Electric Force

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Electric Charge

and Electric Force

Based on a Presentation by

Paul E. Tippens, Professor of Physics

Southern Polytechnic State University

R. Morgan

2009

Objectives: After finishing this

lesson, you should be able to:

• Explain and discuss the fundamental properties

of charge, the charging processes: contact &

induction, and the conservation of electric charge.

• Write Coulomb’s Law and

apply it to problems involving

electric forces.

• Define the electron, the

coulomb, and the microcoulomb

as units of electric charge.

Operational Definition of the

Charging Process



• When amber is rubbed, it obtains the

ability to attract bits of dust and straw.

• Similarly, when other objects are rubbed

they can exhibit an ability to attract or

repel other objects. This process is

called charging of the object.

Operational Definition of the

Charging Process (cont’d)



• When two like objects are rubbed with

like materials, they will have a like

charge (as with two glass rods after

each is rubbed with silk cloth).

Operational Definition of the

Charging Process (cont’d)



• When a charged object touches another

object, the second object can also

become charged. This process is called

conduction.

• When a charged object touches another

object or other objects, these objects will

have like charge.

The Most Fundamental

Operational Property of Charges



• Like charges repel

• Opposite charges attract

The Electroscope

Laboratory devices used to study the

existence of two kinds of electric charge.









Pith-ball Gold-leaf

Electroscope Electroscope

The two parts of an electroscope

will receive a like charge when

touched by a charged object.

They then repel each other.









Pith-ball Gold-leaf

Electroscope Electroscope

When two objects are charged by

rubbing together and then one is used

to charge a pith ball electroscope the

other rubbed object will attract the

pith balls (and then discharge them).









Pith-ball Gold-leaf

Electroscope Electroscope

When two objects are charged by

rubbing together and then one is used

to charge a metal leaf electroscope the

other rubbed object will cause the

leaves to return back to their rest

position.









Pith-ball Gold-leaf

Electroscope Electroscope

Classification of Charges

• Experimentation can show that there are

only two charge conditions.

• By using electroscopes we can now

classify all charges into these two types.

• Benjamin Franklin defined “positive” as

the charge that glass obtains when

rubbed with silk (and all like charges).

• Opposite charges are called “negative”.

Charge at the atomic level.



• Based on J. J. Thomson’s experiments, we

know that the flow of charge is generally the

flow of electrons.

• Atomic protons are generally fixed in place and

do not flow.

• Positive charges can flow too as with positive

ions in a salt water solutions, the acid in lead-

acid batteries, and in living cells.

Charge at the atomic level.

• Charged is “quantized” meaning that it is

“grainy” made up of “building blocks” all

having magnitudes equal to the charge

of a proton or an electron.

• Conservation of charge: charges can not

be created or destroyed. When we

charge an object, we are just moving

charges around.

Charge at the atomic level.

• Conductors are materials that allow

electrons to flow relatively easily

• Insulators hold all electrons tightly

preventing flow.

• Metals are the best conductors because

the metallic bonds allow free movement

of electrons.

Electric Charge: Atomic Level

When a glass rod is rubbed against silk, electrons are

removed from the glass and deposited on the silk.



glass Electrons positive

move from + +

glass to the negative + +

silk silk cloth. - - - -





The glass is said to be positively charged and has a

deficiency of electrons. The silk is said to be

negatively charged and has an excess of electrons.

Electric Charge: Atomic Level

When a rubber rod is rubbed against fur, electrons

are removed from the fur and deposited on the rod.

Electrons negative

move from - -

positive

fur to the - -

++++

rubber rod.







The rod is said to be negatively charged because of an

excess of electrons. The fur is said to be positively

charged because of a deficiency of electrons.

Charging by Rubbing



• Electrons are rubbed off from one

material to the other.

• The material gaining the electrons does

so because it has a stronger attraction to

electrons than does the material losing

the electrons.

Two Negative Charges Repel

1. Charge the rubber rod by rubbing against fur.

2. Transfer electrons from rod to each pith ball.









The two negative charges repel each other.

Two Positive Charges Repel

1. Charge the glass rod by rubbing against silk.

2. Touch balls with rod. Free electrons on the balls

move to fill vacancies on the cloth, leaving each of

the balls with a deficiency. (Positively charged.)









The two positive charges repel each other.

The Two Types of Charge





Rubber glass

Attraction



fur silk



Note that the negatively charged (green) ball is

attracted to the positively charged (red) ball.



Opposite Charges Attract!

The Fundamental (Operational)

Characteristic of Charge

Like charges repel; unlike charges attract.









Neg Neg Neg Pos

Pos Pos

Charging by Contact

1. Take an uncharged electroscope as shown below.

2. Bring a negatively charged rod into contact with knob.



--- - - -

-

-



-- -- -- -

- -

- - -





3. Electrons move down on leaf and shaft, causing them

to separate. When the rod is removed, the scope

remains negatively charged.

Charging Electroscope Positively

by Contact with a Glass Rod:

Repeat procedures by using a positively charged

glass rod. Electrons move from the ball to fill

deficiency on glass, leaving the scope with a net

positive charge when glass is removed.



++++ +

+ + +

+ +

++ + +

+ ++ + ++

Charging Conducting Spheres by

Induction: Not Touching

--- - -

- - Electrons

++

++ -- Repelled



Uncharged Spheres Separation of Charge

--- - -

++

++

-- +

+ + -

-

-

-- + -



Isolation of Spheres Charged by Induction

Induction for a Single Sphere

Induction

--- - -

+ ----

+ --

++ --



Uncharged Sphere Separation of Charge



--- - -

-- - - - -

++ -

++ - + +

+

+



Electrons move Charged by Induction

to ground.

Charge Distribution

on Conductors

• When an excess of electrons is located

on a conductor each electron will repel

every other electron. The result is that

all net negative charge will become

evenly distributed on the very outer

surface of the object.

• In a similar way, if an object is positively

charged, all net charge will exist on the

outer surface of the object.

Charge on Nonconductors

• On non-conductors, the charge (excess/

shortage of electrons) remains in place:

does not distribute across the object.

• When brought near a charged object,

surface charges in a non-conductor can

become polarized (charge polarization).

• The surface can exhibit charged

properties (attraction/repulsion) even

though the net charge of the non-

conductor is zero.

Charge Polarization

in a non-conductor

Charge Polarization

Causing Lightning

The Quantity of Charge

The quantity of charge (q) can be defined in

terms of the number of electrons, but the

Coulomb (C) is a better unit for later work. A

temporary definition might be as given below:



The Coulomb: 1 C = 6.25 x 1018 electrons



Which means that the charge on a single electron is:



1 electron: e- = -1.6 x 10-19 C

Units of Charge

The coulomb (selected for use with electric

currents) is actually a very large unit for static

electricity. Thus, we often encounter a need to

use the metric prefixes.





1 mC = 1 x 10-6 C 1 nC = 1 x 10-9 C





1 pC = 1 x 10-12 C

Example 1. If 16 million electrons are

removed from a neutral sphere, what is

the charge on the sphere in coulombs?

1 electron: e- = -1.6 x 10-19 C

+ +

+ + +

 -1.6 x 10-19 C  + + + +

q  (16 x 106e- )  -  + + +

 1e  + +

q = -2.56 x 10-12 C

Since electrons are removed, the charge

remaining on the sphere will be positive.

Final charge on sphere: q = +2.56 pC

COULOMB’S LAW

The force of attraction or repulsion between two

point charges is directly proportional to the product

of the two charges and inversely proportional to the

square of the distance between them.



F

- q q’ +

qq '

r F 2

F F r

q q’

- -

Coulomb’s Law:

Calculating Electric Force

The proportionality constant k for Coulomb’s

law depends on the choice of units for charge.



kqq ' Fr 2

F  2 where k 

r qq '

When the charge q is in coulombs, the distance r is

in meters and the force F is in newtons, we have:



9 Nm

2 2

Fr

k  9 x 10

qq ' C2

Example 2. A –5 mC charge is placed

2 mm from a +3 mC charge. Find the

force between the two charges.



-5 mC F +3 mC

Draw and label

givens on figure:

q - r + q’

2 mm



kqq ' (9 x 10

9 Nm2

)(5 x 10 C)(3 x 10 C

-6 -6



F 2  C2

r (2 x 10-3m)2

F = 3.38 x 104 N; Attraction



Note: Signs are used ONLY to determine force direction.

Problem-Solving Strategies

1. Read, draw, and label a sketch showing all

given information in appropriate SI units.

2. It is generally easiest to determine the

direction of the force based on attraction/

repulsion and then ignore signs of charges.

3. Superposition Principle: The Resultant force

caused by multiple charges is found by consid-

ering force due to each charge independently

and then adding the force vectors.

4. For forces in equilibrium: SFx = 0 = SFy = 0.

Example 3. A –6 mC charge is placed 4 cm

from a +9 mC charge. What is the resultant

force on a –5 mC charge located midway

between the first charges? 1 nC = 1 x 10-9 C

1. Draw and label. F1

-6 mC F2 +9 mC

2. Draw forces. q3

3. Find resultant;

q1 - r1 - r2 + q2

2 cm 2 cm

right is positive.



kq1q3 (9 x 109 )(6 x 10-6 )(5 x 10-6 )

F1  2  2

; F1 = 675 N

r1 (0.02 m)

kq2 q3 (9 x 109 )(9 x 10-6 )(5 x 10-6 )

F2  2  2

; F2 = 1013 N

r1 (0.02 m)

Example 3. (Cont.) Note that direction

(sign) of forces are found from attraction-

repulsion, not from + or – of charge.



+ -6 mC

F1

q3

F2 +9 mC

F1 = 675 N q1 - r1 - r2 + q2

F2 = 1013 N 2 cm 2 cm



The resultant force is sum of each independent force:



FR = F1 + F2 = 675 N + 1013 N; FR = +1690 N

Example 4. Three charges, q1 = +8 mC,

q2 = +6 mC and q3 = -4 mC are arranged

as shown below. Find the resultant force

on the –4 mC charge due to the others.

+6 mC 3 cm q Draw free-body diagram.

q2 + - -4 mC

3





F2 q3

4 cm 5 cm - -4 mC

53.1o

53.1o

+ F1

q1 +8 mC





Note the directions of forces F1and F2 on q3

based on attraction/repulsion from q1 and q2.

Example 4 (Cont.) Next we find the forces

F1 and F2 from Coulomb’s law. Take data

from the figure and use SI units.

kq1q3 kq2 q3 +6 mC 3 cm q

F1  2 ; F2  2

q2 + F2 - -4 mC

3

r1 r2

(9 x 109 )(8 x 10-6 )(4 x 10 -6 ) 4 cm F1 5 cm

F1 

(0.05 m) 2

53.1o

(9 x 109 )(6 x 10-6 )(4 x 10-6 ) +

F2  q1 +8 mC

(0.03 m) 2

Thus, we need to find resultant of two forces:



F1 = 115 N, 53.1o S of W F2 = 240 N, West

Example 4 (Cont.) We find components of

each force F1 and F2 (review vectors).

F1x = -(115 N) Cos 53.1o F2

= - 69.2 N 240 N F1x q3

- -4 mC

53.1o

F1y = -(115 N) Sin53.1o = F1y

- 92.1 N

Now look at force F2: F1= 115 N



F2x = -240 N; F2y = 0 Rx = SFx ; Ry = SFy



Rx = – 69.2 N – 240 N = -309 N Rx= -92.1 N



Ry = -69.2 N – 0 = -69.2 N Ry= -240 N

Example 4 (Cont.) Next find resultant R from

components Fx and Fy. (review vectors).



Rx= -309 N Ry= -69.2 N

Rx = -309 N q

We now find resultant R,q: f - -4 mC

3







Ry R

R  R  R ; tan f =

2

x

2

y

Rx Ry = -69.2 N



R  (309 N)2  (69.2 N)2  317 N



Thus, the magnitude of R = 317 N

the electric force is:

Example 4 (Cont.) The resultant force is

317 N. We now need to determine the

angle or direction of this force.



R  Rx  Ry  317 N

2 2

-309 N q

f -

Ry 309 N R

tanf   -69.2 N

-62.9 N

R x -69.2 N

The reference angle is: f = 77.40S of W

Or, the polar angle q is: q = 1800 + 77.40 = 257.40



Resultant Force: R = 317 N, q = 257.40

Summary of Formulas:

Like Charges Repel; Unlike Charges Attract.



kqq ' Nm 2

F 2 k  9 x 10 9

2

r C



1 mC = 1 x 10-6 C 1 nC = 1 x 10-9 C





1 pC = 1 x 10-12 C 1 electron: e- = -1.6 x 10-19 C

CONCLUSION: Electric Force


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