Briefly review the concepts of potential energy and work.

•Potential Energy = U = stored work in a system

•Work = energy put into or taken out of system by forces

•Work done by a (constant) force F :

θcos|||| rFrFWv

v

v

v

∆=∆⋅=

∆r

F

θ

Gravitational Potential EnergyLift a book by hand (Fext) at constant velocity.

Fext

mg

initial

final

h

Fext = mg

Wext = Fext h = mgh

Wgrav = -mgh

Define ∆U = +Wext = -Wgrav= mgh

Note that get to define U=0,

typically at the ground.

U is for potential energy, do not confuse with

“internal energy” in Thermo.

Gravitational Potential Energy (cont)

For conservative forces Mechanical Energy

is conserved.

Gravity is a conservative force.

Coulomb force is also a conservative force.

Friction is not a conservative force.

If only conservative forces are acting, then

∆EMech

=0.

0=∆+∆ UEKin

UEE KinMech +=

Electric Potential EnergyCharge in a constant field

- - - - - - - - - - - - - -

+

E

FField=qE

FExt=-qE

+

∆r

Initial position

Final position

∆Uelec = change in U when moving +q from initial to final position.

fieldextif WWUUU −=+=−=∆

rEqU

rFWU fieldfield

v

v

v

v

∆⋅−=∆→

∆⋅−=−=∆

General case

What if the E-field is not constant?

rEqUv

v

∆⋅−=∆

∫ ⋅−=∆f

i

rdEqUv

v Integral over the path

from initial (i) position to final (f)

position.

Electric Potential EnergySince Coulomb forces are conservative, it means that the change

in potential energy is path independent.

∫ ⋅−=∆f

i

rdEqUv

v

Electric Potential EnergyPositive charge in a constant field

Electric Potential EnergyNegative charge in a constant field

Observations• If we need to exert a force to “push” or “pull” against the

field to move the particle to the new position, then U

increases. In other words “we want to move the particle to

the new position” and “the field resists”.

• If we need to exert a force to “hold” the particle so the field

will not move the particle to the new position, then U

decreases. In other words, “the field wants to move the

particle”, and “we resist”.

• Both can be summarized in the following statement:

“If the force exerted by the field opposes the motion of the

particle, the field does negative work and U increases,

otherwise, U decreases”

Potential Energy between two point

charges

r+Q1 +Q

2

Imagine doing work to move the objects from infinitely far apart

(initial) to the configuration drawn above (final).

11

fieldextif WWUUU −=+=−=∆

∫ ⋅−=−=∆f

i

fieldif rdFUUUv

v

∫ ⋅−=−=∆f

i

if rdEqUUUv

v

* Note that force is not

constant over the path!

r+Q1

Consider Q1

fixed and move Q2

from infinity to r.

Potential Energy between two point charges

r+Q1 +Q

2

∫ ⋅−=−=∆f

i

if rdEqUUUv

v E-field generated by Q1.

Moving Q2

through the field.

∫∞

⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=∆

r

if dxx

QQUUU

21

02 4

1

πε

r

QQUUU if

21

04

1

πε=−=∆

Potential Energy between two point charges

We also need to define the zero point for potential energy.

This is arbitrary, but the convention is U=0 when all charged

objects are infinitely far apart.

r

QQUUU if

21

04

1

πε=−=∆

r

QQU f

21

04

1

πε=

Ui=U(∞)= 0 by our convention.

Potential Energy between two

electric charges.

Potential Energy between two point charges

E. Potential energy vs. distance

E. Potential Energy of a charge

distribution

∑=⎟⎟⎠

⎞⎜⎜⎝

⎛+++=

i i

i

r

qq

r

q

r

q

r

qqU

0

0

3

3

2

2

1

1

0

0

4...

4 πεπε

For a continuous charge distribution, replace the sum by

an integral

Potential energy associated to the

field produced by charges qi

r1 r2

r3

q1

q2

q3

q0

E. Potential Energy of a charge

distribution

∑<

=ji ij

jiTotal r

qqU

04

1

πε

For a continuous charge distribution, replace the sum by

an integral

To calculate the TOTAL potential

energy we have to consider all the

fields produced by all the charges qi

on the other(n-1) charges qj

r1 r2

r3

q1

q2

q3

q0

17

r

U(r)

r

U(r)

r

QkQrU 21)( =

Both cases below are for two

point charges separated by a

distance r.

Which graph is correct for two

negative charges? A) Left Plot,

B) Right Plot

CPS Question

Just like a compressed

spring stores potential

energy.

Earlier we found that not only using forces, but also electric fields

was very useful.

rr

QQF ˆ

4

12

21

012 πε=

v

rr

QE ˆ

4

121

01 πε=

v

rr

QQEqF ˆ

4

121

02112 πε

==vv

r+Q1 +Q

2

•Electric Field is the force per unit of charge due to the presence of Q1

•Electric field from Q1

is there even if Q2

is not there.

All of the above are vectors!

Electric Field

We find a similarly useful thing with electric potential energy.

r

QQU 21

012 4

1

πε=

r

QV 1

01 4

1

πε=

r

QQqVU 1

02112 4

1

πε==

r+Q1 +Q

2

•Electric potential is the electric potential energy per unit of charge due

to the presence of Q1

•Electric potential from Q1

is there even if Q2

is not there.

All of the above are scalars!

Electric Potential

+Q1

Electric Field is a vector associated with

a source charge Q1.

Units are [Newtons/Coulomb].

Electric potential is a scalar associated with

a source charge Q1.

Units are [Joules/Coulomb] or [Volts].

rr

QE ˆ

4

121

01 πε=

v

r

QV 1

01 4

1

πε=

Electric Potential

V = Voltage = Electric Potential

Units are [Volts] = [Joules/Coulomb]

U = Electric Potential Energy

Units are [Joules]

Electric Field Units=[Volts]/[meter]=[Newton]/[Coulomb]

Electric Potential of a point charge

r

qV

04

1

πε=

Analogy: Electrical pressure or electrical "height"

Positive charges want to get away from higher voltage towards lower voltage.

Just like a gas wants to move from high to low pressure.

Electric Potential of a charge

distribution

∑=⎟⎟⎠

⎞⎜⎜⎝

⎛+++=

i i

i

r

qq

r

q

r

q

r

qqU

0

0

3

3

2

2

1

1

0

0

4...

4 πεπε

For a continuous charge distribution, replace the sum by

an integral

Potential energy associated to the

field produced by the charges qi

Electric potential due to charges qi

∑=i i

i

r

qV

04

1

πε

r1 r2

r3

q1

q2

q3

q0

“test charge”

∫ ⋅−=∆=∆f

i

rdEq

UV

v

v

Electric Potential from Electric Field

∫∫ ⋅−=⋅−==∆f

i

f

i

Ext rdEqrdFWUv

v

v

r

.

Elevation is a scalar.

Contour lines show paths of

constant elevation.

Gravitational potential

VG=gh

If I stand at a certain

elevation I have a

gravitational potential energy

[Joules] = mgh=mVG

Equipotential surfaces

Electric potential (Voltage) is a

scalar.

Contour lines show paths of

constant Voltage (equipotentials).

If a charge q is at a certain

Voltage, it has an electrical

potential energy [Joules] = qV

Equipotential surfaces

Equipotential surfaces and field lines

•Equi-potential surfaces and field lines are always mutually

perpendicular

•The Field is not necessarily constant on equipotential surfaces

•Larger density of equipotentials means larger variations of V,

and larger values of |E|

E. Field and E. potential

What is Electric Field?

Electric Field is a vector that is

analogous to a steepness

vector.

Steepness cannot have one

number at a given position, it

depends which direction you

look (vector).

∫ ⋅−=−=∆f

i

if rdEVVVv

v

∫∞

⋅−=∞−r

rdEVrVv

v

)()(

∫∞

⋅−==∆r

rdErVVv

v

)(

Given the Electric field vector

as a function of position, we can

compute Voltage.

If we integrate the “steepness” over a path, we

find the change in elevation.

This does not depend on our path taken.

E. Field and E. potential

Vi=V(∞)= 0 by our convention.

dz

dVE

dy

dVE

dx

dVEdzEdyEdxEdV zyxzyx −=−=−=⇒++=− ;;

∫∫ −=⋅−=∆b

a

b

a

dVrdEVv

v

The Electric field vector is the rate of change

of Voltage in a given direction.z

z

Vy

y

Vx

x

VE ˆˆˆ

∂∂−

∂∂−

∂∂−=

v

E. Field and E. potential

VE ∇−=rv The Electric field vector “potential gradient”

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

30

CPS question

Two identical charge, +Q and +Q, are fixed in space. The electric potential (V) at the point X midway between the charges is:A) ZeroB) Non-Zero

+Q +QPoint X

E=0V=?

x

V

+Q +Q

xx

VE ˆ

∂∂−=

v

CPS Question

Drawn are a set of equipotential lines. Consider the electric field at points A and B. Which of the following statements is true?

0V 10V 20V 30VPoint A

Point B

A) |EA| > |EB|B) |EA| < |EB|C)|EA| = |EB|D)Not enough informationgiven.E) None of the above

32

CPS Question

Two charges, +Q and -Q, are fixed in space. The electric field at the point X midway between the charges is:A) ZeroB) Non-Zero

+Q -QPoint X

V=0E=?

x

V

+Q -Q

xx

VE ˆ

∂∂−=

v

33

CPS Question

+Q +Q

-Q-Q

Point P

The magnitude of the electric field at point P is:A) ZeroB) Non-Zero

The magnitude of the voltage at point P is:A) ZeroB) Non-Zero

All points on conductor must be at the same

electrical potential.

∫ ⋅−=∆b

a

rdEVv

v

Imagine point a at Voltage Va.

Since E=0 everywhere inside the conductor (no steepness), integral to

point b is always 0. ∆Vab=0.

E. Potential inside conductors