2 ELECTRIC FORCES AND FIELDS

(1)

ELECTRIC FORCES

(2)

ELECTRIC FORCES AND FIELDS

 THERE ARE TWO TYPES OF THERE ARE TWO TYPES OF CHARGES:

CHARGES:

 POSITIVELY CHARGED PROTONSPOSITIVELY CHARGED PROTONS

 NEGATIVELY CHARGED ELECTRONSNEGATIVELY CHARGED ELECTRONS

(3)

ELECTRIC FORCES AND FIELDS

 THE UNIT OF CHARGE IS THE UNIT OF CHARGE IS

SYMBOLIZED BY THE LETTER “e”

 THE ELECTRON HAS A CHARGE OF THE ELECTRON HAS A CHARGE OF -e. THE PROTON HAS AN EQUAL AND

-e. THE PROTON HAS AN EQUAL AND

OPPOSITE CHARGE OF +e

 THE COULOMB (C) IS THE SI UNIT THE COULOMB (C) IS THE SI UNIT FOR ELECTRIC CHARGE.

(4)

ELECTRIC FORCES AND FIELDS

 THE ELECTRON HAS A MASS OF THE ELECTRON HAS A MASS OF 9.109 X 10

9.109 X 10 -31-31 Kg AND A CHARGE OF Kg AND A CHARGE OF

-1.60 X 10

-1.60 X 10 -19-19 COULOMBS. COULOMBS.

 THE PROTON HAS A MASS OF 1.673 THE PROTON HAS A MASS OF 1.673 X 10

(5)

ELECTRIC FORCES AND FIELDS

 CONDUCTOR: MATERIAL THAT CONDUCTOR: MATERIAL THAT

ALLOWS ELECTRIC CHARGES TO

MOVE FREELY. COPPER, ALUMINUM,

MOST METALS.

 INSULATOR: ELECTRIC CHARGES INSULATOR: ELECTRIC CHARGES DON’T MOVE FREELY. GLASS,

DON’T MOVE FREELY. GLASS,

RUBBER, AND PLASTIC.

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ELECTRIC FORCES AND FIELDS

 TWO CHARGED OBJECTS NEAR ONE TWO CHARGED OBJECTS NEAR ONE ANOTHER MAY EXPERIENCE

ANOTHER MAY EXPERIENCE

ACCELERATION EITHER TOWARD OR

AWAY FROM EACH OTHER BECAUSE

EACH OBJECT EXERTS A FORCE ON

THE OTHER.

THIS IS CALLED

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ELECTRIC FORCES AND FIELDS

 ELECTRIC FORCE:ELECTRIC FORCE:

 DISTANCEDISTANCE BETWEEN THE TWO BETWEEN THE TWO OBJECTS WILL AFFECT THE

OBJECTS WILL AFFECT THE

MAGNITUDE OF ELECTRIC FORCE.

 THE THE AMOUNT OF CHARGEAMOUNT OF CHARGE ON THE ON THE OBJECTS WILL AFFECT THE

OBJECTS WILL AFFECT THE

MAGNITUDE OF ELECTRIC FORCE.

(8)

ELECTRIC FORCES AND FIELDS

 WHAT IS THE RELATIONSHIP WHAT IS THE RELATIONSHIP

BETWEEN DISTANCE, CHARGE, AND

ELECTRIC FORCE?

(9)

ELECTRIC FORCES AND FIELDS

 THE FORMULA SAYS:THE FORMULA SAYS:

 ELECTRIC FORCE EQUALS COULOMB ELECTRIC FORCE EQUALS COULOMB CONSTANT X CHARGE 1 X CHARGE 2

CONSTANT X CHARGE 1 X CHARGE 2

DIVIDED BY DISTANCE SQUARED.

 COULOMBS CONSTANT IS COULOMBS CONSTANT IS 8.99 X 10

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ELECTRIC FORCES AND FIELDS

 PROBLEMPROBLEM: THE ELECTRON AND : THE ELECTRON AND

PROTON OF A HYDROGEN ATOM ARE

SEPARATED, ON AVERAGE, BY A

DISTANCE OF 5.3 X 10

DISTANCE OF 5.3 X 10 -11-11 m. m.

 FIND THE MAGNITUDE OF THE FIND THE MAGNITUDE OF THE ELECTRIC FORCE THAT EACH

ELECTRIC FORCE THAT EACH

PARTICLE EXERTS ON THE OTHER.

(11)

ELECTRIC FORCES AND FIELDS

 GIVEN:GIVEN:

 r = 5.3 X 10 r = 5.3 X 10 -11-11 m m

 kk

C

C = 8.99 X 10 = 8.99 X 10 99

 qq

e

e = -1.60 X 10 = -1.60 X 10 -19-19

 qq

p

(12)

ELECTRIC FORCES AND FIELDS

 FORMULA:FORMULA:

 FF

electric

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ELECTRIC FORCES AND FIELDS

 (8.99 X 10(8.99 X 1099) X (1.60 X 10) X (1.60 X 10-19-19))22/(5.3 X 10/(5.3 X 10-11-11))22

 NOTE: MAGNITUDE IS SCALAR, SO NOTE: MAGNITUDE IS SCALAR, SO WE DISREGARD THE SIGN OF EACH

WE DISREGARD THE SIGN OF EACH

CHARGE AND SQUARE IT.

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ELECTRIC FORCES AND FIELDS

 PROBLEMPROBLEM: A BALLOON RUBBED : A BALLOON RUBBED

AGAINST DENIM GAINS A CHARGE

OF -8.0 uC. WHAT IS THE ELECTRIC

FORCE BETWEEN THE BALLOON AND

DENIM WHEN THE TWO ARE

SEPARATED BY A DISTANCE OF 5.0

CM.?

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ELECTRIC FORCES AND FIELDS

 (8.99 X 10(8.99 X 1099)(-8))(-8)22/(5.0)/(5.0)22

(16)

ELECTRIC FORCES AND FIELDS

 PROBLEMPROBLEM: TWO IDENTICAL : TWO IDENTICAL

CONDUCTING SPHERES ARE PLACED

WITH THEIR CENTERS .30 m APART.

ONE IS GIVEN A CHARGE OF

+12 X 10

+12 X 10 -9-9 C AND THE OTHER IS C AND THE OTHER IS

GIVEN A CHARGE OF -18 X 10

GIVEN A CHARGE OF -18 X 10 -9-9 C. C.

FIND THE ELECTRIC FORCE

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ELECTRIC FORCES AND FIELDS

 (8.99 X 10 (8.99 X 10 99)(12 X 10 )(12 X 10 -9-9)(-18 X 10 )(-18 X 10 -9-9) ) / (.30)

/ (.30)22

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ELECTRIC FORCES AND FIELDS

 SOMETIMES MORE THAN TWO SOMETIMES MORE THAN TWO

CHARGES ARE PRESENT AND IT IS

NECESSARY TO FIND THE NET

ELECTRIC FORCE ON ONE OF THEM.

(19)

ELECTRIC FORCES AND FIELDS

 THE RESULTANT FORCE ON ANY THE RESULTANT FORCE ON ANY SINGLE CHARGE EQUALS THE

SINGLE CHARGE EQUALS THE

VECTOR SUM OF THE INDIVIDUAL

FORCES EXERTED ON THAT CHARGE

BY ALL OF THE OTHER INDIVIDUAL

CHARGES THAT ARE PRESENT.

(20)

ELECTRIC FORCES AND FIELDS

 THIS IS THE THIS IS THE PRINCIPLE OF PRINCIPLE OF SUPERPOSITION

SUPERPOSITION. ONCE THE . ONCE THE

MAGNITUDES OF THE INDIVIDUAL

ELECTRIC FORCES ARE FOUND, THE

VECTORS ARE ADDED TOGETHER.

(21)

ELECTRIC FORCES AND FIELDS

 PROBLEMPROBLEM: THREE POINT CHARGES AT : THREE POINT CHARGES AT

THE CORNERS OF A TRIANGLE. THE CORNERS OF A TRIANGLE.

 qq

1

1 = 6.00 X 10 = 6.00 X 10 -9-9 C C

 qq

2

2 =-2.00 X 10 =-2.00 X 10 -9-9 C C

 qq

3

3 = 5.00 X 10 = 5.00 X 10 -9-9 C C

 FIND THE MAGNITUDE AND DIRECTION FIND THE MAGNITUDE AND DIRECTION

OF THE RESULTANT FORCE OF q

(22)

ELECTRIC FORCES AND FIELDS

 YOU ALREADY KNOW THE THREE CHARGES qYOU ALREADY KNOW THE THREE CHARGES q

1 1 q q22

and q and q₃₃..

 (distance)r(distance)r

2,1

2,1 = 3 m r = 3 m r3,13,1 = 5 m r = 5 m r3,23,2 = 4 m = 4 m

 FF

3,1

3,1 = repulsive + and + = repulsive + and +

 FF

3,2

(23)

ELECTRIC FORCES AND FIELDS

 1) CALCULATE THE MAGNITUDE OF 1) CALCULATE THE MAGNITUDE OF FORCES WITH COULOMBS LAW:

FORCES WITH COULOMBS LAW:

 A.) FA.) F

3

3,F,F11 = =

 (8.99 X 10(8.99 X 1099)(+5.00 X 10)(+5.00 X 10-9-9)(+6.00 X )(+6.00 X 10

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ELECTRIC FORCES AND FIELDS

 B.) FB.) F

3

3FF22 = =

 (8.99 x 10(8.99 x 1099)(5 x 10)(5 x 10-9-9)(-2 x 10)(-2 x 10-9-9) / ) / (4)

(4)22

(25)

ELECTRIC FORCES AND FIELDS

 2) FIND THE X AND Y COMPONENTS 2) FIND THE X AND Y COMPONENTS OF EACH FORCE.

OF EACH FORCE.

 AT THIS POINT, THE DIRECTION OF AT THIS POINT, THE DIRECTION OF EACH COMPONENT MUST BE TAKEN

EACH COMPONENT MUST BE TAKEN

INTO ACCOUNT.

(26)

ELECTRIC FORCES AND FIELDS

 FF

3,1

3,1 WAS 1.08 X 10 WAS 1.08 X 10-8-8 N N

 FOR FFOR F

3,1

3,1::

 FF

X

X = (1.08 X 10 = (1.08 X 10-8-8)(COS 37)(COS 3700))

(27)

ELECTRIC FORCES AND FIELDS

 FF

3,2

3,2 WAS 5.62 X 10 WAS 5.62 X 10-9-9 N N

 FOR FFOR F

3,2

3,2::

 FF

X

X = -F = -F3,23,2 = -5.62 X 10 = -5.62 X 10-9-9

 FF

Y

(28)

ELECTRIC FORCES AND FIELDS

 3) CALCULATE THE MAGNITUDE OF 3) CALCULATE THE MAGNITUDE OF THE TOTAL FORCE ACTING IN BOTH

THE TOTAL FORCE ACTING IN BOTH

DIRECTIONS.

 FF

X

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ELECTRIC FORCES AND FIELDS

 FF

Y

Y tot = 6.50 X 10 tot = 6.50 X 10-9-9 + 0 + 0

 FF

Y

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ELECTRIC FORCES AND FIELDS

 4) USE THE PYTHAGOREAN 4) USE THE PYTHAGOREAN

THEOREM TO FIND THE MAGNITUDE

OF THE RESULTANT FORCE.

 FF

3

3 tot = SQUARE ROOT OF (F tot = SQUARE ROOT OF (FXX tot) tot)22 + +

(F

(31)

ELECTRIC FORCES AND FIELDS

 FF

3

3 tot = THE SQUARE ROOT OF tot = THE SQUARE ROOT OF

 ((3.01 X 10((3.01 X 10-9-9))22 + (6.50 X 10 + (6.50 X 10-9-9))22))

 FF

3

(32)

ELECTRIC FORCES AND FIELDS

 USE THE INVERSE TANGENT USE THE INVERSE TANGENT

FUNCTION TO FIND THE DIRECTION

OF THE RESULTANT FORCE.

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ELECTRIC FORCES AND FIELDS

 A CHARGED OBJECT SETS UP AN A CHARGED OBJECT SETS UP AN ELECTRIC FIELD IN THE SPACE

ELECTRIC FIELD IN THE SPACE

AROUND IT. WHEN A SECOND

CHARGED OBJECT ENTERS THIS

FIELD, THE SECOND OBJECT

INTERACTS WITH THE FIELD OF THE

FIRST PARTICLE

(34)

ELECTRIC FORCES AND FIELDS

 TO CALCULATE ELECTRIC FIELD TO CALCULATE ELECTRIC FIELD

STRENGTH DUE TO A POINT CHARGE: STRENGTH DUE TO A POINT CHARGE:  E = kE = k

C

C x q/rx q/r22

 ELECTRIC FIELD STRENGTH = COULOMBS ELECTRIC FIELD STRENGTH = COULOMBS CONSTANT TIMES CHARGE PRODUCING CONSTANT TIMES CHARGE PRODUCING

(35)

ELECTRIC FORCES AND FIELDS

 PROBLEMPROBLEM: A CHARGE q1 = +7.00uC : A CHARGE q1 = +7.00uC IS AT THE ORIGIN, AND A CHARGE

IS AT THE ORIGIN, AND A CHARGE

q2 = -5 uC IS ON THE X AXIS

0.300 m FROM THE ORIGIN. FIND

THE ELECTRIC FIELD STRENGTH AT

POINT P WHICH IS ON THE Y AXIS

0.400 m FROM THE ORIGIN.

(36)

ELECTRIC FORCES AND FIELDS

 GIVEN:GIVEN:

 qq

1

1 = +7uC = 7.00 x 10 = +7uC = 7.00 x 10-6-6 C C

 qq

2

2 = -5 uC = -5.00 x 10 = -5 uC = -5.00 x 10-6-6 C C

 rr

1

(37)

ELECTRIC FORCES AND FIELDS

 APPLY THE PRINCIPLE OF APPLY THE PRINCIPLE OF

SUPERPOSITION. YOU MUST FIRST

CALCULATE THE ELECTRIC FIELD

PRODUCED BY EACH CHARGE

INDIVIDUALLY AT POINT P AND

THEN THESE FIELDS TOGETHER AS

VECTORS.

(38)

ELECTRIC FORCES AND FIELDS

 1) CALCULATE THE ELECTRIC FIELD 1) CALCULATE THE ELECTRIC FIELD PRODUCED BY EACH CHARGE.

PRODUCED BY EACH CHARGE.

 EE

1

1 = kC x q= kC x q11/r/r112 2

 EE 1

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ELECTRIC FORCES AND FIELDS

 EE

2

2 = kC x q= kC x q22/r/r2222

 EE

2

2 = (8.99 x 10 = (8.99 x 1099)(5.00 x 10)(5.00 x 10-6-6) / (.500 ) / (.500

m)

m)22

 EE

2

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ELECTRIC FORCES AND FIELDS

 2) CALCULATE THE X AND Y 2) CALCULATE THE X AND Y

COMPONENTS OF EACH ELECTRIC

FIELD VECTOR.

 FOR EFOR E

1 1::

 EE

X,1

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ELECTRIC FORCES AND FIELDS

 FOR EFOR E

2

2::

 EE

X,2

X,2 = (E = (E22)(COS 53.10))(COS 53.10)

 (1.80 X 10(1.80 X 1055)(COS 53.1)(COS 53.100))

(42)

ELECTRIC FORCES AND FIELDS

 EE

Y,2

Y,2 = -(E = -(E22)(SIN 53.1)(SIN 53.100))

 -(1.80 X 10-(1.80 X 1055 N/C)(SIN 53.1 N/C)(SIN 53.100))

(43)

ELECTRIC FORCES AND FIELDS

 3) CALCULATE THE TOTAL ELECTRIC 3) CALCULATE THE TOTAL ELECTRIC

FIELD STRENGTH IN BOTH FIELD STRENGTH IN BOTH

DIRECTIONS. DIRECTIONS.

 EE

X,tot

X,tot = E = EX,1X,1 + E + EX,2X,2

(44)

ELECTRIC FORCES AND FIELDS

 EE

Y,tot

Y,tot = E= EY,1Y,1 + E + EY,2Y,2

 3.93 X 103.93 X 1055 N/C – 1.44 X 10 N/C – 1.44 X 1055 N/C N/C

(45)

ELECTRIC FORCES AND FIELDS

 4) USE THE PYTHAGOREAN 4) USE THE PYTHAGOREAN

THEOREM TO FIND THE MAGNITUDE

OF THE RESULTANT ELECTRIC FIELD

STRENGTH VECTOR.

 EE

tot

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ELECTRIC FORCES AND FIELDS

 SQUARE ROOT OFSQUARE ROOT OF

 (1.08 X 10(1.08 X 1055))22 + (2.49 X 10 + (2.49 X 1055))22

 EE

tot

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ELECTRIC FORCES AND FIELDS

 5) USE A SUITABLE 5) USE A SUITABLE

TRIGONOMETRIC FUNCTION TO

FIND THE DIRECTION OF THE

RESULTANT ELECTRIC FIELD

STRENGTH VECTOR.

(48)

ELECTRIC FORCES AND FIELDS

 TAN = ETAN = E

ytot

ytot/E/Extotxtot

 TANTAN-1-1(2.49/1.08)(2.49/1.08)