Chapter 23. Electric ForceChapter 23. Electric ForceA PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of PhysicsPaul E. Tippens, Professor of Physics

Southern Polytechnic State UniversitySouthern Polytechnic State University

© 2007

Objectives: After finishing this Objectives: After finishing this unit, you should be able to:unit, you should be able to:•• Explain and demonstrate the Explain and demonstrate the first law of electrofirst law of electro--

staticsstatics and discuss charging by and discuss charging by contact contact and by and by inductioninduction..

•• Write and apply Write and apply CoulombCoulomb’’s s LawLaw and apply it to problems and apply it to problems involving electric forces.involving electric forces.

•• Define the Define the electronelectron, the , the coulombcoulomb, and the , and the microcoulombmicrocoulomb as units of electric charge.as units of electric charge.

Electric ChargeElectric ChargeWhen a rubber rod is rubbed against fur, electrons When a rubber rod is rubbed against fur, electrons are removed from the fur and deposited on the rod.are removed from the fur and deposited on the rod.

The rod is said to be The rod is said to be negatively chargednegatively charged because of an because of an excessexcess of electrons. The fur is said to be of electrons. The fur is said to be positively positively chargedcharged because of a because of a deficiencydeficiency of electrons.of electrons.

Electrons Electrons move from move from fur to the fur to the rubber rod.rubber rod.

positivenegative

+ + + +

- -

- -

Glass and SilkGlass and SilkWhen a glass rod is rubbed against silk, electrons are When a glass rod is rubbed against silk, electrons are removed from the glass and deposited on the silk.removed from the glass and deposited on the silk.

The glass is said to be The glass is said to be positivelypositively chargedcharged because of because of a a deficiencydeficiency of electrons. The silk is said to be of electrons. The silk is said to be negatively chargednegatively charged because of a because of a excessexcess of electrons.of electrons.

Electrons Electrons move from move from glass to the glass to the silk cloth.silk cloth.

positive

negative- - - -

+ +

+ +

silk

glass

The ElectroscopeThe Electroscope

Pith-ball Electroscope

Gold-leaf Electroscope

Laboratory devices used to study the existence of two kinds of electric charge.

Laboratory devices used to study the existence of two kinds of electric charge.

Two Negative Charges RepelTwo Negative Charges Repel1. Charge the rubber rod by rubbing against fur.1. Charge the rubber rod by rubbing against fur.

2. Transfer electrons from rod to each pith ball.2. Transfer electrons from rod to each pith ball.

The two negative charges repel each other.The two negative charges repel each other.

Two Positive Charges RepelTwo Positive Charges Repel1. Charge the glass rod by rubbing against silk.1. Charge the glass rod by rubbing against silk.

2. Touch balls with rod. Free electrons on the balls 2. Touch balls with rod. Free electrons on the balls move to fill vacancies on the cloth, leaving each of move to fill vacancies on the cloth, leaving each of the balls with a deficiency. (Positively charged.)the balls with a deficiency. (Positively charged.)

The two positive charges repel each other.The two positive charges repel each other.The two positive charges repel each other.

The Two Types of ChargeThe Two Types of Charge

fur

Rubber

Attraction

Note that the negatively charged Note that the negatively charged (green)(green) ball is ball is attractedattracted to the positively charged to the positively charged (red)(red) ball.ball.

Opposite Charges Attract!Opposite Charges Attract!

silk

glass

The First Law of ElectrostaticsThe First Law of Electrostatics

Like charges repel; unlike charges attract.Like charges repel; unlike charges attract.

NegNeg PosNegPosPos

Charging by ContactCharging by Contact1. Take an uncharged electroscope as shown below.1. Take an uncharged electroscope as shown below.

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

3. Electrons move 3. Electrons move downdown on leaf and shaft, causing them on leaf and shaft, causing them to separate. When the rod is removed, the scope to separate. When the rod is removed, the scope remains remains negativelynegatively charged.charged.

--- ---

- - -

------

--- - -

Charging Electroscope Positively Charging Electroscope Positively by Contact with a Glass Rod: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 Spheres by InductionCharging Spheres by Induction--- - -

Uncharged Spheres Separation of Charge--- - -

Isolation of Spheres Charged by Induction

----

++++

----

++++ +

++ +

--

- -

Induction

Electrons Electrons RepelledRepelled

Induction for a Single SphereInduction for a Single Sphere--- - -

Uncharged Sphere Separation of Charge

Electrons move to ground.

Charged by Induction

++

+ +

Induction

----

--- - -

++++

----

- - - -

----

++++

----

The Quantity of ChargeThe Quantity of ChargeThe The quantity of chargequantity of charge (q)(q) can be defined in can be defined in terms of the number of electrons, but the terms of the number of electrons, but the Coulomb (C)Coulomb (C) is a better unit for later work. A is a better unit for later work. A temporarytemporary definition might be as given below:definition might be as given below:

The Coulomb: 1 C = 6.25 x 1018 electronsThe Coulomb: 1 C = 6.25 x 1018 electrons

Which means that the charge on a single electron is:Which means that the charge on a single electron is:

1 electron: e- = -1.6 x 10-19 C1 electron: e- = -1.6 x 10-19 C

Units of ChargeUnits of ChargeThe The coulombcoulomb (selected for use with electric (selected for use with electric currents) is actually a currents) is actually a very large unitvery large unit for static for static electricity. Thus, we often encounter a need to electricity. Thus, we often encounter a need to use the metric prefixes.use the metric prefixes.

1 C = 1 x 10-6 C1 C = 1 x 10-6 C 1 nC = 1 x 10-9 C1 nC = 1 x 10-9 C

1 pC = 1 x 10-12 C1 pC = 1 x 10-12 C

Example 1.Example 1. If If 16 million16 million electrons are electrons are removed from a neutral sphere, what is removed from a neutral sphere, what is the charge on the sphere in coulombs?the charge on the sphere in coulombs?

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

6 --

-1.6 x 10 C(16 x 10 e )1 e

q

q = -2.56 x 10-12 C

Since electrons are Since electrons are removed, removed, the charge the charge remaining on the sphere will be remaining on the sphere will be positive.positive.

Final charge on sphere: q = +2.56 pCq = +2.56 pC

+ + + ++ + ++ + ++ +

+ +

CoulombCoulomb’’s Laws 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.

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.

Fr

FF

q

q q’

q’- +

- -

2

'qqFr

Calculating Electric ForceCalculating Electric ForceThe proportionality constantThe proportionality constant k k for for CoulombCoulomb’’s s lawlaw depends on the choice of units for charge.depends on the choice of units for charge.

2

2

' where '

kqq FrF kr qq

When the charge When the charge qq is in is in coulombscoulombs, the distance , the distance r r is is in in metersmeters and the force and the force F F is in is in newtonsnewtons, we have:, we have:

2 29

2

N m9 x 10' C

Frkqq

Example 2.Example 2. A A ––5 5 CC charge is placed charge is placed 2 mm from a +3 2 mm from a +3 CC charge. Find the charge. Find the force between the two charges.force between the two charges.

- +2 mm

+3 C-5 Cq q’Draw and label

givens on figure: rF

2

29 -6 -6Nm

C2 -3 2

(9 x 10 )( 5 x 10 C)(3 x 10 C'(2 x 10 m)

kqqFr

F = 3.38 x 104 N; AttractionF = 3.38 x 104 N; Attraction

Note: Signs are used ONLY to determine force direction.Note: Signs are used ONLY to determine force direction.

ProblemProblem--Solving StrategiesSolving Strategies

1. Read, draw, and label a sketch showing all 1. Read, draw, and label a sketch showing all given information in appropriate given information in appropriate SI unitsSI units..

2. Do not confuse sign of charge with sign of 2. Do not confuse sign of charge with sign of forces. forces. Attraction/RepulsionAttraction/Repulsion determines the determines the direction (or sign) of the force.direction (or sign) of the force.

3. 3. Resultant forceResultant force is found by considering force is found by considering force due to each charge due to each charge independentlyindependently. Review . Review module on module on vectorsvectors, if necessary., if necessary.

4. For forces in equilibrium: 4. For forces in equilibrium: FFxx = 0 = = 0 = FFyy = 0.= 0.

Example 3.Example 3. A A ––6 6 CC charge is placed charge is placed 4 cm4 cm from a from a +9 +9 CC charge. What is the resultant charge. What is the resultant force on a force on a ––5 5 CC charge located midway charge located midway between the first charges?between the first charges?

- +2 cm

+9 C-6 Cq1 q2r2

2 cm

-r1

1. Draw and label.1. Draw and label.q32. Draw forces.2. Draw forces. F2

F1

1 nC = 1 x 101 nC = 1 x 10--99 CC

3. Find resultant; 3. Find resultant; right is positive.right is positive.

9 -6 -61 3

1 2 21

(9 x 10 )(6 x 10 )(5 x 10 ) ;(0.02 m)

kq qFr

FF11 = = 675 N675 N

9 -6 -62 3

2 2 21

(9 x 10 )(9 x 10 )(5 x 10 ) ;(0.02 m)

kq qFr

FF22 = = 1013 N1013 N

Example 3.Example 3. (Cont.) Note that direction (Cont.) Note that direction (sign) of forces are found from (sign) of forces are found from attractionattraction-- repulsionrepulsion, not from + or , not from + or –– of charge.of charge.

- +2 cm

+9 C-6 Cq1 q2r2

2 cm

-r1

q3F2

F1

FF11 = = 675 N675 NFF22 = = 1013 N1013 N

++

The resultant force is sum of each independent force:The resultant force is sum of each independent force:

FFRR = = FF11 + F+ F22 = = 675 N + 1013 N;675 N + 1013 N; FR = +1690 NFR = +1690 N

Example 4. Example 4. Three charges, Three charges, qq 11 == +8 +8 CC, , qq 22 = = +6 +6 CC and and qq 33 = = --4 4 CC are arranged are arranged as shown below. Find the resultant force as shown below. Find the resultant force on the on the ––4 4 CC charge due to the others.charge due to the others.

Draw Draw freefree--body diagrambody diagram..

-53.1o

-4 Cq3

F1

F2

Note the Note the directionsdirections of forces Fof forces F11 and Fand F22 on on qq33 based on attraction/repulsion from based on attraction/repulsion from qq1 1 and and qq22 . .

+ -

4 cm

3 cm

5 cm

53.1o

+6 C

-4 C

+8 Cq1

q2

q3

+

Example 4 (Cont.)Example 4 (Cont.) Next we find the forces Next we find the forces FF 11 andand FF 22 from Coulombfrom Coulomb’’s law. Take data s law. Take data from the figure and use SI units.from the figure and use SI units.

9 -6 -6

1 2

(9 x 10 )(8 x 10 )(4 x 10 )(0.05 m)

F

F1 = 115 N, 53.1o S of WF1 = 115 N, 53.1o S of W

9 -6 -6

2 2

(9 x 10 )(6 x 10 )(4 x 10 )(0.03 m)

F

F2 = 240 N, WestF2 = 240 N, West

1 3 2 31 22 2

1 2

; kq q kq qF Fr r

Thus, we need to find Thus, we need to find resultant resultant of two forces:of two forces:

+ -

4 cm

3 cm

5 cm

53.1o

+6 C

-4 C

+8 Cq1

q2

q3

F2

F1

+

Example 4 (Cont.)Example 4 (Cont.) We find components of We find components of each force each force FF 11 andand FF 22 (review vectors).(review vectors).

53.1o- -4 C

q3

F1 = 115 N

F1y

F1xFF1x1x = = --(115 N) Cos 53.1(115 N) Cos 53.1oo

= = -- 69.2 N69.2 NFF1y1y = = --(115 N) Sin 53.1(115 N) Sin 53.1oo

= = -- 92.1 N92.1 NNow look at force FNow look at force F22 ::

FF2x2x = = --240 N; F240 N; F2y2y = 0= 0 RRxx == FFxx ; ; RRyy = = FFyy

RRxx = = –– 69.2 N 69.2 N –– 240 N = 240 N = --309 N309 N

RRyy = = --69.2 N 69.2 N –– 0 = 0 = --69.2 N 69.2 N

F2240 N

Rx = -92.1 NRx = -92.1 N

Ry = -240 NRy = -240 N

Example 4 (Cont.)Example 4 (Cont.) Next find resultant R from Next find resultant R from components components FF xx and and FF yy . (review vectors).. (review vectors).

Rx = -309 NRx = -309 N Ry = -69.2 NRy = -69.2 N- -4 C

q3

Ry = -69.2 N

Rx = -309 NR

We now find resultant We now find resultant R,R,::

y2 2

x

R; tan =

Rx yR R R

2 2(309 N) (69.2 N) 317 NR

R = 317 NR = 317 NThus, the magnitude of Thus, the magnitude of the electric force is:the electric force is:

Example 4 (Cont.)Example 4 (Cont.) The resultant force is The resultant force is 317 N317 N. We now need to determine the . We now need to determine the angle or angle or directiondirection of this force.of this force.

2 2 317 Nx yR R R

y

x

R 309 NtanR -69.

2 N

-69.2 N

--309 N

R

-62.9 N

Or, the polar angleOr, the polar angle is: is:

= 1800 + 77.40 = 257.40

The reference angleThe reference angle is: is:

= 77.40S of W

Resultant Force: R = 317 N,

= 257.40Resultant Force: R = 317 N,

= 257.40

Summary of Formulas:Summary of Formulas:

Like Charges Repel; Unlike Charges Attract.Like Charges Repel; Unlike Charges Attract.

1 electron: e- = -1.6 x 10-19 C1 electron: e- = -1.6 x 10-19 C

1 C = 1 x 10-6 C1 C = 1 x 10-6 C 1 nC = 1 x 10-9 C1 nC = 1 x 10-9 C

1 pC = 1 x 10-12 C1 pC = 1 x 10-12 C

29

2

N m9 x 10C

k 2

'kqqFr

CONCLUSION: Chapter 23CONCLUSION: Chapter 23 Electric ForceElectric Force