Electric fields & flux; Gauss’s law - Physics - Notes - JEE

 Page 1


Scott Hughes 3 February 2005
Massachusetts Institute of Technology
Department of Physics
8.022 Spring 2005
Lecture 2:
Electric fields & flux; Gauss's law
2.1 The electric ¯eld
Last time, we learned that Coulomb's law plus the principle of superposition allowed us to
write down fairly simple formulas for the force that a \swarm" of N charges exerts on a test
charge Q:
~
F =
N
X
i=1
Qq
i
^ r
i
r
2
i
:
We found another simple formula for the force that a blob of charged material with charge
density ½ exerts on a test charge:
~
F =
Z
Q½dV ^ r
i
r
2
i
:
And, of course, similar formulas exist if the charge is smeared out over a surface with surface
density ¾, or over a line with line density ¸.
In all of these cases, the force ends up proportional to the test charge Q. We might as
well factor it out. Doing so, we de¯ne the electric ¯eld:
~
E ´
~
F
Q
=
N
X
i=1
q
i
^ r
i
r
2
i
(N Point charges)
=
Z
½dV ^ r
i
r
2
i
(Charge continuum):
Given an electric ¯eld, we can ¯gure out how much force will be exerted on some point
charge q in the obvious way:
~
F =q
~
E.
In cgs units, the electric ¯eld is measured in units of dynes/esu, or esu/cm
2
, or stat-
volts/cm|allofthesechoicesareequivalent. (We'lllearnaboutstatvoltsinaboutaweek.)
In SI units, ¯rst of all, we must multiply all of the above formulae by 1=4¼²
0
. The units of
electric ¯eld are then given in Newtons/Coulomb, or Coulomb/meter
2
, or Volts/meter.
Let's look at a couple of examples.
2.1.1 Example: Electric ¯eld at center of charged ring
Take a ring of radius R with uniform charge per unit length ¸ = Q=2¼R. What is the
electric ¯eld right at its center? Well, if we don't think about it too carefully, we want to do
something like this:
~
E =
Z
¸^ r
r
2
ds;
11
Page 2


Scott Hughes 3 February 2005
Massachusetts Institute of Technology
Department of Physics
8.022 Spring 2005
Lecture 2:
Electric fields & flux; Gauss's law
2.1 The electric ¯eld
Last time, we learned that Coulomb's law plus the principle of superposition allowed us to
write down fairly simple formulas for the force that a \swarm" of N charges exerts on a test
charge Q:
~
F =
N
X
i=1
Qq
i
^ r
i
r
2
i
:
We found another simple formula for the force that a blob of charged material with charge
density ½ exerts on a test charge:
~
F =
Z
Q½dV ^ r
i
r
2
i
:
And, of course, similar formulas exist if the charge is smeared out over a surface with surface
density ¾, or over a line with line density ¸.
In all of these cases, the force ends up proportional to the test charge Q. We might as
well factor it out. Doing so, we de¯ne the electric ¯eld:
~
E ´
~
F
Q
=
N
X
i=1
q
i
^ r
i
r
2
i
(N Point charges)
=
Z
½dV ^ r
i
r
2
i
(Charge continuum):
Given an electric ¯eld, we can ¯gure out how much force will be exerted on some point
charge q in the obvious way:
~
F =q
~
E.
In cgs units, the electric ¯eld is measured in units of dynes/esu, or esu/cm
2
, or stat-
volts/cm|allofthesechoicesareequivalent. (We'lllearnaboutstatvoltsinaboutaweek.)
In SI units, ¯rst of all, we must multiply all of the above formulae by 1=4¼²
0
. The units of
electric ¯eld are then given in Newtons/Coulomb, or Coulomb/meter
2
, or Volts/meter.
Let's look at a couple of examples.
2.1.1 Example: Electric ¯eld at center of charged ring
Take a ring of radius R with uniform charge per unit length ¸ = Q=2¼R. What is the
electric ¯eld right at its center? Well, if we don't think about it too carefully, we want to do
something like this:
~
E =
Z
¸^ r
r
2
ds;
11
where the integral will be take around the ring; ds will be a path length element on the ring,
r is the radius of the ring, and ^ r is a unit vector pointing from each length element on the
ring to the ring's center ...
But wait a minute! You should quickly notice that for each contribution due to a little
length element on the ring, there is an equal contribution from the ring's opposite side
pointing in the opposite direction. Each contribution to the
~
E ¯eld at the center of the ring
is precisely canceled by its opposite contribution:
So the ¯eld at the center of the ring is zero.
2.1.2 Example: Electric ¯eld on ring's axis
Suppose we move along the ring's axis to some coordinate z above the disk's center. The
¯eld does not cancel in this case, at least not entirely: Considering the contribution of pieces
of the ring, we see that there is a component in the radial direction, and a component along
z. The radial component cancels by the same symmetry argument as we used above. The z
component, however, de¯nitely does not cancel:
Radius R
Height above ring z
cancels in radial direction, does not
cancel in z direction.
Vectorial contribution to field
?
So, let's calculate: the contribution coming from a little segment of the ring is
d
~
E =
dQ
r
2
cosµ^ z
12
Page 3


Scott Hughes 3 February 2005
Massachusetts Institute of Technology
Department of Physics
8.022 Spring 2005
Lecture 2:
Electric fields & flux; Gauss's law
2.1 The electric ¯eld
Last time, we learned that Coulomb's law plus the principle of superposition allowed us to
write down fairly simple formulas for the force that a \swarm" of N charges exerts on a test
charge Q:
~
F =
N
X
i=1
Qq
i
^ r
i
r
2
i
:
We found another simple formula for the force that a blob of charged material with charge
density ½ exerts on a test charge:
~
F =
Z
Q½dV ^ r
i
r
2
i
:
And, of course, similar formulas exist if the charge is smeared out over a surface with surface
density ¾, or over a line with line density ¸.
In all of these cases, the force ends up proportional to the test charge Q. We might as
well factor it out. Doing so, we de¯ne the electric ¯eld:
~
E ´
~
F
Q
=
N
X
i=1
q
i
^ r
i
r
2
i
(N Point charges)
=
Z
½dV ^ r
i
r
2
i
(Charge continuum):
Given an electric ¯eld, we can ¯gure out how much force will be exerted on some point
charge q in the obvious way:
~
F =q
~
E.
In cgs units, the electric ¯eld is measured in units of dynes/esu, or esu/cm
2
, or stat-
volts/cm|allofthesechoicesareequivalent. (We'lllearnaboutstatvoltsinaboutaweek.)
In SI units, ¯rst of all, we must multiply all of the above formulae by 1=4¼²
0
. The units of
electric ¯eld are then given in Newtons/Coulomb, or Coulomb/meter
2
, or Volts/meter.
Let's look at a couple of examples.
2.1.1 Example: Electric ¯eld at center of charged ring
Take a ring of radius R with uniform charge per unit length ¸ = Q=2¼R. What is the
electric ¯eld right at its center? Well, if we don't think about it too carefully, we want to do
something like this:
~
E =
Z
¸^ r
r
2
ds;
11
where the integral will be take around the ring; ds will be a path length element on the ring,
r is the radius of the ring, and ^ r is a unit vector pointing from each length element on the
ring to the ring's center ...
But wait a minute! You should quickly notice that for each contribution due to a little
length element on the ring, there is an equal contribution from the ring's opposite side
pointing in the opposite direction. Each contribution to the
~
E ¯eld at the center of the ring
is precisely canceled by its opposite contribution:
So the ¯eld at the center of the ring is zero.
2.1.2 Example: Electric ¯eld on ring's axis
Suppose we move along the ring's axis to some coordinate z above the disk's center. The
¯eld does not cancel in this case, at least not entirely: Considering the contribution of pieces
of the ring, we see that there is a component in the radial direction, and a component along
z. The radial component cancels by the same symmetry argument as we used above. The z
component, however, de¯nitely does not cancel:
Radius R
Height above ring z
cancels in radial direction, does not
cancel in z direction.
Vectorial contribution to field
?
So, let's calculate: the contribution coming from a little segment of the ring is
d
~
E =
dQ
r
2
cosµ^ z
12
=
¸ds
R
2
+z
2
cosµ^ z
=
¸zds
(R
2
+z
2
)
3=2
^ z :
The length ds represents the length of a little segment of the ring. Integrating this up
is trivial: we go around the circumference, so the only \action" as far as the integral is
concerned is that we replace ds with 2¼R. We end up with
~
E =
2¼R¸z
(R
2
+z
2
)
3=2
^ z
=
Qz
(R
2
+z
2
)
3=2
^ z :
2.1.3 Example: Electric ¯eld on disk's axis
Using that result, it is now simple to work out a more complicated
~
E ¯eld, that above a
disk. Suppose we have a charge Q smeared out uniformly over a disk of radius R, so that
the charge per unit area ¾ =Q=¼R
2
. What is
~
E a distance z above the disk?
A simple way to do this is to consider the disk to be made out of an in¯nite number of
in¯nitesimally thin rings:
A single ring of radius r and thickness dr carries a total charge dQ = ¾2¼rdr. The con-
tribution of that ring to the disk's electric ¯eld can be easily ¯gured using our ring electric
¯eld result, replacing R with r and Q with dQ:
d
~
E =¾z^ z
2¼rdr
(z
2
+r
2
)
3=2
:
To get the total ¯eld, we integrate this from r = 0 to r =R:
~
E =
Z
R
0
¾z^ z
2¼rdr
(z
2
+r
2
)
3=2
= 2¼¾^ z
Ã
z
p
z
2
¡
z
p
R
2
+z
2
!
:
13
Page 4


Scott Hughes 3 February 2005
Massachusetts Institute of Technology
Department of Physics
8.022 Spring 2005
Lecture 2:
Electric fields & flux; Gauss's law
2.1 The electric ¯eld
Last time, we learned that Coulomb's law plus the principle of superposition allowed us to
write down fairly simple formulas for the force that a \swarm" of N charges exerts on a test
charge Q:
~
F =
N
X
i=1
Qq
i
^ r
i
r
2
i
:
We found another simple formula for the force that a blob of charged material with charge
density ½ exerts on a test charge:
~
F =
Z
Q½dV ^ r
i
r
2
i
:
And, of course, similar formulas exist if the charge is smeared out over a surface with surface
density ¾, or over a line with line density ¸.
In all of these cases, the force ends up proportional to the test charge Q. We might as
well factor it out. Doing so, we de¯ne the electric ¯eld:
~
E ´
~
F
Q
=
N
X
i=1
q
i
^ r
i
r
2
i
(N Point charges)
=
Z
½dV ^ r
i
r
2
i
(Charge continuum):
Given an electric ¯eld, we can ¯gure out how much force will be exerted on some point
charge q in the obvious way:
~
F =q
~
E.
In cgs units, the electric ¯eld is measured in units of dynes/esu, or esu/cm
2
, or stat-
volts/cm|allofthesechoicesareequivalent. (We'lllearnaboutstatvoltsinaboutaweek.)
In SI units, ¯rst of all, we must multiply all of the above formulae by 1=4¼²
0
. The units of
electric ¯eld are then given in Newtons/Coulomb, or Coulomb/meter
2
, or Volts/meter.
Let's look at a couple of examples.
2.1.1 Example: Electric ¯eld at center of charged ring
Take a ring of radius R with uniform charge per unit length ¸ = Q=2¼R. What is the
electric ¯eld right at its center? Well, if we don't think about it too carefully, we want to do
something like this:
~
E =
Z
¸^ r
r
2
ds;
11
where the integral will be take around the ring; ds will be a path length element on the ring,
r is the radius of the ring, and ^ r is a unit vector pointing from each length element on the
ring to the ring's center ...
But wait a minute! You should quickly notice that for each contribution due to a little
length element on the ring, there is an equal contribution from the ring's opposite side
pointing in the opposite direction. Each contribution to the
~
E ¯eld at the center of the ring
is precisely canceled by its opposite contribution:
So the ¯eld at the center of the ring is zero.
2.1.2 Example: Electric ¯eld on ring's axis
Suppose we move along the ring's axis to some coordinate z above the disk's center. The
¯eld does not cancel in this case, at least not entirely: Considering the contribution of pieces
of the ring, we see that there is a component in the radial direction, and a component along
z. The radial component cancels by the same symmetry argument as we used above. The z
component, however, de¯nitely does not cancel:
Radius R
Height above ring z
cancels in radial direction, does not
cancel in z direction.
Vectorial contribution to field
?
So, let's calculate: the contribution coming from a little segment of the ring is
d
~
E =
dQ
r
2
cosµ^ z
12
=
¸ds
R
2
+z
2
cosµ^ z
=
¸zds
(R
2
+z
2
)
3=2
^ z :
The length ds represents the length of a little segment of the ring. Integrating this up
is trivial: we go around the circumference, so the only \action" as far as the integral is
concerned is that we replace ds with 2¼R. We end up with
~
E =
2¼R¸z
(R
2
+z
2
)
3=2
^ z
=
Qz
(R
2
+z
2
)
3=2
^ z :
2.1.3 Example: Electric ¯eld on disk's axis
Using that result, it is now simple to work out a more complicated
~
E ¯eld, that above a
disk. Suppose we have a charge Q smeared out uniformly over a disk of radius R, so that
the charge per unit area ¾ =Q=¼R
2
. What is
~
E a distance z above the disk?
A simple way to do this is to consider the disk to be made out of an in¯nite number of
in¯nitesimally thin rings:
A single ring of radius r and thickness dr carries a total charge dQ = ¾2¼rdr. The con-
tribution of that ring to the disk's electric ¯eld can be easily ¯gured using our ring electric
¯eld result, replacing R with r and Q with dQ:
d
~
E =¾z^ z
2¼rdr
(z
2
+r
2
)
3=2
:
To get the total ¯eld, we integrate this from r = 0 to r =R:
~
E =
Z
R
0
¾z^ z
2¼rdr
(z
2
+r
2
)
3=2
= 2¼¾^ z
Ã
z
p
z
2
¡
z
p
R
2
+z
2
!
:
13
Beforemovingon, it'susefultoconsidertwolimitsofthisformula. First, considerz large
and positive: this formula can be rewritten
~
E = 2¼¾^ z
0
@
1¡
1
q
1+(R=z)
2
1
A
' 2¼¾^ z
"
1¡
Ã
1¡
1
2
R
2
z
2
!#
=
¼R
2
¾^ z
z
2
=
Q
z
2
^ z :
OnthesecondlinewehaveusedzÀ Rtoexpandthesquarerootviathebinomialexpansion.
The ¯nal result tells us that, at large z the disk has the same
~
E ¯eld as a point charge |
which should not be a surprise!
Next consider z very small and not necessarily positive: putting
p
z
2
=jzj, we have
~
E = 2¼¾^ z
Ã
z
jzj
¡
z
p
R
2
+z
2
!
' 2¼¾^ z
z
jzj
:
The function z=jzj = +1 above the disk, and¡1 below the disk. Thus, our electric ¯eld has
constant magnitude, E = 2¼¾, and points away from the disk: up on the top side, down on
the bottom. Remember this result | we will see it again soon!
2.2 Electric ¯eld lines
A useful concept to illustrate the electric ¯eld is the notion of ¯eld lines. These are ¯ctitious
lines we sketch which point in the direction of the electric ¯eld, and whose density gives us
(at least roughly) an idea of how strong the ¯eld is. Here is a sketch of the electric ¯eld lines
around a pair of point charges:
+ -
The ¯eld lines are bunched up near the charges (where the ¯eld is strongest), and spread
out farther away. They point in towards the minus charge, and away from the plus charge.
Field lines can never cross. If they did, what would the ¯eld direction be at the crossing
point? It wouldn't be de¯ned! Sometimes you have situation where it seems like ¯eld lines
are crossing; it usually turns out the magnitude goes to zero when that happens.
14
Page 5


Scott Hughes 3 February 2005
Massachusetts Institute of Technology
Department of Physics
8.022 Spring 2005
Lecture 2:
Electric fields & flux; Gauss's law
2.1 The electric ¯eld
Last time, we learned that Coulomb's law plus the principle of superposition allowed us to
write down fairly simple formulas for the force that a \swarm" of N charges exerts on a test
charge Q:
~
F =
N
X
i=1
Qq
i
^ r
i
r
2
i
:
We found another simple formula for the force that a blob of charged material with charge
density ½ exerts on a test charge:
~
F =
Z
Q½dV ^ r
i
r
2
i
:
And, of course, similar formulas exist if the charge is smeared out over a surface with surface
density ¾, or over a line with line density ¸.
In all of these cases, the force ends up proportional to the test charge Q. We might as
well factor it out. Doing so, we de¯ne the electric ¯eld:
~
E ´
~
F
Q
=
N
X
i=1
q
i
^ r
i
r
2
i
(N Point charges)
=
Z
½dV ^ r
i
r
2
i
(Charge continuum):
Given an electric ¯eld, we can ¯gure out how much force will be exerted on some point
charge q in the obvious way:
~
F =q
~
E.
In cgs units, the electric ¯eld is measured in units of dynes/esu, or esu/cm
2
, or stat-
volts/cm|allofthesechoicesareequivalent. (We'lllearnaboutstatvoltsinaboutaweek.)
In SI units, ¯rst of all, we must multiply all of the above formulae by 1=4¼²
0
. The units of
electric ¯eld are then given in Newtons/Coulomb, or Coulomb/meter
2
, or Volts/meter.
Let's look at a couple of examples.
2.1.1 Example: Electric ¯eld at center of charged ring
Take a ring of radius R with uniform charge per unit length ¸ = Q=2¼R. What is the
electric ¯eld right at its center? Well, if we don't think about it too carefully, we want to do
something like this:
~
E =
Z
¸^ r
r
2
ds;
11
where the integral will be take around the ring; ds will be a path length element on the ring,
r is the radius of the ring, and ^ r is a unit vector pointing from each length element on the
ring to the ring's center ...
But wait a minute! You should quickly notice that for each contribution due to a little
length element on the ring, there is an equal contribution from the ring's opposite side
pointing in the opposite direction. Each contribution to the
~
E ¯eld at the center of the ring
is precisely canceled by its opposite contribution:
So the ¯eld at the center of the ring is zero.
2.1.2 Example: Electric ¯eld on ring's axis
Suppose we move along the ring's axis to some coordinate z above the disk's center. The
¯eld does not cancel in this case, at least not entirely: Considering the contribution of pieces
of the ring, we see that there is a component in the radial direction, and a component along
z. The radial component cancels by the same symmetry argument as we used above. The z
component, however, de¯nitely does not cancel:
Radius R
Height above ring z
cancels in radial direction, does not
cancel in z direction.
Vectorial contribution to field
?
So, let's calculate: the contribution coming from a little segment of the ring is
d
~
E =
dQ
r
2
cosµ^ z
12
=
¸ds
R
2
+z
2
cosµ^ z
=
¸zds
(R
2
+z
2
)
3=2
^ z :
The length ds represents the length of a little segment of the ring. Integrating this up
is trivial: we go around the circumference, so the only \action" as far as the integral is
concerned is that we replace ds with 2¼R. We end up with
~
E =
2¼R¸z
(R
2
+z
2
)
3=2
^ z
=
Qz
(R
2
+z
2
)
3=2
^ z :
2.1.3 Example: Electric ¯eld on disk's axis
Using that result, it is now simple to work out a more complicated
~
E ¯eld, that above a
disk. Suppose we have a charge Q smeared out uniformly over a disk of radius R, so that
the charge per unit area ¾ =Q=¼R
2
. What is
~
E a distance z above the disk?
A simple way to do this is to consider the disk to be made out of an in¯nite number of
in¯nitesimally thin rings:
A single ring of radius r and thickness dr carries a total charge dQ = ¾2¼rdr. The con-
tribution of that ring to the disk's electric ¯eld can be easily ¯gured using our ring electric
¯eld result, replacing R with r and Q with dQ:
d
~
E =¾z^ z
2¼rdr
(z
2
+r
2
)
3=2
:
To get the total ¯eld, we integrate this from r = 0 to r =R:
~
E =
Z
R
0
¾z^ z
2¼rdr
(z
2
+r
2
)
3=2
= 2¼¾^ z
Ã
z
p
z
2
¡
z
p
R
2
+z
2
!
:
13
Beforemovingon, it'susefultoconsidertwolimitsofthisformula. First, considerz large
and positive: this formula can be rewritten
~
E = 2¼¾^ z
0
@
1¡
1
q
1+(R=z)
2
1
A
' 2¼¾^ z
"
1¡
Ã
1¡
1
2
R
2
z
2
!#
=
¼R
2
¾^ z
z
2
=
Q
z
2
^ z :
OnthesecondlinewehaveusedzÀ Rtoexpandthesquarerootviathebinomialexpansion.
The ¯nal result tells us that, at large z the disk has the same
~
E ¯eld as a point charge |
which should not be a surprise!
Next consider z very small and not necessarily positive: putting
p
z
2
=jzj, we have
~
E = 2¼¾^ z
Ã
z
jzj
¡
z
p
R
2
+z
2
!
' 2¼¾^ z
z
jzj
:
The function z=jzj = +1 above the disk, and¡1 below the disk. Thus, our electric ¯eld has
constant magnitude, E = 2¼¾, and points away from the disk: up on the top side, down on
the bottom. Remember this result | we will see it again soon!
2.2 Electric ¯eld lines
A useful concept to illustrate the electric ¯eld is the notion of ¯eld lines. These are ¯ctitious
lines we sketch which point in the direction of the electric ¯eld, and whose density gives us
(at least roughly) an idea of how strong the ¯eld is. Here is a sketch of the electric ¯eld lines
around a pair of point charges:
+ -
The ¯eld lines are bunched up near the charges (where the ¯eld is strongest), and spread
out farther away. They point in towards the minus charge, and away from the plus charge.
Field lines can never cross. If they did, what would the ¯eld direction be at the crossing
point? It wouldn't be de¯ned! Sometimes you have situation where it seems like ¯eld lines
are crossing; it usually turns out the magnitude goes to zero when that happens.
14
2.3 Flux
The notion of °ux is an important one in physics. It refers to the °ow of some vectorial
quantity through an area. The simplest example to picture is the °ow of °uid. Imagine that
°uid is °owing with velocity ~ v = v^ x. Now imagine that we dip a square wire loop of cross
section area A into this °uid. We describe this loop with a vector
~
A = A^ n, where ^ n is the
normal to the plane of the loop:
Loop, area A
^
Unit normal n
The velocity °ux ©
v
is de¯ned as the overlap between the velocity vector ~ v and the loop
area
~
A:
©
v
=~ v¢
~
A:
For a uniform, plane loop, and constant, uniform velocity ~ v, we have ©
v
= vAcosµ, where
µ is the angle between ^ n and the velocity vector:
F = F = 
F = 
0 v A
v A cos
?
?
For this example, the rate of °uid °ow through this loop is proportional to this °ux, ©
v
.
What if things aren't nice and uniform like this? Then, we break the cross sectional area
of the loop up into little squares d
~
A; we assign each square its own normal ^ n. The °ux then
becomes an integral:
©
v
=
Z
~ v¢ d
~
A:
The integral is taken over the entire surface through which we wish to compute the °ux.
15