coulomb’s law and electric field

Coulomb’s Law and Electric Field

Chapter 24: allChapter 25: all

Physics chapters 24 - 25 2

Electric charge Able to attract other objects Two kinds

Positive – glass rod rubbed with silk Negative – plastic rod rubbed with fur

Like charges repel Opposite charge attract Charge is not created, it is merely

transferred from one material to another


Elementary particles Proton – positively charged Electron – negatively charged Neutron – no charge Nucleus – in center of atom,

contains protons and neutrons Quarks – fundamental particles –

make up protons and neutrons, have fractional charge


ions Positive ions – have lost one or

more electrons Negative ions – have gained one or

more electrons Only electrons are lost or gained

under normal conditions


Conservation of charge The algebraic sum of all the

electric charges in any closed system is constant.


Electrical interactions Responsible for many things

The forces that hold molecules and crystals together

Surface tension Adhesives Friction


Conductors Permit the movement of charge

through them Electrons can move freely Most metals are good conductors


Insulators Do not permit the movement of

charge through them Most nonmetals are good

insulators Electrons cannot move freely


Charging by induction See pictures on pages 539-540


Coulomb’s Law Point charge – has essentially no

volume The electrical force between two

objects gets smaller as they get farther apart.

The electrical force between two objects gets larger as the amount of charge increases


Coulomb’s Law

221

rqq

kF

r is the distance between the charges

q1 and q2 are the magnitudes of the charges

k is a constant 8.99 x 109 N∙m2/C2


Coulombs SI unit of charge, abbreviated C Defined in terms of current – we

will talk about this later


Coulomb’s law constant k is defined in terms of the speed

of light k = 10-7c

k = 1/4pe0

e0 is another constant that will be more useful later

e0 = 8.85 x 10-12 C2/N∙m2


The coulomb Very large amount of charge Charge on 6 x 1018 electrons Most charges we encounter are

between 10-9 and 10-6 C 1 mC = 10-6 C


Examples See pages 543 - 546


Electric Field• A field is a region in space where a

force can be experienced.

• Or: a region in space where a quantity has a definite value at every point.


Electric Field• Produced by a charged particle.• The force felt by another charged

particle is caused by the electric field.

• We can check for an electric field with a test charge, qt. If it experiences a force, there is an electric field.


Electric field• The definite quantity is a ratio of

the electric force experienced by a charge to the amount of the charge.

• Vector quantity measured in N/C.

EF tqtq

FE


Electric field• To determine the field from a point

charge, Q, we place a test charge, qt, at some position and determine the force acting on it.

Q qt

F


Direction of E• If the test charge is positive, E has

the same direction as F.• If the test charge is negative, E

has the opposite direction as F.


Electric Field - Point Charge

2

Q4

1r

Eope

tqFE 2r

Qqk tF

2rQ

kE


Electric Field• The field is there, independent of a

test charge or anything else!• The electric field vector points in

the direction a positive charge would be forced.


Example 1• Two charges, Q1 = +2 x 10-8 C and

Q2 = +3 x 10-8 C are 50 mm apart as shown below.

• What is the electric field halfway between them?

Q1 Q250 mm

E1E2


Example 1• At the halfway point, r1 = r2 = 25

mm.• Magnitudes of fields:

E1 kQ1r1

2 (9 x 109 N • m

2

C2 )(2 x 10 8C)

(2.5 x 10 2 m)2

E2 kQ2r2

2 (9 x 109 N • m

2

C2 )(3 x 10 8 C)

(2.5 x 10 2 m)2


Example 1• E1 = 2.9 x 105 N/C• E2 = 4.3 x 105 N/C• E1 is to the right and E2 is to the

left.• E1 = 2.9 x 105 N/C• E2 = - 4.3 x 105 N/C• E = E1 + E2 = - 1.4 x 105 N/C


Example 2• For the charges in Example 1,

where is the electric field equal to zero?

• Since the fields are in opposite directions between the charges, the point where the field is zero must be between them.

Q1 Q2

E1E2


Example 2E1 E2

kQ1r1

2 kQ2r2

2

Q1r1

2 Q2r2

2

r1 + r2 = s, so r2 = s – r1


Example 2Q1r1

2 Q2r2

2

Q1r1

2 Q2

(s r1)2

(s r1 )2

r12

Q2Q1

s r1r1

Q2Q1

r1 s

1 Q2

Q1

r1 23 mm


Field Diagrams• To represent an electric field we

use lines of force or field lines.• These represent the sum of the

electric field vectors.


Field Diagrams


Field Diagrams• At any point on the field lines, the

electric field vector is along a line tangent to the field line.


Field Diagrams


Field Diagrams• Lines leave positive charges and

enter negative charges.• Lines are drawn in the direction of

the force on a positive test charge.• Lines never cross each other.• The spacing of the lines represents

the strength or magnitude of the electric field.


Point Charges• Lines leave or enter the charges in

a symmetric pattern.• The number of lines around the

charge is proportional to the magnitude of the charge.


Point Charges


Gauss’s Law• Electric flux through a closed

surface is proportional to the total number of field lines crossing the surface in the outward direction minus the number crossing in the inward direction.

0eQEA


Example 25-9 (see page 563)Field of a charged sphere is the

same as if it were a point charge

204

1rqE

pe


Example 25-10 (see page 564)Field of a infinite line of charge is

rE

pe021


Other scenarios • See table on page 567


Example 3• Two parallel metal plates are 2 cm

apart.• An electric field of 500 N/C is placed

between them.• An electron is projected at 107 m/s

halfway between the plates and parallel to them.

• How far will the electron travel before it strikes the positive plate?


Example 3• Two charged parallel plates create

a uniform electric field in the space between them.


Example 3

Evo

This is just like a projectile problem except that the acceleration is not a given value.


Example 3

a =Fm

F = qE = eE

a =eEm

=(1.6 x 10–19C)(500 N/C)

9.1 x 10–31kg

= 8.8 x 1013 m/s2


Example 3• 8.8 x 1013 m/s2 is the vertical

acceleration of the electron.• Horizontally, the acceleration is

zero.• x = vt• v = 1 x 107 m/s & t = ?


Example 3• Back to vertical direction:• y = yo + vot + 1/2at2

• y = 1/2at2

a2yt

2(0.01 m)8.8x1013 m / s2

= 1.5 x 10-8 s


Example 3• Back to horizontal direction:• x = vt• x = (1 x 107 m/s)(1.5 x 10–8 s)

• x = 0.15 m = 15 cm


Dipoles• A pair of charges with equal and

opposite sign.• Induced dipoles, molecular dipoles,

etc.…

coulomb’s law and electric field

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