MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Spring 2013
Problem Set 2 Solutions
Problem 1
(a) Flux Through a Face of a Cube Consider a cube with each side of length a . A pointlike object with charge Q is placed at one corner of a cube shared by three faces. What is
the flux of the electric field emerging from each of the other three square faces of the
cube?
Answer: Consider a cube of side 2a . This cube can be constructed from eight cubes each of side
a.
Place a point-like object with charge Q is placed at the center of this larger cube, then the total
flux through the larger cube, by Gauss’s law, is just Q /  0 . Since each face is identical, the flux
through each face of the larger cube is 1/6 of the total flux or Q / 6 0 . We can divide each larger
face into four equal square faces of side a . Therefore the flux through each of these smaller
square faces is Q / 24 0 . The flux though one of these smaller square faces is the same as the
flux through each face of a cube of side a with the charged object placed at the corner, the
quantity we would like to determine.
(b) Estimation: Excess Charge on a Thundercloud Estimate the magnitude of the excess
charge on a thundercloud just before a lightening strike? (HINT: Dry air breaks down at
an electric field strength of about 3  106 N  C1 ). Explain your reasoning.
Solution. The bottom surface of a thundercloud and the earth can be modeled as a pair of infinite
parallel plate conductors, meaning that the electric field is constant between the cloud and the
earth. We can then use Gauss’ law to find the charge on the thundercloud,
r r Qenc
E
“  dA 
0
which becomes
r
Q
.
E A
0
Therefore the magnitude of the charge on the thundercloud is
Q
r
EA
0
We want to estimate the excess charge on a thundercloud just before a lightening strike. We need
to know that dry air breaks down at an electric field strength of about 3  106 N  C ). So we just
need to estimate the area of the cloud area. Let’s estimate that the linear dimension of the base of
the thundercloud is 1 km and hence an area of about 1 km2, (this is a small cloud). Now we can
estimate the charge from the charge density:
r
Q   0 E A : (8.86  1012 C2  N -1  m -2 )(3  106 N  C-1 )(103 m)2 : 30 C
In reality, thunderclouds look more like electric dipoles (they get a charge at their tops as well)
so this entire calculation is pretty rough, but probably of the right order of magnitude.
Problem 2 Electric Field of a Washer
A very thin washer has outer radius b and inner radius a . The washer is uniformly charged with
a positive surface charge density  . What is the magnitude and direction of the electric field at
an arbitrary point P along the positive z -axis, a distance z from the origin?
Solution. We begin by choosing cylindrical coordinates (r, , z) . Choose a small charged
element a distance r  from the center of the disc. The amount of charge on that element is
dq   r dr d  .
r
We refer to this element as the “source” with position vector r   r rφ.. We are trying to find the
electric field at an arbitrary point a distance z along the positive z-axis. We refer to this as the
r
“field point” with position vector r  z kφ .
The electric field at a point P due to each charged element dq is given by Coulomb’s law:
r
1
dq r r
dE 
(r - r )
r
4 0 r - rr  3
r r
r r
r r
where r - r  is the distance from the source dq to the field point P and (r - r  ) / r - r  is the
unit vector pointing from the source to the field point.
Using our choice of coordinate system and position vectors for the source and the field point we
have that
r r
r - r  zkφ  r rφ
.
r r
r - r  (z 2  r  2 )1/ 2
The electric field along the z-axis due to the distribution is given by the integral expression
r
E(z) 
1
4 0
2 b

0 a
 r dr d (zkφ  r rφ)
(z 2  r  2 )3/ 2
The radial component integrates to zero because
2

0
d r rφ

(z 2  r  2 )3/ 2
2

0
d r (cos φ
i  sin  φ
j) r
 0.
(z 2  r  2 )3/ 2
Thus the expression for the electric field simplifies to
r
z b
r dr 
E(z) 
kφ .

2
2 0 a (z  r  2 )3/ 2
which we can now integrate and find that
b
r
z
1
E(z)  
kφ
2
2 0 (z  r  2 )1/ 2 a
φ
z 
1
1


k
2 0  (z 2  b2 )1/ 2 (z 2  a 2 )1/ 2 
.
Problem 3 Charged Rods
(a) A very thin rod of length L lies along the x-axis with its left end at the origin. The rod has a
non-uniform charge density    x , where  is a positive constant and x is the distance from
the origin.
Calculate the magnitude and direction of the electric field at the point P, shown in the figure
above. Take the limit d ? L . What does the electric field look like in this limit? Is this what
you expect? Explain. Hint: the following mathematical facts may be useful:
x dx
 (x  a)
2

a
 ln(x  a)
xa
ln(1  x)  x 
x2
L
2
(for small x)
Solution:
Choose coordinates as shown in the figure below.
We begin by calculating the total charge on the rod. The charge is the integral of the charge on a
small element dq located at x  is given by
x L
Q

dq
x  0
Using the coordinate system in the figure above with x denotes the distance to the dq ,
dq   ( x  )dx  where  ( x  )   x  is the non-uniform charge density. Note that x  is the
integration variable.
x L
Q

x  0
x L
dq 

 ( x  )dx  
x  0
x L

x  0
 x dx  
 x 2
2
x L

 L2
x  0
2
[ ]  C  m-2 because the units of [ ]  C  m-1 . So the units of
[Q]  [ ][ L2 ]  (C  m-2 )(m2 )  C .
Check: The units of
We will use the integral version of Coulomb’s Law
r
E

source
r
dE  ke
dq
rφdq  , P .
2
r
source dq  , P

r
r
r r
r  d φ
i,
r  x φ
i,
rφdq  , P   φ
i,
rdq  , P  r  r   d  x 
the figure above,
r r
2
 r  r   d  x  , rdq,
 ( x   d)2 , and dq   xdx . Then Coulomb’s Law becomes
P
From
rdq, P
r
E
x L
r
 xdx
 d E  ke  (d  x)2 (φi) .
source
x  0
From the integral table above we find
x L
r
 d

 x dx  φ
E  ke 
(
i)

k


ln(
x

d)

e
2
 x   d

x  0 (d  x  )
x L
( φ
i)
x  0
 d


E  ke  
 ln( L  d )   1  ln(d )   (ˆi )

 L  d

We can combine terms and use our result above for and   (2Q / L2 ) to find that
 d

E  ke (2Q / L2 )  
 1  ln(1  L / d )   (ˆi )

 L  d
In the limit as d  L , we can use the expansion (which can be derived from the Taylor formula)
ln(1  L / d )  L / d 
( L / d )2

2
(for small L / d )
The electric field becomes
E  2ke

Q 
L
L L2


 2     (ˆi )

2 
L   L  d d 2d

This simplifies to
E
2ke
Q 
L2
L2   ˆ


  (i )
L2   ( L  d )d 2d 2  
Since d  L , this finally reduces to
E
2ke
Q L2
(ˆi )
L2 2d 2
ke
Q ˆ
(i )
d2
Check: when d  L , the rod looks like a point charge in agreement with our limit.
(b) Two thin rods of length L carry equal charges Q uniformly distributed over their lengths. The
rods are aligned end to end, with their nearest ends separated by a distance d. What is the
magnitude of the force that one rod exerts on the other?
Solution: The rod is made of a number of charges dq   dx  , where x  is the variable that
moves us through the rod from x  0 to x  L and the linear charge density   Q / L .
We are calculating the electric field at point P, a distance x to the right of the rod. So the
r
i .
distance from dq to P is r  x  x ' φ

r
So, dE 

L
r
r
1 dq r
1
 dx '
r  E   dE  
3
4 0 r
4 0 x  x '
x '0


3
φ
i
x  x' φ
i
4 0


φ
du
i 1


 u2 4 u
u x
0
x L
x L
x
r
  1
1
Q  1
1
E  φ
i
   φ
i
 .


4 0  x  L x 
4 0 L  x  L x 
Note that this points to the right, as we would expect. Now we need to calculate the force on the
right rod due to this electric field.
Force on Right Rod
dL
r
r
r
r dL r
  1
1
φ
dF  dqE  F   dF   dqE   i   dx
 

4 0  x  L x 
xd
xd
Here we are integrating over the x variable, starting at x = d, the left of the right rod, and moving
to d + L, the right. Turning the crank:
r
2 d L  1
1
2
φ
F  φ
i
dx



i
ln x  L  ln x
4 0 x  d  x  L x 
4 0


dL
d
2  x  L 
 φ
i
ln
4 0 
x 
dL
d
where we have used the fact that ln a  ln b  ln a b . Continuing:








2

2
2 
2
r
d  2L d 
1 L d 
 d  2L


d

L

Q
 ln
 φ
 ln

F  φ
i
ln
 ln
 φ
i
i
2
2
4 0 
dL
d 
4 0 



4

L
1

2
L
d
dL 
0



I flipped the sign (and hence the numerator/denominator) to emphasize that the force is positive,
repelling the two rods. Note that the numerator is larger than the denominator.
Problem 4 Non-uniformly Charged Cylinder A solid very long cylinder of radius R has a
charge density   0 (r / R) where  0 is a constant and r is the radial distance from the center
axis of the cylinder. Find the direction and magnitude of the electric field everywhere (both
inside and outside the cylinder).
Solution: We begin by modeling the very long cylinder as infinite so that we may use Gauss’s
Law to calculate the electric field.
There are two regions of space: region I: r  R , and region II: r  R so we apply Gauss’ Law to
each region to find the electric field.
For region I: r  R , we choose a cylinder of radius r and length l as our Gaussian surface.
Because we have assumed that the cylinder is infinite, the electric field by symmetry points
radially outward from the symmetry axis hence the electric flux through this closed surface is
r
r
E

d
A
“ I  EI  2 rl .
Since the charge distribution is non-uniform, we will need to integrate the charge density to find
the charge enclosed in our Gaussian surface. We choose as an integration volume element a
small cylinder shell of length l and radius r  , and thickness dr  , thus dV   2 lr dr  . We use
the integration variable r  in order to distinguish it from the radius r of the Gaussian cylinder.
The associated enclosed charge is dq  ( r  )2 lr dr  . Then
Qenc
0

1
rr
 0 r 0
 2 lr dr  
1
rr
 0 r 0
0 (r  / R)l2 r dr  
0 2 l r   r 2
0 2 lr 3 2 0 lr 3
r
d
r


.


R 0 r 0
3R 0
3 R 0
Notice that the integration variable is primed and the radius of the Gaussian cylinder appears as a
limit of the integral.
Recall that Gauss’ Law equates electric flux to charge enclosed:
r Qenc
r
E

d
A
“ I  .
0
So we substitute the two calculations above into Gauss’ law to arrive at:
2 0 lr
.
EI  2 rl 
3 R 0
3
We can solve this equation for the electric field
r
 r2
EI  EI rφ  0 rφ, 0  r  R .
3R 0
r
 r2
The electric field points radially outward and has magnitude EI  0 , 0  r  R .
3R 0
For region II: r  R : we choose the same cylindrical Gaussian surface of radius r  R .
The electric flux has the same form
r
r
E

d
A
“ II  EII  2 rl
Then
Qenc
0
0 2 l r   R 2
0 2 lR3 2 0 lR2
.

 2 lr dr  
 (r  / R)l2 r dr  
r  dr  

 0 r 0
 0 r 0 0
R 0 r 0
3R 0
3 0
1
r R
1
r R
Gauss’ Law becomes
2 0 lR
.
EII  2 rl 
3 0
2
We can solve this equation for the electric field
r
 R2
EII  EII rφ 0 rφ, r  R .
3 0r
In this region of space, the electric field points radially outward and has magnitude
r
0 R2
EII 
, r  R , so it falls off as 1/ r as we expect since outside the charge distribution, the
3 0r
cylinder acts as if it were a charged wire.
Problem 5: Consider a spherically symmetric charge distribution given by
 0 (1  r / R) ; r  R
0 ; r  R
 (r)  
where  0 is a negative constant. Embedded in the center of the distribution is a point-like
positively charge object with charge Q  0 such that the electric field in the region r  R is
zero.
a) Find an expression for the charge Q of the point-like charged object.
b) Find a vector expression for the electric field in the region r  R .
Solution:
a) For region II: r  R : we choose a spherical Gaussian surface of radius r  R , and we are told
that the electric flux is zero, Therefore the charge enclosed is zero.
The charge is now enclosed inside the Gaussian sphere and is equal to
r R
0  Qenc  Q 

 4 r dr  Q 
2
r  0
Q   0 R3 / 3  0 4
r R

0 (1  r  / R)4 r  2 dr   Q
r  0
r R

( r  2  r  3 / R)dr   Q
r  0
 0 4 ((R3 / 3)  (R 4 / 4))  Q  0 R3 / 3  Q
Therefore
Q   0 R3 / 3
(1.1)
(b) For region I: r  R , we choose a sphere or radius r as our Gaussian surface.
Then, the electric flux through this closed surface is
r
“ E
I
r
 d A  EI 4 r 2 . Becasue the
charge distribution is non-uniform, we will need to integrate the charge density to find the
charge enclosed in our Gaussian surface. In the integral below we use the integration variable
r  in order to distinguish it from the radius r of the Gaussian sphere.
Qenc
0

1
rr

 4 r  dr   Q 
2
 0 r  0
1
rr

 0 r  0

0 4 r   r 2
( r   r  3 / R)dr   Q

 0 r  0

0 4 3
((r / 3)  (r 4 / 4R))  Q
0
0 (1  r  / R)4 r  2 dr   Q
.
Using our result for the charge of the point-like object (eq. 1.1), we have that
Qenc
0
0
0 R3 0  4r 3 r 4 R3 
3
4

((4r / 3)  (r / R)) 

  
0
3 0
 0  3
R
3
Notice that the radius of the Gaussian sphere appears as a limit of the integral. Gauss’ Law
equates electric flux to charge enclosed:
r Qenc
r
E

d
A
“ I  .
0
So we substitute the two calculations above into Gauss’ Law to arrive at:
EI 4 r 2 
0  4r 3 r 4 R3 
  .
 0  3
R
3
We can solve this equation for the electric field
r
  4r r 2 R3 
E I (r)  EI (r)φ
r  0    2  rφ, 0  r  R .
4 0  3 R 3r 
The electric field points radially outward and has magnitude
r
  4r r 2 R3 
EI  0    2  , 0  r  R .
4 0  3 R 3r 
Problem 6: Gauss’s Law for Gravitation
The gravitational force on a point-like object 1 with mass m1 due to the interaction with a second
point-like object 2 of mass m2 that are separated by a distance r12 is given by the expression
r
Gm1m2
F21  
rφ
r212 21
where rφ21 is the unit vector located at object 1 and pointing from object 2 to object 1.
r
a) In your own words, define the gravitational field g of a point-like object of mass m .
Then find a vector expression for the gravitational field at a distance r from the pointlike object.
b) Find an expression for a gravitational version of Gauss’s law.
c) Suppose you are given a spherically symmetric distribution of matter with uniform mass
density  and radius R ? Determine a vector expression for the gravitational field in the
regions (i) r  R and (ii) r  R ?
d) Consider an infinite universe with a uniform mass density  . Use your gravitational
version of Gauss’s Law to show that the gravitational field must be zero everywhere.
Hint: Consider any point P in space. Choose two spherical Gaussian surfaces centered
about different points such that the point P lies on the surface of either Gaussian surface.
What is the gravitational field at the point P ?
Solution:
a) We define the gravitational field at a point P in space due to a distribution of matter as
follows. Place a point-like “test object” with mass mt at the point P . Measure the force
acting on the test object. Note we assume that the gravitational force between the “test
object” and the source of the gravitational field does not distort the distribution of matter
of the source. The gravitational field at P is the defined to be the force on the test object
divided by the mass of that test object,
r
Ft
r
.
g(P) 
mt
b) If we calculate the gravitational field due to a point-like object (source) of mass ms , we
can use the universal law of gravity and find that
r
Fst
Gm m
Gm
r
g s (P) 
  2 s t rφst   2 s rφst .
mt
rst mt
rst
Just like in electrostatic, we now calculate the gravitational flux on a surface of radius r
centered on the point-like source.
“
r
r
gs  d a 
sphere
radius r
“

Gms
phere
radius r
r2
da  
Gms 4 r 2
r2
 G4 ms .
We generalize to any mass distribution because the gravitational force is an inverse square radial
force just as we did for the electrostatic force given by Coulomb’s Law, to get a Gauss’s Law for
gravitation
r
r
g s  d a  G4 menc  G4
“
closed
surface

 dV .
volume
enclosed
c) Suppose you are given a spherically symmetric distribution of matter with mass density
 and radius R ? What is a vector expression for the gravitational field in the regions (i)
r  R and (ii) r  R ?
(i)
for r  R : For our Gaussian surface we choose a sphere of radius r  R , then the
flux of the gravitational field on that surface is
“
r
r
g s  d a  g4 r 2
sphere
radius r
The mass enclosed is
G4

 dV  G4(4 / 3) r 3 .
volume
Applying Gauss’s Law for gravity yields
g4 r 2  G4(4 / 3) r 3
Thus the gravitational field inside the object is
r
g(r)  (G4 / 3)r rφ;
r  R.
(ii)
For the region r  R (outside the object), we still choose a Gaussian sphere of radius
rR .
Gauss’s Law becomes
g4 r 2  G4(4 / 3) R3
Thus the gravitational field outside the object is
G (4 / 3) R3
r
g(r)  
rφ ;r  R .
r2
Note that the mass of the object is M  (4 / 3) R3 so the gravitational field becomes
GM
r
g(r)   2 rφ ; r  R.
r
Outside a spherically symmetric object, the gravitational field looks as if all the matter were
located a point at the center of the sphere and the field falls off as 1 / r 2 . Inside a uniform
spherical matter distribution, the gravitational field is zero at the center and grows as r .
 G4
r rφ;

r
3
g(r)  
 GM rφ ;
 r 2
rR
.
rR
d) Consider any point P in space. Choose two spherical Gaussian surfaces centered about
different points with the same radius r such that the point P lies on the surface of either
Gaussian surface.
By the gravitational version of Gauss’s Law, the gravitation field at P using the Gaussian
r
surface on the left is g L , and the gravitational field due to the Gaussian surface on the right is
r
r
r
g R . By symmetry g L  g R , but that is impossible so the only possible solution is that
r
r
r
g L  g R  0 . Thus the gravitation field must be zero everywhere. This result holds true for a
universe in which we do not consider the effects of general relativity. So in this Newtonian
infinite universe with a uniform mass density, there is no gravitational field.
Problem 7: Infinite Uniformly Charged Slabs
Three infinite uniformly charged thin sheets are shown in the figure below. The sheet on the left
at x  d is positively charged with charge per unit area of 2 , The sheet in the middle at
x  0 is negatively charged with charge per unit area of  , and the sheet on the right at x  d
is positively
charged with charge per unit area of 2 . Determine an expression for the electric
r
field E in each of the regions 1, 2, 3, and 4.
Solution.
We begin by using Gauss’s Law to describe the electric field of an infinite plane with
uniform charge density  . Consider the plane with our choice of Gaussian surface
shown in the figure below.
The electric field at any point on the right and left sides are equal in magnitude but point
r
r
r
r q
in opposite directions. Thus E  E L  E R . Gauss’s Law , “ E  d a  enc , becomes
0
2EAcap   Acap /  0 . Thus Gauss’s Law implies that the magnitude of the electric field
due to the charged plate is uniform and independent of the distance from the charged
plate
E

.
2 0
For the charge configuration given in the problem, we can now use the superposition
principle with the magnitudes of the electric fields E2  2 / 20 , and E   / 20 ,
with the directions shown in the figure below.
The field in region 1 points to the right and has a magnitude equal to the sum
E1  E2  E2  E  2 / 0   / 20  3 / 20 .
The field in region 2 points to the left and has a magnitude equal to the sum
E2  E2  E2  E   / 20 .
The field in region 3 points to the right and has a magnitude equal to the sum
E3  E2  E2  E   / 20 .
The field in region 4 points to the left and has a magnitude equal to the sum
E4  E2  E2  E  2 / 0   / 20  3 / 20 .
Therefore the electric field is the vector sum of these fields and is given by
 3 φ
 2 i ; x  d
0

 
φ
i; d x0

r
 2 0
E(x)  
  φ
i; 0  x  d
 2 0

 3 φ
i; x  d
 2
0

Problem 8 N-P Junction
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’s model the N-P junction 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 a
positive uniform charge density  0 . The adjacent P-type material lies in the range
a  z  0 and has a negative uniform charge density  0 . Thus:
 
 0
 (z)   0

0
0  z a
 a z 0
z a
Find an expression for the electric field in the regions (i) z  a , (ii) a  z  0 , (iii)
0  z  a , and (iv) z  a .
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 semi-conductor )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
r
0
r
r
E 2
ET (z)   r
1
E
r
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
“
closed
surface
r r Q
E  da  enclosed .
0
So for our choice of Gaussian surface, on the cap inside the slab the unit normal for the area
nφ  kφ . So the dot product
vector points in the positive z-direction, thus
r
φ  E da . Therefore the flux is
φ  E2,z kφ  kda
becomes E2  nda
2,z
PS01-21
“
r r
E  da  E2,z Acap
closed
surface
The charge enclosed is
Qenclosed
0

 0 Acap (a  z)
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
vector is
r
  (a  z) φ
E2  0
k.
0
r
r
r
 a
Note when z  a then E 2  0 and when z  0 , E2  0 kφ .
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 –z-direction. So the dot product becomes
r
φ  kda
φ   E da
φ  E1,z ( k)
E1  nda
1,z
Thus Gauss’s Law becomes
 E1,z Acap 
 0 Acap (a  z)
0
.
So the electric field is
PS01-22
E1,z  
0 (a  z)
.
0
The vector description is then
r
  (a  a) φ
E1  0
k
0
r
r
r
 a
Note when z  a then E1  0 and when z  0 , E1  0 kφ .
0
So the resulting field is
r
0
z  a
r
E   0 (a  z) kφ
 2
0
r
ET (z)  
r
E   0 (a  z) kφ
 1
0
r
z a
0
 a  z 0
.
0 z  a
The graph of the electric field is shown below
.
PS01-23