Coulomb law

Coulomb law

Electric Fields of Force• The Force of Electric charges are determined by the type of charge of the objects.

• unlike charges attract

• like charges repel

• A strong charge will attract a weak or neutral charge

Electric Field Patterns

1. Negative and Positive Point Charge: Lines of force radiate from the positive, and inward to the negative

2. Negative and Positive interaction: Forces originate from the Positive and end at the negative using the series of curved lines (attraction)

3. Two Positive charge will deflect force lines away from the charges. (2 Negatives are the same except for direction ( both show repulsion)

Describe 3 ways that Charges interact their force lines.

Coulomb’s law

• SI standard quantity of a charge is a Coulomb (C)• 1 C = 6.0 x 10 28 electrons• 1.9 x 10 -19 C = 1 electron (-) or • 1 Proton (+) or 1 “elementary” charge

• Electric Force : (one of the 4 major forces) varies inversely with the distance between the charge• F 1a d2

Electric Fields

The Electric Field is defined as the Force exerted on a tiny positive test charge at that point divided by the magnitude of the test charge:

E = Fe

q

E is the electric field strength

Fe is the electrostatic force

q is the charge in coulombs

Electric field within conductor

Why is the person in the car protected from the lighting?

The metal “cage” on the inside of the car is uncharged, but the outside surface becomes charged.

Electric Flux• Strength of the electric field depends on the

density of lines in the region.• Electric Flux is the amount of electric field

flowing through a surface• It is the electric field times the area, when the

field is perpendicular to the surface• It is zero if the electric field is parallel to the

surface• Normally denoted by symbol E.• Units are N·m2/C

Electric Field

E = EA

Electric Field

• A uniform electric field can be produced in the space between two parallel metal plates.

• The plates are connected to a battery.• If an Electron moves within the field from the negative

plate to the positive plate, How does the amount of force change?

E

Electric Field Strength-Uniform Field

The field strength at any point in this field is:

E = field strength (Vm-1)V = potential difference (V)d = plate separation (m)

d

VE

The field strength is the same magnitude and

direction at all points in the field

Sample ProblemAn electron (mass m = 9.11×10-31kg) is accelerated in the uniform field E (E = 1.33×104 N/C) between two parallel charged plates. The separation of the plates is 1.25 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate, as seen in the figure. With what speed does it leave the hole?

F = qE = ma

a = qE/m

Vf 2 = vi2 + 2a(d)

Vf 2 = 2ad = 2(qE/m)d

Solution• Vf 2 = 2ad = 2(qE/m)d

= 2 (1.9 x 10 -19C) (1.33×104 N/C) (1.25m) 9.11×10-31kg

= 8.3 x 10 6 m/s

m = 9.11×10-31kg

E = 1.33×104 N/Cd = 1.25 cm

GIVEN:

Coulomb’s law• The magnitude of the force between 2 equal

sized objects • Fe = k q1 q2

r2

k (electrostatic constant =8.99 x 109 N m2/C2)

q (charge in Coulombs)

r ( distance between the charges)

r

Fe

Coulombs Law: 2 Charges• A positive charge of 6.0 x 10 -6C is 0.030m from a second positive

charge of 3.0 x 10 -6C. Calculate the force between the charges.

• Fe = k q1 q2

r2

= (8.99 x 109 N m2/C2 ) (6.0 x 10 -6C) (3.0 x 10 -6C)

( 0.030m )2

= (8.99 x 109 N m2/C2 ) (18.0 x 10 -12C)

(9.0 x 10 -4 m2)

= + 1.8 x 10 -8 N

Coulombs Law: Summary

• The force of electrical charge is an inverse square of the distance between the charge:

• Fe = k q1 q2

r2

The field strength at any point in this field is:

E = field strength (Vm-1)V = potential difference (V)d = plate separation (m)

d

VE

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