ch.7 electric charges & electric fields - science rules! · pdf...

Ch.7 Electric Charges &

Electric Fields

7.3 Electric Fields

“You must feel the Force Luke...”

Introduction to Electric Fields

An x-ray machine uses an

electric field to accelerate

electrons which then

interact with matter to

produce x-rays

Cell phones, LCD screens,

telephones, photocopiers,

and computers all rely on

the principles of

electric fields

Force at a Distance

Force - a push or pull on an object.

Like masses that experience a gravitational

attractive force, charged objects do not need

to be in contact with each other in order to

experience forces of attraction/repulsion.

“action-at-a-distance”

Field of Force

“A field of force exists in a region of space

when an appropriate object placed at any

point in the field experiences a force.”

Properties of Electric Fields

A charge generates an electric field

An electric field (𝜀) causes an electric force

and is the spatial region in which a force is

exerted on any electric charge.

Electric field = the electric force per unit

positive charge (units = N/C).

An electric field exerts attractive/repulsive

forces on other charged objects.

𝜀 is a vector (has both magnitude and

direction); 𝜀 and FE act in the same direction

Electric Fields

Electric field lines ALWAYS start on +ve

charges and end on –ve charges.

Electric field lines NEVER CROSS.

refer to diagrams on pg.339 & 340

http://www.youtube.com/watch?v=laGSICm_agM

Electric Fields

Electric dipole

Field Theory

This diagram illustrates the

generation of an electric field, 𝜺,

by a charge, +q1 . The density of the

shading designates the strength of

the field. If a + test charge, q,

is introduced, it is the field that

interacts with q to produce the

electric force FE (repulsion).

The field strength ↓ as r ↑ FE ↓also;

At infinity 𝜺 and FE become zero

Note: If q1 were negative the electric field and electric

force would be in the opposite direction.

Field Theory

q

+

+

+

+

+

+

+

+

+

+

+ +

positively charged sphere

(which creates electric field)

positive test charge

(which experiences electric field)

field strength

e = FE

q

where e = electric field strength or intensity (N/C)

FE = electric force (N)

q = +ve test charge (C)

NOTE: e and FE

have the same

direction

Comparing Laws

Force exerted

by field (N)

Quantity

affected by field

Field strength

N/C or N/kg

Fg = mg

FE = qe

More Field Theory

FE = qe e = kq1

r2

q1 q

positively charged sphere

(which creates electric field)

positive test charge

(which experiences electric field)

FE

FE = kqq1

r2

Equating both formulas for FE gives an equation dependent upon q1 and r only.

Two “Points” of View

Requires FE and a + test charge q

experiencing the force in the field

(works for charge distributions)

Electric Field

Strength

Requires a master point charge q1,

that is creating the field, and the

distance from this charge (does

NOT work for charge distributions;

only for single point charges)

FE = e = kq1

q r2

Comparison of Gravitational and

Electric Fields and Forces

-

q

FE

FE = kq1q2

r2

FE = qe

e = kq1

r2

m

Fg

Fg = GMEm

r2

Fg = mg

g = GME

r2

Uniform Electric Fields

Instead of point charges, suppose you have two large conducting

plates charged by dry cells. As with the dipole, one plate has a

positive charge and the other plate has a negative charge. In both

cases, the charge spreads uniformly along each plate. The electric

field between the plates of charge extends from the positive plate

to the negative plate and is uniform. These field lines are straight,

parallel to each other, and perpendicular to the plates.

Uniform Electric Fields

In parallel plates because the electric field

lines run straight from 1 plate to the other,

𝜀 does not depend on the separation of

the plates

The magnitude of the electric field b/w 2

plates is directly proportional to the charge

per unit area on the plates. 𝜀 ∝ 𝑞

𝜀 is uniform everywhere in the space b/w

the plates

Example 1a)

Two point charges, q1 = 3.6x10-6 C and

q2 = -2.7x10-6 C, are arranged as shown. If

a test charge is placed at point A, find the

net electric field strength at point A due to

the combined electric fields of both charges.

q1 q2

30 cm 20 cm

A

Example 1b)

What force is exerted on a charge of

4.5x10-6C placed at point A?

q1 q2

30 cm 20 cm

q

A

Example 2

In the diagram, A and C are situated as

shown. What is the magnitude and direction

of the electric field intensity at point B?