Electric field Chapter 22

Week-2

Electric FieldsIn this chapter we will introduce the concept of an electric field. As long as charges are stationary Coulomb’s law described adequately the forces among charges. If the charges are not stationary we must use an alternative approach by introducing the electric field (symbol ). In connection with the electric field, the following topics will be covered:

-Calculate the electric field generated by a point charge.

-Using the principle of superposition determine the electric field created by a collection of point charges as well as continuous charge distributions.

-Once the electric field at a point P is known we will be calculate the electric force on any charge placed at P -Define the notion of an “electric dipole”. Determine the net force , the net torque, exerted on an electric dipole by a uniform electric field, as well as the dipole potential energy.

E

(22-1)

Fields Imagine an object is placed at a particular point in

space. When placed there, the object experiences a force F. We may not know WHY there is a force on the object,

although we usually will. Suppose further that if we double some property of

the object (mass, charge, …) then the force is found to double as well.

Then the object is said to be in a force field. The strength of the field (field strength) is defined as

the ratio of the force to the property that we are dealing with.

Electric Field

If a charge Q is in an electric field E then it will experience a force F.

The Electric Field is defined as the force per unit charge at the point.

Electric fields are caused by charges and consequently we can use Coulombs law to calculate it.

For multiple charges, add the fields as VECTORS.

Two Charges

unitunit rrF

E22

0

00

1

r

qk

r

qqk

qq

Doing it

Q

r

q

A Charge

The spot where we wantto know the Electric Field

unit

unit

r

Qk

q

r

qQk

rF

E

rF

2

2

F

General-

unitjj

jjj

unit

unit

r

Qk

q

General

r

Qk

q

r

qQk

,2

2

2

rF

EE

rF

E

rF

Force Field

q

qo

r

P

E

2

1

4 o

qE

r

Consider the positive charge shown in

the figure. At point P a distance from

we place the test charge . The force

exerted on by is equal to: o

o o

q

r

q q

q q

Electric field generated by a point charge

2

2 2

1

4

1 1

4 4

The magnitude of is a positive number

In terms of direction, points radially

as shown in then figure.

If q were a negative charge the magnitude

o

o

o

o

o o o o

q qF

r

q q qFE

q q r r

E

E

outwards

f remains the same. The direction

of points radially instead

E

E inwards

(22-4)

O O

The net electric electric field generated by a group of point charges is equal

to the vector sum of the electric field vectors ge

E

Electric field generated by a group of point charges. Superposition

1 2 3

1 2 3

1 1 3

1 2

nerated by each charge.

In the example shown in the figure

Here , , and are the electric field vectors

generated by , , and , respectively

Note , , an: d

E E E E

E E E

q q q

E E

1 2 3 1 2 3 1 3

3

2

must be added as vectors

, , x x x x y y y y z z z zE E E

E

E E E E E E E E E

(22-5)

What is the electric field at the center of the square array?

d

+ q

- q

-q

+q/2 +q/2

A system of two equal charges of opposite sign

placed at a distance is known as an "electric

dipole". For every electric dipole we associate

a vector known as "the electric dipol

q d

Electric Dipole

e moment"

(symbol )defined as follows:

The magnitude

The direction of is along the line that connects

the two charges and points from - to .

Many molecules have a built-in electric dip

p

p qd

p

q q

2

ole

moment. An example is the water molecule (H O)

The bonding between the O atom and the two H

atoms involves the sharing of 10 valence electrons

(8 from O and 1 from each H atom)

(22-6)

(22-7)

We will determine the electric field generated by the

electric dipole shown in the figure using the principle of

superposition. The positive charge ge

E

Electric field generated by an electric dipole

( ) 2

( ) 2

( ) ( )

2

nerates at P an electric

1field whose magnitude The negative charge

4

1creates an electric field with magnitude

4

The net electric field at P is:

1

4

o

o

o

qE

r

qE

r

E E E

q qE

r

2 22

2 2

2

2 3 3

1

4 / 2 / 2

1 1 We assume: 14 2 2 2

1 1 =4 22

1

o

o

o o o

q q

r z d z d

q d d dE

z z z z

q d d qdE

z z z z

p

z

21 1 2x x

<<

=

Kinds of continuously distributed charges

Line of charge = the charge per unit length. dq=ds (ds= differential of length along the line)

Area = charge per unit area dq=dA dA = dxdy (rectangular coordinates) dA= 2rdr for elemental ring of charge

Volume =charge per unit volume dq=dV dV=dxdydz or 4r2dr or some other expressions we will

look at later.

Continuous Charge Distribution

Symmetry

Let’s Do it Real Time

Concept – Charge perunit length

dq= ds

The math

)sin(2

)cos(2

)cos()2(

)cos()2(

0

0

0

02

02

0

0

0

r

kd

r

kE

r

rdkE

r

dqkE

E

rdds

x

x

x

y

Why?

CA

dq

Detemine the electric field generated at point P

by a uniformly charged ring of radius and total charge .

Point P lies on the normal to the ring plane that passes through

the ring cente

E

R q

Example :

2 2

r C, at a distance . Consider the charge element

of length and charge shown in the firgure. The distance

between the element and point P is

The charge generates at P an electric fiel

z

dS dq

r z R

dq

2

3/ 23 2 2

d of magnitude

and which points outwards along the line AP.

The z-component of is given by: 4

cos From triangle PACwe have: cos /

4 4

o

z

z zo o

dE

dqdE dE

r

dE dE z r

zdq zdqdE E

r z R

3/ 2 3/ 22 2 2 2

4 4

z

z

o o

dE

z zqE dq

z R z R

(22-9)

The Geometry (the disk)

Define surface charge density=charge/unit-area

dq=dA

dA=2rdr

(z2+r2)1/2

dq= dA = 2rdr

(z2+r2)1/2

R

z

z

rz

rdrzkE

rz

z

rz

drrk

rz

dqkdE

02/322

2/1222222

2

2)cos(

(z2+r2)1/2

Final Result

0z

220

2E

,R

12

When

Rz

zEz

q

q

negative ch

3. Electric field lines extend away from (where they originate)

and towards (where they terminate)

Example 1 : Electric field lines of a nega

a

t

rg

iv

positive

e point c

c

h

harge

arg

e

s

s

e - q

2

1

4 o

qE

r

-The electric field lines point towards the point charge

-The direction of the lines gives the direction of

-The density of the lines/unit area increases as the

distance from decreases.

E

qN

In the case of a positive point charge

the electric field lines have the same

form but they point

outw

ote :

ards

(22-11)

Electric field lines generated by

an electric dipole (a positive

and a negative point charge of

the same size but of opposite

sign)

Example 3.

Electric field lines generated by

two equal positive point charges

Example 4.

(22-13)

Look at the “Field Lines”

F+

F-

x-axis

Consider the electric dipole shown in the figure in

the presence of a uniform (constant magnitude and

direction) electric field

Forces and torques exerted on electric dipoles

by a uniform electric field

along the -axis

The electric field exerts a force on the

positive charge and a force on the

negatice charge. The net force on the dipole

0net

E x

F qE

F qE

F qE qE

The net torque generated by and about the dipole center is:

sin sin sin sin2 2

In vector form:

The electric dipole in a uniform electric field does not move

bu

F F

d dF F qEd pE

p E

t can rotate about its center p E

0netF

(22-14)

A

B

U

180˚

90 90

90

sin

sin cos

U d pE d

U pE d pE p E

Potential energy of an electric dipole

in a uniform electric field

p E

E

p

min

At point ( 0) has a minimum

value

It is a position of equilibriums

A

table

U

U pE

max

At point ( 180 ) has a maximum

value

It is a position of equilibriumunst le

B

ab

U

U pE

U p E

cosU pE

(22-15)

p

Ef

Fig.b

p

Ei

Fig.a Consider the electric dipole in Fig.a. It has an electric

dipole moment and is positioned so that is at an p p

Work done by an external agent to rotate an electric

dipole in a uniform electric field

i

f

angle

with respect to a uniform electric field

An external agent rotates the electric dipole and brings

it in its final position shown in Fig.b. In this position

is at an angle with respect to

E

p

The work W done by the external agent on the dipole

is equal to the difference between the initial and

final potential energy of the dipole

cos cos

cos cos

f i f i

i f

E

W U U pE pE

W pE

(22-16)

What did we learn in this chapter??

We introduced the concept of the Electric FIELDFIELD. We may not know what causes the field. (The

evil Emperor Ming) If we know where all the charges are we can

CALCULATE E. E is a VECTOR. The equation for E is the same as for the force

on a charge from Coulomb’s Law but divided by the “q of the test charge”.

What else did we learn in this chapter? We introduced continuous distributions of

charge rather than individual discrete charges.

Instead of adding the individual charges we must INTEGRATE the (dq)s.

There are three kinds of continuously distributed charges.

Kinds of continuously distributed charges

Line of charge = the charge per unit length. dq=ds (ds= differential of length along the line)

Area = charge per unit area dq=dA dA = dxdy (rectangular coordinates) dA= 2rdr for elemental ring of charge

Volume =charge per unit volume dq=dV dV=dxdydz or 4r2dr or some other expressions we will

look at later.

The Sphere

thk=dr

dq=dV= x surface area x thickness= 4r2 dr

dq

Summary

222

,2

2

2

)()()(

r

rdsk

r

rdAk

r

rdVk

r

Qk

q

General

r

Qk

q

r

qQk

unitjj

jjj

unit

unit

E

rF

EE

rF

E

rF

(Note: I left off the unit vectors in the lastequation set, but be aware that they should

be there.)

To be remembered …

If the ELECTRIC FIELD at a point is E, then E=F/q (This is the definition!)

Using some advanced mathematics we can derive from this equation, the fact that:

EF q