MTE1 results

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Mean 75% = 90/120

From last times

Electric charges and forceElectric Field and ways to calculate itMotion of charges in E-fieldGauss’ Law

Today:

More on Gauss’ law and conductors in electrostatic equilibrium

Work, potential energy Electric potential

PHYS208, SPRING 2008

What is the Electric Flux?

(Component of E-field ⊥ to surface) x (surface area)

So you need to understand how the E-field is directed respect to each piece of surface through which you have to calculate it

Why do we need to calculate it?to use Gauss’ law to determine E

PHYS208, SPRING 2008

How to calculate Electric Flux?

Electric flux through a surface: (component of E-field ⊥ to surface) x (surface area)

!

"E

!

ˆ s

Ecos! is component of E-fieldperpendicular to surface

PHYS208, SPRING 2008

GAUSS’ LAW for any closed surface

Flux thru closed surface depends ONLY on the charge enclosed by surface!!

But you need to calculate the Flux to determine EThe Flux depends on the normal component of E that

crosses the surfacebecause of the scalar product in the surface integral

PHYS208, SPRING 2008

Quiz 1

Rank fluxes through the blue dashed spherical surfaces for the 3 cases:

+Q

+Q

+Q+2Q

2)3)

1)

A) 1)<2)=3)B) 3)<1)<2)C) 1) = 3) < 2)D) 1) < 2) < 3)

!

Qr3

a3"0

<Q

"0

<2Q

"0

Flux thru spheres:

!

4"r2E(r)

PHYS208, SPRING 2008

Conductors in equilibrium electrostatic equilibrium

In a conductor in electrostatic equilibrium there is no net motion of charge

Property 1: E=0 everywhere inside the conductor

Conductor slab in an external field E:

• if E-field not null inside the conductor, free electrons would be accelerated

• These electrons would not be in equilibrium.

• When the external field is applied, the electrons redistribute until the magnitude of the internal field equals the magnitude of the external field

•The total field inside the conductor is zero

Etot = E+Ein= 0

Ein

Etot =0

PHYS208, SPRING 2008

Property 2: Charge on only conductor surface

• chose a gaussian surface inside the conductor (as close to the surface as desired)

•There is no net flux through the gaussian surface (since E=0)

•Any net charge must reside on the surface (cannot be inside!)

E=0

PHYS208, SPRING 2008

Magnitude and direction of E-Field

Field lines

E-field perpendicular to the surface:parallel component to E would put force on chargescharges would accelerate along the surface NO equilibrium

Applying Gauss’s law

Remember from P207

• 1st Newton’s law: An object at rest will remain at rest and a moving object will continue to move at constant velocity on a straight line if F=0.

• 2nd Newton’s law(superposition principle applies):

• A moving particle has kinetic energy

• Work:

• Total work and work-energy theorem:

!

r F

net= m

r a

!

F

ds

>0: Force is in direction of motion

<0: Force is opposite to direction of motion

=0: Force is perpendicular to direction of motion

!

dW =r F " d

r s = Fdscos#

Infinitesimal displacement

!

W = F " dsA

B

# = KB$K

A

!

K=1

2mv

2

Potential energy

Fg = mg

F

yf

yi

Work to lift the block done by external agent (F): W = mgh = mg(yf-yi) > 0 (the agent has to make an effort to move the block)Work done by gravitational field: Wg = -mgh

y

h

Conservative Forces: work done by the force is independent on the path and depends only on the starting and ending locations (if the initial and final point coincide => W = 0)

It is possible to define the potential energy U

Wconservative = -! U = Uinitial - Ufinal = !K

"Ug = Ug - U0 = -mghUg = U0 + mgy for gravitational force and U0 =U(y=0) = 0

Electric force workConsider bringing two positive charges together

They repel each other Force is conservative

Pushing them together requires work

Stop after some distance

How much work was done?

+ +

!

r F " d

r x =

xinitial

x final

# ke

Q1Q2

R $ x( )2

dx =xinitial

x final

# ke

Q1Q2

R $ x xinitial

x final

=

= ke

Q1Q2

R $ x final

$ ke

Q1Q2

R $ xinitial

== ke

Q1Q2

rfinal

$ ke

Q1Q2

rinitial

> 0

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

! Net force on Q1 ?

! Force from hand on Q1 ?

Zero

!

1

4"#o

Q1Q2

R2

! Total work done by hand

Force in

direction of motion

+ +

Q1 Q2R

xinitial xfinal

+

rfinalrinitial

!

rfinal < rinitial "1

rfinal>

1

rinitial for like charges

!

r F

!

r F

Coulomb

x

Potential energy of 2 point charges

The external agent makes positive work (because charges repel) and changes the potential energy of the system

Wext = !U = Ufinal - Uinitial >0 => U increases.

The energy is stored in the electric field as electric potential energy.

We can set: Uinitial = U(!) = 0 (at infinite distance force becomes null). Hence, electric potential energy of 2 charges at distance r:

!

Wext = keQ1Q2

rfinal" ke

Q1Q2

rinitial= ke

Q1Q2

r12

Q1

Q2

If we initially put Q1 very far away (rinitial = !) and move Q1 at distance rfinal =r12

from Q2:

!

Uelec = keQ1Q2

r

More about U of 2 charges

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

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

Quick QuizTwo balls of equal mass and equal charge are held fixed a distance r12 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

!

Wel = "#U =Uinitial "Ufinal = keQ1Q2

r12

!

Wel = "K = K final #Kinitial = K final =1

2mv

2= ke

Q1Q2

r12

=0 very far away

=0 initially fixed

if r’12 = 1/4 r12 =>K’final = 4 Kfinal

and v’final = 2 vfinal

U with multiple charges! If there are more than two

charges, then find U foreach pair of charges andadd them

! For three charges:

Quick QuizHow much work would it take YOU to assemble 3 negative charges?

A. W = +19.8 mJ

B. W = 0 mJ

C. W = -19.8 mJ

!1µC

!3µC

!2µC5 m

5 m5 m

Electric Potential

!

"U /q #V = Electric potential

Work done by E-field produced by some source charge on a test charge q to move it along a path:

!

Wel = "#U =r F Coulomb $ d

r s = q

r E $%% d

r s = q

r E $ d

r s %

Define:

V scalar quantity and units volts 1 V = 1 J/C V created by some source charge or charge distribution and

independent on test charge V can be used to determine potential energy of charge q and

source charge system V connected to E due to source charge (more in this week

laboratory)

Equipotential lines

Lines of constant potentialIn 3D, surfaces of constant potential(your Lab)

Quick QuizIn the Figure, q1 is a negative source charge and q2 is a test charge. If q2 is initially positive and then is changed to a negative charge of the same magnitude, the potential at the position of q2 due to q1

(A.) increases (B.) decreases (C.) remains the same

+ ! -

Answer: (c). The potential is established only by the source

charge and is independent of the test charge.

Electric potential at point charge

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

q

Q

!

UQq r( ) = keQq

r

!

Vq r( ) =UQq r( )q

= keQ

r

Electric Potential of point charge

Ele

ctr

ic p

ote

ntial energ

y=qoV

A

B

qo > 0

Every point in space has a numerical value for the electric potential

Potential energy: U = q0V (q0 test charge)VB > VA

Means that work must be done to move the test charge q0 from A to B to overcome the Coulomb repulsive force.

y

x+Q

!

V =kQ

r

Distance from ‘source’ charge +Q

Work done = qoV

B-q

oV

A =

!

"r F

Coulomb( ) • dr l

A

B

#

Differential form:

!

qodV = "r F Coulomb • d

r l

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 +100"C charge from A to B?

A. +2 mJ

B. -20 mJ

C. +8 mJ

D. +100 mJ

E. -100 mJ

Superposition: electric potential of dipole

+Q -Q

x=+ax=-a

Superposition of

• potential from +Q

• potential from -Q

+ =

y

Eg: V(y=0) = kQ/a-kQ/a=0In general: for a group of point charges

!

V = keqi

rii

"

Electric Potential of a continuous charge distribution

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

Total potential:

This value of V used the reference that V = 0 at infinite distance

Van de Graaff generator

!V

• The high-voltage electrode is a hollow metal dome mounted on an insulated column

• Charge is delivered continuously to dome by a moving belt of insulating material

•Large potentials can be developed by repeated trips of the belt

•Protons can be accelerated through such large potentials

•Sparks: electrons from rod to dome excite air molecules that emit light

!

"U = q"V = #"K = K f