WHY?

=

Electromagnetism I

Lecture 2

Coulomb’s Law and the Concept of Electric Field

Learning Objectives

•To present a mathematical expression to the experimentally observed force

•To describe Coulomb’s Law Quantifying the Forces between Charges – superposition principle

•To introduce the concept of the Electric Field

- a great way to deal with forces!

Charles Augustin de CoulombBorn: 14 June 1736 in Angoulême, France

Died: 23 Aug 1806 in Paris, France

He worked on applied mechanics, but he is best known for his work on electricity and magnetism, and in particular the Coulomb Law

Coulomb studied Friction

Friction is commonly studied by rubbing two surfaces.

And rubbing two surfaces sometimes

can create charges.

Create is the wrong word.

Charge transfer

Coulomb’s Law•Like charges repel, unlike charges attract

•The force acts along the line joining two charged particles

•The force between two charged particles is proportional to the magnitude of each charge

•The force between two charged particles is inversely proportional to the square of the distance between the particles

21221

1221 r̂

r

qqF ∝

Mathematically

21221

12

021 ˆ

4

1r

r

qqF

πε=

•q1 and q2 are measured in Coulombs

•r21 is in metres

•F21 is measured in Newtons

In SI Units

ε0 is called the permittivity of free space = 8.854 10-12 Coulomb2 Newton-1 m-2 (simplifies equations)

For two protons this is 1034 smaller than the repulsive force between two proton. Gravitational force is only significant because it is always attractive!

r̂r

mGmF grav 2

21−=

Gravitational Force

Worked Example: assume that the positive and negative charge within a 1 pence coin do not cancel, but there is an excess of 0.0001% of electrons. What force would two such pennies exert on each other if they are placed 1 m apart?

1p 1p

1 m

1p 1p

1 m

One pence 3 g copperT&M lectures: Molar mass of Cu 63.5 g

One pence = (3/63.5) 6x1023 = 2.84 1022 Cu atoms

One Cu atoms contains 29 electrons. Hence one pence

contains 8.25 1023 electrons

This corresponds to 131956 C. 0.0001% of that is 0.132 C

0.132 C at 1 m gives F = 157,000,000 N (17,662 Tons)

So when we say that macroscopic objects are neutral, we mean it!

Compute the ratio of the electric force to the gravitational

force exerted by a proton on an electron at a separation

of 106 m.

mp = 1.67x10-27 kg me = 9.11x10-31 kg

e = 1.60x10-19 C G = 6.67x10-11 N m2 kg-2

Classwork

39

0

2

220

2

102724

and 4

×==∴

==

.epg

E

epgE

mGme

FF

r

mGmF

re

F

πε

πε

(a)Did you really need to know how far

apart the two particles are?

(a)Why is gravity the dominating force between large objects such as planets and stars?

Forces between many charges - the principle of superposition

What if you have more than two positive charges?

q2

q3

F12

F21

q1

F31

F13

Ftotal

•Consider the charges one pair at a time

•Vector sum of forces

Procedure

In general the force on a charge q due to a collection of charges is the vector sum of all the individual forces due to all the others.

∑=

+++=

ii

i

i r̂r

qq

FFFF

20

321

4

πε

L

This is known as the principle of superposition of forces

The Electric Field

•Concept of the Electric Field

•Electric Field Strength E

•Electric Field Lines

•Calculation of E at a point

•The Electric Dipole

Definition of Electric FieldConsider a small test charge, q0, placed at a point in a distribution of point charges. The force on this test charge is:

∑=i

ii

i r̂r

qqF

20

00 4πεDefine electric field E as

∑==i

ii

i r̂r

q

q

FE

200

0

4πε

∑=i

i

i

i rr

qE ˆ

2

04πε

Thus, the E at a certain point is equal to the electric force per unit charge at that point

F = qE

Once we know E at a certain point we can calculate the force on any charge q placed there, because E is the force acting on a unit charge.

A better known example of a field; the gravitational force Fg that the earth exerts on a mass m0:

gmF g 0=

0m

Fg g=

We can interpret g as the gravitational field

These are lines drawn in space so that the tangent to them at any point gives the direction of E at that point.

Visualisation of EUse field lines of force

The E field near to a point charge

r̂r

qE

204πε

=

•The density of the field lines is proportional to the strength of E

The absolute number of lines

is not important. You cannot

obtain the field strength by

counting the number of lines.

Adding lines must follow rules.

•Field lines only start on positive charge, they can only end on negative charge

Example: E field lines around an electric dipole

The tangent at any point gives

the direction of E at that point.

Field lines do not cross one another!

Field lines are continuous,only stop at charges!

Field can be measured without knowing

the detailed configuration of the

charges which produce the field

E

Test charge

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎠⎞

⎜⎝⎛ +

−

⎟⎠⎞

⎜⎝⎛ −

= 220

22

4

1

ar

q

ar

qE

πε

The E-field due to an Electric Dipole - Calculation

To simplify the calculation, we will only compute the field along the axis

r

E

-q +q

a

a/2

€

E =q

4πε0r2

2a /r

1−a2

r2

⎛

⎝ ⎜

⎞

⎠ ⎟

This simplifies to:

304

2

r

qaE

πε≈

For r >> a

The quantity qa is the electric moment of the dipole, p

For r >>>>>>>> a

€

E ≈2qa

4πε0r3

=2q

4πε0r2

.a

r≈ 0

-q +q

ar

For large r, r+a r

P

To the distant point P, the field looks like that from -q and +q from

the same point.

304

2

r

pE

πε≈

electric dipole moment = aq=p;

note r-3 dependence

Review and Summary

•Coulomb’s law gives the force on a charged particle placed in a distribution of charges

•Forces must be superposed as vectors - superpositon principle

•Field lines provide a graphical representation of electric fields

• An electric dipole is a pair of electric charges of equal magnitude q but opposite sign separated by a distance a

•Attention must be paid to direction, it depends on the signs of charges and geometry of configuration

∑=i

ii

i rr

qqF ˆ

4

20πε

Coulomb’s Law

Definition of an Electric Field

∑=i

i

i

i rr

qE ˆ

2

04πε

The effects of gravitational attraction are negligible, and hence do not need to be considered when discussing atomic (or molecular) interactions. 

Gravity is the dominating force between large objects because these objects are electrically neutral – containing equal numbers of positive and negative electrical charges – the net electrical force is zero because there are attractive and repulsive forces. The gravitational force is attractive only.

Next Lecture

Calculating E-fields of continuous charge distributions