Physics 122 September 7, 2010Thi i h d!This course is hard!Don’t expect to cram all the material before exams.

k h h l d It is important to keep up with the material, and not slip behind. You should definitely look at the homework problems before coming to class.m w p m f m g .Labs start next week (Sept. 13-15).You should read the lab materials before coming to You should read the lab materials before coming to lab, and take the pre-lab quiz. Lab writeup and pre-lab-quiz will be on your section of PHY124 in Bl kb d s s s s ft lit h is s l dBlackboard as soon as a software glitch is resolved.There will generally be clicker questions each Thurday based on that week’s labbased on that week s lab.Question 1

Here is how the WebAssign gradebook looked yesterday at g y y~1:30 PM.

ND = Not Downloaded Students ND = Not Downloaded. Students who haven’t even looked at the problems.

NS = No Submissions.Students who downloaded, but haven’t submitted any of the problems for grading.

Original policy of 10% credit subtracted for wrong answers h b l d Y h 10 has been canceled. You have 10 “free” tries for each problem

As of 3:00 PM today, more than 80% of you have not yet downloaded the Chapter 16 homework problems Not good!downloaded the Chapter 16 homework problems. Not good!

In the first homework assignment, 10% credit was lost for each incorrect answer We have changed that so there is each incorrect answer. We have changed that so there is no penalty for incorrect answers.

1 2

Question #2.

Charges 1 and 2 have 1 2what signs?

Answer: 1 is +, 2 is -

Question 3: If there are two large planes of charge, negative g p g , gon the left and positive on the right, for an observer between the sheets of charge, what is the direction of the electric field?the direction of the electric field?-- +-- +-- +-- +-- +

↑

P-- +-- +-- +

← P →

↓ Answer: field +-- +-- +

points to the left

-- +-- +

Conductor: a material in which charges are free to move.*

Put a charge next to a conductor: How do the field lines look?

?? Question 4:

What’s wrong with this picture?

Electric field lines meet a conductor perpendicular to the perpendicular to the surface.

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WRONG. Charges on the d ld

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conductor would move in response to the field.

If the conductor carries a net negative If the conductor carries carries a net negative charge zero net charge

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Remember some concepts about potential energy from Ph i 121 I thi th i it ti l fi ld Physics 121. In this room, there is a gravitational field that exerts a force F = mg (vector) pointing downward on an object with mass m. j

The amount of work required to carry the bj t f A t B i

B

F

object from A to B isW = m g (hB-hA),independent

F = m gp

of the path.

hBThere is a (scalar) gravitational B

AThere is a (scalar) gravitational potential energy U = m g h.

Only differences in potential hAOnly differences in potential energy are meaningful.

If you carry a particle with charge +q around in a place where there is a uniform electric field E, say pointing in the x direction the charge feels a force F = q e in the x the x direction, the charge feels a force F = q e in the x direction.

EE

xBB

E

If the particle moves from A to B, the amount of work done by the field is

Edone by the field isW = E q (xB-xA),independent of the path.x

ExA

AF = e E

Neglecting gravity, the particle has a E g g g y ppotential energy Uelectric = –q E x.

E

If you carry a particle with charge -q around in a place where there is a uniform electric field E, say pointing in the x direction the charge feels a force F = -q e in the x the x direction, the charge feels a force F = -q e in the x direction.

EE

xBB

ENote sign

If the particle moves from A to B, the amount of work done by the field is

Edone by the field isW = q E (xA-xB) = q E Δx,independent of the path.x

ExA

AF = e E

Neglecting gravity, the particle has a E g g g y ppotential energy Uelectric = +q E x.

E

Note sign

D fi l t i t ti l

Electric Potential energy of a charge q

Define electric potential = gy g q

qUnits: Joule / Coulomb,

hi h ill i V lt*

E

which we will give a new name, Volt*

Higher potential Lower potential f henergy for + charge

Lower potential f h

energy for + chargeHigher potential energy for – charge

EEEenergy for – charge

Higher potential

energy for – chargeLower potentialE

E

El t i fi ld li s i t f m i s Electric field lines point from regions of higher potential to lower potential.

* Alessandro Volta (1745-1827), inventor of the first battery.

Homework problem from Chapter 15:Each of the protons in a particle beam has a kinetic p penergy of 3.25x10-15 J. What is the magnitude and direction of an electric field that will stop these protons in a distance of 1 25 m?in a distance of 1.25 m?

KEi + PEi = KEf + PEfΔPE KE KE 3 25 10 15 JΔPE = KEi – KEf = 3.25x10-15 J

= q E Δx = 1.6x10-19C x E x 1.25m

E = 3.25x10-15 J / (1.6x10-19C x 1.25m) = 1.63x104 N/C.

(J = Nm so J/m = N and the dimensions are OK )(J = Nm, so J/m = N, and the dimensions are OK.)

Higher potential energy at the point where the proton t th fi ld i t t d th f th stops, so the field points towards the source of the

proton.

Question 5:QThere is a uniform electric field E of magnitude 150 N/C, pointing South. An electron is released from rest, and starts to move in response to the field What is its and starts to move in response to the field. What is its kinetic energy when it has travelled 30 cm?

There is a uniform electric field E of magnitude 150 gN/C, pointing South. An electron is released from rest, and starts to move in response to the field. What is its kinetic energy when it has travelled 30 cm?kinetic energy when it has travelled 30 cm?

As before, ΔKE = ─ΔPE = ─q E (xf – xi) 1 6 10 19C 150 N/C 0 3 7 2 10 18 J= 1.6x10-19C x 150 N/C x 0.3m = 7.2x10-18 J

Which direction does it move?

Whichever way it goes, it is moving from higher to lower potential energy so ΔPE is negativepotential energy, so ΔPE is negative.

If the particle were a proton, what would be the t ?outcome?

A BATTERY is a device that produces a fixed potential difference between two conductors.p

This terminal is at an electric potential 1.5 pVolts more positive than this terminal.

Thi i l i l i This terminal is at an electric potential 12 Volts more positive than this terminal.p

Q sti 6:Question 6:If I carry a proton from the – to the + terminal of a 12 Volt battery, how does its potential energy change?

If I carry a proton from the – to the + terminal of a 12 Volt battery how does its potential energy change?Volt battery, how does its potential energy change?

Proton has a positive charge, so the change in potential energy in going to a more positive potential is positiveenergy in going to a more positive potential is positive.

Change in potential energy = charge x change in potential.

ΔU = q ΔV = 1.6x10-19 Coul. x 12 Volt = 1.92x10-18 Joule(Volt = J/C, so Coulomb x Volt = Joule) (Volt J/C, so Coulomb x Volt Joule)

A ski lift raises the potential energy of its passenger by m g Δh.

A battery raises the potential energy of y p gycharge that flows through it by q ΔV.

We have expressed electric field in units of N/C (force per unit charge). Consider the uniform electric ( p g )field between two large conducting planes.

The work done by the field in V2V1

y fmoving a charge q from V1 to V2 is q E Δx.

EThe change in potential energy of the charge is –q E Δx which i l (V V ) is also q (V2 – V1).

So V2 < V1 as drawn, and

Δx

2 1 ,E = (V1-V2)/Δx.

Units: Units: Newton/Coulomb = Volt/meter

Gravitational Potential Energy (Phy 121) of an isolated massgy ( y )

F = G M m / R 2 is the magnitude of the force between two objects F is a vectorobjects. F is a vector.

U = ─G M m / R is the gravitational potential energy. Work d h ll M d t t th

Note sign

you do when you allow masses M and m to come together from a great distance. There is no vector U.

Work you do to move a mass m from position x1 to x2 isU2 ─U1 (means U(x2) – U(x1)).

In the uniform gravitational potential near Earth’s surface, U = m g h where h is distance measured up from some

f i treference point.

Electric potential

E = ke Q / R 2 is the magnitude of the electric field due to a charge Q. E is a vector.a charge Q. E is a vector.

V = ke Q / R is the electric potential at a distance R from an isolated (point) charge Q There is no vector Van isolated (point) charge Q. There is no vector V.

Work you do to move a charge q from position x1 to x2 is q (V2 - V1) (means V(x2) – V(x1)).

V (x) is a function of position x, even if there is a V (x) s a funct on of pos t on x, even f there s a complicated arrangement of charges.

V (x) becomes more positive near positive charges more V (x) becomes more positive near positive charges, more negative near negative charges.

Electric potential as a function of distance from a 1μC chargedistance from a 1μC charge.

90,000 Volts at 0.1 meter

9000 Volts at 1 meter

900 Volts at 10 meters

Potential drops to zero very far from the charge.

Potential energy of two +1μC charges as a function of their separation. U = q V (r) = ke q1 q2 / r 2.

0.09 Joules at 0.1 meter

Remember that the zero of potential energy can be chosen

bit il It i ti l t arbitrarily. It is conventional to chose the zero at great separation.

Positive potential energy – you have to do work in order to bring those two charges together.

Potential energy of a +1μC charge and a -1μC charge as a function of their separationcharge, as a function of their separation.

Distance (meters)

The force is attractive, so you could get some work out of letting th h t ththe charges come together.

Question 7:T h f 1 2 10 6 C d 3 5 10 6 C i iti ll Two charges of 1.2x10-6 C and 3.5x10-6 C are initially separated by 1 meter. How much work do I have to do to move them to a separation of 0.3 meter?p

(Hint. Think of one of them as fixed, and find the difference of potential energy when I move the difference of potential energy when I move the other one closer.)

Two charges of 1.2x10-6 C and 3.5x10-6 C are initially t d b 1 t H h k d I h t d separated by 1 meter. How much work do I have to do

to move them to a separation of 0.3 meter?

Think of the 1.2x10-6 C charge as fixed.

The potential a distance of 1 m away isThe potential a distance of 1 m away is9x109 Nm2C-2 x 1.2x10-6 C / 1 m = 10.8 kV0.3 m away: 9 109 N 2C 2 1 2 10 6 C / 0 3 36 0 kV9x109 Nm2C-2 x 1.2x10-6 C / 0.3 m = 36.0 kV

Moving a charge of 3.5x10-6 C from a point of 10.8 kV g g ppotential to a point of 36.0 kV potential means increasing its potential by 25.2 kV, which raises its potential energy by 3 5x10-6 C x 25 200 V = 0 088 Jpotential energy by 3.5x10 C x 25,200 V 0.088 J.

Potential due to several charges –add potentials of each chargeadd potentials of each charge

P l P h f h l Potential at point P is the sum of the potentials produced by q1 and q2. q1 is +5μC, 4 m away from the observation point, q2 is -2μC, 5 m away.p , q2 μ , m y

6 691 2 5 10 2 10( ) 9 10 7650 V

4 5e eq qV P k k

− −⎛ ⎞× ×= + = × − =⎜ ⎟

1 2

( )4 5e er r ⎜ ⎟

⎝ ⎠

10 cmQuestion 8:

3 1 C

QThree charges of +1x10-6 C are at the corners of a square, 10 cm on a side What is the 3 ea. +1 μCcm on a side. What is the potential at the fourth corner?

10 cmThree charges of +1x10-6 C are

3 1 C

gat the corners of a square, 10 cm on a side. What is the potential at the fourth corner? 3 ea. +1 μCpotential at the fourth corner?

Two of the charges are a distance 0.10 m, and the third is a distance 0.1414 m (0.10 x √2).

31 2 qq qV k⎛ ⎞

= + +⎜ ⎟⎜ ⎟1 2 3

6 6 69 1 10 1 10 1 109 10

0 1 0 1 0 1414

eV kr r r

− − −

+ +⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞× × ×= × + +⎜ ⎟

⎝ ⎠5

0.1 0.1 0.14142.44 10 V

⎝ ⎠= ×

For any distribution of charge the

- - --

For any distribution of charge, the potential V(x) can be found by adding contribution from each h

+++- -

charge.

++ +

+ x

One special case:

A spherically symmetric charge distribution produces V (r) = ke Q / rwhere r is measured from the center where r is measured from the center of the sphere, and Q is the total charge on (in) the sphere. It looks l k ll f h h d like all of the charge is concentrated at the center.

True for solid sphere or spherical shell.

You already knew this because the field is the same outside a sphere as sam outs a sph r as for a point charge at center.

What is the relation between Electric Field lines and Equipotentials?Equipotentials?

Recall work done by a force which moves an object:

W F dW = F d cosθ

Replace work W by change of potential V, and force Fb l t i fi ld E nd u s th t ΔV E d sθ hi h by electric field E, and you see that ΔV = E d cosθ, which will be zero (constant V) if cosθ=0, i.e., θ = 90°.

Surfaces of constant potential are perpendicular to electric field lines.

Field lines (black) and equipotentials (red) of an electric dipole.

From http://hyperphysics.phy-astr.gsu.edu/hbase/electric/equipot.html

The surface of a conductor is an equipotential.

Restatement of the fact that electric field lines are perpendicular to the surface of a conductor where they meet itmeet it.

Labs start next week (Sept. 13-15).You should read the lab materials before coming to lab and take the before coming to lab, and take the pre-lab quiz. Info on your section of PHY124 in Blackboard as soon as a

f l h l dsoftware glitch is resolved.

Your first lab is on tracing out f gelectric field lines and equipotentials.

There will be clicker questions every There will be clicker questions every Thurday based on that week’s lab.