Electric PotentialSymmetry, Equipotentials and Energy

What is an expert: three possibilities?

• Knows answer without waking up• Has worked through most

problems during a previous life• Is armed with a) logic b) intuition

and c) ability to use math

• Knows answer without waking up• Has worked through most

problems during a previous life• Is armed with a) logic b) intuition

and c) ability to use math

You are all well on the way to being experts

Enjoy and learn from the challenges

• Equipotentials and conductors• Electric Potential for other symmetries

A problem of infinities• We got V from E, now how can we get 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: – Calculate electric field of dipole from

potential

Today…

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.

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

AB

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

Preflight 6:

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…

( ) 0V (where )

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

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

Along an equipotential surface, there is NO change in V . Pick A and B on itThen,

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.

0 VldEB

A

ldE

The electric field is always perpendicular to an equipotential

surface!

Why??

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!!

+ + + +

+ +

+ + +

+ + +

+ +

Conductors

• ClaimE-field lines are always locally perpendicular to a conducting surface

• Why??

If the E-field had a component parallel to the surface, charges would move until the surface charge distribution acquired a pattern that had no component of E parallel to the surface. Remember the charge on the surface, together with all other charges, generates E-field.

+ + + +

+ +

+ + +

+ + +

+ +

Conductors

• ClaimThe local E-field magnitude at a point on the surface of a conductor is related to the local surface charge density by E=/0. (this is E right at the surface)

• Why??

Just zoom in.

+ + + +

+ +

+ + +

+ + +

+ +

A B

3) The two conductors are now connected by a wire. How do the potentials at the conductor surfaces compare now ?

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

4) 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

Preflight 6:

Two 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 dielectric (air) breakdown first?(a) Ball 1 (b) Ball 2 (c) Same Time

r1

Ball 1

r2

Ball 2

surface 2

QE kr

QV kr

Smaller r higher E closer to breakdown

surfaceVEr

Ex. 100 kVV 3

6

100 10 V 0.03m 3cm3 10 V/m

r

High Voltage Terminals must

be big!

Followup Question:

• What about cylindrical and planar symmtries?• E-fields are easily obtained using techniques from

last two lectures – Gauss’ Law, etc• To get potential, integrate -E from reference point

to any point (x,y,z). Where to start?• Charge distribution extends to infinity, so we can

not pick rinfinity for “place” where V0.• Must pick some other point• This point and other points form an equipotential

surface where V=0• Discuss planar symmetry, then cylindrical

symmetry

Other symmetries

• Let’s assume is positive• Pick x=0 plane to be V=0• Integrate –E from this plane

to get V(x)• Potential drops on both sides

since E // dl on both sides

planar symmerty

x

x=0

0

00

( ) ( )

( )2

x

x

V x dxE x

dx

+ for x>0- for x<0

x

x=0

V(x)

0

slope = 2

0

V(x) = 2

x

• With more planes, just use superposition• at x=0, - at x=d• Still pick x=0 plane to be V=0• Second plane has offset wedge and opposite charge• Add potential graphs from both source planes to get final result

planar symmerty

x

x=0

V(x)

x

x=0

x=d

x

x=0

V(x)

0

( )V d d

• Consider an infinite line charge on the z-axis• Again, we can’t set V at infinity = 0, since line

charge extends to infinity• Equipotential surfaces will be cylinders, V has cyl

sym• Pick one of these cylinders (r=a) to be V=0

cylindrical symmetry

Segment ofinfinite

line charge C/m

End on view

V=0 here

a

cylindrical symmetry

End on view

V=0 here

a

0 0

( ) ( )

ln2 2

r

a

r

a

V r drE r

dr rr a

V(r)

r

aCurve shows V(r) for pos

What happens when neg?Still have zero at a, and curve flips through that point on x-axis…

• More complicated cylindrically symmetrical charge distributions can also be solved by superposition Vtotal(r) = V1(r)+ V2(r)+ V3(r)+ V4(r)+….

• But…Must make sure you keep V(a)=0 for all

of the potential functions you add up.

cylindrical symmetry

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.

1

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

Conductors versus InsulatorsCharges move to Charges cannot cancel electric field move at allin the conductor

E=0 equipotential Charge distributionsurface on insulator

unaffected by external fields

All charge on surface Charge can sit “inside”

(Appendix B describes method of “images” to find the surface charge distribution on a conductor [only for your reading pleasure!])

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

Vdxxr 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

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

2

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 < 0We know V(x) “everywhere”

To obtain Ex “everywhere”, use

2Lecture 6, ACT 2

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

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.

• 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...

3

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

A

qrQ

BQ

2q

2r

• 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

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):

QAqr

QB2q2r

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

3A

• 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 UnitsMKS: 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 for sperically symmetric

charges)

• At some specific place for charge distributions that extend to infinity

• Conductors are equipotentials and E is always perpendicular to the surface of a conductor

• Find E from V:

• Potential Energy

• Next time capacitors

qVUVE

Appendix A: Electric Dipole

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

aa

+q

-q

r

r1r2

Appendix A: Dipole Field

x =

y =

0

0

Etot

E

E

r

/ /

z

aa

+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: FYI: 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 so that the total 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

+++++

+++++

-----

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

o

surfacesurfacesurfacenormal

rrErE

)(

)()(

Appendix B: (FYI) 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

+ ---

--

-