Lecture06 Electric Potentials of Charge Distributions; Equipotentials Electric Field and Electric Potential Gradient

8/3/2019 Lecture06 Electric Potentials of Charge Distributions; Equipotentials Electric Field and Electric Potential Gradient

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http://www.lab-initio.com(nz183.jpg)


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Todays lecture is unofficially brought to you by

The Museum of Electricity


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Announcements

y Ex

am 1 is Tuesday, September 20, 5:00-6:15 pm.Exam rooms (on next slide) will be posted on the Physics 24web site under Course Information.

y One of the homework problems for tomorrow is Special

Homework #3. You can find it on the web here.Dont forget to work this problem!

y Wednesday ofthis week is deadline to submit the

appropriate memo or e-mail regarding an exam conflict. Followthis web link for instructions on what to do.


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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 from5:00-6:15 pm!

y Physics 24 Test Room Assignments, Fall 2011:

Instructor Sections Room

Dr. Hagen K, L 295 ToomeyDr. Hale E, G 104 PhysicsDr. Hor J, M 227 FultonDr. Parris B, H 125 Butler-Carlton (Civil)Dr. Peacher D, F G-5 HSSDr. Schmitt A, C 120 Butler-Carlton (Civil)

4:30 and 5:30 Exams 202 PhysicsSpecial Needs Testing Center


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More Announcements

y Exam 1 special arrangements:7 Test Center students. You need to also make anappointment with Test Center. By before now.

Five 4:30 exam students (so far).

Three 5:30 exam students (so far).

y If any of tomorrows homework problems involve a ring ofcharge: there is no starting equation for V on the axis or at thecenter of ring of charge. You must derive any equation you use.

I sent these 7 students an e-mail Friday. If you did not

receive the e-mail, you areNO

T on the Test Center list!


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Todays 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, andelectric potential to understand and apply several useful facts about conductors inelectrostatic equilibrium.


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Electric Potential of a Charge Distribution

Example: potential and electric field between two parallelconducting plates.

Assume V0


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V0

V1

plate 1

1 0plate 0

V V V E d( ! ! &&"

E

d"

d d

0 0V E dx E dx Ed( ! ! !

d

x

y

z

VE , or V Ed

d

(! ( !

The famous Mr.Ed equation!*

Ill discuss in lecture why theabsolute value signs are needed.

*2004, Prof. R. E. Olson.

|(V|=Ed


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V Ed( !

Important note: the derivation of

did not require rectangular plates, or any plates at all. Itworks as long as E is uniform and parallel or antiparallel

to d.

In general, E should be replaced by the component of alongthe displacement vector .

E&

d&" V E d( !

&&


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Example: A rod of length L located along the x-axis has a totalcharge Q uniformly distributed along the rod. Find the electricpotential at a point P along the y-axis a distance d from theorigin.

Thanks to Dr. Waddill for this fine example.

y

x

P

d

L

dq

xdx

r

P=Q/L

dq=Pdx

2 2

dq dxdV k k

r x d

P! !

L

0V dV!

What are we assuming when we use this equation?


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y

x

P

d

L

dq

xdx

r

L L

2 2 2 20 0

dx Q dxV k k

Lx d x d

P! !

A good set of math tables willhave the integral:

2 22 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.


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Example: Find the electric potential due to a uniformly chargedring 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 thering is the same distance

from the point P.

2 2

dq dqdV k k

r x R

! !

2 2ring ring

dqV dV k

x R! !


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P

R

dQ

r

xx

2 2 ringkV dq

x R!

2 2

kQV

x R

!

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

You must derive an equation for thepotential at the center of a ring if you

need it for homework! In lecture I willshow you how easy the derivation is.

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

Homework hint: derivethis equation in

tomorrows homework!


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Quiz time (maybe for points, maybe just for practice!)


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Example: A disc of radius R has a uniform charge per unit areaW and total charge Q. Calculate V at a point P along the central

ax

is of the disc at a distancex

from its center.

P

r

dQ

xx

R

The disc is made ofconcentric rings. Thearea of a ring at a

radius r is 2Trdr, andthe charge on each ringis W(2Trdr).

We *can use the equation for the potential due to a ring,replace R by r, and integrate from r=0 to r=R.

ring2 2

k 2 rdrdV

x r

W T!

*I just derived it, so I get to use it.


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P

r

dQ

xx

R

R

2 2 2 2ring ring 00 0

1 2 rdr rdr V dV4 2x r x r

W T W! ! !TI I

R

2 2 2 2 2 2

2

0 0 00

QV x r x R x x R x2 2 2 R

W W

! ! ! I I TI

2

Q

R

W !T


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2 220

QV x R x

2 R!

TI

P

r

dQ

xx

R

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


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See your text for other examples of potentials calculated fromcharge distributions, as well as an alternate discussion of theelectric field between charged parallel plates.

Remember: worked examples in the te

xt are testable.

Homework Hint!

Problems 23.35 and 23.36: you must derive an expression forthe potential outside a long conducting cylinder. *Seeexample 23.10. V is not zero at infinity in this case. Use

f

iV E d .( !

&&"

*Caution: in example 23.10 the text is not careful with signs.


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Special Dispensations

1. In homework problem 22.14, it is shown that the electric fieldoutside a spherically symmetric charge distribution is the same asthe electric field due to a point charge (having the total charge ofthe distribution) at the center of the charge distribution.You mayuse this fact in future homework and on exams without

further proof. If you do so, you should state what you are doing.

2. In chapter 23, your text shows that the electric potential outsidea spherically symmetric charge distribution is the same as the

potential of a point charge (of the same total charge) at the centerof the charge distribution.You may use this fact in futurehomework and on exams without further proof. If you do so,you should state what you are doing.


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Special Dispensations

3. For tomorrows homework only: you may use the equation forthe electric field of a long straight wire without first proving it:

line

0

E .2 r

P!

TI

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


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Homework Hint!

In energy problems involving potentials, you may know thepotential but not details of the charge distribution thatproduced it (or the charge distribution may be complex). In

that case, you dont want to attempt to calculate potentialenergy. Instead, use U q V .( ! (


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PRACTICALAPPLICATION

For some reason you think practical applications are important.

Well, I found one!


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More Cooking with High Voltage

Application: Deep Space Propulsion Systems

Dr. Joshua Rovey, MAE Dept.


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Demo: Faraday Cage


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Todays agenda:




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 inelectrostatic equilibrium.


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Equipotentials

Equipotentials are contour maps of the electric potential.

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


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The electric field must be perpendicular to equipotential lines.Why?

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

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

Otherwise charge would flow and it wouldnt be a static case.

Equipotential lines are another visualization tool. Theyillustrate where the potential is constant. Equipotential lines

are actually projections on a 2-dimensional page of a 3-dimensional equipotential surface. (Just like the contourmap.)


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Here are some electric field and equipotential lines I generatedusing an electromagnetic field program.

E

quipotential lines are shown in red.

Ill discuss in lecture someimplications this figure hasfor charged particle motion.


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Toy

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


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Todays agenda:







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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 higherequipotential surface to a lower equipotential surface.

You can use E to calculate V:b

b aa

V V E d . ! &&"

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

dVdV E d E d E

d! ! !

" "

&&" "

"

E


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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

x x x! ! !

x x x

V V V or E i j k Vx y z

x x x! !

x x x

& &


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More *^#$*& signs!

r

dVE

dr!

x y z

V V VE , E , E .

x y z

x x x! ! !

x x x

dV

E d!

" "

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


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Example (from a Fall 2006 exam problem): In a region ofspace, 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 areconstants. Find the electric field at the origin

2x(0,0,0)

(0,0,0)

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

x

x! ! !

x

y (0,0,0)

(0,0,0)

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

y

x! ! !

x

z(0,0,0)

V

E (0, 0, 0) 0z

x

! !x

V E(0,0,0) 400 im

!

&


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Todays agenda:







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

Potentials and Fields Near Conductors

When there is no net flow of charge, or no flow at all (theelectrostatic case), then a number of conclusions can bereached using Gauss Law and the concepts of electric fieldsand potentials


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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 everywhereinside, is the same.

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

Equipotential surfaces just outside the conductor must beparallel to the conductors surface.


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Another key point: the charge density on a conductor surfacewill vary if the surface is irregular, and surface charge collects

at sharp points.Therefore the electric field is large (and can be huge) nearsharp points.

Another Practical Application

To best shock somebody, dont touch them with your hand;touch them with your fingertip.

Better yet, hold a small piece of bare wirein your hand and gently touch them withthat.

Documents

Lecture06 Electric Potentials of Charge Distributions; Equipotentials Electric Field and Electric Potential Gradient