ELECTRIC POTENTIALELECTRIC POTENTIAL

September 13, 2006September 13, 2006

Goings On For the Next Few Goings On For the Next Few DaysDays

Quiz Today – Gauss/Electric FieldQuiz Today – Gauss/Electric Field Today – Begin Chapter 25 – PotentialToday – Begin Chapter 25 – Potential Monday - PotentialMonday - Potential Wednesday – EXAMINATION #1Wednesday – EXAMINATION #1 Friday – More PotentialFriday – More Potential

Picture a Region ofPicture a Region ofspace Where there is an space Where there is an

Electric FieldElectric Field

Imagine there is a particle of charge Imagine there is a particle of charge q at some location.q at some location.

Imagine that the particle moves to Imagine that the particle moves to another spot within the field.another spot within the field.

Work must be done in order to Work must be done in order to accomplish this.accomplish this.

Electric Potential

We will be dealing with Work Energy

We do work if we try to move a charge in an electric field. Does the FIELD do work? Does the FIELD have a brain? Muscles?

Energy Methods

Often easier to apply than to solve directly Newton’s law equations.

Only works for conservative forces. One has to be careful with SIGNS.

VERY CAREFUL!

I need some help.

THINK ABOUT THIS!!!THINK ABOUT THIS!!!

When an object is moved from one point to another in an Electric Field, It takes energy (work) to move it. This work can be done by an external force (you).

You can also think of this as the FIELDFIELD

doing the negativenegative of this amount of work on the particle.

Move It!

Move the charge at constant velocity so it is in mechanical equilibrium all the time.

Ignore the acceleration at the beginning because you have to do the same amount of negative work to stop it when you get there.

And also remember:

The net work done by a conservative (field)force on a particle moving

around a closed path is

ZERO!

A nice landscape

mg

h

Work done by external force = mgh

How much work here by gravitational field?

The gravitational case:

Someone else’s path

IMPORTANT

The work necessary for an external agent to move a charge from an initial point to a final point is INDEPENDENT OF THE PATH CHOSEN!

The Electric Field

Is a conservative field. No frictional losses, etc.

Is created by charges. When one (external agent) moves a test

charge from one point in a field to another, the external agent must do work.

This work is equal to the increase in potential energy of the charge.

It is also the NEGATIVE of the work done BY THE FIELD in moving the charge from the same points.

A few things to remember…

A conservative force is NOT a Republican. An External Agent is NOT 007.

Electric Potential EnergyElectric Potential Energy

When an electrostatic force acts between two or more charged particles, we can assign an ELECTRIC POTENTIAL ENERGY U to the system.

Example: NOTATION U=PEU=PE

A B

dd

E

q

F

Work done by FIELD is Fd

Negative of the work done by the FIELD is -Fd

Change in Potential Energy is –Fd.Change in Potential Energy is –Fd.The charge sort-of “fell” to lower potential energy.

HIGH U LOWER U

Another way to look at it:

Consider an external agent. The external agent moves a charge from one

point to the other. The work done by the external agent in

moving the charge between these points is the change in potential energy.

This is the negative of the work that the FIELD does!

Electrons have those *&#^ negative signs. Electrons sometimes seem to be more

difficult to deal with because of their negative charge.

They “seem” to go from low potential energy to high.

They DO! They always fall AGAINST the field! Strange little things. But if YOU were

negative, you would be a little strange too!

AN IMPORTANT DEFINITION

Just as the ELECTRIC FIELD was defined as the FORCE per UNIT CHARGE:

We define ELECTRICAL POTENTIAL as the POTENTIAL ENERGY PER UNIT CHARGE:

q

FE

q

UV

VECTOR

SCALAR

UNITS OF POTENTIAL

VOLTCoulomb

Joules

q

UV

Let’s move a charge from one point to another via an external force. The external force does

work on the particle. The ELECTRIC FIELD

also does work on the particle.

We move the particle from point i to point f.

The change in kinetic energy is equal to the work done by the applied forces. Assume this is zero for now.

q

WV

WUUU

also

WW

K

if

WWKKK

applied

appliedif

fieldapplied

fieldappliedif

0

Furthermore…

VqW

so

q

W

q

UV

applied

applied

If we move a particle through a potential difference of V, the work from an external

“person” necessary to do this is qV

Example

Electric Field = 2 N/C

1 C d= 100 meters

Joules

mCN4102

100)/(2C1qEdPE

Energy. potentialin Change

agent EXTERNALby doneWork

One Step More

Joules

mCN4102

100)/(2C1qEdPE

Energy. potentialin Change

agent EXTERNALby doneWork

Volts 200200101

102

q

PE POTENTIALin Change

6

4

C

J

C

JoulesV

Consider Two Plates

OOPS …

Look at the path issue

The difference in potential between the accelerating plates in the electron gun of a TV picture tube is about 25 000 V. If the distance between these plates is 1.50 cm, what is the magnitude of the uniform electric field in this region?

An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 × 10–17 J. Calculate the charge on the ion.

ImportantImportant

We can define an absolute level of potential. To do this, we need to define a REFERENCE

or ZERO level for potential. For a uniform field, it doesn’t matter where

we place the reference. For POINT CHARGES, we will see shortly

that we must place the level at infinity or the math gets very messy!

An Equipotential Surface is defined as a surface on which the potential is constant.

0VIt takes NO work to move a charged particlebetween two points at the same potential.

The locus of all possible points that require NO WORK to move the charge to is actually a surface.

Example: A Set of Equipotenital Surfaces

Back To YesteryearBack To Yesteryear

Field Lines and Equipotentials

EquipotentialSurface

ElectricField

Components

EquipotentialSurface

ElectricField

Enormal

Eparallel

x

Work to move a charge a distancex along the equipotential surfaceIs Q x Eparallel X x

BUT

This an EQUIPOTENTIAL Surface No work is needed since V=0 for such a

surface. Consequently Eparallel=0 E must be perpendicular to the equipotential

surface

Therefore

E

E

E

V=constant

Field Lines are Perpendicular to the Equipotential Lines

Equipotential

)(0 ifexternal VVqWork

Consider Two EquipotentialSurfaces – Close together

V

V+dV

dsab

Work to move a charge q from a to b:

VVectords

dVE

and

dVEds

qdVVdVVqdW

also

qEdsdsFdW

external

appliedexternal

E...

)(E

Where

zyx

kji

Over a certain region of space, the electric potential is V = 5x – 3x2y + 2yz2. Find the expressions for the x, y, and z components of the electric field over this region. What is the magnitude of the field at the point P that has coordinates (1, 0, –2) m?

Typical Situation

dF W

Keep in Mind

Force and Displacement are VECTORS!

Potential is a SCALAR.

UNITS

1 VOLT = 1 Joule/Coulomb For the electric field, the units of N/C can be

converted to: 1 (N/C) = 1 (N/C) x 1(V/(J/C)) x 1J/(1 NM) Or

1 N/C = 1 V/m So an acceptable unit for the electric field is now

Volts/meter. N/C is still correct as well.

In Atomic Physics

It is sometimes useful to define an energy in eV or electron volts.

One eV is the additional energy that an proton charge would get if it were accelerated through a potential difference of one volt.

1 eV = e x 1V = (1.6 x 10-19C) x 1(J/C) = 1.6 x 10-19 Joules.

Nothing mysterious.

Coulomb Stuff: A NEW REFERENCE: INFINITY

204

1

r

qE

Consider a unit charge (+) being brought from infinity to a distance r from a Charge q:

q r

To move a unit test charge from infinity to the point at a distance r from the charge q, the external force must do an amount of work that we now can calculate.

x

Just Do It!

OK, doing it!

AB

r

r

unit

AB

rrkq

r

drkq

dr

qk

dVV

B

A

112

2sr

sE

Set the REFERENCE LEVEL OF POTENTIALat INFINITY so (1/rA)=0.

For point charges

i i

i

r

qV

04

1

For a DISTRIBUTION of charge:

volume r

dqkV

Ponder –

What is the potential a distance d from an infinite plane whose charge per unit area is ?

Given two 2.00-μC charges, as shown in Figure P25.16, and a positive test charge q = 1.28 × 10–18 C at the origin, (a) what is the net force exerted by the two 2.00-μC charges on the test charge q? (b) What is the electric field at the origin due to the two 2.00-μC charges? (c) What is the electrical potential at the origin due to the two 2.00-μC charges?

The three charges in Figure P25.19 are at the vertices of an isosceles triangle. Calculate the electric potential at the midpoint of the base, taking q = 7.00 μC.

A disk of radius R has a non-uniform surface charge density σ = Cr, where C is a constant and r is measured from the center of the disk. Find (by direct integration) the potential at P.

Example: Find potential at Pq1 q2

q3 q4

d

rP

md

r

md

qqqqr

V

919.02

3.1

)(1

4

14321

0

q1=12nC q2=-24nC q3=31nC q4=17nC q=36 x 10-9C

V=350 Volts (check the arithmetic!!)

An Examplefinite line of charge

d

r

x

dx

d

xLLV

and

xd

dxV

xd

dxdV

L

2/122

0

02/122

0

2/1220

)(ln

4

1

)(4

1

)(4

1

P

At P Using table of integrals

What about a rodthat goes from –L to +L??

Example

zR

220

22

0

22

0

12

2

2

Rz

zE

zRzdz

d

z

VE

zRzV

z

z

Which was the result we obtained earlier

disk=charge per unit area

In the figure, point P is at the center of the rectangle. With V = 0 at infinity, what is the net electric potential in terms of q/d at P due to the six charged particles?

Continuing

dd

s

ddd

dds

12.152

4

5

42

222

222

s

1 2 3

45 6

d

qxV

d

qk

d

qk

d

q

d

qkV

d

qqqq

d

qqk

r

qkV

i i

i

91035.8

93.093.8812.1

108

12.1

5533

2/

22

HRW gets 8.49 … one of us is right!

Derive an expression in terms of q2/a for the work required to set up the four-charge configuration in the figure, assuming the charges are initially infinitely far apart.

1 2

3 4

aadiagonal 71.12