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The Electric Field
Coulomb’s law presents us with several problems.
Calculations become difficult as we increase the number of
charges. Visualization becomes difficult.
What do we do about continuous (approximately)
distributions of charges?
The force of gravity acts at a distance, without contact. For a
long time, this was difficult to for scientists to accept.
Now we have electrostatic forces acting at a distance,
without contact.
To help us visualize how electric forces can act at a distance,
we “invent” the electric field.
The electric field “connects” the charges—we can draw it,
and invoke it is the “thing” that provides contact.
An electric field extends outward from every charge and
permeates all of space. Other charges experience a force due
to the electric field.
We define the electric field by specifying the force it exerts
on an infinitesimally small* positive test charge q.
OSE:
E=F/q
*Why infinitesimally small? If it weren’t infinitesimally small, q itself would contribute to the
electric field, exert forces on the other charges, possibly causing them to move, and in general foul
up the measurement of electric field.
We can use our test charge to determine the electric field
due to another charge Q located a distance r away. From
Coulomb’s law, the magnitude of the force on q is
Qq
F k 2 ,
Qq
r
so that the magnitude of the electric field due to Q is

Q
r 
k
k .
Qq
E
Q
q
2
r2
This is almost an OSE, but E is a vector, so we need to specify
a direction.
Our test charge was positive, so the force exerted on it by
another positive charge is repulsive. The electric field
direction is the same as the force direction. The electric
field points away from a positive charge.
OSE:
Q
E  k 2 , away from 
Q
r
The principle of superposition applies to the electric field:
the field due to a collection of charges is the vector sum of
the fields due to the individual charges.
Example: Calculate the total electric field at point A.
y
30 cm
A
=30º
Q1=-50C
Q2=+50C
x
52 cm
Yes, this is almost the same setup as the example in the previous lecture.
Step 0: Think!
This is an electric field problem (that’s what the problem
statement says!).
We want the electric field at point A due to charges Q1 and Q1.
The electric field is additive, so we can calculate EQ1 and EQ2
and add the two.
If we do our vector addition using components, we must
resolve our electric fields into their x- and y-components.
y E
Q2
Step 1: Diagram
Draw axes, showing
origin and directions—
done.
Draw and label
relevant quantities—
show the r’s.
r2= 30 cm
Draw a representative
sketch—done.
A
EQ1
=30º
Q1=-50C
Q2=+50C
52 cm
If there is no possibility of confusion, it is
acceptable to write E1 instead of EQ1. If you do
it in your diagram, do it in your equations too!
Draw and label electric fields due to individual charges.
Draw components of fields which are not along axes.
x
y E
Q2
Step 2: OSE
Q
E k 2 ,
Q
r
away from 
r2= 30 cm
A
EQ1
=30º
Q1=-50C
Q2=+50C
52 cm
Step 3: Replace Generic Quantities by Specifics
Q2
E
 k 2 , up
Q2
r2
E
0
Q 2 ,x
Q2
E
 k 2
Q 2 ,y
r2
x
y E
Q2
Step 3 (continued)
Q1
E
 k 2 cos
Q 1,x
r1
Q1
E
 k 2 sin
Q 1,y
r1
A
r2= 30 cm
Q1
E  k 2 , towards Q 1
Q1
r1
EQ1
=30º
Q1=-50C
Q2=+50C
52 cm
Remember, a vector has both magnitude and direction. A vector component is also a
vector. Our convention is that subscripts x and y indicate vector components. The sign
gives the direction if the component is along an axis, so a separate indication of
direction is not needed.
x
Step 4: Complete the Math
E E E
Q1
Q2
If there were two points, A and B, at which E was to be
calculated, you would need to use the subscript A on the E’s
here. I don’t use the subscript in this example because I am
only doing the calculation at point A.
E E
E
x
Q1,x
Q 2 ,x
E E
E
y
Q1,y
Q 2 ,y
Q1
E  k 2 cos   0
x
r1
Q1
Q2
E  k 2 sin   k 2
y
r1
r2
Q1
E  k 2 cos 
x
r1
The boxed answers on the preceding slide are sufficient for a
symbolic answer.
Plugging in numbers gives Ex = 1.1x106 N/C and Ey = 4.4x106
N/C. It may look like my solution took more steps than
necessary, but beware of skipped steps. Skipping steps is
asking for trouble!
How could you calculate the force on an electron placed at point A? How
would you calculate the electric field at a point halfway between the two
charges?
One could use symmetry arguments to calculate the electric
field at any point halfway between the two charges. You are
welcome to use symmetry arguments in your calculations.
Just make sure it is valid to do so!
Electric Field Lines
Although the electric field helps us understand how electric
forces can act at a distance, we may still have a visualization
problem.
We could draw an arrow at every point in space. The
direction would show the direction of E, and the length
would show the magnitude.
Too many arrows!
Too confusing!
-
+
A better approach is to use “field lines” to help us visualize
the variation of the field in space. Field lines show the
direction of the field, but not its magnitude. Field lines are
also called “lines of force.”
Field lines start at positive charges
and end at negative charges.
The electric field is tangent to the
field lines.
Make the number of lines
proportional to the charge.
The closer together the lines, the
stronger the electric field.
The electric field in between two charged plates is
(approximately) constant.
Electric Fields and Conductors
If charges in a conductor at rest, the electric field inside must
be zero. Why?
If the electric field were not zero, the charges would
experience a force and would accelerate.
Any excess charge in a conductor spreads itself out on the
surface. Why?
Like charges repel. The charges move as far away from each
other as possible.
A charge inside a spherical
conducting shell gives rise to an
electric field outside the shell,
even though no field exists inside
the metal.
The electric field is always
perpendicular to the surface
outside of a conductor. Why?
If there were a component of the electric field parallel to the
surface, electrons would move along the surface until they
came to a location where there was no parallel component.
Other Items of Interest
E&M Fields. Little program that lets you explore fields and
forces. An old dos program, runs fine in Windows XP.
The fine print: click on the link, select “open,” and follow the
prompts. This will copy a small (1.2 megabyte) set of files in the
folder c:\Program Files\emfield. You can change the install folder if
you want.
After the files are copied, open windows explorer and run the file
START.BAT in the emfield folder.
To “uninstall” the emfield program, just delete the emfield folder
and all the files in it.
The program has been scanned for viruses, but you are always
welcome to run your own scan.
More Items of Interest
Electric Field Hockey. Little program that lets you play
hockey, using charges to move the “puck.” An old dos
program, runs fine in Windows XP.
The fine print: click on the link, select “open,” and follow the
prompts. This will copy a small (0.8 megabyte) set of files in the
folder c:\Program Files\hockey. You can change the install folder if
you want.
After the files are copied, open windows explorer and run the file
START.BAT in the hockey folder.
To “uninstall” the hockey program, just delete the hockey folder
and all the files in it.
The program has been scanned for viruses, but you are always
welcome to run your own scan.
Even More Goodies!
Some potentially useful links:
The Physics Zone at Science Joy Wagon.
Lessons on Electrostatics at the Physics Zone.
JC Physics.com physics for A-level and Junior College
Students (in other words, algebra-based physics).
If there were nothing more to electric fields than the material
I have presented so far, they would not be much of a “big
deal.”
At best, they would give us a method slightly different than
Coulomb’s law for calculating forces between electrical
charges.
At worst, they would confuse us by introducing the abstract
concept of “fields” that act on particles even though the
fields aren’t made of anything we can touch.
The abstraction we are about to make is very powerful.
Electric field lines extend outward from positive charges,
towards negative charges, and the density of lines depends on
the magnitude of the charge.
We define electric flux as the number of electric field lines
that pass perpendicularly through a surface.
Actually, we just use the vector dot product to find how much
of the electric field is parallel to the surface normal, and
multiply by the area to get the flux: =BA, or d=BdA.
The total flux through any surface is proportional to the
charge inside the surface (remember, the number of electric
field lines is proportional to the charge).
Remember our constant k from Coulomb’s law? It is related to
0 by k=1/40. We could write  using k, if we wish.
But the total flux through a surface is the integral of d over
the surface…
This gives us Gauss’ law:
Q
 E dA  0
Gauss’ law is a BIG DEAL.
It is one of Maxwell’s equations—the
four fundamental equations that
describe Electricity and Magnetism.
You should think of Maxwell’s equations as being on the same
footing as our conservation laws.
Sometimes you see physics majors walking around with t-shirts with a
bunch of funny-looking equations on them. Maxwell’s equations!
That little circle on the integral sign means integrate over a closed surface.
For the mathematicians in the audience,
Gauss’ law can be written in an equivalent
differential form.
I’m not going to carry this any further here.
If you took physics 24 or 25, you’d
learn to love the integrals you’d do
for Gauss’ law calculations.
You can buy a Maxwell’s equation
from the UMR Society of Physics
Students and impress everybody
with your knowledge of physics.