Chapter 23. Electric Chapter 23. Electric ForceForce

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity© 2007

Objectives: After finishing Objectives: After finishing this unit, you should be this unit, you should be able to:able to:

• Explain and demonstrate the Explain and demonstrate the first law of electro- first 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 Coulomb’s Coulomb’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.

positive

negative

+ + + +

--

--

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 Two Negative Charges RepelRepel

1. 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 Two Positive Charges RepelRepel

1. 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 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 The First Law of ElectrostaticsElectrostatics

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 Charging Electroscope Positively by Contact with a Positively by Contact with a

Glass Rod: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 Charging Spheres by InductionInduction

--- - -

Uncharged Spheres Separation of Charge--- - -

Isolation of Spheres Charged by Induction

----

++++

----

++++ +

++ +

-

-- -

Induction

Electrons Electrons RepelledRepelled

Induction for a Single Induction for a Single SphereSphere

--- - -

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 can be defined in terms of the number of electrons, but in terms of the number of electrons, but the the Coulomb (C)Coulomb (C) is a better unit for later is a better unit for later work. A work. A temporarytemporary definition might be as definition might be as given below: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 Which means that the charge on a single electron is:electron is:

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

Units of ChargeUnits of Charge

The The coulombcoulomb (selected for use with (selected for use with electric currents) is actually a electric currents) is actually a very large very large unitunit for static electricity. Thus, we often for static electricity. Thus, we often encounter a need to use the metric encounter a need to use the metric prefixes.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 electrons are removed from a neutral sphere, are removed from a neutral sphere, what is the charge on the sphere in what is the charge on the sphere in coulombs?coulombs?

1 electron: e- = -1.6 x 10-

19 C -196 -

-

-1.6 x 10 C(16 x 10 e )

1 eq

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

+ + + + + + +

+ + ++ +

+ +

Coulomb’s LawCoulomb’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.

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

r

FF

q

q q’

q’- +

- -

2

'qqF

r 2

'qqF

r

Calculating Electric ForceCalculating Electric ForceThe proportionality constantThe proportionality constant k k for for Coulomb’s lawCoulomb’s law depends on the choice of depends on the choice of units for charge.units for charge.

2

2

' where

'

kqq FrF k

r qq

2

2

' where

'

kqq FrF k

r 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

Frk

qq

2 29

2

N m9 x 10

' C

Frk

qq

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

- +2 mm

+3 C-5 Cq q’

Draw and label givens on figure: r

F

2

2

9 -6 -6NmC

2 -3 2

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

(2 x 10 m)

kqqF

r

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.

Problem-Solving Problem-Solving StrategiesStrategies

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 4 cmcm from a from a +9 +9 CC charge. What is the charge. What is the resultant force on a resultant force on a –5 –5 CC charge charge located midway between the first located midway between the first charges?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 10-9-9 C C

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 qF

r FF11 = = 675 N675 N

9 -6 -62 3

2 2 21

(9 x 10 )(9 x 10 )(5 x 10 );

(0.02 m)

kq qF

r 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 attraction- repulsionattraction- repulsion, 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 N

FF22 = = 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 675 N + 1013 N;N;

FR = +1690 NFR = +1690 N

Example 4. Example 4. Three charges, Three charges, qq11 = = +8 +8 CC, , qq22 = = +6 +6 CC and and qq33 = -4 = -4 CC are are arranged as shown below. Find the arranged as shown below. Find the resultant force on the resultant force on the –4 –4 CC charge charge due to the others.due to the others.

Draw Draw free-body diagramfree-body diagram..

-53.1o

-4 Cq3

F1

F2

Note the Note the directionsdirections of forces F of forces F11and 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 Next we find the forces forces FF11 andand F F22 from Coulomb’s law. from Coulomb’s law. Take data from the figure and use SI Take data from the figure and use SI units.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 q

F 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 We find components of each force components of each force FF11 andand F F22 (review vectors).(review vectors).

53.1o- -4 C

q3

F1= 115 N

F1y

F1x

FF1x1x = - = -(115 N) Cos 53.1(115 N) Cos 53.1oo

= - 69.2 N= - 69.2 N

FF1y1y = - = -(115 N) Sin 53.1(115 N) Sin 53.1oo = =

- 92.1 N- 92.1 NNow look at force FNow look at force F22::

FF2x2x = = -240 N; F-240 N; F2y2y = 0 = 0 RRxx = =FFxx ; R ; Ry y = = FFyy

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

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

F2

240 N

Rx= -92.1 NRx= -92.1 N

Ry= -240 NRy= -240 N

Example 4 (Cont.)Example 4 (Cont.) Next find resultant R Next find resultant R from components Ffrom components Fxx and F and Fyy. (review . (review vectors).vectors).

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

- -4 Cq3

Ry = -69.2 N

Rx = -309 N

R

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

y2 2

x

R; tan =

Rx yR 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 The resultant force is force is 317 N317 N. We now need to . We now need to determine the angle or determine the angle or directiondirection of of this force.this force.

2 2 317 Nx yR R R 2 2 317 Nx yR R R

y

x

R 309 Ntan

R -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 10

Ck

29

2

N m9 x 10

Ck

2

'kqqF

r 2

'kqqF

r

CONCLUSION: Chapter 23CONCLUSION: Chapter 23Electric ForceElectric Force