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Integrating Problem Solving in Physics
In the treatment of fields in physics, there are several integrals which pop
up and look frightfully different from their tamer cousins you encountered
in Calculus class. The contain vector products and are decorated with
circles, double integral signs and odd differentials like ds and dA.
Fear not! We need only point out what these embellishments represent and
restrict ourselves to the simpler applications and we can find out quite a lot
about vector fields. In this tour, I hope to show you how to navigate the
fields of vector physics.
We have two new kinds of integrals here, line integrals and area integrals.
We also will have a variety of geometries in which to employ our tools.
The most powerful tools we have in using these relations are symmetry and
the ability to choose any path or surface we wish for the integral.
To get a real flavor for what mathematical physics is all about and why it is
so powerful, you need to get the big picture of how situations are analyzed
and modeled to gain understanding. To solve a problem you haven’t seen
before, you need to gain an understanding of the situation being described,
decide what ideas and principles apply to the problem, then work out the
geometry and finally use the calculus and algebra to produce the new
solution. For more involved problems, some of these steps may be revisited
for modification and clarification.
Schematically:
The geometry is often the hardest part of the process. In order to employ
calculus, the geometry from the diagram must lead to a way of adding
contributions from small parts or modeling curves with short, straight
segments. This is what calculus does. One of the most powerful examples of
this process is that of taking a result for a point particle and extending it to
a distributed source, such as a spherical mass modeled as a collection of
small point masses each with its own gravity.
As examples of this process, let’s look at the electric and magnetic fields
and their various integrals. Several of these integrals appear with strange
decorations on them and involve vector products and multiple dimensions, but
not to worry! With the power of symmetry and cheating properly by
choosing simple surfaces and paths, we can reduce these to tamer integrals
of the standard sort, and may not even need to do the integral to find the
answer!
There are a few basic principles which science has adopted and those are
given here.
Principle of Universality: Correct physical laws hold everywhere in the
universe at all times, past and future, subject to local conditions.
Example: Newton’s law of gravitation holds for planetary systems anywhere in the
universe as well as for falling objects on any planet or star.
Principle of Isometery: There are no special directions with different laws.
Differences are the result of the distribution of matter and energy in the
local region. This is often the source of the phrase “By symmetry”.
Example: The inverse square law for light, gravity, etc.
Principle of Causality: (So far, this one’s taking some uncomfortable hits at
the quantum level.) Every event has a specific physical cause which precedes
it in time.
Example: An electric field is produced by an electric charge and radiates out from the
charge at the speed of light.
Principle of Superposition: If there are more than one source of the same
type for an effect, the effects from each source are added to get the total
effect. (This principle is not valid for nonlinear situations where the effects
are not independent)
Example: Waves from two different sources add to give interference.
The Electric Field
Principle: The electric field from a point charge extends in all directions
equally.
This principle says that the electric field should have the same strength at
any point on a sphere centered on the point charge. By symmetry, the
direction of the field must be radially out (+ charge) or in (- charge). That is
because any tilt in the field vector would indicate a difference between
points on the sphere the vector is tilting away from versus the points it
leans toward, which violates the Principle of Isotropy.
This idea can be applied to two spheres of different radii. For both, the
field points radially outward and is the same at all points on the sphere. But
this means that the field is more spread out over the larger sphere and so
must follow the inverse square law. Including a constant for the effect of
the medium on the field (1/o for free space), this gives
E =
___q___
(4r2) o
This expression represents the electric field produced in terms of a source
(the charge q), geometry (4r2 surface of a sphere), and the property of the
medium (o). We will see this with all our expressions.
Having a model for the electric field due to a point charge, we can go on to
construct models for distributed charges, such as lines, sheets and spheres.
To do so, the distribution is modeled as a collection of point charges and the
fields produced by the point charges are added up. This is what leads to
these integrals with all the vector products and decorations. Let’s look at
the electric field for a few situations and then broaden our discussion to the
magnetic field.
Electric field for a disk of charge
To keep our calculations simple, we will limit our discussion to a circular disk
of radius R and look for the electric field at points along the axis of the
disk. More general problems can be done, but then you are in the full-blown
calculus treatment and that is beyond our scope in this course.
Diagram:
Principles:
Superposition,
Electric field from a point charge
Geometry:
Electric field from a point charge is the same at any point which is the
same distance away, so find portions of the disk which are equidistant from
the arbitrary point on the axis. These are rings.
Consider a single ring within the disk, radius a and thickness dr which is
concentric with the disk. Work this problem, then add up results for all
rings.
The field from a point charge dq on the ring of radius a is
E = ___dq___
in the direction of r in the diagram.
2
(4r ) o
Superposition of the fields from the points on the ring leads to a
cancellation of the field components parallel to the plane of the ring (points
at opposite ends of a diameter will have equal and opposite parallel
components). This leaves only the perpendicular components to add, so we
get from each point
E =
_dq cos()___
(4r2) o
along the axis.
Calculus:
And for the whole ring, just add up the area of the ring
and multiply by charge density . dq now becomes
2ada, setting up the integral for the whole disk.
dE =
_2ada cos()_
(4r2) o
along the axis.
This gives
E = _  a cos() da
2o 
r2
Now we need to use some more geometry to eliminate some of the
dependent variables. The variable a has simple limits of 0 and R, the radius
of the disk, so let’s try to get r and cos() in terms of a. We also expect the
distance d from the disk to be in the result.
Geometry:
From the initial diagram, r2 = a2 + d2 and cos() = d/r
Calculus:
Substituting these in gives the integral:
E = _d R _a da___
2o 0 (a2 + d2)3/2
E =
=
_  1 - ___d___ 
2o 
(R2 + d2)1/2 
_d __-1___ |R
2o (a2 + d2)1/2 |0
Check: at R=0 (no disk), E=0
at R = , E = /2o
the value for an infinite sheet of charge
The form of the electric field for a point charge suggests another approach,
which is strengthened by a visual image. Think about the field lines pointing
straight out radially from the charge and crossing a spherical surface. If
you count up all the lines crossing the surface, you get the same number,
regardless of the size of the sphere, and the number of lines is
conventionally proportional to the charge at the center.
This leads to Gauss’s Law. For the electric field, Gauss’s Law states that the
surface integral of EdA around a closed surface is equal to the enclosed
charge/o. This says the flux of the electric field through a closed surface
is equal to the enclosed charge/o.
 EdA = q/o
The integral on the left is a surface integral and the differential dA is a two
dimensional differential of the area perpendicular to the surface and
pointing outward. Let’s talk a bit about surface integrals in the limited
context we will encounter them.
Surface integrals:
Surface integrals we will use involve integrating over a surface we choose
the dot product of a vector with a differential, dA. The differential of
area dA is perpendicular to the surface and pointing outward.
We have three such surface integrals, Gauss’s Law, the integral of the
current density through the surface bounded by the curve in Ampere’s Law
and the integral for the flux through a surface. All these are essentially
flux integrals, summing the flux, or flow, of some vector quantity through
the surface.
Let’s look at Gauss’s law for three different geometries. For the electric
field, Gauss’s Law states that the surface integral of EdA around a closed
surface is equal to the enclosed charge q/o. This says the flux of the
electric field through a closed surface is equal to q/o.
 EdA = q/o
Applying this to a point charge gives a right-hand side of Q/o . In
choosing the surface to use for the integral, it is important to use our two
principles of problem solving :
(a) be creatively lazy and (b) cheat properly
The integral will be simplest if the electric field is perpendicular to the
surface everywhere. Since the field lines are radially outward, they are
perpendicular to a spherical surface and, even better, the value of the
electric field should be constant everywhere on the surface. The integral
then becomes
 EdA =  EdA = E dA = E A
since dA = A
For a spherical surface, this gives
E 4r2 = Q/o
E = Q/(4r2 o)
That’s the power of symmetry and vector calculus!
Let’s try an infinite conducting sheet with a surface charge density of
 C/m2.
Of three possible directions near the sheet, the vector perpendicular to the
surface is the unique direction. The other two are both parallel to the
surface. So our surface of integration should have portions parallel to and
perpendicular to the surface. Think of a tuna can with one end inside the
sheet and both ends parallel to the surface of the sheet.
Since the electric field radiates out from electric charge, the field must be
perpendicular to the surface of the sheet. (If it were not, there must be
something different about the direction it tilts; since there isn’t any
difference, there must not be any tilt.)
Given this, the dot product EdA , with dA perpendicular to the surface,
equals zero for the cylindrical part of the integration surface and is simply
EdA for the top and bottom of the integration surface.
The electric field inside a conductor is zero (see problem 2 below). This now
makes the integral simple to evaluate.
 EdA = q/o
surface of integration = can with sides  sheet and
top and bottom parallel to surface of the sheet.
Integral = 0 from the cylindrical side (EdA=0) + 0 from bottom surface in
sheet (E=0) +  EdA =  EdA for the top surface. On the top surface, E
must be the same everywhere, by symmetry; every point is equivalent to
every other point. This then gives
 EdA = EdA = EA since dA = A, the area.
The enclosed charge within the surface of integration is (/2)A, where A is
the area of the sheet within the surface of integration. The (/2) is the
charge density on the top surface of the sheet, the rest being on the
bottom surface.
This gives EA = q/o = A/2o so E = /2o .
[1] Apply the same approach to show that the electric field inside a
conductor is zero. [Hint: move the top surface of integration.]
[2] For a linear charge distribution of density  C/m, what surface of
integration would you pick to use for Gauss’s law so that the field is either
parallel to or perpendicular to the surface? Can you evaluate the integral?
Solve for E.
The other surface integrals were the flux integrals. These are often not
closed surfaces, but surfaces bounded by some closed curve. For these, the
idea is to pick a surface for which the vector field is either parallel to or
perpendicular to the surface everywhere.
The two types we encounter frequently are
Flux, B = BdA and later, E = EdA
Ampere’s Law I enclosed = jdA ; j = current density.
The proper way to cheat on these integrals is to examine the vector field
over the area of integration and find what part of the area has a constant
value and direction for that field. Slice the area into these parts and then
add them up, thus avoiding one integration. Even sneakier, you can stretch
and bend the surface as needed so that the vector field is either parallel or
perpendicular to each part of the surface. We shouldn’t need that here.
Some common ways of slicing up the area of integration are listed in the
table below. These choices avoid most cases of having to do a double integral
with limits which depend on the second variable (which we don’t assume you
know how to do).
Field dependence
Surface
dA______
linear variable x
rectangle axb
a dx
radius , r
circle
2r dr
dependent on xy
square
dxdy
angle 
circle
Rd (rare)
We will generally restrict ourselves to fields which intersect a flat surface
at right angles, so the dot product is just multiplication. For a field at right
angles to the surface, B = BdA = BdA , with no vector product left.
Then choose shape of surface to fit dependence of vector field (B) on
variables (we already do this for problems we pitch at you). The problems
involving conducting loops, coils, etc. and the surface they enclose are chosen
with the above in mind so they can be done more simply without full – blown
Calculus methods.
Try these:
[3] A wire of radius R carries a current density given by a(R-r)e-r .
Calculate the current I flowing through the wire.
[4] A square loop of wire extending from x=0 to x=L in the xy plane has a
magnetic field through it given by B = ax + b(L-x) in the z direction.
Calculate the magnetic flux through the loop.
[5] A second square loop in the xy plane is in a magnetic field given by
B = ax + b/y in the z direction. Calculate the magnetic flux through the loop
with coordinates x = 0 to X by y = 1 to Y. Hint: use superposition for the X
and Y parts of the field.
For Ampere’s Law, the other side of the equation is a line integral. You have
actually done some of these before, we just didn’t tell you! A line integral
involves integrating the dot product of some vector quantity with a vector
tangent to the line of integration. The example you have done in the past is
the work integral:
W =  Fds where ds is the differential element of path length.
You didn’t notice because we used simple paths. The most obvious case of
this being a vector line integral is in the case of friction where the exact
shape and length of the path matters.
Line Integrals
The line integrals we will encounter have an integrand with a dot product of
some vector and a differential ds. They often have a circle around the
center of the integral sign meaning the integral is evaluated over a closed
path. I’ll omit the circle on the integral;
 v ds .
In these integrals,
the differential is a vector of differential length tangent to the curve.
In order to simplify the integral, we will pick the integration path so that
the vector v is either parallel to it or perpendicular to it, reducing the dot
product to either simply multiplying the two magnitudes or zero.
Example:
The B field around a current-carrying wire is a set of concentric circles
around the wire. Ampere’s Law is one of the common line integrals we will
work with:
Bds = o Ienclosed
Since B is in the shape of circles, we will pick a circle of radius r centered
on the wire as our path of integration. Everywhere on such a circle B is
parallel to ds, so Bds is simply Bds. The value of B is the same everywhere
on the path, so
o Ienclosed = Bds = Bds = B 2r since ds = path length.
The rest of the process is just solving for B :
o Ienclosed = B 2r ; B = o Ienclosed /(2r)
Much easier than the hard way!
Here’s one for you to try:
[6] In a region of uniform magnetic field, can there be any current
perpendicular to the field? Let the B field be in the +x direction.
(a) Find a path for which the B field is either along the path or
perpendicular to it.
Path :
(b) Now evaluate Bds for the various parts of your path. Remember that
ds proceeds around the path in one direction, clockwise or counterclockwise.
(c) Add the values for Bds to find the value of the integral.
Sum =
(d) Given this result, can there be any such current?
There is a law for the electric and gravitational fields which is a line
integral, but we will not be using it. It mainly gives a constraint on the field
shape, modeling the fact that the field lines in these cases cannot circle the
source like the magnetic field does.
page 8
[1] Apply the same approach to show that the electric field inside a
conductor is zero. [Hint: move the top surface of integration.]
[2] For a linear charge distribution of density  C/m, what surface of
integration would you pick to use for Gauss’s law so that the field is either
parallel to or perpendicular to the surface? Can you evaluate the integral?
Solve for E.
Page 9
[3] A wire of radius R carries a current density given by a(R-r)e-r .
Calculate the current I flowing through the wire.
[4] A square loop of wire extending from x=0 to x=L in the xy plane has a
magnetic field through it given by B = ax + b(L-x) in the z direction.
Calculate the magnetic flux through the loop.
[5] A second square loop in the xy plane is in a magnetic field given by
B = ax + b/y in the z direction. Calculate the magnetic flux through the loop
with coordinates x = 0 to X by y = 1 to Y. Hint: use superposition for the X
and Y parts of the field.
Page 11
[6] In a region of uniform magnetic field, can there be any current
perpendicular to the field? Let the B field be in the +x direction.
(a) Find a path for which the B field is either along the path or
perpendicular to it.
Path :
(b) Now evaluate Bds for the various parts of your path. Remember that
ds proceeds around the path in one direction, clockwise or counterclockwise.
(c) Add the values for Bds to find the value of the integral.
Sum =
(d) Given this result, can there be any such current?