ELECTRICITY AND MAGNETISM
P10D
Coulomb’s Law
The
force of attraction or repulsion between two point charges q1
and q2 is directly proportional to the product of their
charges and inversely proportional to the square of the distance between them.
where
F12 is the force exerted on point charge q1
by point charge q2 when they are separated by a distance r12.
The
unit vector 
is directed from q2 to q1
along the line between the two charges.
The
constant 
is called the permitivity of
free space. In SI units
where force is in Newtons (N), distance in meters (m) and charge in coulombs
(C),
Coulomb’s
law and the Principle of Superposition
constitute the physical input for electrostatics.
The
force on any one charge due to a collection of other charges is the vector sum
of the forces due to each individual charge.
Here
Problem
Solving Strategies:
1.
Draw a clear diagram of the situation. Be sure to distinguish
between the fixed external charges and the charges which the forces must be
found. The diagram should contain the coordinate axes for reference.
2.
Electric force is a vector quantity; when many forces are present
the net force is a vector sum.
3.
Search for symmetries in the distribution of charges that give
rise to the electric force. When symmetries are present, the net force along
certain directions will be zero.
Example:
Consider three point charges q1 = q2 = 2.0 mC and q3 = -3 mC which are placed as shown. Calculate the
net force on q1 and q3.
The
force on q1 is F1 = F12 +
F13.
Similarly,
F3 = F31 + F32
Electric
Field.
Electric field is defined as the electric force per unit charge.
The direction of the field is taken to be the direction of the force it would
exert on a positive test charge. The electric field is radially outward from a
positive charge and radially in toward a negative point charge.
The
electric field can be defined by measuring the magnitude and direction of the
electric force F on a test charge q0. A small test charge is
used so as not to interfere with the field distribution of the other charges.
Thus we define the electric field as
.
The
SI unit of electric field is NC-1.
For
a point charge, 
Electric Field Concepts
Field Line Diagrams:
A
convenient way to visualize the electric field due to any charge distribution
is to draw a field line diagram. At any point the
field line has the same direction as the electric field vector
Electric field lines diverge from positive charges and converge
into negative charges.
Rules
for constructing filed lines:
1.
Field lines begin at positive charge and end at negative charge
2.
The number of field lines shown diverging from or converging into
a point is proportional to the magnitude of the charge.
3.
Field lines are spherically symmetric near a point charge
4.
If the system has a net charge, the field lines are spherically
symmetric at great distances
5.
Field lines never cross each other.
The Electric Dipole and their Electric Fields
An
electric dipole consists of two charges + q and –q, of equal magnitude but
opposite sign, that are separated by a distance L.
From
the diagram, E = Ex i = (E1x
+ E2x)i = 2E1x i
where
.
But
Hence
I
Define
the product p = qL as the electric
dipole moment. We make p = qL a vector by
defining L to be directed from –q to +q.
The
vector p points from the negative charge to the positive charge.
The
electric field decreases with r as 1/r3.
Finally,
.
If r
>>L, then 
and 
Electric
Dipole Field:
Charge Distributions:
The
simplest kind of charge distribution is an isolated point charge (i.e.,
an amount of charge covering such a small region of space that we need not be
concerned about its dimensions).
When
the finite size of the space occupied by a collection of charges must be
considered, it is useful to consider the density of charge. The word
density is used is used in three different ways.
(rho) for the charge
per unit volume, the volume density
(sigma) for the charge
per unit area, the area density
(lambda) for the charge
per unit length, the linear density
Thus we can write
Cm-3, 
Cm-2, 
Cm-1.
Find the electric
field a distance z above the midpoint of a straight line segment of
length 2L which carries a uniform line charge l.
(Model of a transmission line)
It is advantageous to chop the line up into symmetrically
placed pairs
(at ± x).
Here 
where q is the total charge on the rod
The horizontal components of the two fields cancel.
The net field of the pair is
Here 
, and x runs from 0 to L
How to evaluate this integral
Consider
;
Let 
, then 
and
. Substitute these into the integral:
From diagram, 
Hence 
Substitute the limits x = 0 and x = L
to get the desired answer.
For points far from the line ( z >> L),
this result simplifies to
.
The line “looks” like a point charge 
, so the field reduces to that of a point charge 
.
Find the electric field a distance z
above one end of a straight line segment of length L, which carries a
uniform line charge l.
(Model of a transmission line)
Net electric field E = Ex + Ez
For z >> L and 
, 
Find the electric field a distance z above the
center of a circular loop of radius R, which carries a uniform line
charge l.
Here, 
where q is the total charge of the loop
Horizontal components cancel, leaving:
(both constants)
But 
Find the electric field a distance z above
the center of a flat circular disk of radius a, which carries a uniform
surface density, 
.
(Models an
electrostatic microphone)
A typical element is a ring of radius r and
thickness dr, which has an area 
.
The charge density on the disk is 
. The charge on the element is:
The electric field at P produced by this ring
is
Since we have expressed a positive ring thickness as
dr, we sum the rings from the center toward the edge. That is, the
radius ranges from 0 to a.
.
Let 
, then du = 2r dr, and
So
When z << a, 
When z >> a, 
which is the expected
result for a point charge located at the origin.
Gauss’ Law: The net flux through any closed surface is
proportional to the net charge enclosed by that surface, i.e., 
The total of the electric flux out of a closed surface is equal to the
charge enclosed divided by the permittivity.
The
electric flux through an area is defined as the electric field multiplied by
the area of the surface projected in a plane perpendicular to the field.
Gauss's
Law is a general law applying to any closed surface. It is an important tool
since it permits the assessment of the amount of enclosed charge by mapping the
field on a surface outside the charge distribution.
For
geometries of sufficient symmetry, it simplifies the calculation of the
electric field.
Gauss' Law, Integral Form
The
area integral of the electric field over any closed surface is equal to the net
charge enclosed in the surface divided by the permittivity of space. Gauss' law
is a form of one of Maxwell's equations, the four fundamental equations for
electricity and magnetism.
Gauss'
law permits the evaluation of the electric field in many practical situations
by forming a symmetric Gaussian surface surrounding a charge distribution and
evaluating the electric flux through that surface.
Flux and Gauss's Law
Example: Water Flow
Suppose water flows through a square surface of area A with
a uniform velocity 
.
We want to define flux as the amount of water that flows
through the surface per second. First, notice that the flux depends on the
direction of the velocity vector relative
to the surface.
Questions
Compare
the water flux in the following three situations. Rank the flux in cases (a), (b),
and (c), from largest to
smallest.
(a)
is
perpendicular to the surface
(b)
is
at a 45 degree angle to the surface
(c)
is
parallel to the surface
If the biggest flux is 1,
what is the relative number corresponding to the flux in the other two cases?
Answer
(a): 1
(b): 
(c): 0
Electric Flux
The
concept of electric flux is useful in association with Gauss' law. The electric
flux through a planar area is defined as the electric field times the component
of the area perpendicular to the field.
If
the area is not planar, then the evaluation of the flux generally requires an
area integral since the angle will be continually changing.
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When
the area A is used in a vector operation like this, it is understood that the
magnitude of the vector is equal to the area and the direction of the vector is
perpendicular to the area.
Gaussian Surface:
Part
of the power of Gauss' law in evaluating electric fields is that it applies to
any surface. It is often convenient to construct an imaginary surface called a
Gaussian surface to take advantage of the symmetry of the physical situation.
Conductor at Equilibrium
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For
a conductor at equilibrium: |
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1. The net electric charge of a conductor resides entirely on its
surface. (The mutual repulsion
of like charges from Coulomb's Law demands that the charges be as far apart as
possible, hence on the surface of the conductor.)
2. The electric field inside the conductor is zero.
(Any net electric field in the conductor would cause charge to move since it is
abundant and mobile. This violates the condition of equilibrium: net force =0.)
3. The external electric field at the surface of the conductor is
perpendicular to that surface.
(If there were a field component parallel to the surface, it would cause mobile
charge to move along the surface, in violation of the assumption of
equilibrium.)
Electric Field: Conductor Surface
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The
fact that the conductor is at equilibrium is an important constraint in this
problem. It tells us that the field is perpendicular to the surface, because
otherwise it would exert a force parallel to the surface and produce charge
motion. Likewise it tells us that the field in the interior of the conductor
is zero, since otherwise charge would be moving and not at equilibrium. |
Examining
the nature of the electric field near a conducting surface is an important
application of Gauss' law. Considering a cylindrical Gaussian surface
oriented perpendicular to the surface, it can be seen that the only
contribution to the electric flux is through the top of the Gaussian surface.
The flux is given by
and
the electric field is simply
While
strictly true only for an infinite conductor, it tells us the limiting value
as we approach any conductor at equilibrium. |
Gauss's Law for a Single Point Charge
Gauss's Law applies to any charge contribution, but let us apply
it now to the simplest case, that of a single point charge q. 
Construct a spherical Gaussian surface of radius r around
the charge q.
Take a small area on the Gaussian surface. The area vector dA
points radially outward, as does
the electric field vector 
at
this point.
Therefore, the electric flux through this small area is
From the spherical symmetry, all of such small area elements
contribute equally to the total.
According to Gauss's Law,
because 
=
q. Solving for the electric field gives
which is just Coulomb's Law!
A Uniformly Charged
Solid Sphere
An insulating sphere of radius R has a total charge q
uniformly distributed throughout its volume.
We want to find the electric field everywhere, that is, at an inside
point r < R, and also at an outside point r > R.
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We already know that for a spherically symmetric case, 
.
We have to be careful about qenc, though. qenc is the net charge within the
Gaussian surface, not the total charge q of the entire sphere. For this
reason, we need to multiply the total charge by the ratio of the volume of the
Gaussian surface to the volume of the entire sphere:
Therefore,
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For the case where r > R:
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The graph below is of the magnitude of the electric field due to a
uniformly charged sphere, plotted as a function of distance from the center of
the sphere.
Electric Field for a Charged Sheet
What is the electric field due to a large, thin charged sheet made
of some nonconducting material?
We construct a cylindrical Gaussian surface, and place it
symmetrically in the charged sheet as shown in the above figure. Let the area
of each top surface of the cylinder be A, and the surface charge density
on the sheet be s.
Applying Gauss's Law to this cylindrical Gaussian surface gives
and we note qenc = sA.
Therefore,
Electric Field between Capacitor Plates
Consider two large, parallel conducting plates, one with a charge
density + s
and the other with charge density - s.
The two opposite charges are attracted to the inside of the
plates, as shown. We construct a cylindrical Gaussian surface, with one base
area within a conducting plate and another base between the plates. If the area
of each base is A, then we have
where E is the electric field at the base located between
the parallel plates, since E = 0 on the base inside the conductor plate.
Then, we have
Electric Field Due to a Line Charge
Consider an infinitely long straight line charge with linear
charge density l,
which is in units of Coulombs per meter.
Choose a cylinder because it mimics the symmetry of the wire.
We construct a cylindrical Gaussian surface of radius r and
length L. The electric flux about this Gaussian surface is
Therefore,
Question
Find
the electric field at all points due to a long, solid cylinder of radius R
and uniform charge density l.
AnswerB
Begin by choosing an appropriate Gaussian surface: one that mimics
the symmetry of the case. Therefore, since we are concerned with a long
cylinder, we choose a Gaussian surface that is also a long cylinder.
To find the field everywhere, we need to examine two cases: when the radius r
of the Gaussian surface is less than R, and when r > R.
Case
I: r < R
We note that the electric field will be constant everywhere on the cylindrical
Gaussian surface; we also note that this surface has a surface area of 2 prL. Thus, Gauss's Law becomes
The
uniform charge density of the cylinder is r. Therefore, the enclosed charge qenc is simply r times the volume enclosed by the
Gaussian surface, pr2L. This gives us
Therefore,
for r < R,
Case
II: r > R
Again, the electric field will be constant everywhere on the cylindrical
Gaussian surface, and the surface area of that surface is still 2 prL. With r > R, the entire cylinder is enclosed, so
qenc is equal to the uniform charge density r times the volume of the charged
cylinder, pR2L. This gives us
Therefore,
for r > R,
To
summarize, here is a plot of the electric field versus
the radius r of the Gaussian surface for this problem:
A Uniformly Charged Sphere
Question
Use Gauss's Law to find the electric field everywhere due to a
uniformly charged insulator shell, like the one shown below. The shell has a
total charge Q, which is uniformly distributed throughout its volume.
(a)
What is the charge on the inner surface of the conductor?
(b) What is the charge on the outer surface of the conductor?
(c) Use Gauss's Law to find the electric field for radius r < a.
(d) Use Gauss's Law to find the electric field for radius a < r
< b.
(e) Use Gauss's Law to find the electric field for radius r > b.
Answer
We need to
look at this problem in three parts: one, for when the radius 
;
for when a < r < b, and for when 
.
I. :
No charge is enclosed in this case that is, qenc = 0. Therefore, the flux
through the Gaussian surface must be zero, and so the electric field =
0 everywhere in this region.
II. 
:
Here, the charge enclosed is found by multiplying the total charge Q by
the ratio of the volume of charge enclosed by the Gaussian surface to the
volume of the entire charged shell:
We recall that
and as the electric field will be
constant everywhere on the spherical Gaussian surface, we can substitute as
follows:
Therefore,
III. :
Here, qenc is Q, so we have
Again, since the electric field will be
constant at every point on the spherical Gaussian surface, we have
Which becomes
To
summarize our findings, here is a plot of the electric field as a function of
the radius of the Gaussian surface:
Potential Energy, Work, and the Electric
Field:
The potential energy difference can be defined as
the negative of the work 
done by a conservative force F on an object moved from point
A to point B.
where dl is a small element of the
path from A to B and l is a vector from A to B.
Potential Difference:
The electric potential difference from point A to
point B is the potential energy change per unit charge in moving from A to B.
In the case of a uniform field,
The Volt and the Electron Volt:
The unit of potential difference is the volt (V). 1
V = 1J/C. To say that a car has a 12V battery means that the battery does 12 J
of work on every coulomb the moves between its two terminals.
The electron volt (eV) is the energy gained by a
particle carrying one elementary charge when it moves through a potential
difference of one volt.
1eV = 1.602 ´ 10-19 J.
Calculating Potential Difference:
The potential of a point charge.
The electric field of a point charge q is given by 
.
Consider two points A and B at distances rA
and rB from a positive point charge.
The distance between them is rB - rA
but we cannot just multiply this distance by the electric field because the
field varies with position. Instead we must integrate as follows:
What if the points do not lie on the same radial line?
The potential difference is independent of path and
one path between A and B consists of a radial segment and a circular arc. Since
E is perpendicular to the arc, it takes no work to move a charge along
the arc. The potential difference 
therefore arises only
from the radial segment.
Defining the potential to be zero at some point
allows us to speak of the “potential at a point”, meaning the potential
difference from the reference point to the point in question. For isolated
point charges, a convenient zero is
infinitely far from the charge; then the potential at an arbitrary point
a distance r from the point q is
Potential
difference in the field of a line charge
We know from before that 
for field of a line charge.
Hence
Finding potential differences using superposition:
where the ri’s are the distances
from each of the charges to the point P.
Electric potential is a scalar quantity, so the sum
above is a scalar sum, and there is no need to consider angles or vector
components.
Continuous charge distributions:
We can calculate the potential of a continuous
charge distribution by considering it to be made up of infinitely many
infinitesimal charge elements dq. Each acts like a point charge and
therefore contributes to the potential at some point P an amount dV
given by
where the zero potential is at infinity. The
potential at P is the sum of all the contributions dV from all the
charge elements.
.
A charged ring
A total charge Q
is uniformly distributed over a thin ring of radius a. What is the
potential on the axis of the ring?
At the center of the ring,
A charged disk:
A charged disk of radius a
carries a total charge Q distributed uniformly over its surface. What is the
potential at a point P on the disk’s axis, a distance x from the disk?
Divide the disk into charge elements dq.
If a ring shaped element has charge dq and radius r, then from above,
Then
We must relate r and dq.
Unwinding
the ring gives a strip of area 
. The surface density 
is the total charge divided by the disk’s area: 
.
The charge dq on our
infinitesimal ring of area 
is
then
