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Today’s lecture is unofficially brought to you by

The Museum of Electricity

Announcements

Exam 1 is Tuesday, September 22, 5:00-6:00 pm.

Exam rooms (on next slide) will be posted on the Physics 2135 web site under “Course Information.”

One of the homework problems for tomorrow is Special Homework #4. You can find it on the web here.

Wednesday of this week is deadline to submit the appropriate memo or e-mail regarding an exam conflict. Follow this web link for instructions on what to do.

Know the exam time!

Find your room ahead of time!

If at 5:00 on test day you are lost, go to 104 Physics and check the exam room schedule, then go to the appropriate room and take

the exam there.

Exam is from 5:00-6:00 pm!

Physics 2135 Test Rooms, Fall 2015:

Instructor Sections RoomDr. Hagen room assignments coming soonDr. HaleDr. HorDr. KurterDr. MadisonDr. ParrisMr. UpshawDr. Vojta

Special Accommodations Testing Center

More Announcements

Exam 1 special arrangements:Test Center students: you need to make an appointment with Testing Center. By before now.

I will send you a confirming e-mail by the end of the Tuesday before the exam. If you do not receive the e-mail, you are NOT on the Test Center list!

More Announcements

LEAD Schedule (http://lead.mst.edu/)

Physics Learning Center, 2-4:30 pm and 6-8:30 pm Monday, Wednesday, rooms 129/130 Physics.

Tutor available, 7-9 pm Monday through Thursday; check the LEAD schedule.

Other LEAD Tutors available; check the schedule here. (You can get one-on-one tutoring if you don’t like the group environment of the Physics Learning Center).Burns & McDonnell Student Success Center (B&MSSC) Tutors available; check the schedule here. (Another place to get one-on-one tutoring).

More Announcements

If any of tomorrow’s homework problems involve a ring of charge: there is no starting equation for V on the axis or at the center of ring of charge. You must derive any equation you use.

If any of tomorrow’s homework problems involve a spherically-symmetric charge distribution: do not use the starting equation for V due to a point charge. You must derive any equation you use.

Today’s agenda:

Electric potential of a charge distribution.You must be able to calculate the electric potential for a charge distribution.

Equipotentials.You must be able to sketch and interpret equipotential plots.

Potential gradient.You must be able to calculate the electric field if you are given the electric potential.

Potentials and fields near conductors.You must be able to use what you have learned about electric fields, Gauss’ law, and electric potential to understand and apply several useful facts about conductors in electrostatic equilibrium.

Example 1: potential and electric field between two parallel conducting plates.

Assume V0<V1 (so I have a direction to draw the electric field). Also assume the plates are large compared to their separation, so the electric field is constant and perpendicular to the plates.

V0 V1

EAlso, let the plates be separated by a distance d.

d

V0 V1

plate 1

1 0 plate 0V V V E d

E

d

d d

0 0V E dx E dx Ed

d

xy

z

VE , or V Ed

d

The famous “Mr. Ed equation!*”

I’ll discuss in lecture why the absolute value signs are needed.

*2004, Prof. R. E. Olson.

|V|=Ed

d is dx

V Ed

Important note: the derivation of

did not require rectangular plates, or any plates at all. It works as long as E is uniform and parallel or antiparallel to d.

If is uniform but not parallel or antiparallel to , then use.

d

V E d

E

Example 2: A rod* of length L located along the x-axis has a total charge Q uniformly distributed along the rod. Find the electric potential at a point P along the y-axis a distance d from the origin. (*“rod” means “really thin”)

y

x

P

d

L

dq

xdx

r

=Q/L

dq=dx

2 2

dxk

x d

L

0V dV

*What are we assuming when we use this equation?

*dq

dV kr

Q

y

x

P

d

L

dq

xdx

r

L L

2 2 2 20 0

dx Q dxV k k

Lx d x d

A good set of math tables will have the integral:

2 2

2 2

dxln x x d

x d

2 2kQ L L dV ln

L d

Include the sign of Q to get the correct sign for V.

What is the direction of V?

Q

Example 3: Find the electric potential due to a uniformly charged ring of radius R and total charge Q at a point P on the axis of the ring.

P

R

dq

r

xx

Every dq of charge on the ring is the same distance from the point P.

dqdV k

r

2 2 2 2ring ring ring

kdq kV dV dq

x R x R

2 2

dqk

x R

Q

P

R

dq

r

xx

2 2 ring

kV dq

x R

2 2

kQV

x R

Could you use this expression for V to calculate E? Would you get the same result as I got in Lecture 3?

You must also derive an equation for the potential at the center of a ring if you need it for homework! In the next slide I will show you how

easy the derivation is.Include the sign of Q to get the correct sign for V.

Homework hint: you must derive this

equation in tomorrow’s homework!

Q

Example 4: Find the electric potential at the center of a uniformly charged ring of radius R and total charge Q.

Every dq of charge on the ring is the same distance from the point P.

dqdV k

r

ring ring ring

kdq k kQV dV dq

R R R

dqk

R

Q

Rdq

Example 4: A disc of radius R has a uniform charge per unit area and total charge Q. Calculate V at a point P along the central axis of the disc at a distance x from its center.

P

r′

dq

xxR

The disc is made of concentric rings. The area of a ring at a radius r′ is 2r′dr′, and the charge dq on each ring is (2r′dr′), where =Q/R2.

Q

dqdV k .

r

2 2r x r

Each ring is a distance from point P.2 2r x r

Start with

P

r′

dq

xxR

Q

2 2

k 2 r dr

x r

This is the (infinitesimal) potential for an (infinitesimal) ring of radius r′.

dqdV k

r2 2r x r

On the next slide, just for kicks I’ll replace k by 1/40.

R

2 2 2 2ring ring 00 0

1 2 r dr r drV dV

4 2x r x r

R

2 2 2 2 2 22

0 0 00

QV x r x R x x R x

2 2 2 R

2

Q

R

P

r′

dq

xxR

Q

2 2r x r

2 22

0

QV x R x

2 R

Could you use this expression for V to calculate E? Would you get the same result as I got in Lecture 3?

P

r′

dq

xxR

Q

2 2r x r

Example 5: calculate the potential at a point outside a very long insulating cylinder of radius R and positive uniform linear charge density .

To be worked at the blackboard in lecture…

I would prefer to not start with and integrate. Why?

dqdV k

r

Reason #1: a mathematical pain. What do I pick for my charge element dq? Little cubes? Ugh. Circular rings that I have to integrate from 0 to R and from - to along the axis? Long cylindrical shells that I have to integrate from 0 to R? There must be an easier way.Reason #2: I suspect that we would eventually find (read your text) that V= at any finite distance from the cylinder. Not very useful.Reason #3: we have previously derived a simple expression for outside a cylinder of charge. This can easily be integrated to find V.

E

r R0

λ R V V ln

2πε r

If we tried to use V=0 at r= then

r0

λ V V ln

2πε r

(V is infinite at any finite r).

That’s why we can’t start withdq

dV k .r

Things to note:

V is zero at the surface of the cylinder and decreases as you go further out. This makes sense! V decreases as you move away from positive charges.

0

λ R V r ln .

2πε r

Things to note:

r R0

λ R V V ln

2πε r For >0 and r>R, Vr – VR

<0.

Our text’s convention is Vab = Va – Vb. This is explained on page 759. Thus VrR = Vr – VR is the potential difference between points r and R and for r>R, VrR < 0.

In Physics 1135, Vba = Va – Vb. I like the Physics 1135 notation because it clearly shows where you start and end. But Vab has mathematical advantages which we will see in Chapter 24.

See your text for other examples of potentials calculated from charge distributions, as well as an alternate discussion of the electric field between charged parallel plates.

Remember: worked examples in the text are “testable.”

Make sure you know what Vab means, and how it relates to V.

Vif = Vf – Vi so Vif = -Vif

Special Dispensation

For tomorrow’s homework only: you may use the equation for the electric field of a long straight wire without first proving it:

line0

E .2 r

Of course, this is relevant only if a homework problem requires you to know the electric field of a long straight wire.

You can also use this equation for the electric field outside a long cylinder that carries charge.

Homework Hint!

Problems like 23.32 and 23.33: you must derive an expression for the potential outside a long conducting cylinder. See example 23.10. V is not zero at infinity in this case. Use

f

iV E d .

If 23.32 and 23.33 are not assigned, don’t be disappointed. We can still get you on this in problems in chapter 24!

Homework Hints!

In energy problems involving potentials, you may know the potential but not details of the charge distribution that produced it (or the charge distribution may be complex). In that case, you don’t want to attempt to calculate potential energy using . Instead, use

U q V .

If the electric field is zero everywhere in some region, what can you say about the potential in that region? Why?

1 2

12

q qU k

r

PRACTICAL APPLICATION

For some reason you think practical applications are important.Well, I found one!

The recipes on the Museum’s web site are sometimes slow to load, so I have copied a couple here…

Charred Cheese Sandwich

Take two pieces of yummy white bread, and place a piece of good ol' American cheese in between them.

Place the uncooked sandwich in the broilng & frying unit on your DL-1000 Electro Range (as yet unavailable).

Put on your fire retardant safety suit.

Make sure you have a fully charged fire extinguisher handy!

Set the cooking voltage to 19,000 volts at 3,600 watts, hit the timer switch and dive for cover!

The recipes on the Museum’s web site are sometimes slow to load, so I have copied a couple here…

Jacob’s Ladder Kebabs

This recipe will taste best if you can get the freshest ingredients. We use red & green peppers, and Italian sausage. However feel free to add mushrooms, onions, or any of your other favorites!

Cut your veggies and sausages into small cubes, and place them on the optional Kebab-O-Matic cooker attachment on your DL-1000 Electro Range (as yet unavailable).

Crank up the power and run like heck!

More Cooking with High Voltage

Application: Deep Space Propulsion Systems

Dr. Joshua Rovey, MAE Dept.

Today’s agenda:

Electric potential of a charge distribution.You must be able to calculate the electric potential for a charge distribution.

Equipotentials.You must be able to sketch and interpret equipotential plots.

Potential gradient.You must be able to calculate the electric field if you are given the electric potential.

Potentials and fields near conductors.You must be able to use what you have learned about electric fields, Gauss’ law, and electric potential to understand and apply several useful facts about conductors in electrostatic equilibrium.

Equipotentials

Equipotentials are contour maps of the electric potential.

http://www.omnimap.com/catalog/digital/topo.htm

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.

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.

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.)

Here are electric field and equipotential lines for a dipole.

Equipotential lines are shown in red.

I’ll discuss in lecture some implications this figure has for charged particle motion.

“Toy”

http://www.falstad.com/vector2de/

Today’s agenda:

Electric potential of a charge distribution.You must be able to calculate the electric potential for a charge distribution.

Equipotentials.You must be able to sketch and interpret equipotential plots.

Potential gradient.You must be able to calculate the electric field if you are given the electric potential.

Potentials and fields near conductors.You must be able to use what you have learned about electric fields, Gauss’ law, and electric potential to understand and apply several useful facts about conductors in electrostatic equilibrium.

Potential Gradient(Determining Electric Field from Potential)

The electric field vector points from higher to lower potentials.More specifically, E points along shortest distance from a higher equipotential surface to a lower equipotential surface.You can use E to calculate V:

b

b a aV V E d .

You can use the differential version of this equation to calculate E from a known V:

dVdV E d E d E

d

E

For spherically symmetric charge distribution:

r

dVE

dr

In one dimension:

x

dVE

dx

In three dimensions:

x y z

V V VE , E , E .

x y z

V V Vˆ ˆ ˆor E i j k Vx y z

r

dVE

dr

x y z

V V VE , E , E .

x y z

dVE

d

Calculate -dV/d(whatever) including all signs. If the result is +, points along the +(whatever) direction. If the result is -, E points along the –(whatever) direction.

E

E

Example (from a Fall 2006 exam problem): In a region of space, the electric potential is V(x,y,z) = Axy2 + Bx2 + Cx, where A = 50 V/m3, B = 100 V/m2, and C = -400 V/m are constants. Find the electric field at the origin

2x

(0,0,0)(0,0,0)

VE (0,0,0) Ay 2Bx C C

x

y (0,0,0)(0,0,0)

VE (0,0,0) (2Axy) 0

y

z(0,0,0)

VE (0,0,0) 0

z

V ˆE(0,0,0) 400 im

Today’s agenda:

Electric potential of a charge distribution.You must be able to calculate the electric potential for a charge distribution.

Equipotentials.You must be able to sketch and interpret equipotential plots.

Potential gradient.You must be able to calculate the electric field if you are given the electric potential.

Potentials and fields near conductors.You must be able to use what you have learned about electric fields, Gauss’ law, and electric potential to understand and apply several useful facts about conductors in electrostatic equilibrium.

When there is a net flow of charge inside a conductor, the physics is generally complex.

Potentials and Fields Near Conductors

When there is no net flow of charge, or no flow at all (the electrostatic case), then a number of conclusions can be reached using Gauss’ Law and the concepts of electric fields and potentials…

The electric field inside a conductor is zero.

Summary of key points (electrostatic case):

Any net charge on the conductor lies on the outer surface.The potential on the surface of a conductor, and everywhere inside, is the same.

The electric field just outside a conductor must be perpendicular to the surface.

Equipotential surfaces just outside the conductor must be parallel to the conductor’s surface.

Another key point: the charge density on a conductor surface will vary if the surface is irregular, and surface charge collects at “sharp points.”

Therefore the electric field is large (and can be huge) near “sharp points.”

Another Practical Application

To best shock somebody, don’t touch them with your hand; touch them with your fingertip.

Better yet, hold a small piece of bare wire in your hand and gently touch them with that.