+
+0 ; 1.0 m < z < 2.0 m
+2
#2
-3 & 3
+ V ! % V " m ( z ; 0 m < z < 1.0 m
3
3
$
'
+
V (z) = *
+ 2 V+ # 2 V " m -3 & z 3 ; ! 1.0 m < z < 0 m
%$ 3
('
+3
+
+0 ; ! 2.0 m < z < ! 1.0 m
+(2V " m -1 )z + 4V ; z < ! 2.0 m
,
The graph below shows the variation of an electric potential V ( z ) as a function of z .
!
a) Give the electric field vector E for each of the six regions in (i) to (vi) below?
!
!
Solution: We shall the fact that E = !"V . Since the electric potential only depends on
the variable z , we have that the z -component of the electric field is given by
Ez = !
The electric field vector is then given by
dV
.
dz
!
dV ˆ
E = Ez ˆi = !
k
dz
(i) z > 2.0 m :
!
dV ˆ
d
E=!
k = ! ((!2V " m -1 ) z + 4V )kˆ = 2V " m -1 kˆ
dz
dz
(ii) 1.0 m < z < 2.0 m :
(iii) 0 m < z < 1.0 m :
!
dV ˆ !
E=!
k =0
dz
!
dV ˆ
d !2
!2
" "
E=#
k = # % V # % V $ m -3 & z 3 & kˆ = (2 V $ m -3 )z 2kˆ
dz
dz ' 3
'3
( (
Note that the z 2 has units of [m 2 ], so the value of the electric field at a point just inside
z − = 1.0 m − ε m where ε > 0 is a very small number is given by
!
V
E! = (2 V " m -3 )(1 m)2 k̂ = 2 k̂
m
Note that the z-component of the electric field, 2 V ! m "1 , has the correct units.
(iv) −1.0 m < z < 0 m :
!
#2
& &
dV
d #2
E=!
k̂ = ! % V+ % V " m -3 ( z 3 ( k̂ = ! 2 V " m -3 z 2 k̂
dz
dz $ 3
$3
' '
(
)
The value of the electric field at a point just inside z + = !1.0 m + " m is given by
(v) −2.0 m < z < − 1.0 m :
(vi) z < ! 2.0 m :
!
V
E! = !(2 V " m -3 )(1 m)2 k̂ = !2 k̂ .
m
!
dV ˆ !
E=!
k =0
dz
!
dV ˆ
d
E=!
k = ! ((2V " m -1 ) z + 4V )kˆ = !2V " m -1 kˆ
dz
dz
b) Make a plot of the z-component of the electric field, Ez , as a function of z . Make
sure you label the axes to indicate the numeric magnitude of the field.
c) Qualitatively describe the distribution of charges that gives rise to this potential
landscape and hence the electric fields you calculated. That is, where are the charges,
what sign are they, what shape are they (plane, slab…)?
In the region !1.0 m < z < 1.0 m there is a non-uniform (in the z-direction) slab of
positive charge. Note that the z-component of the electric field is zero at z = 0 m ,
negative for the region !1.0 m < z < 0 m , and positive for 0 m < z < 1.0 m as it should
for a positive slab that has zero field at the center.
In the region 1.0 m < z < 2.0 m there is a conductor where the field is zero.
On the plane z = 2.0 m , there is a positive uniform charge density ! that produces a
constant field pointing to the right in the region z > 2.0 m (hence the positive
component of the electric field).
On the plane z = 1.0 m , there is a negative uniform charge density "! .
In the region !2.0 m < z < ! 1.0 m there is a conductor where the field is zero.
On the plane z = ! 2.0 m , there is a positive uniform charge density ! that produces a
constant field pointing to the left in the region z < ! 2.0 m (hence the positive
component of the electric field).
On the plane z = ! 1.0 m , there is a negative uniform charge density "! .
Problem 2: Electric Field from Electric Potential
The electric potential V (x) for a planar charge distribution is given by:
)
+
+
+
+
V (x) = *
+
+
+
+,
for x < !d
0
"
x%
!V0 $ 1+ '
# d&
2
for ! d ( x < 0
"
x%
!V0 $ 1+ 2 '
d&
#
for 0 ( x < d
!3V0
for x > d
where !V0 is the potential at the origin and d is a distance.
This function is plotted to the right, with d = 2 cm and V0 = 2 V , the x-axis with units in
cm, the y-axis in units of Volts.
!
(a) What is the electric field E(x) for this problem?
Region I: x < -d
!
#V
E = !"V = !
î = 0
#x
Region II: !d " x < 0
2
!
V #
#
"#
x& &
x&
E = ! % !V0 % 1 + ( ( î = 2 0 % 1 + ( î
"x %$
d ' ('
d$
d'
$
Region III: 0 ! x < d
!
V
#
"#
x&&
E = ! % !V0 % 1 + 2 ( ( î = 2 0 î
"x $
d ''
d
$
Region IV: x > d
!
#V
E = !"V = !
î = 0
#x
(b) Plot the electric field that you just calculated on the graph below. Be sure to properly
label the y-axis (top and bottom) to indicate the limits of the magnitude of the E field!
This makes sense. Where the potential is flat, the E field is 0.
Problem 3 Dielectric
A dielectric rectangular slab has length s , width w , thickness d , and dielectric constant
! . The slab is inserted on the right hand side of a parallel-plate capacitor consisting of
two conducting plates of width w , length L , and thickness d . The left hand side of the
capacitor of length L ! s is empty. The capacitor is charged up such that the left hand
side has surface charge densities ±! L on the facing surfaces of the top and bottom plates
respectively and the right hand side has surface charge densities ±! R on the facing
surfaces of the top and bottom plates respectively. The total 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 EL on the left hand side
in terms of ! L , ! R , ! , s , w , L , ! 0 , and d as needed.
b) Find an expression for the magnitude of the electric field ER on the right hand
side in terms of ! L , ! R , ! , s , w , L , ! 0 , and d as needed.
c) Find an expression that relates the surface charge densities ! L and ! R in terms of
! , s , w , L , ! 0 , and d as needed.
d) What is the total charge +Q on the entire top plate? Express your answer in terms
of ! L , ! R , ! , s , w , L , ! 0 , and d as needed.
e) What is the capacitance of this system? Express your answer in terms of ! , s ,
w , L , ! 0 , and d as needed.
f) Suppose the dielectric is removed. What is the change in the stored potential
energy of the capacitor? Express your answer in terms of Q , ! , s , w , L , ! 0 ,
and d as needed.
Solutions:
a) Find an expression for the magnitude of the electric field EL on the left hand side
in terms of ! L , ! R , ! , s , w , L , ! 0 , and d as needed.
!
Using Gauss’s Law E L = L
"o
b) Find an expression for the magnitude of the electric field ER on the right hand
side in terms of ! L , ! R , ! , s , w , L , ! 0 , and d as needed.
Using Gauss’s Law for dielectrics ER =
σR
κε o
c) Find an expression that relates the surface charge densities ! L and ! R in terms of
! , s , w , L , ! 0 , and d as needed.
σ d
The potential difference on the left side is EL d = L . On the right hand side it is
εo
ER d =
σ Rd
σ
. Since these must be equal we must have R = σ L
κε o
κ
d) What is the total charge +Q on the entire top plate? Express your answer in terms
of ! L , ! R , ! , s , w , L , ! 0 , and d as needed.
Q = σ L ( L − s ) w + σ R sw
e) What is the capacitance of this system? Express your answer in terms of ! , s ,
w , L , ! 0 , and d as needed.
The potential difference is EL d =
C=
σ Ld
, so the capacitance is
εo
σ ( L − s ) w + σ R sw ε o w ⎡
σ ⎤ ε w
Q
= L
=
L − s ) + R s ⎥ = o ⎡⎣( L − s ) + κ s ⎤⎦ = C
(
⎢
σ Ld
∇V
d ⎣
σL ⎦ d
εo
f) Suppose the dielectric is removed. What is the change in the stored potential
energy of the capacitor? Express your answer in terms of Q , ! , s , w , L , ! 0 ,
and d as needed.
Since the battery has been removed, the charge on the capacitor does not change
when we do this. So the change in the energy stored is
1 Q2 1 Q2 Q2 ⎛ 1 1 ⎞
−
=
⎜ − ⎟
2 Co 2 C
2 ⎝ Co C ⎠
⎞ Q2d ⎡ 1
⎤
Q2 ⎛ d
d
1
=
−
−
⎜
⎟=
⎢
⎥
2 ⎜⎝ ε o Lw ε o w ⎡⎣( L − s ) + κ s ⎤⎦ ⎟⎠ 2ε o w ⎣ L ( L − s ) + κ s ⎦
⎤
(κ − 1) s
Q2d ⎡ ( L − s ) + κ s − L ⎤ Q2d ⎡
⎢
⎥=
⎢
⎥.
change in energy =
2ε o w ⎢ ⎡ L ⎡⎣( L − s ) + κ s ⎤⎦ ⎤ ⎥ 2ε o w ⎢ ⎡ L ⎡⎣( L − s ) + κ s ⎤⎦ ⎤ ⎥
⎦⎦
⎦⎦
⎣⎣
⎣⎣
change in energy =
Problem 4: Capacitors and Dielectrics
(a) Consider a plane-parallel capacitor completely filled with a dielectric material of
dielectric constant κ . What is the capacitance of this system?
The capacitance is
C=
!" 0 A
= ! C0
d
(b) A parallel-plate capacitor is constructed by filling the space between two square
plates with blocks of three dielectric materials, as in the figure below. You may assume
that ℓ >> d. Find an expression for the capacitance of the device in terms of the plate area
A and d, ! 1 , ! 2 , and ! 3 .
The
capacitor can be regarded as being consisted of three capacitors,
! " A2
!" A2
!" A2
, C2 = 2 0
and C3 = 3 0
, with C2 and C3 connected in series, and
C1 = 1 0
d/2
d/2
d
the combination connected in parallel with C1. Thus, the equivalent capacitance is
! 1
1$
C = C1 + #
+ &
" C2 C3 %
=
'1
= C1 +
) 0 A ! ( 1 ( 2( 3 $
+
d #" 2 ( 2 + ( 3 &%
C2C3
( ) A 2 ) 0 A ! ( 2( 3 $
= 1 0
+
C2 + C3
d
d #" ( 2 + ( 3 &%
Problem 5: Conductors and Potential
A charge !Q is at the center of a neutral metal (conducting) sphere of radius R1 that is in
turn centered in a larger metal (conducting) sphere of radius R2 , which carries a net
charge of +Q . The potentials of the inner and outer spheres with respect to infinity are
respectively:
Inner
1
1
A. keQ( ! )
R2 R1
Q
B. !ke
R1
C. 0 ,
D. keQ(
Outer
Q
ke
R2
0
keQ(
1
1
! )
R2 R1
1
1
! )
R1 R2
0
Explain your reasoning.
Reason: The induced charge distributions are shown in the figure below. The overall
charge is zero, so there is not charge on the outer surface of the outer sphere.
The electric field outside is therefore zero. The potential is therefore constant outside the
outer sphere and it is zero at infinity, it is also zero on the surface of the outer sphere,
V (R2 ) = 0 .
The electric field between the spheres is given just the field due to a negative charge !Q
so
! !k Q
E = 2e r̂
r
So the potential difference between the two spheres is given by
V (R1 ) ! V (R2 ) = !ke #
R1
R2
$ 1
!Q
1'
r̂ " dr! r̂ = !keQ & ! )
2
r
% R1 R2 (
Since V (R2 ) = 0 , the potential on the inner surface is
" 1
1%
V (R1 ) = keQ $
! '.
# R2 R1 &
Problem 6: Faraday Cage
Consider
two
nested,
spherical
conducting shells. The first has inner
radius a and outer radius b. The second
has inner radius c and outer radius d.
In the following four situations, determine the total charge on each of the faces of the
conducting spheres (inner and outer for each), as well as the electric field and potential
everywhere in space (as a function of distance r from the center of the spherical shells).
In all cases the shells begin uncharged, and a charge is then instantly introduced
somewhere.
a) Both shells are not connected to any other conductors (floating) – that is, their net
charge will remain fixed. A positive charge +Q is introduced into the center of
the inner spherical shell. Take the zero of potential to be at infinity.
b) The inner shell is floating but the outer shell is grounded – that is, it is fixed at
V = 0 and has whatever charge is necessary on it to maintain this potential. A
negative charge !Q is introduced into the center of the inner spherical shell.
c) The inner shell is grounded but the outer shell is floating. A positive charge +Q
is introduced into the center of the inner spherical shell.
d) Finally, the outer shell is grounded and the inner shell is floating. This time the
positive charge +Q is introduced into the region in between the two shells. In
!
this case the questions “What are E(r) and V (r) ?” cannot be answered
analytically in some regions of space. In the regions where these questions can be
answered analytically, give answers. In the regions where they can’t be answered
analytically, explain why, but try to draw what you think the electric field should
look like and give as much information about the potential as possible (is it
positive or negative, for example).
Solution
a) Both shells are not connected to any other conductors (floating) – that is, their net
charge will remain fixed. A positive charge +Q is introduced into the center of
the inner spherical shell. Take the zero of potential to be at infinity.
There is no electric field inside a conductor. Also, the net charge on an isolated conductor
is zero (i.e. Qa + Qb = Qc + Qd = 0 ).
Qa = !Q , Qb = +Q , Qc = !Q , Qd = +Q
Using the Gauss’s law,
# Q
r̂,
%
2
4
!"
r
0
%!
%0,
%
!
% Q
E(r) = $
r̂,
2
% !4!" 0 r
%0,
%
% Q
% 4!" r 2 r̂,
0
&
r>d
c<r<d
b<r<c
a<r<b
r<a
r
Since V (r) ! V (") = ! # E(r)dr , the potential difference is then
"
+ Q
,
r>d
- 4#$ 0 r
- Q
,
c<r<d
4
#$
d
0
-- Q % 1 1 1 (
V (r) ! V (") = ,
'& ! + *) , b < r < c
4
#$
r c d
0
- Q % 1 1 1(
'& ! + *) , a < r < b
4
#$
b c d
0
- Q % 1 1 1 1 1(
' ! + ! + * ,r < a
-. 4#$ 0 & r a b c d )
b) The inner shell is not connected to ground (floating) but the outer shell is
grounded – that is, it is fixed at V = 0 and has whatever charge is necessary on it
to maintain this potential. A negative charge !Q is introduced into the center of
the inner spherical shell.
!
!
Since the outer shell is now grounded, Qd = 0 to maintain E(r) = 0 outside the outer
shell. We have.
Qa = Q , Qb = !Q , Qc = +Q , Qd = 0 .
Again using Gauss’s Law yields
!
$0 ,
&
&! Q r̂,
&& 4"# 0 r 2
!
E(r) = % !
&0,
&
Q
r̂,
&!
2
&' 4"# 0 r
r>c
b<r <c
a<r<b
r<a
The potential difference is then
+0,
r>c
-! Q % 1 ! 1 ( ,
b<r <c
- 4#$ 0 '& r c *)
V (r) ! V (") = , Q % 1 1 (
a<r<b
-! 4#$ '& b ! c *) ,
0
- Q % 1 1 1 1(
-!
' ! + ! *, r < a
. 4#$ 0 & r a b c )
c) The inner shell is grounded but the outer shell is not connected to ground. A
positive charge +Q is introduced into the center of the inner spherical shell.
!
!
The inner shell is grounded and Qb = 0 to maintain E(r) = 0 outside the inner shell.
Because there is no electric field on the outer shell, Qc = Qd = 0 .
Qa = !Q , Qb = Qc = Qd = 0
Gauss’s Law then yields
!
#0,
r>a
!
%
E(r) = $ Q
% 4!" r 2 r̂, r < a
0
&
The potential difference is then
r>a
+0,
V (r) ! V (") = , Q % 1 1 (
- 4#$ '& r ! a *) , r < a
0
.
d) Finally, the outer shell is grounded and the inner shell is floating. This time the
positive charge +Q is introduced into the region in between the two shells. In
!
this case the questions “What are E(r) and V (r) ?” cannot be answered
analytically in some regions of space. In the regions where these questions can be
answered analytically, give answers. In the regions where they can’t be answered
analytically, explain why, but try to draw what you think the electric field should
look like and give as much information about the potential as possible (is it
positive or negative, for example).
The electric field within the cavity is zero. If there is any field line that began and ended
! !
on the inner wall, the integral " E ! d s over the closed loop that includes the field line
would not be zero. This is impossible since the electrostatic field is conservative, and
therefore the electric field must be zero inside the cavity. The charge Q between the two
conductors pulls minus charges to the near side on the inner conducting shell and repels
plus charges to the far side of that shell. However, the net charge on the outer surface of
the inner shell (Qb) must be zero since it was
uncharged (floating). Since the
! initially
!
outer shell is grounded, Qd = 0 to maintain E(r) = 0 outside the outer shell. Thus,
!
!
Qa = Qb = Qd = 0 , Qc = !Q and E(r) = 0 , r < b or r > c
!
For, E(r) is in fact well defined but it is very complicated. The field lines are shown in
the figure below.
What can we say about the electric potential? V (r) = 0 for r > c , and V (r) = constant
for r < b but the potential is very complicated defined between the two shells.
Problem 7 N-P Semiconductors Revisited Potential Difference
When two slabs of n -type and p -type semiconductors are put in contact, the relative
affinities of the materials cause electrons to migrate out of the n -type material across the
junction to the p -type material. This leaves behind a volume in the n -type material that
is positively charged and creates a negatively charged volume in the p -type material.
Let us model this as two infinite slabs of charge, both of thickness a with the junction
lying on the plane z = 0 . The n -type material lies in the range 0 < z < a and has uniform
charge density + !0 . The adjacent p -type material lies in the range !a < z < 0 and has
uniform charge density ! "0 . Thus:
# + !0
%
!(x, y, z) = !(z) = $" !0
%
&0
0 < z< a
" a< z< 0
z >a
a) Find the electric field everywhere.
b) Find the potential difference between the points P1 and P2. . The point P1. is located
on a plane parallel to the slab a distance z1 > a from the center of the slab. The
point P2. is located on plane parallel to the slab a distance z2 < !a from the center
of the slab.
Solution:
In this problem, the electric field is a superposition of two slabs of opposite charge density.
!
Outside both slabs, the field of a positive slab E P (due to the p -type semi-conductor ) is
!
constant and points away and the field of a negative slab E N (due to the n -type semiconductor) is also constant and points towards the slab, so when we add! both
contributions we find that the electric field is zero outside the slabs. The fields E P are
!
shown on the figure below. The superposition of these fields ET is shown on the top line
in the figure.
The electric field can be described by
!
!0
#!
!
#E 2
ET ( z ) = $ !
#E1
#!
%0
z < "a
"a < z< 0
0< z < a
.
x >d
We shall now calculate the electric field in each region using Gauss’s Law:
For region !a < z < 0 : The Gaussian surface is shown on the left hand side of the figure
below. Notice that the field is zero outside. Gauss’s Law states that
!
!
"
"" E ! d a =
closed
surface
Qenc
.
#0
So for our choice of Gaussian surface, on the cap inside the slab the unit normal for the
area vector points in the positive z-direction, thus nˆ = +kˆ . So the dot product becomes
!
E2 ! nˆ da = E2, z kˆ ! kˆ da = E2, z da . Therefore the flux is
"
""
closed
surface
The charge enclosed is
! !
E ! d a = E2,z Acap
Qenc " #0 Acap (a + z)
=
!0
!0
where the length of the Gaussian cylinder is a + z since z < 0 .
Substituting these two results into Gauss’s Law yields
E2, z Acap =
# !0 Acap (a + z )
"0
Hence the electric field in the n -type is given by
E2, x =
# !0 (a + z )
.
"0
The negative sign means that the electric field point in the –z direction so the electric
field is
!
# ! (a + z ) ˆ
E2 = 0
k.
"0
!
!
!
#! a
Note when z = !a then E2 = 0 and when z = 0 , E2 = 0 kˆ .
"0
We make a similar calculation for the electric field in the p -type noting that the charge
density has changed sign and the expression for the length of the Gaussian cylinder is
a ! z since z > 0 . Also the unit normal now points in the negative z-direction. So the dot
product becomes
!
E1 ! nˆ da = E1, z (" kˆ ) ! kˆ da = " E1, z da
Thus Gauss’s Law becomes
# E1, z Acap =
+ !0 Acap (a # z )
"0
So the electric field is
E1, z = #
The vector description is then
!0 (a # z )
.
"0
!
# ! (a # a) ˆ
E1 = 0
k
"0
.
!
! !
#! a
Note when z = a then E1 = 0 and when z = 0 , E1 = 0 kˆ .
"0
So the resulting field is
!
#0
z < $a
%!
%E = $ !0 (a + z ) kˆ
%% 2
!
"0
ET ( z ) = &
!
%E1 = $ !0 (a $ z ) kˆ
%
"0
%!
z >a
%'0
$a < z< 0
.
0< z < a
The graph of the electric field is shown below
.
(ii) Find the potential difference between the points P1 and P2. . The point P1. is located
on a plane parallel to the slab a distance z1 > a from the center of the slab. The
point P2. is located on plane parallel to the slab a distance z2 < !a from the center of
the slab.
The electric potential difference is given by the integral
P2
!
!
V ( P2 ) ! V ( P1 ) = ! # ET " dr
P1
We first break this line integral into four pieces covering each region
z = z2
z =0
z =! a
" z =a !
!
!
!
!
!
!
!#
V ( z2 ) ! V ( z1 ) = ! % ) ET $ dr + ) ET $ dr + ) ET $ dr + ) ET $ dr & .
% z=z
&
z =a
z =0
z =! a
' 1
(
Since the fields are zero outside the slab, the only non-zero pieces are
z =! a
!
" z =0 !
!
!#
V ( z2 ) ! V ( z1 ) = ! % ) ET $ dr + ) ET $ dr & .
z =0
' z =a
(
!
We now use our explicit result for the electric field in each region and that dr = dz kˆ ,
z =# a
$ z =0
%
V ( z2 ) # V ( z1 ) = # ' + E1, z kˆ & dz kˆ + + E2,a kˆ & dz kˆ (
z =0
) z =a
*
z =# a
$ z =0 # !0 (a # z )
# !0 (a + z ) %
= #' +
dz + +
dz (
"0
"0
z =0
) z =a
*
We now calculate the two integrals
V ( z2 ) # V ( z1 ) =
!0 ( za # z 2 / 2)
"0
z =0
z =a
+
!0 ( za + z 2 / 2)
"0
z =# a
z =0
!0 (a 2 # a 2 / 2) !0 ((#a )a + (#a ) 2 / 2)
!0 (a 2 / 2) !0 (a 2 / 2) .
=#
+
=#
#
"0
"0
"0
"0
=#
!0 a 2
"0
The potential difference is negative because we are moving along the direction of the
field. So the type pn -semiconductor slab established a small potential difference across
the slab.
Problem 8: Conceptual Questions
Part (a) A parallel plate capacitor has capacitance C. It is connected to a battery of
EMFe until fully charged, and then disconnected. The plates are then pulled apart an
extra distance d, during which the measured potential difference between them changed
by a factor of 4.
Below are a series of questions about how other quantities changed. Although they are
related you do not need to rely on the answers to early questions in order to correctly
answer the later ones.
a) Did the potential difference increase or decrease by a factor of 4?
INCREASE
DECREASE
Answer. Increase.
b) By what factor did the electric field change due to this increase in distance?
Make sure that you indicate whether the field increased or decreased.
Answer.
Since the charge cannot change (the battery is disconnected) the electric field cannot
change either. No Change!
c) By what factor did the energy stored in the electric field change?
Make sure that you indicate whether the energy increased or decreased.
Answer.
The electric field is constant but the volume in which the field exists increased, so the
energy must have increased. But by how much? The energy U = 12 QV . The charge
doesn’t change, the potential increased by a factor of 4, so the energy: Increased by a
factor of 4
d) A dielectric of dielectric constant k is now inserted to completely fill the volume
between the plates. Now by what factor does the energy stored in the electric field
change? Does it increase or decrease?
Answer.
Inserting a dielectric decreases the electric field by a factor of k so it decreases the
potential by a factor of k as well. So now, by using the same energy formula U = 12 QV ,
Energy decreases by a factor of k
e) What is the volume of the dielectric necessary to fill the region between the plates?
(Make sure that you give your answer only in terms of variables defined in the
statement of this problem, fundamental constants and numbers)
Answer.
How in the world do we know the volume? We must be able to figure out the crosssectional area and the distance between the plates. The first relationship we have is from
knowing the capacitance:
! A
C= 0
x
where x is the original distance between the plates. Make sure you don’t use the more
typical variable d here because that is used for the distance the plates are pulled apart.
Next, the original voltage V0 = E x, which increases by a factor of 4 when the plates are
moved apart by a distance d, that is, 4 V0 = E (x+d). From these two equations we can
solve for x:
4V0 = 4Ex = E x + d
! x=d 3
(
)
Now, we can use the capacitance to get the area, and multiply that by the distance
between the plates (now x + d) to get the volume:
(
)
Volume = A x + d =
% 4d 2C
xC
dC " d
x+d =
+
d
'& = 9!
!0
3! 0 $# 3
0
(
)
Problem 9: Spherical Capacitor and Stored Energy
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?
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 ! 4# r
The enclosed charge is Qenc = +Q , and the electric field is everywhere perpendicular to
the surface. Thus Gauss’s Law becomes
E ! 4" r 2 =
Q
Q
$E=
#0
4"# 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 = $ 4!" 0 r 2
%0
elsewhere
&
We know the positively charged inner sheet is at a higher potential so we shall calculate
a
! !
!V = V (a) " V (b) = " $ E # d s = " $
a
b
b
a
Q
Q
Q ' 1 1*
dr =
=
"
>0
2
4%& 0 r b 4%& 0 )( a b ,+
4%& 0 r
which is positive as we expect.
We can now calculate the capacitance using the definition
C=
C=
4"# 0
4"# 0 ab
Q
Q
=
=
=
b$ a
!V
Q % 1 1( % 1 1(
$
$
4"# 0 '& a b *) '& a b *)
4!" 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 ! 0 times an area ab divided by a length b ! a ,
exactly the same units as the formula for a parallel-plate capacitor C = ! 0 A / d . Also note
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%
A ! 4"
! 4"
= 4"a 2 ! 4"ab
$ 2 &
$ 2 &
2
2
So, in this limit, the spherical formula is the same at the plate one
4"# 0 ab # 0 4" a 2 # 0 A
.
!
=
b! a b $ a
d
d
C = lim
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:
Q
The electric field E(a) =
. Therefore the maximum charge is
4!" 0 a 2
Qmax = 4!" 0 Emax (a)a 2 =
(3.0 # 106 V $ m -1 )(0.1 m)2
= 3.3 # 10%6 C
9
2
2
(9 # 10 N $ m $ C )
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
2C
(2)(2.2 ! 10"11 F)
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
!V =
2
Q 4"# 0 E(a)a (b $ a) E(a)a(b $ a)
=
=
C
4"# 0 ab
b
(3.0 % 106 V & m -1 )(0.1 m)(0.1 m)
!V =
= 1.5 % 105 V
(0.2 m)
.
We already calculated the potential difference in part a):
!V =
Q % 1 1(
.
$
4"# 0 '& a b *)
Q
Q
or
= E(a)a 2 . Substitute this into our expression for
2
2
4!" 0 a
4!" 0
potential difference yielding
Recall that E(a) =
# 1 1&
(b " a)
(b " a)
!V = E(a)a 2 % " ( = E(a)a 2
= E(a)a
ab
b
$ a b'
in agreement with our result above.
Problem 10: Cylindrical Capacitor
Consider two nested cylindrical conductors of height h and radii a & b respectively. A
charge +Q is evenly distributed on the outer surface of the pail (the inner cylinder), -Q on
the inner surface of the shield (the outer cylinder). You may ignore edge effects.
a)
Calculate the electric field between the two cylinders (a < r < b).
b)
Calculate the potential difference between the two cylinders:
c)
Calculate the capacitance of this system, C = Q/DV
d)
cm.
Numerically evaluate the capacitance, given: h ≅ 15 cm, a ≅ 4.75 cm and b ≅ 7.25
e)
Find the electric field energy density at any point between the conducting
cylinders. How much energy resides in a cylindrical shell between the conductors of
radius r (with a < r < b ), height h , thickness dr , and volume 2! rh dr ? Integrate your
expression to find the total energy stored in the capacitor and compare your result with
that obtained using U E = (1 / 2)C(!V )2 .
(a) Calculate the electric field between the two cylinders (a < r < b).
For this we use Gauss’s Law, with a Gaussian cylinder of radius r, height l
! !
Q
1 Q
Q
E
"
"" ! d A = 2# rlE = $inside = $ h l % E(r)a <r <b = 2# r$ h
0
0
0
(b) Calculate the potential difference between the two cylinders:
The potential difference between the outer shell and the inner cylinder is
a
' b*
Q
Q
Q
dr $ = "
ln r $ b =
ln ) ,
2# r $% 0 h
2#% 0 h
2#% 0 h ( a +
b
a
!V = V (a) " V (b) = " &
(c) Calculate the capacitance of this system, C = Q/DV
C=
2"# o h
|Q |
|Q |
=
=
| !V |
$ b'
|Q | $ b '
ln & ) ln & )
2"# 0 h % a (
% a(
(d) Numerically evaluate the capacitance given:
h ≅ 15 cm, a ≅ 4.75 cm and b ≅ 7.25 cm
C=
2!" o h
1
15 cm
=
+ 20 pF
9
-1
# b & 2 ) 9 * 10 m F
# 7.25 cm &
ln % (
ln %
$ a'
$ 4.75 cm ('
e)
Find the electric field energy density at any point between the conducting
cylinders. How much energy resides in a cylindrical shell between the conductors of
radius r (with a < r < b ), height h , thickness dr , and volume 2! rh dr ? Integrate your
expression to find the total energy stored in the capacitor and compare your result with
that obtained using U E = (1 / 2)C(!V )2 .
The total energy stored in the capacitor is
1
1 # Q &
uE = ! 0 E 2 = ! 0 %
2
2 $ 2" r! 0 h ('
Then
2
2
1 # Q &
Q 2 dr
dU = uE dV = ! 0 %
2
"
rh
dr
=
2 $ 2" r! 0 h ('
4"! 0 h r
Integrating we find that
U=
!
b
a
dU =
!
b
a
Q 2 dr
Q2
=
ln(b / a) .
4"# 0 h r
4"# 0 h
From part d) C = 2!" o h / ln(b / a) , therefore
U=
!
b
a
dU =
!
b
a
Q 2 dr
Q2
Q2 1
=
ln(b / a) =
= C$V 2
4"# 0 h r
4"# 0 h
2C 2
which agrees with that obtained above.
Problem 11 Capacitance of Multiple Plates
A capacitor is made of three sets of parallel plates of area A , with the two outer plates on
the left and the right connected together by a conducting wire. The outer plates are
separated by a distance d . The distance from the middle plate to the left plate is z . The
distance from the inner plate to the right plate is d ! z . You may assume all three plates
are very thin compared to the distances d and z . Neglect edge effects.
a) The positive terminal of a battery is connected to the outer plates. The negative
terminal is connected to the middle plate. The potential difference between the
outer plates and inner plate is !V = V (z = 0) " V (z) . Find the capacitance of this
system.
b) Find the total energy stored in this system.
Solution:
When the battery is connected positive charges QL and QR appear on the outer plates (on
the inner facing surfaces) and negative charges !QL and !QR respectively appear on
each side of the inner plate.
The plates on the left in the above figure act as a capacitor arranged as show in the figure
below.
plates on left
plates on right
This is parallel plate capacitor with capacitance C L = QL / !V . The plates on the right
also act as a capacitor with capacitance C R = QR / !V . All the plates have the same area
A so neglecting edge effects we can use Gauss’s Law to show that the magnitude of the
electric field on the left is E L = QL / ! 0 A for parallel plates on the left. Because the
electric field is uniform the potential difference is
!V = E L z = QL z / " 0 A .
Therefore the positive charge on the left satisfies
QL =
!V " 0 A
z
(1)
A similar argument applied to the plates on the right show that
!V = E R (d " z) = QR (d " z) / # 0 A .
Hence the positive charge on the right satisfies
QR =
!V " 0 A
d#z
The total positive charge at the higher potential is the sum QL + QR and so the
capacitance of the capacitor is
Q + QR
C= L
!V
(2)
(3)
We can now substitute Eqs. (1) and (2) into Eq. (3) and solve for the capacitance
C=
!V " 0 A
!V " 0 A
" Ad
$1
1 '
.
+
= "0 A& +
= 0
)
!Vz
!V (d # z)
% z d # z ( z(d # z)
Note: Because the outer plates are at the same potential you can think of these capacitors
as connected in parallel. Therefore the equivalent capacitance is
C = CL + CR =
QL QR
+
!V !V
agreeing with our result above.
b) Find the total energy stored in this system.
The stored energy is just
U=
" Ad!V 2
1
C!V 2 = 0
.
2
2z(d # z)
Problem 12: Capacitance, Stored Energy, and Electrostatic Work
Two flat, square metal plates have sides of length L , and thickness s 2 , are
arranged parallel to each other with a separation of s , where s << L so you may ignore
fringing fields. A charge Q is moved from the upper plate to the lower plate. Now a
force is applied to a third uncharged conducting plate of the same thickness s 2 so that it
lies between the other two plates to a depth x , maintaining the same spacing s 4
between its surface and the surfaces of the other two.
a) What is the capacitance of this system?
b) How much energy is stored in the electric field?
c) If the middle plate is released, it starts to move. Will it move to the right or left?
Hint: If the middle plate moves to the left by a small positive amount !x , the
change in potential energy is approximately !U ! (dU / dx)!x . Will the stored
potential energy increase or decrease?
d) For a small displacement !x in the direction you determined in part c), find the
horizontal force exerted by the charge distribution on the outer plates acting on
the charges on the middle plate that cause it to move.
Solution:
a) What is the capacitance of this system?
Divide the plates into left and right sides. The area on the left is AL = L( L ! x) and the
area on the right is AR = Lx . The charge densities on the two sides are shown in the
figure below. The charge on the left is QL = ! L L( L " x) . The charge on the right is
QR = ! LR (Lx) .
The electric field on the two sides is shown in the figure below.
We can calculate the electric field on both sides using Gauss’s Law (neglecting edge
effects. We find
EL = ! L / " 0 , ER = ! R / " 0
Because the electric field is uniform on both sides (neglecting edge effects)
!VL = E L s =
" Ls
#0
(1)
Note that on the right side the electric field is zero in the middle conductor so
!VR =
" R (s / 4) " R (s / 4) " R s
.
+
=
#0
#0
2# 0
(2)
The potential difference on the two sides are the same because the upper and lower plates
are held at the same potential difference
!V " !VL = !VR .
Therefore we can solve for a relationship between the charge densities on the two sides
by setting Eq. (1) equal to Eq. (2)
! Ls ! Rs
=
"0
2" 0
Hence
!L =
!R
.
20
(3)
The capacitance of the system is the total charge divided by the potential difference
C=
QL + QR
.
!V
Using our results for the charges and the potential difference we have that
C=
! L L( L " x) + ! R Lx
! Ls / #0
(4)
We can now substitute Eq. (3) into Eq. (4) and find that the capacitance is
C=
(! R / 2)L( L " x) + ! R Lx # 0 (L( L " x) + 2Lx) # 0 L( L + x)
.
=
=
! R s / 2# 0
s
s
b) How much energy is stored in the electric field?
The energy stored in the capacitor is then
U=
Q2
Q 2s
.
=
2C 2! 0 L( L + x)
c) If the middle plate is released, it starts to move. Will it move to the right or left?
Hint: If the middle plate moves to the left by a small positive amount !x , the
change in potential energy is approximately !U ! (dU / dx)!x . Will the stored
potential energy increase or decrease?
If the middle plate moves a positive distance !x to the left resulting in a larger value of
x , then the stored potential energy changes by an amount
!U =
dU
Q 2s
!x = "
!x .
dx
2# 0 L( L + x)2
This change is negative hence the middle plate will move to the left decreasing the stored
potential energy of the system. Because the change in potential energy is negative, the
work done by the electric forces is positive
Q 2s
W=
"x .
2! 0 L( L + x)2
d) For a small displacement !x in the direction you determined in part c), find the
horizontal force exerted by the charge distribution on the outer plates acting on
the charges on the middle plate that cause it to move.
For a small displacement in the positive x -direction, the work done is equal to
W = F!x . Therefore the horizontal force exerted by the charges on the outer plates
acting on the charges on the middle plate is given by
F=
Q 2s
.
2! 0 L( L + x)2
Problem 13 Energy and Force in a Capacitor
A flat conducting sheet A is suspended by an insulating thread between the surfaces
formed by the bent conducting sheet B as shown in the figure on the left. The sheets are
oppositely charged, the difference in potential, in volts, is !V . This causes a force F , in
addition to the weight of A , pulling A downward.
a) What is the capacitance of this arrangement of conductors as a function of y , the
distance that plate A is inserted between the sides of plate B ?
b) How much energy is needed to increase the inserted distance by !y ?
c) Find an expression for the difference in potential !V in terms of F and relevant
dimensions shown in the figure.
Solution:
a) We begin by assuming the plates are very large and use Gauss’s Law to calculate the
electric field between the plates
!
!
1
"
"" E ! d a = #
qenc .
0
Our choice of Gaussian surface is shown in the figure below.
Then
!
!
"
"" E ! d a = EA
cap
and
1
1
qenc = " Acap .
!0
!0
Thus Gauss’s Law implies that
! 1
E = " î , between the plates .
!0
Note that the potential difference between the positive and negative plates is
x =0
V (+) ! V (!) = !
"
Ex dx =
x=s
1
$s .
#0
So the surface charge density is equal to
!=
" 0 (V (+) # V (#))
s
The area between the plates is yb . Note that there is a charge on inner surface of the
surrounding sheet equal to ! yb , so the total charge on the outer sheets is Q = 2! yb .
Therefore the capacitance is
C( y) =
2" yb# 0 2# 0 yb
Q
=
=
.
V (+) ! V (!)
"s
s
b) How much energy is needed to increase inserted distance by !y ?
When the inserted distance is increased by
!C =
2" 0 !yb
s
Because the charge on the plates is fixed, the energy stored in the capacitor is given by
U ( y) =
Q2
2C( y)
When the inserted distance is increased by !y , the energy stored between the plates
decreases by
dU
Q2
(V (+) " V ("))2 2# 0 !yb
!U =
!C = " 2 !C = "
dC
2
s
2C
c) This decrease in energy is used to pull the hanging plate in between the two positive
charged plates. The work done in pulling the hanging plate a distance !y is given by
!W = Fy !y .
By conservation of energy
0 = !U + !W = "
(V (+) " V ("))2 2# 0 !yb
+ Fy !y .
2
s
We can solve this equation for the y-component of the force
Fy =
(! (+) " ! ("))2 2# 0 b
2
s
or
V (+) ! V (!) =
sFy
" 0b
.