Coulombs law

Electric Charge and

Coulomb’s Law

Fundamental Charge: The charge on one electron.

e = 1.6 x 10 -19 C

Unit of charge is a Coulomb (C)

Two types of charge:

Positive Charge: A shortage of electrons.

Negative Charge: An excess of electrons.

Conservation of charge – The net charge of a closed system remains constant.

+

n + +

+ +

+

n

nn

n n

-

-

-

-

-

-

Neutral Atom

Number of electrons = Number of protons

Nucleus

Negative Atom

Number of electrons > Number of protons

-2e = -3.2 x 10-19C

-

-

Positive Atom

Number of electrons < Number of protons

+2e = +3.2 x 10-19C

Electric Forces

Like Charges - Repel

Unlike Charges - Attract

- +F F

+ +FF

Coulomb’s Law – Gives the electric force between two point charges.

221

r

qqkF =

k = Coulomb’s Constant = 9.0x109 Nm2/C2

q1 = charge on mass 1

q2 = charge on mass 2

r = the distance between the two charges

The electric force is much stronger than the gravitational force.

Inverse Square

Law

221

r

qqkF =

If r is doubled then F is :

If q1 is doubled then F is :

If q1 and q2 are doubled and r is halved then F is :

¼ of F

2F

16F

Two charges are separated by a distance r and have a force F on each other.

q1q2

r

F F

Example 1

Example 2

Two 40 gram masses each with a charge of 3μC are placed 50cm apart. Compare the gravitational force between the two masses to the electric force between the two masses. (Ignore the force of the earth on the two masses)

3μC40g

50cm

3μC40g

221

r

mmGFg =

211

)5.0(

)04)(.04(.1067.6 −×= N131027.4 −×≈

221

r

qqkFE =

2

669

)5.0(

)103)(103(100.9

−− ×××= N324.0≈

The electric force is much greater than the gravitational force

5μC

- 5μC

Three charged objects are placed as shown. Find the net force on the object with the charge of -4μC.

- 4μCF2

F1

F1 and F2 must be added together as vectors.

20cm

20cm

cm282020 22 ≈+

45º

45º

221

r

qqkF =

NF 5.4)20.0(

)104)(105(109

2

669

1 =×××=−−

NF 30.2)28.0(

)104)(105(109

2

669

2 =×××=−−

Example 3

F1

F2

45º

2.3cos45≈1.6

2.3sin45≈1.6

F1 = < - 4.5 , 0.0 >

F2 = < 1.6 , - 1.6 >+

Fnet = < - 2.9 , - 1.6 >

NFnet 31.36.19.2 22 ≈+=

- 2.9

- 1.63.31

θ

299.2

6.1tan 1 ≈

−−= −θ

3.31N at 209º

29º

Example 4

Two 8 gram, equally charged balls are suspended on earth as shown in the diagram below. Find the charge on each ball.

qq

20º

L = 30cmL = 30cm

FEFE

r =2(30sin10º)=10.4cm

2

2

221

r

qk

r

qqkFE ==

10º10º

30sin10º

r

Draw a force diagram for one charge and treat as an equilibrium problem.

FE

Fg = .08N

T

q

Tsin80º

NT

T

081.80sin

08.

08.80sin

≈=

=

Tcos80º

Cq

kq

qk

TFE

7

22

2

2

103.1

)104(.014.

80cos)081(.104.

80cos

−×=

=

=

=

80º