Electrical Forces and Fields - Drexel Universityjoetrout/phys102/electric_forces...gravitational and...

1

Electrical Forces and Fields

Dr. Joseph J. [email protected]/~joetrout

mailto:[email protected]

2

Benjamin Franklin

Two type of charges: positive and negative.

Like charges repel one another.

Unlike charges attract one another.

3

++

4

++

w=mg

F T

F E

Static Equilibrium

5

++

w=mg

F T

F E

FT cos

FT sin

∑ F y=F T cos−mg=0F T cos=mg

∑ F x=F E−F T sin =0F E=FT sin

6

++

w=mg

F T

F E

FT cos

FT sin

∑ F y=FT cos−mg=0F T cos=mg

F T=mg

cos

∑ F x=F E−F T sin=0F E=F T sin

F E= mgcos sin=mg tan

7

--

w=mg

F T

F E

FT cos

FT sin

∑ F y=FT cos−mg=0F T cos=mg

F T=mg

cos

∑ F x=F E−F T sin=0F E=F T sin

F E= mgcos sin=mg tan

8

““Two Type of Charges.”Two Type of Charges.”

Electron

Protron

Neutron

−1.6 X 10−19C

1.6 X 10−19C

0C

9.11 X 10−31kg

1.67 X 10−27kg

1.67 X 10−27kg

Particle Charge Mass

9

6 electrons 6 protons

6 X −1.6 X 10−19C 6 X 1.6 X 10−19C

0C

10


16 X −1.6 X 10−19C 6 X 1.6 X 10−19C

10 X −1.6 X 10−19C =−1.6 X 10−18C

11


4 X −1.6 X 10−19C 6 X 1.6 X 10−19C

2 X 1.6 X 10−19C =3.2 X 10−18C

12

ConductorConductor – Materials in which charges move freely.

InsulatorInsulator – Materials in which charges cannot move freely.

Semi-ConductorSemi-Conductor - Materials in which charges move freely under certain conditions.

13

Charging by InductionCharging by Induction

+ - + - +- + - + -+ - + - +- + - + -

14


+ - + - +- + - + -+ - + - +- + - + -

+ + + + + + +

15


- -- - -- - - - -

+ + + + + + +

+ + + + ++ + ++ +

16


+ - + - +- + - + -+ - + - +- + - + -

+ + + + + + +

17


- -- - -- - - - -

+ + + + + + +

+ + + + ++ + ++ +

18


- -- - -- - - - -

+ + + + + + +

+ + + + ++ + ++ +

“Ground”

“Switch”

19


- -- - -- - - - -

+ + + + + + +

- - - - -- - -- -

20


- -- - -- - - - -

+ + + + + + +

- - - - -- - -- -

21


- -- - -- - - - -

- - - - -- - -- -

Negatively Charged.

22

q1

q2

F 21

F 12

23

q1

q2

F 21

F 12

r12

r 21

24

q1

q2

F 21

F 12

r12

r 21

25

q1

q2

F 21 F 12

r12

r 21

26

F e=k∣q1∣∣q2∣r12

2 r12

k= 14o

=9 X 109 N m2

C 2

o=8.85 X 10−12 C 2

N m2

o=Permittivity of Free Space

Charles Coulomb

Charles Coulomb

● Electric Force between two stationary particles:– inversely proportional to the square of the the

distance ( r ) between the two particles and directly along the line connecting them.

– proportional to the product of the charges ( q1 and q2 )

– Attractive if the charges are of opposite signs and repulsive if the charges are of like signs.

Charles Coulomb

2

212

2

29

221

108542.8

1098.84

1

NmCX

CNmXkk

rqq

kF

o

oe

ee

−=

===

=

ε

π ε

permittivity of free space

Gravitational vs. Electrical

Consider two electrons 2 angstrom apart. What is thegravitational and electrical forces between them?


2rmGmF ee

G =


( )( )( )

=

=

−

−−−

210

3131

2

211

2

1000.21011.91011.9107.6

mXkgXkgX

kgNmX

rmGmF ee

G


( )( )( )

NX

mXkgXkgX

kgNmX

rmGmF ee

G

51

210

3131

2

211

2

1039.1

1000.21011.91011.9107.6

−

−

−−−

=

=

=


221

rqqk

Fe =


( )( )( )

=

=

−

−−

210

1919

2

29

221

1000.21060.11060.11099.8

mXCXCX

CNmX

rqqk

Fe


( )( )( )

NX

mXCXCX

CNmX

rqqk

Fe

27

210

1919

2

29

221

1075.5

1000.21060.11060.11099.8

−

−

−−

=

=

=


2451

27

101039.11075.5 ≈= −

−

NXNX

FF

G

e

The electrical force is much greater than the gravitational force at this distance.

Don’t forget force is a vector!

rrqqkF e ˆ2

2112 =

Force exerted by charge 1 on charge 2, with a unitvector that points from charge 1 to charge 2.

Forces obey Newton’s third law!

2112 FF −=

40

4.0m

3.0m

q1=10C

Find the net force on qo.

q2=−10C

q0=10C

41

4.0m

3.0m

q1=10C


q2=−10C

q0=10C

∣F 10∣

∣F 20∣

42

4.0m

3.0m

q1=10C


q2=−10C

q0=10C

∣F 10∣

∣F 20∣ ∣F 0∣

0

43

F 0=F 0x

iF 0 y

j=0.056 N i−0.10 N j

∣F 0∣=F 0 x

2 F 0 y

2 =0.11N

0= tan−1F 0 y

F 0x=−60.75o

F 0 x=∣F 10∣=k

q1q0

r102 =0.056 N

F 0 y=−∣F 20∣=k

q2q0

r202 =−0.10 N

44

4.0m

3.0m

q1=10mC

q2=20mC

q0=10mC


45

4.0m

3.0m

q1=10mC

q2=20mC

q0=10mC

F 20

F 10

46

4.0m

3.0m

q1=10mC

q2=20mC

q0=10mC

F 20

F 10

47

∣F 10∣=kq1q0

r102 =2.25 X 107N

∣F 20∣=kq2q0

r202 =1.125 X 105N

F 0x=F 10 cos=1.35 X 107N

F 0 y=F 10 sinF 20=1.81 X 107N

F 0=F 0x

2 F 0 y

2 =2.258 X 107N

o=tan−1F 0 y

F 0x=53.28o

48

q1

q2q

0

q1=20C q2=50Cq0=10C

5.0 m

F 10F 20

49

q1

q2q

0

q1=20C q2=50Cq0=10C

5.0 m

F 10 F 20

50

q1

q2q

0

q1=20C q2=50Cq0=10C

5.0 m

Find place where the net force on charge qo is equal to zero.

51

q1

q2q

0

q1=20C q2=50Cq0=10C

5.0 m


F 10F 20

F 0= F 20− F 10=0F 20=F 10

k∣q2∣∣q0∣r20

2 =k ∣q1∣∣q0∣r10

2

∣q2∣r20

2 =∣q1∣r10

2

52

q1

q2q

0

q1=20C q2=50Cq0=10C

5.0 m


F10F 20

F 0= F 20− F 10=0F 20=F 10

k∣q2∣∣q0∣r20

2 =k ∣q1∣∣q0∣r10

2

∣q2∣r20

2 =∣q1∣r10

2

∣q2∣5m−x 2

=∣q1∣x2

5.0 m - xx

53

q1

q2q

0

q1=20C q2=50Cq0=10C

5.0 m


F10F 20

F 0= F 20− F 10=0F 20=F 10

k∣q2∣∣q0∣r20

2 =k ∣q1∣∣q0∣r10

2

∣q2∣r20

2 =∣q1∣r10

2

∣q2∣5m−x 2

=∣q1∣x2

5.0 m - xx

x2

5m−x 2=∣q1∣∣q2∣

x5m− x

=∣q1∣∣q2∣

x=∣q1∣∣q2∣

5m−x

x=5m∣q1∣∣q2∣

−x∣q1∣∣q2∣

54

q1

q2q

0

q1=20C q2=50Cq0=10C

5.0 m


F10F 20

F 0= F 20− F 10=0F 20=F 10

k∣q2∣∣q0∣r20

2 =k ∣q1∣∣q0∣r10

2

∣q2∣r20

2 =∣q1∣r10

2

∣q2∣5m−x 2

=∣q1∣x2

5.0 m - xx

x 2

5m−x 2=∣q1∣∣q2∣

x5m−x

=∣q1∣∣q2∣

x=∣q1∣∣q2∣

5m−x

x=5m∣q1∣∣q2∣

− x∣q1∣∣q2∣

xx∣q1∣∣q2∣

=5m∣q1∣∣q2∣

x 1∣q1∣∣q2∣=5m∣q1∣

∣q2∣

x=5m ∣q1∣∣q2∣

1∣q1∣∣q2∣

55

q1

q2q

0

q1=20C q2=50Cq0=10C

5.0 m


F10F 20

5.0 m - xx

x=5m ∣q1∣∣q2∣

1∣q1∣∣q2∣

x=5m ∣20C∣∣50C∣

1∣20C∣∣50C∣

x=1.94m

56

q1

q2

q0

Find net force on charge q0.

xa

a

57

q1

q2

q0


r10

r20

xa

a

F 20

F 10

q1=q2=qr10=r20=r

58


F 20 y

F 10 y

F 10x, F 20x

Y components cancel. X-components have equal magnitude.

q1=q2=qr10=r20=r

59

q1

q2

q0


r10

r20

xa

a

F 20

F 10

F 10=F 20=kq qor2

F o=2F 10 x=2F 10 cos=2k

qq0

r2 xr =2kxq q0

r3 =2kxq q0

x2a2 3 /2

F o=2kxq q0

x2a2 3 /2i

q1=q2=qr10=r20=r

60

Electric Field – The force per charge. Use a small test charge to determine the pattern of the force.

+

+

q

qoSmall Test Charge

61


+

+

q

qo

E=Fqo

62


+

+

q

qo

E=Fqo

63


+q qo

E=Fqo

+

64


+q

qo

E=Fqo

+

65


+q

qo

E=Fqo+

66


+q

qo

E=Fqo+

67


+q

68


+q

-Symmetrical

- Points away from positive charges.

69


-q

-Symmetrical

- Points towards negative charges.

70


- ++

71


- ++

72


- +

+

73


- +

+

74


- ++

75

1. Number of lines proportional to the charge.

2. Lines leave the charge perpendicular to the surface of the charge.

3. Lines point in the direction of the Electrical Force felt by a positive charge (i.e. away from positive charges and towards negative charges).

4. Lines NEVER cross one another.

5. Lines are symmetric around the charge.

6. Density of lines is proportional to the Electric Field in that area.

Drawing Electric Field Lines.Drawing Electric Field Lines.

76

- +32C−32C

1 line per 4C

77

- +32C−32C

1 line per 4C

78

79

80

q1 q2

81q1 q2

82q1 q2

83q1 q2

84

E=Fq0

E point charge =kqr2 r

E=k∑i

qir i

2 r i

Electric Field

85

http://physics.weber.edu/schroeder/software/EField/

http://www.its.caltech.edu/~phys1/java/phys1/EField/EField.htmlhttp://hibp.ecse.rpi.edu/~crowley/java/Efield/App.htmhttp://web.mit.edu/jbelcher/www/java/vecnodyncirc/vecnodyncirc.html

86

q1

q2


r10

r20

xa

a

E20

E10

F o=2kxqq0

x2a2 3 /2i

E=2kx qx2a2 3 /2

i

q1=q2=qr10=r20=r

87

E=k∫ dqr 2 r

Volume Charge Density =QV= dqdV

Surface Charge Density =QA= dqdA

Linear Charge Density =Ql=dqdl

Electric Field

E=k∫ dqr 2 r

Volume Charge Density =QV= dqdV

Surface Charge Density =QA= dqdA

Linear Charge Density =Ql=dqdl

Electric Field

Electric Field of a ring of charge:

R

P


R

P

dq

d E


R

P

dq

d E

dq

Y – components cancel.


dE=k dqr2

dE x=kdqr2 cos=k dq

r2xr= kxr3 dq

E x=∫dE x=∫ kxr3 dq=

kxr3∫dq=Q total

kxr 3 =

kxQx 2R2 3 /2


E x=kxQ

x2R2 3 /2R=0.1mx=1.2mQ=8 nC

E= kxQx2R2 3 /2

i

E=9 X 109 Nm2

C2 1.2m 8 X 10−9C

1.2m 20.1m 23 /2i=49.48 N

Ci


E x=kxQ

x2R2 3 /2R=0.1mx=1.2mQ=8nC

E= kxQx2R2 3 /2

i

E=9 X 109 Nm2

C2 1.2m 8 X 10−9C

1.2m 20.1m 23 /2i=49.48 N

Ci

F 0=q0E=−5C 49.48 N

Ci=−247.4 N i

qo=−5C

Electric Field of a charged sheet ( another look ):

R


RP

=QA= dqdA

xr

dq

dr E x=kxQ

x2R2 3 /2

dE= kxx2R2 3 /2

dq

Recall for a ring:


RP

=QA= dqdA

xr

dq

drE x=

kxQx2R2 3 /2

dE= kxx2R2 3 /2

dq

Recall for a ring:

dq= dA= 2r dr

d E


=QA= dqdA

dE= kxx2r2 3/2

2 r dr

E=∫ kxx2r23/2

2 r dr

E=2 x k ∫ x2r2 −3 /2r dr

∫2 x2r2 −3 /2r dr

u=x2r2

du=2r dr

∫u−3 /2du=[ u−1 /2

−1/2 ]=[−2 x2r2 −1 /2]oR


=QA= dqdA

dE= kxx2r2 3 /2

2 r dr

E=∫ kxx2r23 /2

2 r dr

E=2 x k ∫ x2r2 −3 /2r dr

E=−2 x k [x 2r2 −1 /2]oR

E=2 x k [ 1∣x∣

− 1xR2 ]


=QA= dqdA

E=2 x k [ 1∣x∣

− 1xR2 ]

What if R ∞ ?

E=2k =2 14o =

2o

=2 nCm2

=−2 nCm2 E=

2o

2o=o

=2 nCm2

=−2 nCm2 E=

2o

2o=o

q0

=2 nCm2

=−2 nCm2 E=

2o

2o=o

q0

w=mg

FT

F E=q0E

=2 nCm2

=−2 nCm2 E=

2o

2o=o

q0

w=mg

FT

F E=q0E

∑ F y=F T cos−mg

F T=mg

cos

=2 nCm2

=−2 nCm2 E=

2o

2o=o

q0

w=mg

FT

F E=q0E

∑ F y=FT cos−mg=0

F T=mg

cos

∑ F x=q0E−F T sin =0

q0=F T sinE

q0=mg tan /o

107

Gauss's Law

108

Flux:

=Electrical Field X Area=E A

A=5m2

E=100 NC

=100 NC 5m2

=500 N m2

C

109

Flux:

=Electrical Field X Area

A=5m2

E=100 NC

=100 NC 5m2

=500 N m2

C

110

Flux:

=Electrical Field X Area

E

A

111

Flux:

=E⋅A

E

A

A cos

112

Flux:

=E⋅A=∣E∣∣A∣cos

E

A

A cos

113

Flux:

=E⋅A=∣E∣∣A∣cos=10 NC

5m2 cos 30o=43.30 N m2

C

E=10 NC

A=5m2

=30o

A cos

114

q1

2

115

E=Q inside

o

Gauss's Law

116

q

r

Gauss's LawGauss's Law

E=E A

E=k qr2 AE= 1

4o

qr2 A

117

q

r


E=E A

E=k qr2 AE= 1

4o

qr2 A

E= 14o

qr2 4 r2

E=qo=Q inside

o

118

q

r


E=Qinside

o

119

q

r


=∫ E⋅ dA

∫ E⋅dA

120

q r


=∫ E⋅ dA

∫ E⋅dA=∫∣E∣∣dA∣cos

A

121

q r


=∫ E⋅ dA

∫ E⋅dA=∫∣E∣∣dA∣cos

∫ E⋅dA=∫∣E∣∣dA∣cos 0o

∫ E⋅dA=∫∣E∣∣dA∣∫ E⋅dA=∣E∣∫∣dA∣∫ E⋅dA=k q

r 2 4 r2

A

122

q r


=∫ E⋅ dA

∫ E⋅dA=k qr2 4r 2

∫ E⋅dA= 14o qr2 4 r2

∫ E⋅dA= qo

A

123

q r


=∫ E⋅ dA=qenco

A

124

E=Qinside

o

Volume Charge Density =QV=

qV

Surface Charge Density =QA=

q A

Linear Charge Density =Ql=q l

125

Infinite Line of Charge

q

EE

126

Infinite Line of ChargeE

=Ql=q l

= constant

l

A1

A2

No electrical field passes through A1 or A2 .

127


l

r

l

2r

A=2 r l

=Ql=q l

= constant

128


l

r

l

2r

A=2 r lqinside= l

=Ql=q l

= constant

129


l

r

l

2r

A=2 r l

=Ql=q l

= constant

qinside= l

E=q insideo

E A= lE 2 r l = l

E=

2o r

130


l

r

l

2r

A=2 r l

=Ql=q l

= constant

qinside= l

E=qinsideo

E A= lE 2 r l = l

E=

2o r

E=22

2o r

E=2

4o r=2k

r

131

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

Infinite Sheet of Charge

132

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


AA

qenc=A

133

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


AA

EAE A= Ao

2EA=Ao

E=

2o

E=Qinside

o

134

=2 nCm2

=−2 nCm2 E=

2o

2o=o

135

=2 nCm2

=−2 nCm2 E=

2o

2o=o

q0

136

=2 nCm2

=−2 nCm2 E=

2o

2o=o

q0

w=mg

FT

F E=q0E

137

=2 nCm2

=−2 nCm2 E=

2o

2o=o

q0

w=mg

FT

F E=q0E

∑ F y=F T cos−mg

F T=mg

cos

138

=2 nCm2

=−2 nCm2 E=

2o

2o=o

q0

w=mg

FT

F E=q0E

∑ F y=FT cos−mg=0

F T=mg

cos

∑ F x=q0E−F T sin =0

q0=F T sinE

q0=mg tan /o

139

Hollow Sphere with total charge Q.

E=Qinside

oE A=Q

o

E 4 r2=Qo

E= Q4r2o

=k Qr2

140

Solid Conducting Sphere with total charge Q.

E=Qinside

o

+ +

141


E=Qinside

o

+ +

142


E=Qinside

o

+ ++ +

143


E=Qinside

o

+ +

+

+

144


E=Qinside

o

+ +

+

+

+ +

145


E=Qinside

o

+ +

+

+

+

+

146


E=Qinside

o

+ +

+

+

+

+

+

+

+

+

+

+

All charge will lie on the surface of the conducting sphere.

E A=Qo

E 4 r2=Qo

E= Q4 r2o

=k Qr2

r >= R

R

r

147


E=Qinside

o

+ +

+

+

+

+

+

+

+

+

+

+

All charge will lie on the surface of the conducting sphere.

E A=Qo

E 4 r2 =0E=0

r < R

Rr

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