From last time(s)…

Oct. 4, 2007 1

From last time(s)…

Work, energy, and (electric) potential Electric potential and charge Electric potential and electric field.

Electric charges, forces, and fields Motion of charged particles in fields.

Today…

No honors lecture this week

Oct. 4, 2007 2

Forces, work, and energy Particle of mass m at rest Apply force to particle - what happens? Particle accelerates

Stop pushing - what happens? Particle moves at constant speed Particle has kinetic energy

Oct. 4, 2007 3

Work and energy Work-energy theorem:

Change in kinetic energy of isolated particle = work done

€

dW =r F ⋅d

r s = Fdscosθ

Total work

€

ΔK = dWstart

end

∫ =r F • d

r s

start

end

∫

Oct. 4, 2007 4

Electric forces, work, and energy

Consider bringing two positive charges together They repel each other

Pushing them together requires work Stop after some distance How much work was done?

+ +

Oct. 4, 2007 5

Calculating the work E.g. Keep Q2 fixed, push Q1 at constant velocity

Net force on Q1 ?

Force from hand on Q1 ?

+ +Q1 Q2

Zero

€

1

4πεo

Q1Q2

R2

Total work done by hand

€

rF • d

r s

start

end

∫ = ke

Q1Q2

R − x( )2 dx

xinitial

x final

∫

= ke

Q1Q2

R − x xinitial

x finalForce in direction of motion

R

+

xinitialxfinal

Oct. 4, 2007 6

Conservation of Energy

Work done by hand

€

=ke

Q1Q2

R − x xinitial

x final

= ke

Q1Q2

R − x final

− ke

Q1Q2

R − x initial

= ke

Q1Q2

rfinal

− ke

Q1Q2

rinitial

> 0

Where did this energy go?

Energy is stored in the electric field as electric potential energy

for pos charges

Oct. 4, 2007 7

Electric potential energy of two charges

Define electric potential energy U so that

€

Wexternal = ΔK + ΔUWork done on system

Change in kinetic energy

Change in electric potential energy

Works for a two-charge system if

€

U r( ) = ke

Q1Q2

r+ const

Define: potential energy at infinite separation = 0

€

U r( ) = ke

Q1Q2

rThen

Units of Joulesfor two charges

Oct. 4, 2007 8

Quick QuizTwo balls of equal mass and equal charge are held

fixed a distance R apart, then suddenly released. They fly away from each other, each ending up moving at some constant speed.

If the initial distance between them is reduced by a factor of four, their final speeds are

A. Two times bigger

B. Four times bigger

C. Two times smaller

D. Four times smaller

E. None of the above

Oct. 4, 2007 9

More About U of 2 Charges

Like charges U > 0 and work must be done to bring the charges together since they repel (W>0)

Unlike charges U < 0 and work is done to keep the charges apart since the attract one the other (W<0)

Oct. 4, 2007 10

Electric Potential Energy of single charge

Work done to move single charge near charge distribution. Other charges provide the force, q is charge of interest.

+++

+q

€

rF =

r F q1

+r F q2

+r F q3

q1

q2

q3

€

U =r F • d

r s ∫ =

r F q1

+r F q 2 +

r F q3( ) • d

r s ∫

= Uq1r1( ) + Uq 2 r2( ) + Uq 3 r3( )

= kq1q

r+ k

q2q

r+ k

q3q

rSuperposition of individual interactionsGeneralize to continuous charge distribution.

Oct. 4, 2007 11

Electric potential

Electric potential V usually created by some charge distribution.

V used to determine electric potential energy U of some other charge q

V has units of Joules / Coulomb = Volts

€

ΔU =r F Coulomb • d

r s ∫ = q

r E • d

r s ∫ = q

r E • d

r s ∫

Electric potential ΔU energy proportional to charge q

€

ΔU /q ≡ V = Electric potential

Oct. 4, 2007 12

Electric potential of point charge

Consider one charge as ‘creating’ electric potential, the other charge as ‘experiencing’ it

Q

q

€

Vq r( ) =UQq r( )

q= ke

Q

r

€

UQq r( ) = ke

Qq

r

Oct. 4, 2007 13

Electric Potential of point charge

Potential from a point charge Every point in space has a

numerical value for the electric potential

y

x

€

V =kQ

r

Distance from ‘source’ charge +Q

+Q

Oct. 4, 2007 14

Potential energy, forces, work

U=qoV Point B has greater potential

energy than point A Means that work must be done

to move the test charge qo

from A to B. This is exactly the work to

overcome the Coulomb repulsive force.

Ele

ctric

pot

entia

l ene

rgy=

qoV

A

B

qo > 0Work done = qoVB-qoVA =

€

− r

F Coulomb( ) • dr l

A

B

∫

Differential form:

€

qodV = −r F Coulomb • d

r l

Oct. 4, 2007 15

Quick QuizTwo points in space A and B have electric potential

VA=20 volts and VB=100 volts. How much work does it take to move a +100µC charge from A to B?

A. +2 mJ

B. -20 mJ

C. +8 mJ

D. +100 mJ

E. -100 mJ

Oct. 4, 2007 16

V(r) from multiple charges

Work done to move single charge near charge distribution. Other charges provide the force, q is charge of interest.

q

q1

q2

q3

€

U =r F • d

r s ∫ =

r F q1

+r F q 2 +

r F q3( ) • d

r s ∫

= Uq1r1( ) + Uq 2 r2( ) + Uq 3 r3( )

= kq1q

r+ k

q2q

r+ k

q3q

r

= q kq1

r+ k

q2

r+ k

q3

r

⎛

⎝ ⎜

⎞

⎠ ⎟

= q Vq1 r( ) + Vq2 r( ) + Vq 3 r( )( )Superposition of individual electric potentials

€

V r( ) = Vq1 r( ) + Vq 2 r( ) + Vq 3 r( )

Oct. 4, 2007 17

Quick Quiz 1 At what point is the electric potential zero for this

electric dipole?

+Q -Q

x=+ax=-aA

B

A. A

B. B

C. Both A and B

D. Neither of them

Oct. 4, 2007 18

Superposition: the dipole electric potential

+Q -Q

x=+ax=-aSuperposition of• potential from +Q• potential from -Q

Superposition of• potential from +Q• potential from -Q

+ =

V in plane

Oct. 4, 2007 19

Electric Potential and Field for a Continuous Charge Distribution

If symmetries do not allow an immediate application of the Gauss’ law to determine E often it is better to start from V!

Consider a small charge element dq The potential at some point due to this charge element is

To find the total potential, need to integrate over all the elements

This value for V uses the reference of V = 0 when P is infinitely far away from the charge distribution

Oct. 4, 2007 20

Quick QuizTwo points in space have electric potential VA=200V & VB=150V.

A particle of mass 0.01kg and charge 10-4C starts at point A with zero speed. A short time later it is at point B.

How fast is it moving?

A. 0.5 m/s

B. 5 m/s

C. 10 m/s

D. 1 m/s

E. 0.1 m/s

Oct. 4, 2007 21

E-field and electric potential If E-field known,

don’t need to know about charges creating it. E-field gives force From force, find work to move charge q

+++

+

€

rE

q

€

rF elec = q

r E

€

ΔU =r F me • d

r s ∫ = −q

r E • d

r s ∫ = −q

r E • d

r s ∫

€

ΔV = ΔU /q = −r E ⋅d

r s ∫Electric potential

Non-constant potential Non-zero E-field

Oct. 4, 2007 22

Potential of spherical conductor

Zero electric field in metal -> metal has constant potential Charge resides on surface,

so this is like the spherical charge shell. Found E = keQ / R2 in the radial direction. What is the electric potential of the conductor?

€

V R( ) −V ∞( ) = − E∞

R

∫ • ds

= ER

∞

∫ • ds

Integral along some path, from point on surface to inf.

difficult path

easy path

Easy because is same direction as E,

€

E • ds = E dr = Edr

Oct. 4, 2007 23

Electric potential of sphere

Conducting spheres connected by conducting wire.Same potential everywhere.

€

V R( ) −V ∞( ) = − E∞

R

∫ dr = kQ

r2R

∞

∫ dr

= −kQ

r R

∞

= kQ

R

So conducting sphere of radius R carrying charge Q is at a potential

€

kQ /R

R1R2

Q1 Q2

But not same everywhere

But not same everywhere

Oct. 4, 2007 24

Connected spheres

Since both must be at the same potential,

€

kQ1

R1

=kQ2

R2

⇒Q1

Q2

=R1

R2

Surface charge densities?

€

σ =Q

4πR2⇒

σ 1

σ 2

=R2

R1

Charge proportional to radius

Charge proportional to radius

Surface charge density proportional to 1/R

Surface charge density proportional to 1/R

Electric field?

Since

€

E =σ

2εo

Local E-field proportional to 1/R (1/radius of curvature)

Local E-field proportional to 1/R (1/radius of curvature)

Oct. 4, 2007 25

Varying E-fields on conductor

Expect larger electric fields near the small end. Can predict electric field proportional to local radius of curvature.

Large electric fields at sharp points, just like square Fields can be so strong that air is ionized and ions accelerated.

Oct. 4, 2007 26

Quick Quiz

Four electrons are added to a long wire. Which of the following will be the charge distribution?

A)

B)

C)

D)

Oct. 4, 2007 27

Conductors: other geometries Rectangular conductor (40

electrons) Edges are four lines Charge concentrates at

corners Equipotential lines closest

together at corners So potential changes faster

near corners. So electric field is larger at

corners.

Oct. 4, 2007 28

E-field and potential energy

Oct. 4, 2007 29

Zero

What is electric potential energy of isolated charge?

Oct. 4, 2007 30

The Electric Field

is the Electric Field It is independent of the test charge,

just like the electric potential It is a vector, with a magnitude and direction, When potential arises from other charges,

= Coulomb force per unit charge on a test charge due to interaction with the other charges.

€

qodV = −r F Coulomb • d

r l ⇒

dV =−

r F Coulomb

qo

⎛

⎝ ⎜

⎞

⎠ ⎟• d

r l

€

=− r

E • dr l

€

rE

€

rE

We’ll see later that E-fields in electromagnetic waves exist w/o charges!

Oct. 4, 2007 31

Electric field and potential

Electric field strength/direction shows how the potential changes in different directions

For example, Potential decreases in direction of local E field at rate Potential increases in direction opposite to local E-field at rate potential constant in direction perpendicular to local E-field

€

dV = −r E • d

r l Said before that

€

rE • d

r l = 0( )

€

∝ r

E

€

∝ r

E

Oct. 4, 2007 32

Potential from electric field

Electric field can be used to find changes in potential

Potential changes largest in direction of E-field.

Smallest (zero) perpendicular to E-field

€

dV = −r E • d

r l

€

dV = −r E • d

r l

€

dr l

€

rE

V=Vo

€

V = Vo −r E d

r l

€

V = Vo +r E d

r l

€

dr l

€

dr l

€

V = Vo

Oct. 4, 2007 33

Quick Quiz 3

Suppose the electric potential is constant everywhere. What is the electric field?

A) Positive

B) Negative

C) Zero

Oct. 4, 2007 34

Electric Potential - Uniform Field

Constant E-field corresponds to linearly increasing electric potential

The particle gains kinetic energy equal to the potential energy lost by the charge-field system

€

ΔV = −r E • d

r s ⇒ VB −VA = −

r E

A

B

∫ • dr s

€

=− r

E A

B

∫ dr s = −E d

r s

A

B

∫ = −Ex

€

rE || d

r s

€

rE || d

r s E cnstE cnst

A B

x

€

E = −ΔV

d= −

VB −VA

d

+

Oct. 4, 2007 35

Electric field from potential

Said before that Spell out the vectors:

This works for

€

dV = −r E • d

r l

€

dV = − Exdx + Eydy + E zdz( )

€

Ex = −dV

dx, Ey = −

dV

dy, E z = −

dV

dz

Usually written

€

r E = −

r ∇V = −

dV

dx,dV

dy,dV

dz

⎛

⎝ ⎜

⎞

⎠ ⎟

€

r E = −

r ∇V = −

dV

dx,dV

dy,dV

dz

⎛

⎝ ⎜

⎞

⎠ ⎟

Oct. 4, 2007 36

Equipotential lines

Lines of constant potential In 3D, surfaces of constant potential

Oct. 4, 2007 37

Electric Field and equipotential lines for + and - point charges

The E lines are directed away from the source charge

A positive test charge would be repelled away from the positive source charge

The E lines are directed toward the source charge

A positive test charge would be attracted toward the negative source charge

Blue dashed lines are equipotential

Oct. 4, 2007 38

Quick Quiz 1

C

C

C m

m

m

1. W = +19.8 mJ2. W = 0 mJ3. W = -19.8 mJ

Question: How much work would it take YOU to assemble 3 negative charges?

Likes repel, so YOU will still do positive work!

Oct. 4, 2007 39

Work done to assemble 3 charges

W1 = 0

1C

3C

2C m

m

m

• W2 = k q1 q2 /r

• W3 = k q1 q3/r + k q2 q3/r (9109)(110-6)(310-6)/5 + (9109)(210-6)(310-6)/5

=16.2 mJ

• W = +19.8 mJ• WE = -19.8 mJ• UE = +19.8 mJ

=(9 109)(1 10-6)(2 10-6)/5 =3.6 mJ

q3

q2q1

Similarly if they are all positive:

Oct. 4, 2007 40

Quick Quiz 2

Q

Q

Q m

m

m

1. positive2. zero3. negative

The total work required for YOU to assemble the set of charges as shown below is:

€

W1 = 0

W2 = kQ(−Q)

d

W3 = kQQ

d+ k

Q(−Q)

d

Total work = −kQ2

d

Oct. 4, 2007 41

Why ΔU/qo ?

Why is this a good thing?

ΔV=ΔU/qo is independent of the test charge qo

Only depends on the other charges. ΔV arises directly from these other charges, as described last time.

Last week’s example: electric dipole potential

-Q +Q

x=+ax=-a

Superposition of• potential from +Q• potential from -Q

Superposition of• potential from +Q• potential from -Q

Oct. 4, 2007 42

Dipole electric fields

Since most things are neutral, charge separation leads naturally to dipoles.

Can superpose electric fields from charges just as with potential

But E-field is a vector, -add vector components

+Q -Q

x=+ax=-a

Oct. 4, 2007 43

Quick Quiz 2

In this electric dipole, what is the direction of the electric field at point A?

A) Up

B) Left

C) Right

D) Zero +Q -Q

x=+ax=-a

A

Oct. 4, 2007 44

Dipole electric fields

+Q -Q

Note properties of E-field lines

Oct. 4, 2007 45

Conservative forces

Conservative Forces: the work done by the force is independent on the path and depends only on the starting and ending locations.

It is possible to define the potential energy U Wconservative = Δ U = Uinitial - Ufinal = = -

(Kfinal - Kinitial) = -ΔK

Fg

Oct. 4, 2007 46

Potential Energy of 2 charges Consider 2 positive charged particles. The electric force between them is

The work that an external agent should do to bring q2 at a distance rf from q1 starting from a very far away distance is equal and opposite to the work done by the electric force. Charges repel W>0!

r12

F

€

W = F ⋅dr∞

r12∫ = −WE

Oct. 4, 2007 47

Potential Energy of 2 charges

Since the 2 charges repel, the force on q2 due to q1

F12 is opposite to the direction of motion The external agent F = -F12 must do positive work!

W > 0 and the work of the electric force WE < 0

r12

Fdr

€

W = −keq1q2 −1

r

⎡ ⎣ ⎢

⎤ ⎦ ⎥

∞

r12

= ke

q1q2

r12

€

W = −WE = − F ⋅dr∞

r12

∫ = − Fdr∞

r12∫ = − ke

q1q2

r2dr

∞

r12∫

Oct. 4, 2007 48

Potential Energy of 2 charges

Since WE = - U = Uinitial - Ufinal = = -W W = U

We set Uinitial = U( ) = 0 since at infinite distance the force becomes null

The potential energy of the system is

Oct. 4, 2007 49

More than two charges?

Oct. 4, 2007 50

U with Multiple Charges

If there are more than two charges, then find U for each pair of charges and add them

For three charges: