Equipment

• Clear Tape

Forces and Fields (6)

In the most fundamental equations about the universe, we find fields.

Black holes, the Aurora Borealis, and microwave ovens all are understood in terms of fields.

Fields are abstract, but quite real.

Mr. KlapholzShaker Heights

High School

Electricity

Fun with clear tape that you can try at home

• Make two pieces of tape have the same type of charge

• See what like charges do.• See what an uncharged object does to a

charged object.• Make two pieces with unlike charge.• See what unlike charges do.

( Coulomb’s Law )

Electric Force

Meet the Electron

Meet the Proton

QP = -Qe MP 2000•Me

Meet the Neutron

QN = 0 MN MP 2000 Me

0

Alpha ‘Particle’

0

0

Why does an alpha attract an electron?

Which is more (or are they equal): • the force on the alpha, or the force on the electron? • the acceleration of the alpha, or the acceleration of the electron?

The Electric Force• Have we ever actually seen evidence for

this force?• Can it be repulsive? How do you know?• Can this force be attractive? • Where is this force greatest (close or far)?• This force is strongest for highly charged

objects (why do we think this is true?).• How is this force like gravity, and how is it

different?

Electric Force Basics

• The force on object One equals the force on object Two. Recall the third law.

• The force is proportional to the charge of object One. F Q1

• The force is proportional to the charge of object Two. F Q2

• The force is proportional to the product of the charges: F Q1•Q2

Electric Force and Distance (R)

The farther apart the objects,the less the force.

€

F∝ 1

R2

Put all our ideas together

€

F ∝ Q1Q2R2

Change the proportionality to an equality by including a constant.

€

F = kQ1Q2R2

Hey, this looks familiar!

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F = GM1M2

R2

Gravitation

k = 9.00 x 109 Nm2/C2

Let’s try an example:

One object has a charge of +0.5 Coulombs,and the other object has a charge of +2 C.

The objects are 3 meters apart.How much force repels these objects?

Show your buddy how to start this problem:

€

F = kQ1Q2R2

= (9 ×109)(0.5)(2)32

How much does this turn out to be, and what are the units?F = _____ Newtons. That’s a billion Newtons.

€

F = kQ1Q2R2

= (9 ×109)(0.5)(2)32

How much does this turn out to be, and what are the units?F = 109 Newtons. That’s a billion Newtons. This force is _ _ _ _.

€

F = kQ1Q2R2

= (9 ×109)(0.5)(2)32

How much does this turn out to be, and what are the units?F = 109 Newtons. That’s a billion Newtons. This force is huge. And a Coulomb is a _ _ _ _ amount of charge.

€

F = kQ1Q2R2

= (9 ×109)(0.5)(2)32

How much does this turn out to be, and what are the units?F = 109 Newtons. That’s a billion Newtons. This force is huge. And a Coulomb is a huge amount of charge.What would have been different if one of the charges was negative?

Electric Field (E)

• The electric field shows the direction that a charged object would be forced (if there was a positive charge at that spot).

• More intense fields are indicated by ‘lines’ that are drawn more densely…

Electric Field due to a positive point charge.

http://facstaff.gpc.edu/~pgore/PhysicalScience/electric-charge.html

Shape of the Electric Field

Electric Field due to a negative charge.

http://www.tutorvista.com/content/physics/physics-iv/electric-charges/electric-field-lines.php

Electric Field due to two positive charges.

http://www.tutorvista.com/content/physics/physics-iv/electric-charges/electric-field-lines.php

Electric Field due to a positive point charge and a negative point charge.

http://electron9.phys.utk.edu/phys136d/modules/m4/efield.htm

Shape of the Electric Field

http://web.ncf.ca/ch865/englishdescr/3DEFldDipole.html

Electric Field due to a oppositely charged parallel plates (edge view).

http://en.wikibooks.org/wiki/FHSST_Physics/Electrostatics/Electric_Fields

Shape of the Electric Field

http://web.ncf.ca/ch865/englishdescr/EFld2Plates.html

Basics of Electric Field Strength• To test the electric field at a point in space, put a

small charged object at that location. Likely there will be a force on the object (due to the other charges in the universe). If there is a force on the object, then there is an electric field at that point.

• If you doubled the charge of our “test charge”, the force on it would double. The strength of the field does not depend on the charge of the test object.

• If you didn’t bother to put a charge at the point, there would still be a field there (but no force).

There is no electric field inside a conductor.

http://electron9.phys.utk.edu/phys136d/modules/m5/conductor.htm

This happens because electrons in a conductor are free to move, so if there was an electric field

then charges would be forced to move to aposition until there was no more field.

The Strength of the Electric Field

E = F / q

Field = Force / Charge

The uptight definition of the Electric Field requires that q be positive:

E = limq0 F / q

One.How much force acts on ?

+4 C+2 C

3 meters

How much force acts on ?

• F = kQq ÷ D2 • F = (9x109)(2)(4) ÷ 32 • F = 8 x 109 N• In what direction is the force on ?

What is the value of theElectric field at ?

• E = F / q• E = (8 x 109 N) ÷ ?• E = (8 x 109 N) ÷ (4 C)• E = 2 x 109 units? • E = 2 x 109 N / C

What is the direction of theElectric field at ?

• Vital Definition: The direction of E at a point is the same as the direction that a positive charge would be forced if it was at that point.

• In this case, since is a positive charge that is forced to the right, the Electric field is to the Right.

Two.Find the force on and the Electric field at .

-3 C+2 C

3 meters

Find F and E.

• F = kQq/D2 • F = (9x109)(2)(-3) ÷ 32 • F = -6 x 109 N (towards the left)

• E = F/q = (-6x109) ÷ (-3) • E = 2x109N/C (towards right)• Compare to the first question (!) …

Why do the E’s have the same magnitude and direction?

+4 C+2 C

3 meters

One

Two

-3 C+2 C

3 meters

In the previous question, we found out that:

• The force depends on both charges, but• The Electric field depends on the charge that is

not at the spot that we are looking at.• E = F ÷ q • E = (kQq/D2) ÷ q • E = kQ ÷ D2 (the q at the spot cancels).

Three. What is the E at theempty spot (x) ?

x+5 C

6 meters

Two ways to find E.

• Slow: make up a value for q (at x) and calculate force. Then use E = F ÷ q.

• Faster, use E = kQ ÷ D2 • E = (9x109)(5) ÷ 62 • E = 1.25 x 109 N/C

Four. What is the E at theempty spot (x) ?

xQ

D

E = kQ ÷ D2