The Electric Field

January 13, 2006

Calendar

Quiz Today Continuation with Coulomb’s Law and

the concept of the Electric Field MLK Holiday on Monday Next week we continue with same

topics – see the schedule on the website

Quiz on Friday

Last Time We found two kinds of charge. Like charges repel, unlike charges

attract. Discussed induction … a bit Fundamental unit of charge is the

COULOMB. Coulomb’s Law Assignment: read text about

induction.

One example of induction

Polarize

Ground

Remove Ground

Positive !

Coulomb’s Law

229

0

221

0

/1094

1

4

1

CNmxk

r

qqunit

rF

The Unit of Charge is calledTHE COULOMB

Smallest Charge: e ( a positive number) 1.6 x 10-19 Coul.

electron charge = -eProton charge = +e

What would be the magnitude of the electrostatic force between two 0.500 C charges separated by the following distances, if such point charges existed (they do not) and this configuration could be set up?

(a) 1.25 m__________N

(b) 1.25 km __________N

Three point charges are located at the corners of an equilateral triangle as shown in Figure P23.7. Calculate the resultant electric force on the 7.00-μC charge.

Two small beads having positive charges 3q and q are fixed at the opposite ends of a horizontal, insulating rod, extending from the origin to the point x = d. As shown in Figure P23.10, a third small charged bead is free to slide on the rod. At what position is the third bead in equilibrium? Can it be in stable equilibrium?

This is WAR

• You are fighting the enemy on the planet Mongo.

• The evil emperor Ming’s forces are behind a strange green haze.

• You aim your blaster and fire … but ……

Ming themerciless

this guy is

MEAN!

Nothing Happens! The Green thing is a Force Field!

The Force may not be with you ….

Side View

TheFORCE FIELD

Force

Positiono

|Force| Big!

Properties of a FORCE FIELD

It is a property of the position in space.

There is a cause but that cause may not be known.

The force on an object is usually proportional to some property of an object which is placed into the field.

EXAMPLE: The Gravitational Field That We Live In.

m Mmg

Mg

Mysterious Force

F

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 itQ

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

Two ChargesWhat is the Electric Field at Point P?

The two S’s

•Superposition

•Symmetry

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

Kinds of continuously distributed charges Line of charge

or sometimes = 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

ymmetry

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?

A Harder Problem

A line of charge=charge/length

setupsetup

dx

L

r

x

dE dEy

2/

02/322

2/

02/322

22

2

2

22

)(2

)(2

)()cos(

)(

)cos(

L

x

L

x

L

Lx

xr

dxkrE

xr

dxrkE

xr

r

xr

dxkE

(standard integral)

Completing the Math

r

kL

r

kLE

Lr

L

Lrr

kLE

x

x

2

2

4

:line long VERY a oflimit In the

4

:nintegratio theDoing

22

22

1/r dependence

Dare we project this??

Point Charge goes as 1/r2

Infinite line of charge goes as 1/r1

Could it be possible that the field of an infinite plane of charge could go as 1/r0? A constant??

The Geometry

Define surface charge density=charge/unit-area

dq=dA

dA=2rdr

(z2+r2)1/2

dq= x 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

Look at the “Field Lines”

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 or sometimes = 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

dqr

thk=dr

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

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