Electric Potential

We introduced the concept of potential energy in mechanics

Let’s remind to this concept and apply it to introduce electric potential energyWe start by revisit the work done on a particle of mass m -by a force in general-by a conservative force such as the gravitational force In general, work done by a force F moving a particle from point ra to rb

b

a b

a

W F dr

In general we have to specify how we get from a to b, e.g., for friction force

x2

x1

0( )ar t

( )b fr t

However,for a conservative force such as gravity we rememberF

where is the potential energy

b

b a

a

d Independent of the pathbetween ( )b fr t and 0( )ar t

With this we obtainb

a b

a

W dr

Gravity as an exampleGravitational force derived from ( )

Mr G

r

h

Pot. energydependson h, not howto get there.

Can we find a function U =U (r) such that ( )U r F is the force exerted by a point charge q on a test charge q0 ?We expect the answer to be yes, due to the similarity between Coulomb force and gravitational force

02

0

1ˆ

4Coulomb qqF r

r0

2ˆgravity m m

F G rr

00( ) , /

mmr G m

r

Potential energy

Potential

Let’s try 0

0

1( )

4

qqU r

r

In fact we see

02

0

ˆ( )

1ˆ

4

Coulomb dUF U r r

drqq

rr

simple because of radial symmetry where U(r)=U(r)

We conclude

0

0

1( )

4

qqU r

r

Electric potential energy of electrostatically interacting point charges q and q0

r

U

qq0<0attractive potential

r

U

qq0>0repulsive potential

As always, potential defined only up to an arbitrary constant. Expression above uses U(r)=0 as reference point

We know already the superposition principle for electric fields and forces, can we find a net potential energy for q0 interacting with several point charges?

1 2 3( ) ( ) ( ) ( ) ...F r F r F r F r

Net force q0 experiencesForce exerted on q0 by charge q1 at r1

Force exerted on q0 by charge q2 at r2

Force exerted on q0 by charge q3 at r3

r

x

y

q0

r1

q1

r 1-r

r3-r

q2

r2

r2-r

q3r3

Note: textbook on p. 785 defines I prefer to keep r-dependence explicitly visible

1 01 1

10

2 02 2

20

1( ) ( )

4

1( ) ( )

4

...

q qF r U r

r r

q qF r U r

r r

1 2 3( ) ( ) ( ) ( ) ...F r F r F r F r 1 1 1( ) ( ) ( ) ...U r U r U r

1 2 3( ) ( ) ( ) ... ( )U r U r U r U r

0 1 2 3 0

1 2 30 0

( ) ...4 4

i

i i

q q q q q qU r

r r r r r r r r

iir r r

The last expression answered the question about the potential energy of the charge q0 due to interaction with the other point charges q1, q2, …,

r

x

y

q0

r1

q1

r 1-r

r3-r

q2

r2

r2-r

q3r3

Those point charges q1, q2, …, interact as well.Each charge with all other charges

If we ask for the total potential energy of the collection of charges we obtain

0

1

4i j

i j i j

q qU

r r

This is the energy it takes to bring the charges from infinite separation to their respective fixed positions ri

makes sure that we count each pair only once

What is the speed of charge q after moving in the field E from the positive to the negative plate.Neglect gravity.

Clicker question

+++++++++++++++++++++++++++++++

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

+

d

1)

2 /v qEd m

2 /v mEd q2)

3) / 2v qEd m

4) 0/ 4v qEd m

5) None of the above

Goal: Making the potential energy a specific, test charge independent quantityWe are familiar by now with the concept of creating specific quantities, e.g.,Force on a test charge F qE

Electric field: test charge independent, specific quantity /E F q

0( )Mm

r Gr

Gravitational potential energy

0

MG

m r

test mass independent, specific potential

Electric potential V

0

UV

q Specific, test charge independent potential energy.

The SI unit of the potential is volt (V) .

Meaning of a potential difference

Point b

Point a

Wa->b work done by electric force during displacement of charge q0 from a to b.

a b b aW U U U

0 0 0 0

a b b ab a a b

W U U UV V V V

q q q q

Voltage of the battery

Alternatively we can ask:What is the work an external force, F, has to do to move charge q0 from b to aThis force is opposite to the electric force, Fel, above.Hence:

00 0 0 0

1 1 1 1/

a a b

el el a b a b

b b a

W q Fdr F d r F d r W V Vq q q q

We know these two alternative interpretations already from mechanics

za

b

Fg=-mg

( ) ( )b

a b b a a b a b

a

W mg dz mg z z mg z z

a

b

F=mg To slowly (without adding kinetic energy) move mass from b to a we need an external force acting against gravity

( )a

a b a b

b

W mg dz mg z z

Fromb

a b

a

W F dr 0

b

a

q E d r

0F q E

and 0

a ba b

WV V

q

b

a b

a

V V E dr We obtain the potential difference (voltage) from the path independent line integral taken between points a and b

0 1 2 3 4

0.0

0.5

1.0

E(r

)4

0R2 /Q

r/R

0.0

0.5

1.0

V(r

)4

0R

/Q

point a becomes variable point in distance r

Let’s calculate the potential of a charged conducting sphere by integrating the E-field

R

r

b

a b

a

V V E dr We start from

point b becomes reference point at r

20

( ) ( )4 r

Q drV r V r

r

for r>R

:=0

0 0

1( )

4 4r

Q QV r

r r

For r<R

20

( ) ( ) 04

R

r R

Q drV r V r dr

r

0

( )4

QV r

R

An important application of our “potential of a conducting sphere”- problem

R

According to our considerations above we find at the surface of the conducting sphere:

04surface surface

QV E R

R

There is a dielectric breakdown field strength, Em, for all insulating materials including air For E>Em air becomes conducting due to discharge

max potential of a sphere before discharge in air sets independs on radius

mm

VE

RFrom

http://en.wikipedia.org/wiki/File:Plasma_wheel_2_med_DSIR2018.jpg

Wartenberg pinwheel charged to a very high voltageNote, that discharge sets in at regions of small R

Demonstration: Surface Charge Density

How do we actually measure the charge on the proof plane ?