Physics 1502: Lecture 2Today’s Agenda

• Announcements:– Lectures posted on: www.phys.uconn.edu/~rcote/– HW assignments, solutions etc.

• Homework #1:Homework #1:– On Masterphysics this FridayOn Masterphysics this Friday

• Homeworks posted on Masteringphysics – You need to register (included in cost of book)– Go to masteringphysics.com and register– Course ID: MPCOTE33308

• Labs: Begin in two weeks

Today’s Topic :

• End of Chapter 20– Define Electric Field in terms of force on "test charge"

– Electric Field Lines

– Example Calculations

– Continuous charge distributions => integrate

– Moving charges: Use Newton’s law

• Demonstration of Mastering Physics

Coulomb's Law

SI Units:

• r in meters

• q in Coulombs

• F in Newtons

F12= 1

4q1q2

r2

14

= 8.987 109 N m2/C2

r

q2

r F1

2

q1

F21

r

Charles Coulomb (1736-1806)

Electric Fields

Introducing the Electric Field:

a quantity, which is independent of that charge q, and depends only upon its position relative to the collection of charges.

The force, F, on any charge q due to some collection of charges is always proportional to q:

A FIELD is something that can be defined anywhere in space

it can be a scalar field (e.g., a Temperature Field)

it can be a vector field (as we have for the Electric Field)

Lecture 2, ACT 1

• Two charges, Q1 and Q2 , fixed along the x-axis as shown, produce an electric field E at a point (x,y) = (0,d) which is directed along the negative y-axis.

– Which of the following statements is true?

(a) Both charges Q1 and Q2 must be positive.

Q2Q1 x

y

Ed

(b) Both charges Q1 and Q2 must be negative.

(c) The charges Q1 and Q2 must have opposite signs.

How Can We Visualize the E Field?

• Vector Maps:

arrow length indicates vector magnitude

+ chg

+O

• Graphs:

Ex, Ey, Ez as a function of (x, y, z)Er, E, E as a function of (r, , )

x

Ex

Example• Consider a point charge fixed at the origin of

a co-ordinate system as shown.

– The following graphs represent the functional dependence of the Electric Field.

Q x

y

r

r0

Er

0 2

Er

• As the distance from the charge increases, the field falls off as 1/r2.• At fixed r, the radial component of the field is a constant, independent of !!

E

Lecture 2, ACT 2

x

• Consider a point charge fixed at the origin of a co-ordinate system as shown.

– Which of the following graphs best represents the functional dependence of the Electric Field at the point (r,)?

Q

y

r

Fixedr>0

0 2

Ex

0 2

Ex

0 2

Ex

Another Way to Visualize E ...• The Old Way:

Vector Maps

• Lines leave positive charges and return to negative charges

• Number of lines leaving/entering charge = amount of charge

• Tangent of line = direction of E

• Density of lines = magnitude of E

+O

• A New Way: Electric Field Lines

+ chg - chg

+O O

x

y

a

a

+Q

-Q r

Electric Dipole

EE

Symmetry

Ex = ?? Ey = ??

Calculate for a pt along x-axis: (x,0)

What is the Electric Field generated by this charge arrangement?

Electric Dipole: Field Lines• Lines leave positive charge

and return to negative charge

• Ex(x,0) = 0

What can we observe about E?

• Ex(0,y) = 0

• Field largest in space betweenthe two charges

• We derived:

... for r >> a,

Field Lines from 2 Like Charges• Note the field lines from 2

like charges are quite different from the field lines of 2 opposite charges (the electric dipole)

• There is a zero halfway between charges

• r>>a: looks like field of point charge (+2q) at origin.

Lecture 2, ACT 3

• Consider a dipole aligned with the y-axis as shown.

– Which of the following statements about Ex(2a,a) is true?

+Q

x

y

a

a-Q

a

2a

(a) Ex(2a,a) < 0 (b) Ex(2a,a) = 0 (c) Ex(2a,a) > 0

q1 q2

Electric Dipole

x

y

a

a

+Q

-Q

Coulomb ForceRadial

E y

Q

y a y ay 0

41 1

02 2

,

E yQ ay

ya

y

y 04

4

10 4

2

2

2,

E

Now calculate for a pt along y-axis: (0,y)

Ex = ?? Ey = ??

What is the Electric Field generated by this charge arrangement?

€

Ex 0, y( ) = 0

Electric Dipole

x

y

a

+Q

-Q

Case of special interest:

(antennas, molecules)

r > > a

r

For pts along x-axis: For pts along y-axis:

E rQa

ry ,0 2

14 0

3

For r >>a, For r >>a,

E rQa

ry 0 4

14 0

3,

a

E rQ ar

ra

r

y 04

4

10 4

2

2

2,

€

Ex 0,r( ) = 0

€

Ey r,0( ) = −21

4πε0

Qa

r2 + a2( )

3/2

€

Ex r,0( ) = 0

Electric Dipole Summary

x

y

a

a

+Q

-Q

Case of special interest:

(antennas, molecules)

r > > a

r

• Along y-axis

E rQa

ry 0 4

14 0

3,

€

Ex 0,r( ) = 0

• Along x-axis

E rQa

ry ,0 2

14 0

3

€

Ex r,0( ) = 0

• Along arbitrary angle

€

E∝Qa

E∝1

r 3

dipole moment

with

Electric Fields fromContinuous Charge

Distributions• Principles (Coulomb's Law + Law of Superposition) remain

the same.

Only change:

 

  

See Examples in text

P

rq

E

r

∫∑

→V

ei

i

i

i

qe

e

r

dqk

r

qk

r

qk

rrE

rE

ˆˆlim

ˆ

220

2

r

r

Charge Densities• How do we represent the charge “Q” on an extended object?

total chargeQ

small piecesof charge

dq

• Surface of charge:

= charge per unit area

dq = dA

• Line of charge:

= charge per unit length

dq = dx

• Volume of charge:

= charge per unit volume

dq = dV

++++++++++++++++++++++++++

rE(r) = ?

Example: Infinite line of charge

How do we approach this calculation?

++++++++++++++++++++++++++

rE(r) = ?

In words:

“add up the electric field contribution from each bit of charge, using superposition of the results to get the final field”

In practice:

• Use Coulomb’s Law to find the E field per segment of charge

• Plan to integrate along the line…

• x: from to OR : from to

+++++++++++++++

• Any symmetries ? This may help for easy cancellations.

Infinite Line of Charge

++++++++++++++++ x

y

dx

r'r

dE

Charge density =

We need to add up the E field contributions from all segments dx along the line.

Infinite Line of Charge

++++++++++++++++ x

y

dx

r'r

dEWe use Coulomb’s Law to find dE:

But x and are not independent!

x = r tandx = r sec2 d

And what is r’ in terms of r?

But what is dq in terms of dx?

€

′ r =r

cosθ€

dq = λ dx€

dE =1

4πε0

dq

′ r 2

Therefore,

€

dE =1

4πε0

λdx

r / cosθ( )2

€

dE =1

4πε0

λ cos2 θdx

r2

€

dE =1

4πε0

λdθ

r

Infinite Line of Charge• Components:

++++++++++++++++ x

y

dx

r'r

dE

• Integrate:

E dE drx x∫ ∫

14 02

2

sin

/

/

E dE dry y∫ ∫

14 02

2

cos

/

/

Ex

€

dEx = −1

4πε0

λdθ

rsinθ

Ey

€

dEy = +1

4πε0

λdθ

rcosθ

Infinite Line of Charge• Solution:

• Conclusion:

The Electric Field produced by an infinite line of charge is:– everywhere perpendicular to the line

– is proportional to the charge density

– decreases as 1/r.

++++++++++++++++ x

y

dx

r'r

dE

0sin2/

2/∫

d

2cos2/

2/

d

€

Ex = 0

Ey =1

4πε0

2λ

r

Lecture 2, ACT 4• Consider a circular ring with a uniform charge

distribution ( charge per unit length) as shown. The total charge of this ring is +Q.

• The electric field at the origin is

(a) zero

R

x

y

+++

++ +

+++

++

+ + + + ++

++++

+

(b)

€

1

4πε0

2πλ

R(c)

€

1

4πε0

πRλ

R2

SummaryElectric Field Distibutions

Dipole ~ 1 / R3

Point Charge ~ 1 / R2

Infinite Line of Charge ~ 1 / R

Motion of Charged Particles in

Electric Fields • Remember our definition of the Electric Field,

• And remembering Physics 1501,

Now consider particles moving in fields. Note that for a charge moving in a constant field

this is just like a particle moving near the earth’s surface.

ax = 0 ay = constant

vx = vox vy = voy + at

x = xo + voxt y = yo + voyt + ½ at2

€

rF = q

r E

€

rF = m

r a ⇒

r a =

qr E

m

Motion of Charged Particles in

Electric Fields • Consider the following set up,

++++++++++++++++++++++++++

- - - - - - - - - - - - - - - - - - - - - - - - - - e-

For an electron beginning at rest at the bottom plate, what will be its speed when it crashes into the top plate?

Spacing = 10 cm, E = 100 N/C, e = 1.6 x 10-19 C, m = 9.1 x 10-31 kg

Motion of Charged Particles in

Electric Fields ++++++++++++++++++++++++++

- - - - - - - - - - - - - - - - - - - - - - - - - - e-

vo = 0, yo = 0

vf2 – vo

2 = 2ax

Or,

€

v f2 = 2

qE

m

⎛

⎝ ⎜

⎞

⎠ ⎟Δx

€

v f2 = 2

1.6x10−19C( ) 100N /C( )

9.1x10−31kg

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥0.1m( )

€

v f =1.9x106 m / s

• Homework #1 on Mastering Physics– From Chapter 20

Recap of today’s lecture

• Define Electric Field in terms of force on "test charge"

• Electric Field Lines

• Example Calculations

• Continuous charge distributions => integrate

• Moving charges: Use Newton’s law