If q1
and q2 both have the same
polarity q2 will be repelled by
q1. If they have opposite
charges, q2 will be attracted to
q1.
The strength of the force depends on the inverse square of the
distance between the two charges.
A few comments about notation. The "squiggly" line under a variable ( ~ ) denotes a vector. The caret above a variable
( ^ ) indicates a unit vector. The indices r12 indicate that the vector points from 1 toward 2. We'll also mostly be using the mks system of units in this course.
A few comments about notation. The "squiggly" line under a variable ( ~ ) denotes a vector. The caret above a variable
( ^ ) indicates a unit vector. The indices r12 indicate that the vector points from 1 toward 2. We'll also mostly be using the mks system of units in this course.
I'd encourage you to put
all the constants and units together on one page so that
you can refer to for homework and exam questions.
Both the midterm and final exams will be open notes.
The principle of superposition applies with Coulomb's law. The figure below wasn't shown in class.
When multiple charges are present, the force exerted on
one of the charges is the vector sum of the forces exerted
on the charge by all the other charges.
If a charge q is placed in the vicinity of a charge Q, we could use Coulomb's law to determine the force that Q exerts on q. We can imagine a vector field, the electric field, existing around Q even before q is brought into the picture. Multiplying q times E would give the force that Q exerts on q. The expression for electric field is shown above.
We'll try an example problem: calculating the electric field a distance r away from an infinitely long line of charge (charge is distributed uniformly along the line). The symbol λ will stand for the line charge density (careful: λ is often used for conductivity as well)
We'll write down expressions for the contributions to the field dE from short segments above and below z = 0. The z components of the E field at Point r point in opposite directions and cancel out. The radial components point in the r-direction (away from line of charge) and add.
The principle of superposition applies with Coulomb's law. The figure below wasn't shown in class.
If a charge q is placed in the vicinity of a charge Q, we could use Coulomb's law to determine the force that Q exerts on q. We can imagine a vector field, the electric field, existing around Q even before q is brought into the picture. Multiplying q times E would give the force that Q exerts on q. The expression for electric field is shown above.
We'll try an example problem: calculating the electric field a distance r away from an infinitely long line of charge (charge is distributed uniformly along the line). The symbol λ will stand for the line charge density (careful: λ is often used for conductivity as well)
We'll write down expressions for the contributions to the field dE from short segments above and below z = 0. The z components of the E field at Point r point in opposite directions and cancel out. The radial components point in the r-direction (away from line of charge) and add.
To determine the
total field we must include all the segments from
z = 0 to infinity. We must integrate the
expression for dE from 0 to infinity. The
solution is shown at the bottom of the figure.
We'll come back to this problem later in this lecture and solve it in a much simpler way.
Next we'll consider a charge completely surrounded and enclosed by a surface, S.
The vector r points outward from q and n is a unit vector normal (perpendicular) to the small increment of area dA.
We will evaluate the following surface integral
The surface integral of electric field over a
surface enclosing a charge q is just q divided by
εo the
integral form of Gauss' Law. The shape of
the surface is irrelevant.
To see how useful this expression can be we'll return to our earlier problem involving the infinite line of charge and use Gauss' Law to determine the electric field. We'll see that it is a much easier process.
The crucial part of the problem is the choice
of surface. We draw a cylinder centered on
the line of charge. This is the area that we
will integrate E over in the Gauss Law
expression. It is also important to realize
that the E field has a radial component only.
There is no contribution to the surface
integral from the ends of the cylinder (E is
perpendicular to the normal vector, the dot
productand E and n is therefore zero). E is
parallel to the normal vector along the side of
the cylinder. E is also constant on the side
of the cyliner (E is a function of r and r stays
constant as you integrate over the surface of the
cylinder side). In the end we obtain the
same expression for E as we did in the earlier
example.
It's relatively easy to show that the surface integral of the E is zero when charge is located outside the volume. We didn't do this in class.
This might be a little hard to visualize because the figure isn't drawn very well. But along the front portion of the surface the r unit vector points inward, n points outward. Along the back surface r and n both point outward. The two increments of solid angle have the same value but opposite signs so they cancel.
We'll go back to the integral form of Gauss' Law.
A surface S contains a volume V. We can use the divergence theorem to write the surface integral of electric field as a volume integral of the dot product of electric field. The charge enclosed by the surface is just the volume integral of the volume charge density. Combining these two expressions
In order for the two volume integrals to be
equal for an arbitrary volume, the terms inside
the integrals must be indentical.
It's been a long lecture, we're almost done.
Next we imagine placing a conductor in an electric field
Free electrons inside the conductor will quickly move around and redistribute themselves in such a way that they will cancel out the field in the conductor (as the electrons move they leave behind positively charged atoms). Very quickly the field inside the conductor will become zero. There is also no space charge inside the conductor, all the charge is distributed along the surface on the conductor. The electric field lines in the vicinity of the conductor will be distorted and will strike the surface of the conductor perpendicular (if there were any tangential component, charges would move and cancel it out).
Next we'll use Gauss' Law to determine a relationship between the surface charge density and the electric field strength at the surface of a conductor.
A cylindrical volume, half inside and half
outside the conductor is used. You must
integrate the electric field, E, over the surface
of the cylinder.
1. The E field is zero inside the conductor. So you get no contribution to the surface integral from the bottom end of the cylinder.
2. Both the sides of the cylinder
and the E field lines are perpendicular to the
surface of the conductor. The dot product of
E and the normal to the cylinder surface is zero
(E and n are perpendicular) So there is no
contribution to the surface integral from the
sides of the cylinder.
3. The only contribution comes from
the top end of the cylinder. E and the
normal vector are parallel. So the
contribution from the top end is just E A, where A
is the area of the top end of the
cylinder. The charge enclosed is just
A σ, where σ is the surface charge
density. The electric
field at the surface of the conductor is
perpendicular to the conductor and has the value
highlighted in yellow.
We plugged a value for the fair weather electric field into this expression on the first day of class to determine the charge density on the surface of the earth.
We'll come back to this problem later in this lecture and solve it in a much simpler way.
Next we'll consider a charge completely surrounded and enclosed by a surface, S.
The vector r points outward from q and n is a unit vector normal (perpendicular) to the small increment of area dA.
We will evaluate the following surface integral
Substituting
in the expression for electric field
surrounding a point charge.
This could be
a difficult integral to evaluate for a surface
of arbitrary shape. However we can first
note that the surface integral can be
rewritten as an integral of solid angle over a
closed surface.
The increment
of solid angle dΩ is really just a
3-dimensional version of an increment of angle
dθ. The following figure might make this
clearer.
I
didn't show the following figure in class
(too hard to draw on the whiteboard).
In the upper
figure a line of length r sweeps through some
angle increment dθ traces out a segment of
length rdθ. In the bottom figure r times
an increment of solid angle dΩ maps out an
area r2
dΩ.
It is relatively easy to show that the integral of solid angle over a surface is 4 π.
It is relatively easy to show that the integral of solid angle over a surface is 4 π.
To do this we
just consider a sphere.
Now let's go
back to the surface integral of the electric
field.
To see how useful this expression can be we'll return to our earlier problem involving the infinite line of charge and use Gauss' Law to determine the electric field. We'll see that it is a much easier process.
It's relatively easy to show that the surface integral of the E is zero when charge is located outside the volume. We didn't do this in class.
This might be a little hard to visualize because the figure isn't drawn very well. But along the front portion of the surface the r unit vector points inward, n points outward. Along the back surface r and n both point outward. The two increments of solid angle have the same value but opposite signs so they cancel.
We'll go back to the integral form of Gauss' Law.
A surface S contains a volume V. We can use the divergence theorem to write the surface integral of electric field as a volume integral of the dot product of electric field. The charge enclosed by the surface is just the volume integral of the volume charge density. Combining these two expressions
It's been a long lecture, we're almost done.
Next we imagine placing a conductor in an electric field
Free electrons inside the conductor will quickly move around and redistribute themselves in such a way that they will cancel out the field in the conductor (as the electrons move they leave behind positively charged atoms). Very quickly the field inside the conductor will become zero. There is also no space charge inside the conductor, all the charge is distributed along the surface on the conductor. The electric field lines in the vicinity of the conductor will be distorted and will strike the surface of the conductor perpendicular (if there were any tangential component, charges would move and cancel it out).
Next we'll use Gauss' Law to determine a relationship between the surface charge density and the electric field strength at the surface of a conductor.
1. The E field is zero inside the conductor. So you get no contribution to the surface integral from the bottom end of the cylinder.
We plugged a value for the fair weather electric field into this expression on the first day of class to determine the charge density on the surface of the earth.