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Quiz 1

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Quiz 2

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“I've been giving this lecture to first-year classes for over twenty-five years. You'd think they would begin to understand it by now.”—English mathematician John Littlewood

Graphic “borrowed” from http://www.howstuffworks.com. Go there!

Example 17-1: An electron in the picture tube of a TV is accelerated from rest through a potential difference Vca = +5000 V.

0 V

+5000 V

(a) What is the change in potential energy of the electron?

How do you remember which is anode and which is cathode? I remember that electrons are “cathode rays,” because a monitor is a CRT and has a “gun” that “shoots” electrons. Cathode rays must come from the cathode. If electrons are exiting the cathode, it must be negative. Therefore, the anode is positive. See text, page 518This is the “opposite” of anions and cations. *!?$ chemists!

(a) What is the change in potential energy of the electron?

Reminder: the steps to reach the solution are draw a fully-labeled diagram,* OSE, replace generic quantities by specifics in OSE, solve algebraically, numerical answer only at very end.

*Kind of done—see previous slide.

OSE: PEif= q Vif

PEca= q Vca = q (Va – Vc)

PEca= (-1.6x10-19 C)(+5000 V – 0 V)

PEca= -8x10-16 J

PEca= -8x10-16 J

What is the meaning of the – sign? The electron’s potential energy has decreased. Total energy is conserved. What happened to the electron’s kinetic energy?

(b) The potential energy lost by the electron becomes kinetic energy. What is the speed of the electron as a result of its acceleration?

OSE: Ef – Ei = (Wother ) if

“Nooooo… that was last semester. This is this semester. You can‘t…”

Yes I can!

OSE: Ef – Ei = (Wother ) if

Wother refers to work done by non-conservative forces. A conservative force is one for which you can define a potential.* We have defined a potential for Coulomb’s Law forces, which tells you they must be conservative.

*A good way to think of a conservative force is that it doesn’t “rob” you of useful energy. Energy is “conserved” as you move from initial state to final state. Of course, energy never disappears, but nonconservative forces are especially wasteful of useful energy.

Work done by conservative forces gets accounted for in the potential energies, which in this equation are contained in the E’s.

PEca= -8x10-16 J

Ef – Ei = (Wother ) if

Kf + Uf – Ki – Ui = 0

0 no non-cons. forces present

0

a

0

b

the electron is initially at resta

setting V=0 at the cathode means electron has PE=0 thereb

½mvf2 = -Uf

vf2 = -2 Uf /m

vf = ( -2 Uf /m)½

bad idea - I combined two algebra steps here

vf = ( -2 Uf /m)½

vf = ( -2 PEf /m)½ PE and U mean the same thing

vf = ( -2 q Vf /m)½

vf = [ -2 (-1.6x10-19 C) (+5000 V) / 9.11x10-31 kg ]½

vf = 4.2x107 m/s

This answer is (slightly) in error compared to the result you would get from a measurement because a relativistic calculation is needed for the velocity of a fast-moving electron.

A note on potential energies:

Before Chapter 17, the potential energies that went into Ef – Ei = (Wother ) if were spring and gravitational.

In Chapter 17, we have discovered a new (to us) potential energy which results from the electric force (a conservative force).

Now when we write Ef – Ei = (Wother ) if, we must include Ugrav, Uspring, and Uelectric.

Why have you been writing PE for electrical potential energy, instead of Uelec?

Because that’s the way the author of our College Physics textbook does it.

17.2 Electric Potential and Electric Field

- -

- -

- -

- -

- -

- -

- -

+

E

+ +

+ +

+ +

+ +

+ +

a b

The magnitude of the work done in moving a charge from a to b in a uniform electric field E is

Wab = q Vab

= qEd

Vab = Ed

d

E = Vab/d

This gives us another equation for the electric field.

i fΔVOSE E = , away from +

d

D

= FD = +Fd F

Example 17-2: Two parallel plates are charged to a voltage of 50 V. If the separation between the plates is 0.050 m, calculate the electric field between them.

- -

- -

- -

- -

- -

- -

- -

E

+ +

+ +

+ +

+ +

+ +

a b

d=0.05 m

V = 50 V

i fΔVE = , away from +

d

E = |V| / d

E = 50 V / 0.05 m

E = 1000 V/m, to the right

“This is just more stuff to learn. There’s already enough stuff. Why bother?”We started out with charges. That led us to forces due to the charges – Coulomb’s Law. We introduced the electric field to help us visualize the forces throughout all of 3D space.The electric field is a vector quantity. Vectors are a pain to deal with. Wouldn’t you rather work with scalars?

We needed to introduce energy, hence potentials and potential differences. A cool bonus is that potentials are scalars, and are related to the electric field.

Wouldn’t you rather work with scalar potentials instead of electric field vectors?

Sure you would.

Yes, you have to.Sure you would.

If we were doing this “right,” we would define

x x 0

V(xyz)E = - lim

x

and similarly for the y and z-components of E.

Calculus would then tell us

x

dV(xyz)E = -

dx

and similarly for the y and z-components of E.

We could then elegantly write

E(xyz) = - V(xyz) .

just a fancy way of writing all 3 derivative components at once

I’m not sure that the way I have presented this material, which is the way our College Physics text presents it, is the best way to do it.

It is not the way you would present it to a physics major, who has already had enough calculus so that it is like a second language.

On the other hand, the presentation here tries to make sense out of the relationships between electric force, electric field, and electric potential. The traditional physics approach is more like a decree from above.You can see how hyperphysics does it, if you are curious. Electric potential energy. Work and voltage.

17.3 Equipotential LinesEquipotentials in 2 dimensions are like contour maps.

Equipotential lines are another visualization tool. They illustrate where the potential is constant. Equipotential lines are actually projections on a 2-dimensional page of a 3-dimensional equipotential surface. (“Just like” the contour map.)The electric field must be perpendicular to equipotential lines. Why?

Otherwise work would be required to move a charge along an equipotential surface, and it would not be equipotential.

I got the map on the previous page fromhttp://www.omnimap.com/catalog/digital/topo.htm.

In the static case (charges not moving) the surface of a conductor is an equipotential surface. Why?

Otherwise charge would flow and it wouldn’t be a static case.

Here are some electric field lines I generated using the emfield program.

Equipotential lines are shown in red.

17.4 The Electron Volt, a Unit of Energy

The joule represents too large an energy scale when we discuss individual electrons or atoms.

An electron volt (eV) is the amount of energy an electron acquires in being accelerated through a potential difference of 1 volt.

1 eV = 1.6x10-19 joules

The eV is not a unit of energy in the SI (mks) system. If you are doing calculations involving mks, you should* convert eV to joules before you calculate, otherwise there is a good chance you will introduce an error.

*i.e., must, unless you really know what you are doing

At the beginning of chapter 17, we defined electric potential in terms of work and potential energy:

Va = (PE)a /q

Wif = q Vif

PEif= q Vif

We also learned that electric field and electric potential are related:

|E| = |Vab|/d

But an electric field can arise from a collection of point charges. Could we calculate the electric potential due to a collection of point charges?

17-5 Electric Potential for Point Charges

So far, we have been talking about the energy of charges in an electric field.

But electric fields can be related back to point charges.

In this section we introduce the electric potential for point charges.

A few slides back we saw that Ex = -V / x. In calculus-based physics, this is a derivative, and because E is actually a vector, this is a 3 dimensional vector derivative.

We can use calculus (see next page) to integrate to get the expression for the potential for a point charge, which is

OSE: VQ = kQ/r = Q/40r.

As you might expect, the collection due to a sum of point charges is equal to the sum of potentials. I’ll make this an OSE.

OSE: Vnet = Vi.

As with all equations involving charges, the sign on Q is important.

We have taken V=0 at r= from the point charge Q. Note the 1/r dependence (for F and E the dependence was 1/r2). V is a scalar and F and E are vectors, so working with V should be much easier.

x

dV(xyz)E = -

dx

r

dV(r)E = -

dr

rE dr= -dV(r) noooo… (math majors understand)

ZZZZZZZZZZZZZZ

rE dr= -dV(r) correctly including vector nature

We can use calculus to derive the expression (previous page) for the potential for a point charge…

ZZZZZZZZZZZZZZ

rdV(r)= -E dr

You aren’t responsible for this derivation, but I want to “expose” you to it.

ZZZZZZZZZZZZZZb b

ra adV(r)= - E dr

ZZZZZZZZZZZZZZb

b a raV V = - E dr

b

b a 2a

kqV V = - dr

rfor a point charge

r

r 2

kqV V = - dr

rremember, the r in the integral isa “dummy” variable

r

r

kqV 0=

r

kq

V rr

in this derivation, I switch from using Q to q for our point charge—no particular reason

For a continuous distribution of charges, replace the sum by an integral.

k dq

V rr

Example 17-3: What minimum work is required by an external force to bring a charge q=3.00 C from a great distance away (take r = ) to a point 0.500 m from a charge Q = 20.0 C?

This problem can be solved without a diagram, although you may make one if it helps you.

OSE: Wif = q Vif

Wif = q (Vf - Vi)

Wif = q (kQ/rf - kQ/ri)

Wif = q (kQ/rf - kQ/ri)

Wif = kqQ (1/rf - 1/ri)

Wif = (9x109)(3x10-6)(20x10-6) / (0.5)

0, because ri =

Wif = kqQ /rf

Wif = 1.08 J

Example 17-4: Calculate the electric potential at point A in the figure below due to the two charges shown.

x

y

Q2=+50CQ1=-50C

52 cm

60 cm

30 c

m

=30º

A

OSE: Vnet = Vi.

VA = V1 + V2.

OSE: VQ = kQ/r = Q/40r

VA = V1 + V2

VA = kQ1/r1 + kQ2 /r2

x

y

Q2=+50CQ1=-50C

52 cm

60 cm

30 c

m

=30º

A

r 1 =

r2 =

All the numbers are in SI units, making the calculation easy.

VA = (9x109)[(-50x10-6/0.6) + (+50x10-6/0.3)]

VA = kQ1/r1 + kQ2 /r2

Which would you rather work: this example, or example 16-8, which calculated the electric field, used vectors, and took 6 slides?

VA = 7.5x105 V.

Conceptual example. All charges in the figure have the same magnitude.

+ -

+ -

+ +

(i)

(ii)

(iii)

Which set has a positive potential energy?

Which set has the most negative potential energy?

Which set requires the most work to separate the charges to ?

17-6 Electric DipolesAn electric dipole is two charges +Q and –Q separated by a distance ℓ.

The figure shows electric field lines and equipotential lines for an electric dipole.

Electric dipoles appear frequently in physics, chemistry, and biology.

Potential due to a dipole.

ℓ

r

r

r

P

-Q

+Q

P

kQ k(-Q) 1 1 ΔrV = + = kQ - = kQ

r (r +Δr) r (r+Δr) r r+Δr

If P is far from the charges, so that r>>ℓ, then r = ℓcos and r >> r.

P

ΔrV = kQ

r r+Δr

becomes

2

kQ cosθV= .

r

that’s a script lowercase letter l

ℓ

r

r

r

P

-Q

+Q

The product Qℓ is called the dipole moment of the dipole.

dipole 2

kp cosθOSE V = for r>> .

r

ℓ

r

r

r

P

-Q

+Q

Example: dipole moment of C==O at point P (see text for numbers).

C+ O-P

9 30

2 210

9 10 8 10 cos180kp cosθV= 0.089 V .

r 9 10

The potential at P is much greater if you remove one of the charges (makes sense; charges almost “cancel”).