1 Norah Ali Al-moneef king Saud unversity 23.1 Properties of Electric Charges 23.2 Charging Objects By Induction 23.3 Coulomb’s Law 23.4 The Electric Field

1Norah Ali Al-moneef king Saud unversity

23.1 Properties of Electric Charges23.2 Charging Objects By Induction23.3 Coulomb’s Law23.4 The Electric Field23.6 Electric Field Lines23.7 Motion of Charged Particles in a Uniform Electric Field

1/10/2006

Electric Charge• Types:

– Positive• Glass rubbed with silk • Missing electrons

– Negative• Rubber/Plastic rubbed with fur• Extra electrons

• Arbitrary choice – convention attributed to ?

• Units: amount of charge is measured in [Coulombs]

• Empirical Observations:– Like charges repel– Unlike charges attract

2Norah Ali Al-moneef king Saud unversity1/10/2006

Charge in the Atom

• Protons (+)• Electrons (-)• Ions• Polar Molecules


23.1 Properties of Electric Charges• Conservationelectricity is the implication that electric charge is always conserved.• That is, when one object is rubbed against another, charge is not created in the

process. The electrified state is due to a transfer of charge from one object to the other.

• One object gains some amount of negative charge while the other gains an equal amount of positive charge.

• Quantization– The smallest unit of charge is that on an electron or proton. (e = 1.6 x 10-19 C)

• It is impossible to have less charge than this• It is possible to have integer multiples of this charge

Q Ne4Norah Ali Al-moneef

king Saud unversity1/10/2006

Conductors and Insulators

• Conductor transfers charge on contact• Insulator does not transfer charge on contact• Semiconductor might transfer charge on contact


23.2 Charging Objects By Induction

1/10/2006

Charge Transfer Processes

• Conduction• Polarization• Induction


23-3 Coulomb’s Law• Empirical Observations

• Formal Statement

1 2F q q2

1F

r

Direction of the force is along the line joining the two charges

1 212 212

21

kq qˆF r

r


Norah Ali Al-moneef king Saud unversity 8

• Consider two electric charges: q1 and q2

• The electric force F between these two charges separated by a distance r is given by Coulomb’s Law

• The constant k is called Coulomb’s constant and is given by

221

r

qkqF

229 /CNm109k

• The coulomb constant is also written as

• 0 is the “electric permittivity of vacuum”

– A fundamental constant of nature

2

212

00 Nm

C 1085.8 where

4

1

k

2

21

0

4

1

r

qqF

1/10/2006


• Double one of the charges– force doubles

• Change sign of one of the charges– force changes direction

• Change sign of both charges– force stays the same

• Double the distance between charges– force four times weaker

• Double both charges– force four times stronger

1/10/2006


Example:What is the force between two charges of 1 C separated by 1 meter?

Answer: 8.99 x 109 N,

1/10/2006

Coulomb’s Law Example• What is the magnitude of the electric force of

attraction between an iron nucleus (q=+26e) and its innermost electron if the distance between them is 1.5 x 10-12 m

• The magnitude of the Coulomb force is• F = kQ1Q2/r2

• = (9.0 x 109 N · m2/C2)(26)(1.60 x 10–19 C)(1.60 x 10–19 C)/(1.5x10–12 m)2

• = 2.7 x 10–3 N.



Example - The Helium NucleusPart 1: The nucleus of a helium atom has two protons and two neutrons. What is the magnitude of the electric force between the two protons in the helium nucleus?

Answer: 58 NPart 2: What if the distance is doubled; how will the force change?

Answer: 14.5 N

Inverse square law: If the distance is doubled then the force is reduced by a factor of 4.

1/10/2006

13

• Consider two charges located on the x axis• The charges are described by

– q1 = 0.15 C x = 0.0 m

– q2 = 0.35 C x = 0.40 m

• Where do we need to put a third charge for that charge to be at an equilibrium point?

At the equilibrium point, the forces from the two charges will cancel.

x1 x2

Example - Equilibrium Position

Norah Ali Al-moneef king Saud unversity1/10/2006

mx

x

qqk

x

qqk

16.0

)4.0()( 232

231

Here the forces from q1 and q2 can balance.

q3

xx

x x4.0

Zero Resultant Force, Example

– The magnitudes of the individual forces will be equal

– Directions will be opposite– Will result in a quadratic– Choose the root that gives the forces in

opposite directions

Two fixed charges, 1mC and -3mC are separated by 10cm as shown in the figure (a) where may a third charge be located so that no force acts on it?

cmx

xx

x

qqk

x

qqk

7.13

)10(103

)(101

)10()(

2

6

2

6

232

231

1/10/2006 14Norah Ali Al-moneef king Saud unversity

1/10/2006 Norah Ali Al-moneef king Saud unversity 15

two charges are located on the positive x-axis of a coordinate system, as shown in the figure. Charge q1=2nC is 2cm from the origin, and charge q2=-3nC is 4cm from the origin. What is the total force exerted by these two charges on a charge q3=5nC located at the origin?

The total force on q3 is the vector sum of the forces due to q1 and q2 individually.

The total force is directed to the left, with magnitude 1.41x10-4N.

Example:


Example - Charged Pendulums• Consider two identical charged balls hanging

from the ceiling by strings of equal length 1.5 m (in equilibrium). Each ball has a charge of 25 C. The balls hang at an angle = 25 with respect to the vertical. What is the mass of the balls?

Step 1: Three forces act on each ball:

Coulomb force, gravity and the tension of the string.

mgTFd

kqTF

y

x

cos

sin

:lefton Ball

2

2

x

y

1/10/2006


Example - Charged Pendulums (2)Step 2: The balls are in equilibrium positions. That means the sum of all forces acting on the ball is zero!

tan

/

cos

sin

2

2

22

d

kqmg

mg

dkq

T

T

Answer: m = 0.76 kgA similar analysis applies to the ball on the right.

d=2 l sin

1/10/2006

mgTd

kqT

cos

sin2

2


Electric Force and Gravitational Force

• Coulomb’s Law that describes the electric force and Newton’s gravitational law have a similar functional form

• Both forces vary as the inverse square • of the distance between the objects.• Gravitation is always attractive.• k and G give the strength of the force.

Felectric kq1q2

r2Felectric k

q1q2

r2 Fgravity Gm1m2

r2Fgravity G

m1m2

r2

1/10/2006


Example: An electron is released above the surface of the Earth. A second electron directly below it exerts an electrostatic force on the first electron just great enough to cancel out the gravitational force on it. How far below the first electron is the second?

e

e

mg

Fe

r = ?

)8.9)(1011.9(

)106.1()109(

31

2199

212

21

x

xr

mg

qqkrmg

r

qqk

mgFE

5.1 m

1/10/2006

N102.8

m103.5

C1060.1

C

N.m109.

4

1

8

211

219

2

29

2

2

0

r

eFe

Compare the electrostatic and gravitational the forces

N106.3

m103.5

k1067.1k1011.9

k

N.m107.6

47

211

2731

2

211

2g

gg

gr

mmGF pe

Fe/Fg = 2 x 1039 The force of gravity is much weaker than the electrostatic force

The electron and proton of a hydrogen atom are separated (on the average) by a

distance of approximately 5.3 x10-11 m. Find the magnitudes of the electric force

and the gravitational force between the two particles.


Norah Ali Al-moneef king Saud unversity 211/10/2006

Electric Forces and VectorsElectric Fields and Forces are ALL vectors, thus all rules

applying to vectors must be followed.Consider three point charges, q1 = 6.00 x10-9 C (located at the origin),q3 = 5.00x10-9 C, and q2 = -2.00x10-9 C, located at the corners of a RIGHT triangle. q2 is located at y= 3 m while q3 is located 4m to the right of q2. Find the resultant force on q3.

q1

q2 q3

3m

4m

5m

q3

Which way does q2 push q3?Which way does q1 push q3?

Fon 3 due to 2

Fon 3 due to 1

= 37

= tan-1(3/4)


2,3

2

999

2,3 4

)102)(100.5()1099.8(

F

xxxF

1,3

2

999

1,3 5

)105)(106()1099.8(

F

xxxF

)(tan

)()(F

1062.6)37sin(

1018.3

)37cos(

1

22resultant

91,3

9

2,31,3

x

y

res

yx

y

x

x

F

FDirection

F

FF

NxFF

NxF

FFF

q1

q2 q3

3m

4m

5m

q3

Fon 3 due to 2

Fon 3 due to 1

= 37= tan-1(3/4)

5.6 x10-9 N

1.1x10-8 N

F3,1cos37

F3,1sin37

7.34x10-9 N

64.3 0 above the +x


Coulomb’s Law Example • Q = 6.0 mC• L = 0.10 m• What is the magnitude and

direction of the net force on one of the charges?

F1

+

F2

F3

Q

x

y

L

++

+

L

QQ

Q


We find the magnitudes of the individual forces on the charge at the upper right corner:

F1= F2 = kQQ/L2 = kQ2/L2

= (9 x109 N · m2/C2)(6 x10–3 C)2/(0.100 m)2 = 3.24 x107 N.F3= kQQ/(L√2)2 = kQ2/2L2 = (9 x109 N · m2/C2)(6 x10–3 C)2 /2(0.100 m)2

= 1.62 x107 N.

1/10/2006


along the diagonal, or away from the center of the square. From the symmetry, each of the other forces will have the same magnitude and a direction away from the center: The net force on each charge is= 6.20 107 ء N away from the center of the square.

.

45 tan

10 2.6107 385.4107 385.42 2

10 385.410 1.14510 24.3

10 385.410 1.14510 24.3

10 1.145707.010 1.62 sin

10 1.145707.010 1.62 cos

45 so 1L

L tan

0

01

7

777

777

7733y

7733x

0

FF

45 so 1LL

tan

FF

FF

FF

x

y

y

x

NF

N

N

N

N

1/10/2006


Example - Four ChargesConsider four charges placed at the corners of a square with sides of length 1.25 m as shown on the right. What is the magnitude of the electric force on q4 resulting from the electric force from the remaining three charges?

Answer:

F (on q4) = 0.0916 N

… and the direction?

1/10/2006

23-7; Three point charges are located at the corners of an equilateral triangle. Calculate the net electric force on the 7.00 uC charge.

HOMEWORK :

23-8: Two small beads having positive charges 3q and q are fixed at the opposite ends of a horizontal insulating rod extending from the origin to the point x =d. a third small charged bead is free to slide on the rod. At what position is the third bead in equilibrium? Can it be in stable equilibrium?


23-12; An object having a net charge of 24.0 C is placed in a uniform electric field of 610 N/C that is directed vertically. What is the mass of this object if it “floats” in the field?

3-18; Two 2.00uC point charges are located on the x axis. One is at x = 1.00 m, and the other is at x =- 1.00 m. (a) Determine the electric field on the y axis at y =0.500 m. (b) Calculate the electric force on a - 3.00uC charge placed on the y axis at y = 0.500 m.

23-41; An electron and a proton are each placed at rest in an electric field of 520 N/C. Calculate the speed of each particle 48.0 ns after being released.

23-44; The electrons in a particle beam each have a kinetic energy of 1.60 x 10-

17 J. What are the magnitude and direction of the electric field that stops these electrons in a distance of 10.0 cm?


Documents

1 Norah Ali Al-moneef king Saud unversity 23.1 Properties of Electric Charges 23.2 Charging Objects By Induction 23.3 Coulomb’s Law 23.4 The Electric Field