ELECTRIC POTENTIALELECTRIC POTENTIAL

Spring , 2008Spring , 2008

Chapter 24 Electric PotentialIn this chapter we will define the electric potential ( symbol V ) associated with the electric force and accomplish the following tasks:

Calculate V if we know the corresponding electric field Calculate the electric field if we know the corresponding potential V Determine the potential V generated by a point charge Determine the potential V generated by a continuous charge distribution Determine the electric potential energy U of a system of charges Define the notion of an equipotential surface Explore the geometric relationship between equipotential surfaces and electric field lines Explore the potential of a charged isolated conductor

(24 - 1)

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 must be Imagine that the particle must be moved to another spot within the moved to another spot within the field.field.

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

What (or who) must do this What (or who) must do this work?work?

An external agent An external agent (person)(person)

The Field itselfThe Field itself Either of the aboveEither of the above

Electric Potential

We will be dealing with Work Energy & Conservation

Work must be done to move a charge in an electric field. Let’s do a demo ….

I need some help.

What we will do ….

For the moment, assume the charge has MASS. (It may not.) Assume the charge is initially stationary. The charge is to be moved to the left. The charge is to be moved at CONSTANT velocity.

+charge

E

Mr. ExternalMrs. Fields

During this process, who is pushing?

Mr. External Mrs. Fields

ENERGY is required to bring the charge up to speed (if it has mass).

ENERGY is required to bring the particle back to rest (if it has mass).

The sum of these two is ZERO.

During this process, who is actually doing work?

Mr. External Mrs. Fields Both of them Neither of them.

Clearly

Both are doing work. BOTH are applying a force through a

distance. BOTH get tired!

About the work that they do .. Mrs. Fields does more work than Mr.

External. Mr. External does more work than Mrs.

Fields. Both do the same amount of work. Each does the negative amount of work

than the other does.

Each does the negative amount of work than the other does.

So, when we move a charge in an Electric Field ..

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.

Summary--Summary--

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.

And also remember:

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

around a closed path is

ZERO!

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.

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.

The change in potential energy of a charge is the amount of work that is done by an external force in moving the charge from its initial position to its new position.

It is the negative of the work done by the FIELD in moving the particle from the initial to the final position.

Definition – Potential Energy

PE or U is the work done by an external agent in moving a charge from a REFERENCE POSITION to a different position.

A Reference ZERO is placed at the most convenient position Like the ground level in many gravitational

potential energy problems.

Example:

E

Zero Levelq

F

d

Work by External Agent

Wexternal = F d = qEd= U

Work done by the Fieldis:Wfield= -qEd = -Wexternal

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

Consider Two Plates

OOPS …

Look at the path issue

ImportantImportant

We defined an absolute level of potential. To do this, we needed to define a

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

placed 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.

A collection of points that have the same

potential is known as an equipotential

surface. Four such surfaces are shown in

the figure. The work done by as it moves

a charge be

E

q

Equipotential surfaces

tween two points that have a

potential difference is given by:

V

W q V

2 1

2 1

For path I : 0 because 0

For path II: 0 because 0

For path III:

For path IV:

When a charge is moved on an equipotential surface 0

The work don

I

II

III

IV

W V

W V

W q V q V V

W q V q V V

V

Note :

e by the electric field is zero: 0W

W q V

(24 - 10)

SA B

V

q

F

E

r

Consider the equipotential surface at

potential . A charge is moved

by an electric field from point A

to point B along a path

V q

E

The electric field E is perpendicular

to the equipotential surfaces

.

Points A and B and the path lie on S

r

Lets assume that the electric field forms an angle with the path .

The work done by the electric field is: cos cos

We also know that 0. Thus: cos 0

0, 0,

E r

W F r F r qE r

W qE r

q E r

0 Thus: cos 0

The correct picture is shown in the figure belo

9

w

0

(24 - 11)S

V

E

Field Lines are Perpendicular to the Equipotential Lines

Equipotential

)(0 ifexternal VVqWork

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.

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

r1

r2

r3

P

q1

q3

q2

Consider the group of three point charges shown in the

figure. The potential generated by this group at any

point P is calculated using the principle of super

V

Potential due to a group of point charges

1

31 2

1 2

2 3

31 21 2 3

1 2 3

1 2 3

position

We determine the potentials , and generated

by each charge at point P.

1 1 1, ,

4 4 4

We add the thr

1

ee terms:

1 1

4 4 4

o o

o o o

o

V ,

qq qV

r r

V V

qq qV V V

r r r

V V V V

1.

2.

3r

1 2

11 2

The previous equation can be generalized for charges as

1 1 1 1...

follo

4 4 4 4

ws:n

n i

o o o n o i

q qq q

r r

n

Vr r

(24 - 5)

1 2 3V V V V

For a DISTRIBUTION of charge:

volume r

dqkV

dqO A

Potential created by a line of charge of

length L and uniform linear charge density λ at point P.

Consider the charge element at point A, a

distance from O. From triangle OAP we

dq dx

x

Example :

2 2

2 2

2 20

2 2

2

2

0

2

2

2

2

have:

Here is the distance OP

The potential created by at P is:

1 1

4 4

4

ln4

l

l

n ln4

n

o o

L

o

L

o

o

r d x d

dV dq

dq dxdV

r d x

dxV

d x

V x d x

V L L x

dxx

d

x dd x

(24 - 8)

A

B

VV+dV

The work done by the electric field is given by:

(eqs.1)

also cos cos (eqs.2)

If we compare these two equations we have:

cos cos

From triangle PAB we see that

o

o

o o

W q dV

W Fds Eq ds

dVEq ds q dV E

ds

cos is the

component of along the direction s.

Thus:

s

s

E

E E

VE

s

s

VE

s

(24 - 13)Calculating the electric field E from the potential V

A

B

VV+dV

We have proved that: s

VE

s

s

VE

s

The component of in any direction is the negative of the rate

at which the electric potential changes with distance in this direction

E

If we take s ro be the - , -, and -axes we get:

If we know the function ( , , )

we can determine the components of

and thus the vector itself

x

y

z

x y z

V x y z

E

VE

xV

Ey

VE

z

E

ˆˆ ˆV V VE i j k

x y z

(24 - 14)

Potential energy of a system of point charges

We define as the work required to assemble the

system of charges one by one, bringing each charge

from infinity to its final position

Using the above de

U

U

finition we will prove that for

a system of three point charges is given by:U

2 3 1 31 2

12 23 134 4 4o o o

q q q qq qU

r r r x

y

O

q1

q2

q3

r12

r23

r13

(24 - 15)

, 1

each pair of charges is counted only once

For a system of point charges the potential energy is given by:

Her1

e is the separation between and 4

T

i

ij

ni j

i jo iji j

i j

n q U

r q qq q

Ur

Note :

he summation condition is imposed so that, as in the

case of three point charges, each pair of charges is counted only once

i j

1

1

2

2 2

1 1 22

12 12

3

3 3

1 2

13 23

1 3 23

13

Bring in

0

(no other charges around)

Bring in

(2)

(2)4 4

Bring in

(3)

1(3)

4

1

4

o o

o

o

q

W

q

W q V

q q qV W

r r

q

W q V

q qV

r r

q q qW

r

Step2 :

Step 1 :

e

St p3 :

3

23

1 2 3

2 3 1 31 2

12 23 134 4 4o o o

q q q qq

q

r

W W W W

qW

r r r

(24 - 16)

O

x

y

q1

Step 1

x

y

O

r12q1

q2

Step 2

Step 3

x

y

O

r12

r13

r23

q1

q2

q3

path

A

B

conductor

0E

We shall prove that all the points on a conductor

(either on the surface or inside) have the same

potential

Potential of an isolated conductor

Consider two points A and B on or inside an conductor. The potential difference

between these two points is give by the equation:

We already know that the electrostatic field

B A

B

B A

A

V V

V V E d S

>>>>>>>>>>>>>>

inside a conductor is zero

Thus the integral above vanishes and for any two points

on or inside the conductor.B A

E

V V

A conductor is an equipotential surface

(24 - 17)

We already know that the surface of a conductor

is an equipotential surface. We also know that

the electric field lines are perpendicular to the

equipote

Isolated conductor in an external electric field

ntial surfaces.

From these two statements it follows that the electric field vector is

perpendicular to the conductor surface, as shown in the figure.

All the charges of the conductor reside on the surface and a

E

rrange

themselves in such as way so that the net electric field inside the

conductor

The electric field just out side the conductor is:

0.

in

outo

E

E

(24 - 18)

Electric field and potential

in and around a charged

conductor. A summary

All the charges reside on the conductor surface.

The electric field inside the conductor is zero 0

The electric field just outside the conductor is:

The electric field

in

outo

E

E

1.

2.

3.

4. just outside the conductor is perpendicular

to the conductor surface

All the points on the surface and inside the conductor have the same potential

The conductor is an eequipotential surf

5.

ace

0inE

ˆouto

E n

E

E

n̂

n̂(24 - 19)

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