Electric PotentialEquipotentials and Energy

• Equipotentials and conductors• E from V• Calculate electric field of dipole from potential• Electric Potential Energy

– of charge in external electric field– stored in the electric field itself (next time)

• Appendix: – Example calculation of a spherical charge configuration– Calculate electric field of dipole from potential

Today…

Text Reference: Chapter 24.3,5 and 25.1 examples: 24.7,11,13,15 and 25.1

Sparks• High electric fields can ionize nonconducting materials

(“dielectrics”)

• Breakdown can occur when the field is greater than the “dielectric strength” of the material.– E.g., in air,

6 6max 3 10 N/C 3 10 V/m 30 kV/cm E

Insulator ConductorDielectric

Breakdown

Ex.

doorknobV

fingerV

2mmd

Arc discharge equalizes the potential

What is ΔV?

max V E d

30 kV/cm • 0.2 cm

6 kV

Note: High humidity can also bleed the charge off reduce ΔV.

r1

Ball 1

r2

Ball 2

Question 1Two charged balls are each at the same potential V. Ball 2 is

twice as large as ball 1.

As V is increased, which ball will induce breakdown first?

(a) Ball 1 (b) Ball 2 (c) Same Time

Question 1Two charged balls are each at the same potential V. Ball 2 is

twice as large as ball 1.

As V is increased, which ball will induce breakdown first?

(a) Ball 1 (b) Ball 2 (c) Same Time

r1

Ball 1

r2

Ball 2

surface 2

QE k

r

QV k

r

Smaller r higher E closer to breakdownsurface V

Er

Ex. 100 kVV3

6

100 10 V0.03m 3cm

3 10 V/m

r

High Voltage Terminals must be big!

Lightning!

Factoids: ~ 200 M voltsV

~ 40,000 ampI

~ 30mst12~ 10 WP

_

+

__

+ +

Collisions produce charged particles.

The heavier particles (-) sit near the bottom of the cloud; the lighter particles (+) near

the top.

Stepped Leader

Negatively charged electrons

begin zigzagging downward.

AttractionAs the stepped

leader nears the ground, it

draws a streamer of

positive charge upward.

Flowing Charge

As the leader and the

streamer come together, powerful

electric current begins flowing

Contact!Intense wave of positive charge, a “return stroke,” travels upward

at 108 m/s

Two spherical conductors are separated by a large distance.They each carry the same positive charge Q. Conductor A has a larger radius than conductor B.

Compare the potential at the surface of conductor A with the potential at the surface of conductor B.

A B

a) VA > VB b) VA = VB c) VA < VB

Question 2

Potential from a charged sphere

•The electric field of the charged sphere has spherical symmetry.

•The potential depends only on the distance from the center of the sphere, as is expected from spherical symmetry.

•Therefore, the potential is constant along a sphere which is concentric with the point charge. These surfaces are called equipotentials.

•Notice that the electric field is perpendicular to the equipotential surface at all points.

Er

Equipotential

Last time…

(where ) ( ) 0V

EquipotentialsDefined as: The locus of points with the same potential.

• Example: for a point charge, the equipotentials are spheres centered on the charge.

0 VldEB

A

ldE

Along the surface, there is NO change in V (it’s an equipotential!)

Therefore,

We can conclude then, that is zero.

If the dot product of the field vector and the displacement vector is zero, then these two vectors are perpendicular, or the electric field is always perpendicular to the equipotential surface.

The electric field is always perpendicular to an equipotential

surface!

Why??

From the definition of potential

Conductors

• ClaimThe surface of a conductor is always an equipotential surface (in fact, the entire conductor is an equipotential).

• Why??

If surface were not equipotential, there would be an electric field component parallel to the surface and the charges would move!!

+ +

+ +

+ +

+ + +

+ + +

+ +

A B

The same two conductors that were in Question 2 are now connected by a wire, before they each carried the same positive charge Q. How do the potentials at the conductor surfaces compare now ?

a) VA > VB b) VA = VB c) VA < VB

Question 3

A B

The same two conductors that were in Question 2 are now connected by a wire, before they each carried the same positive charge Q. What happens to the charge on conductor A after it isconnected to conductor B ?

a) QA increases

b) QA decreases

c) QA doesn’t change

Question 4

Charge on Conductors?

• How is charge distributed on the surface of a conductor? – KEY: Must produce E=0 inside the conductor and E normal to the

surface .

Spherical example (with little off-center charge):

- ---

- --

-

-

-

-

--

-

-

+

+

+

++

+

+

++

+ +

+ +

+

+

+

+q

E=0 inside conducting shell.

charge density induced on outer surface uniform

E outside has spherical symmetry centered on spherical conducting shell.

charge density induced on inner surface non-uniform.

An uncharged spherical conductor has a weirdly shaped cavity carved out of it. Inside the cavity is a charge -q.

How much charge is on the cavity wall?

(a) Less than< q (b) Exactly q (c) More than q

1A

How is the charge distributed on the cavity wall?

(a) Uniformly

(b) More charge closer to –q

(c) Less charge closer to -q

1B

Lecture 6, ACT 1

How is the charge distributed on the outside of the sphere?

(a) Uniformly

(b) More charge near the cavity

(c) Less charge near the cavity

1C

-q

Lecture 6, ACT 1

By Gauss’ Law, since E=0 inside the conductor, the total charge on the inner wall must be q (and therefore -q must be on the outside surface of the conductor, since it has no net charge).

An uncharged spherical conductor has a weirdly shaped cavity carved out of it. Inside the cavity is a charge -q.

How much charge is on the cavity wall?

1A -q

(a) Less than< q (b) Exactly q (c) More than q

How is the charge distributed on the cavity wall?

(a) Uniformly

(b) More charge closer to -q

(c) Less charge closer to -q

1B

Lecture 6, ACT 1

The induced charge will distribute itself nonuniformly to exactly cancel everywhere in the conductor. The surface charge density will be higher near the -q charge.

E

-q

How is the charge distributed on the outside of the sphere?

(a) Uniformly

(b) More charge near the cavity

(c) Less charge near the cavity

1C

Lecture 6, ACT 1

As in the previous example, the charge will be uniformly distributed (because the outer surface is symmetric). Outside the conductor the E field always points directly to the center of the sphere, regardless of the cavity or charge.

Note: this is why your radio, cell phone, etc. won’t work inside a metal building!

-q

Charge on Conductor Demo

• How is the charge distributed on a non-spherical conductor?? Claim largest charge density at smallest radius of curvature.

• 2 spheres, connected by a wire, “far” apart

• Both at same potential

But: rS

rL

Smaller sphere has the larger surface charge density !

Equipotential Example

• Field lines more closely spaced near end with most curvature – higher E-field

• Field lines to surface near the surface (since surface is equipotential).

• Near the surface, equipotentials have similar shape as surface.

• Equipotentials will look more circular (spherical) at large r.

Electric Dipole Equipotentials

•First, let’s take a look at the equipotentials:

Electric FishSome fish have the ability to produce & detect electric fields

• Navigation, object detection, communication with other electric fish

• “Strongly electric fish” (eels) can stun their prey

Black ghost knife fish

Dipole-like equipotentialsMore info: Prof. Mark Nelson,

Beckman Institute, UIUC

-Electric current flows down the voltage gradient-An object brought close to the fish alters the pattern of current flow

• We can obtain the electric field E from the potential V by inverting our previous relation between E and V:

r

V

dxxr ˆ

V+dVdxEdxxEdV x ˆ

• Expressed as a vector, E is the negative gradient of V

• Cartesian coordinates:

• Spherical coordinates:

E from V?

This graph shows the electric potential at various points along the x-axis.

8) At which point(s) is the electric field zero?

A B C D

Preflight 6:

• Consider the following electric potential:

• What electric field does this describe?

... expressing this as a vector:

• Something for you to try:

Can you use the dipole potential to obtain the dipole field? Try it in spherical coordinates ... you should get (see Appendix):

E from V: an Example

allows us to calculate the potential function V everywhere (keep

in mind, we often define VA = 0 at some convenient place)

If we know the electric field E everywhere,

allows us to calculate the electric field E everywhere

If we know the potential function V everywhere,

• Units for Potential! 1 Joule/Coul = 1 VOLT

The Bottom Line

2

A point charge Q is fixed at the center of an uncharged conducting spherical shell of inner radius a and outer radius b. – What is the value of the potential Va at

the inner surface of the spherical shell?

(c)(b)(a)

a

b

Q1A

The electric potential in a region of space is given by

The x-component of the electric field Ex at x = 2 is

(a) Ex = 0 (b) Ex > 0 (c)Ex < 0

1B

Lecture 6, ACT 2

A point charge Q is fixed at the center of an uncharged conducting spherical shell of inner radius a and outer radius b. – What is the value of the potential Va at

the inner surface of the spherical shell?

a

b

Q

(c)(b)(a)

1A

Eout

• How to start?? The only thing we know about the potential is its definition:

• To calculate Va, we need to know the electric field E• Outside the spherical shell:

• Apply Gauss’ Law to sphere:

• Inside the spherical shell: E = 0

Lecture 6, ACT 2

The electric potential in a region of space is given by

The x-component of the electric field Ex at x = 2 is

(a) Ex = 0 (b) Ex > 0 (c)Ex < 0

We know V(x) “everywhere”

To obtain Ex “everywhere”, use

1BLecture 6, ACT 2

Electric Potential Energy

• The Coulomb force is a CONSERVATIVE force (i.e. the work done by it on a particle which moves around a closed path returning to its initial position is ZERO.)

• The total energy (kinetic + electric potential) is then conserved for a charged particle moving under the influence of the Coulomb force.

3

• Therefore, a particle moving under the influence of the Coulomb force is said to have an electric potential energy defined by:

this “q” is the ‘test charge” in other examples...

E

A

BC

6) If a negative charge is moved from point A to point B, its electric potential energy

a) increasesb) decreasesc) doesn’t change

Preflight 6:

Lecture 6, ACT 3

(a) UA < UB (b) UA = UB (c) UA > UB

Two test charges are brought separately to the vicinity of a positive charge Q.

– charge +q is brought to pt A, a distance r from Q.

– charge +2q is brought to pt B, a distance 2r from Q.

– Compare the potential energy of q (UA) to that of 2q (UB):

3A

• Suppose charge 2q has mass m and is released from rest from the above position (a distance 2r from Q). What is its velocity vf as it approaches r = ?

(a) (b) (c)

3B

A

qrQ

BQ

2q

2r

Lecture 6, ACT 3• Two test charges are brought

separately to the vicinity of positive charge Q. – charge +q is brought to pt A, a

distance r from Q.– charge +2q is brought to pt B, a

distance 2r from Q.– Compare the potential energy of q (UA)

to that of 2q (UB):

Q

Aq

r

Q

B2q2r

(a) UA < UB (b) UA = UB (c) UA > UB

3A

•Look familiar? •This is ALMOST the same as ACT 2 from the last lecture.

• In that ACT, we discovered that the potential at A was TWICE the potential at B. The point was that the magnitudes of the charges at A and B were IRRELEVANT to the question of comparing the potentials.• The charges at A and B are NOT however irrelevant in this ACT!!• The potential energy of q is proportional to Qq/r.• The potential energy of 2q is proportional to Q(2q)/(2r).• Therefore, the potential energies UA and UB are EQUAL!!!

Lecture 6, ACT 3• Suppose charge 2q has mass m and is released from

rest from the above position (a distance 2r from Q). What is its velocity vf as it approaches r = ?

(a) (b) (c)

3B

• What we have here is a little combination of 111 and 112.• The principle at work here is CONSERVATION OF ENERGY.• Initially:

• The charge has no kinetic energy since it is at rest. • The charge does have potential energy (electric) = UB.

• Finally:• The charge has no potential energy (U 1/R) • The charge does have kinetic energy = KE

Energy Units

MKS: U = QV 1 coul-volt = 1 joule

for particles (e, p, ...) 1 eV = 1.6x10-19 joules

Accelerators• Electrostatic: Van de Graaff

electrons 100 keV ( 105 eV)

• Electromagnetic: Fermilab

protons 1TeV ( 1012 eV)

Summary

• Physically, V is what counts

• The place where V=0 is “arbitrary” (at infinity)

• Conductors are equipotentials

• Find E from V:

• Potential Energy

• Next time, capacitors:

qVU

VE

Reading assignment: 25.2, 4

Examples: 25.2,3,5,6 and 7

Appendix A: Electric Dipole

Now we can use this potential to calculate the E field of a dipole (after a picture)

(remember how messy the direct calculation was?)

• Rewrite this for special case r>>a:

The potential is much easier to calculate than the field since it is an algebraic sum of 2 scalar terms.

z

a

a

+q

-q

r

r1r2

r2-r1

Appendix A: Electric Dipole

• Calculate E in spherical coordinates:

the dipole moment

z

a

a

+q

-q

r

r1

r2

Appendix A: Dipole Field

x =

y =

0

0

Etot

E

E

r

/ /

z

a

a

+q

-q

r

Sample Problem• Consider the dipole shown at the

right. – Fix r = r0 >> a– Define max such that the polar component of the electric field has its

maximum value (for r = r0).

z

a

a

+q

-q

r

r1

r2

(a) max = 0 (b) max = 45 (c) max = 90

What is max?

• The expression for the electric field of a dipole (r >> a) is:

• The polar component of E is maximum when sinis maximum.

• Therefore, E has its maximum value when = 90.

Appendix B: Induced charge distribution on conductor via

“method of images”• Consider a source charge

brought close to a conductor:

+--

-

-

-

+

++

++

• Charge distribution “induced” on conductor by source charge:

• Induced charge distribution is “real” and sources E-field that is zero inside conductor!

– resulting E-field is sum of field from source charge and induced charge distribution

– E-field is locally perpendicular to surface

– just like the homework problem.

+++++

+++++

-----

• With enough symmetry, can solve for on conductor– how? Gauss’ Law

o

surfacesurfacesurfacenormal

rrErE

)(

)()(

Appendix B: Induced charge distribution on conductor via

“method of images”• Consider a source charge brought

close to a planar conductor:

• Charge distribution “induced” on conductor by source charge– conductor is equipotential

– E-field is normal to surface

– this is just like a dipole

o

surfacesurfacesurfacenormal

rrErE

)(

)()(

• Method of Images for a charge (distribution) near a flat conducting plane:– reflect the point charge through the surface and put a

charge of opposite sign there– do this for all source charges

– E-field at plane of symmetry - the conductor surface determines

+ --

-

-

-

-