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					                  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 C and q3 = -3 C 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 .

(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 .

(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 .
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.




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
For a conductor at equilibrium:



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
                                           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
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         and the electric field is simply
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.           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.


    For the case where r < R:

    We construct a spherical Gaussian
    surface of radius r and apply
    Gauss's Law.




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,
    For the case where r > R:

    We construct a spherical Gaussian
    surface through an outside point r >
    R, and apply Gauss's Law. In this
    case,




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 .

Applying Gauss's Law to this cylindrical Gaussian surface gives




and we note qenc = A.
Therefore,
Electric Field between Capacitor Plates
Consider two large, parallel conducting plates, one with a charge density + 
and the other with charge density - .




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 ,
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 .

                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 rL. Thus, Gauss's Law becomes




      The uniform charge density of the cylinder is . Therefore, the
      enclosed charge qenc is simply  times the volume enclosed by the
      Gaussian surface, r2L. 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 rL. With r > R, the entire cylinder is enclosed, so qenc is
equal to the uniform charge density  times the volume of the
charged cylinder, R2L. 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

				
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