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.

An electric field extends outward from every charge and permeates all of space. Other charges experience a force due to the electric field.

*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.

OSE: E = F / q

We define the electric field by specifying the force it exerts on an infinitesimally small* positive test charge q.

The electric field “connects” the charges—we can draw it, and invoke it is the “thing” that provides contact.

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

2

QqF k ,Qq r

so that the magnitude of the electric field due to Q is

2

2

QqQrE k k .Q rq

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.

2

QOSE: E k , 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.

Yes, this is almost the same setup as the example in the previous lecture.

x

y

Q2=+50C

A

Q1=-50C

52 cm

60 cm

30

cm

=30º

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.

Step 1: Diagram

Draw a representative sketch—done.

Draw and label electric fields due to individual charges.Draw components of fields which are not along axes.

Draw and label relevant quantities—show the r’s.

Draw axes, showing origin and directions—done. x

y

Q2=+50C

A

Q1=-50C

52 cm

60 cm

30

cm

=30º

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!

r 2=

r1 =EQ1

EQ2

Step 2: OSE

2

QE k , Q raway from

Step 3: Replace Generic Quantities by Specifics

22

2 2

QE k , upQ r

22

2 2

QE kQ ,y r

2

E 0Q ,x

x

y

Q2=+50C

A

Q1=-50C

52 cm

60 cm

30

cm

=30ºr 2=

r1 =EQ1

EQ2

Step 3 (continued)

112

1 1

QE k , towards QQ r

12

1 1

QE k sinQ ,y r

12

1 1

QE k cosQ ,x r

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

y

Q2=+50C

A

Q1=-50C

52 cm

60 cm

30

cm

=30ºr 2=

r1 =EQ1

EQ2

Step 4: Complete the Math

1 2E E EQ Q

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.

,1 2

E E Ex Q ,x Q x ,1 2

E E Ey Q ,y Q y

121

QE k cos 0x r 1 2

2 21 2

Q QE k sin ky r r

121

QE k cosx r

The boxed answers on the preceding slide are sufficient for a symbolic answer.

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!

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?

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!

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.

Too many arrows!Too confusing!

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.

- +

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.

The electric field is always perpendicular to the surface outside of a conductor. Why?

A charge inside a spherical conducting shell gives rise to an electric field outside the shell, even though no field exists inside the metal.

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: =BA, or d=BdA.

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=1/40. 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:

0

QE dA

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.