Electric Potential

qVU =

Equipotentials and Energy

Today: Mini-quiz + hints for HWK

Should lightening rods have a small

or large radius of curvature ?

V is proportional to 1/R. If you want a high voltage to pass through the rod then use a small radius of curvature.

Air is normally an insulator, however for large E fields (E>3 x 106 V/m) it starts conducting.

Empire State Building, NYC

Electrical PotentialReview:

Wa → b = work done by force in going from a to b along path.

∫∫ •=•=→

b

a

b

aba ldEqldFW

rrrr

a

b

dl

F

θ∫ •−=−=−=∆ →

b

abaab ldEqWUUU

rr

∫ •−=−=−

=∆

=−=∆→ b

a

baabab ldE

q

W

q

UU

q

UVVV

rr

U = potential energy

V = electric potential

• Potential difference is minus the work done per unitcharge by the electric field as the charge moves from a to b.• Only changes in V are important; can choose the zero at any point.

Let Va = 0 at a = infinity and Vb → V, then:

∫∞ •−=r

ldEVrr

allows us to calculate Veverywhere if we know E

Potential from charged spherical shell

R

R

q

4πεπεπεπε0000 R

R r

V

q

4πεπεπεπε0000 r

• Potential

• r > R:

• r < R:

04

1

0

++++====−−−−−−−−====−−−−====••••−−−−==== ∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∞∞∞∞∞∞∞∞

====

∞∞∞∞====a

Q)dr(E)dr(E)dr(EldE)r(V

r

a

r

a

r

r

r

rr

rπεπεπεπε

rr

r

Q)dr(EldE)r(V

r

r

rr

r 04

1

πεπεπεπε====−−−−====••••−−−−==== ∫∫∫∫∫∫∫∫

∞∞∞∞

====

∞∞∞∞====

rr

• E-field (from Gauss' Law)

Er = 0• r < R:

Er = 1

4πε 0

q

r2

• r >R:

R

RR

q

q

ELECTRIC POTENTIAL for Charged Sphere (Y&F, ex.23.8)

Suppose we have a charged

metal sphere with charge q.

What is the electric potential

as a function of radius r?

q

N.B. V is continuous but E is not.

A point charge Q is fixed at the center of an uncharged conductingspherical shell of inner radius aand outer radius b.

– What is the value of the potential Va at the inner surface of the spherical shell?

(c)b

QV a

04

1

πε=(b)

a

QV a

04

1

πε=(a) 0=aV

a

b

Q1

Clicker exercise

A point charge Q is fixed at the center of an uncharged conducting spherical shell of inner radius aand outer radius b.

– What is the value of the potential Va at the inner surface of the spherical shell?

a

b

Q

(c)b

QV a

04

1

πε=(b) a

QV a

04

1

πε=(a) 0=aV

1

Eout

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

∫∞

∞ •−=−≡a

aa ldEVVVrr

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

• Apply Gauss’ Law to sphere:

∫ •−∫∞

•−=a

bldE

bldEaV

rrrr0

4

1

0

+=b

QVa

πε b

QV a

04

1

πε=

20

ˆ

4 r

rQE

πε=

r

• Inside the spherical shell: E = 0

Clicker exercise

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:

Electrical PotentialTwo ways to find V at any point in space:

• Use electric field:

• Sum or Integrate over charges:

i

i

i r

qV ∑=

04

1

πε

∫=r

dqV

04

1

πε

Pq1

q3

q2r1

r2r3

P

dq

r

Examples of integrating over a distribution of charge:• line of charge (review this one)• ring of charge• disk of charge

You should be able to do these.

∫∞ •−=r

ldEVrr

Infinite line charge or conducting cylinder.

Linear charge density λr

02E

r

λ

πε=

0 0

ln( )2 2

b

ba b

aa

rV V Edr

r r

λ λ

πε πε− = = =∫ ∫

Suppose we set rbto infinity, potential is infinite

Instead, set ra=r and rb=r0 at some fixed radius r0.

Cover homework hints

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 on a sphere which is

concentric with the charged sphere. These surfaces are called equipotentials.

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

Er

Equipotential

( ) 0V ∞ ≡

0

1( )

4

qV r

rπε=

(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 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.

∫ •−=−B

A

AB ldEVVrr

0 VldE

B

A

=∆=•− ∫rr

ldErr

•

The electric field is always perpendicular to an equipotential surface!

Why??

EXAMPLES of Equipotential Lines

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

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 is

connected to conductor B ?

a) QA increases

b) QA decreases

c) QA doesn’t change

Preflight 6:

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.

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.

Conservation of 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:

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

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)mr

Qqv f

04

1

πε= (b)

mr

Qqv f

02

1

πε= (c) 0=fv

3B

Clicker Problem• 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

A

qr

Q

B

2q2r

(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!!!

Clicker Problem

• 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)mr

Qqv f

04

1

πε= (b)

mr

Qqv f

02

1

πε= (c) 0=fv

3B

• What we have here is a little combination of 170 and 272.• 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

KEUB =2

0 2

1

2

2

4

1fmv

r

)q(Q=

πε mr

Qqv f

0

2

2

1

πε=

How to obtain vector E field from V

dl

E

θ

∫∞ •−=r

ldEVrr

∫∫ =•−=−=∆b

a

b

aab dVldEVVV

rr

θcosdlEldEdV −=•−=rr

For an infinitesimal step:

• Determine V from E:

Determining E from V:

Cases:• θ = 0: dV = E dl (maximum)• θ = 90o: dV = 0• θ = 180o: dV = -E dl

Can write:

)(

)ˆˆˆ()ˆˆˆ(

dzEdyEdxE

kdzjdyidxkEjEiEldEdV

zyx

zyx

++−=

++•++−=•−=rr

directional derivative

dV depends on direction

Example: V due to spherical charge distribution.

Potential GradientTake step in x direction: (dy = dz = 0)

dxEdzEdyEdxEdV xzyx −=++−= )(

x

V

dx

dVE

constzy

x∂

∂−=−=

.,

y

VEy

∂

∂−=

z

VEz

∂

∂−=

Similarly:

And:

Vkz

Vj

y

Vi

x

VE ∇−=

∂

∂+

∂

∂+

∂

∂−=

rr)ˆˆˆ(

)ˆˆˆ( kz

jy

ix ∂

∂+

∂

∂+

∂

∂=∇

rgradient operator

Gradient of V points in the direction that V increases the fastestwith respect to a change in x, y, and z.

E points in the direction that V decreases the fastest.E perpendicular to equilpotential lines.

partial derivative

Potential Gradient

Example: charge in uniform E field

y

o

q

EU = qEy

V = U/q = Eywhere V is taken as 0 at y = 0.

jEkjEi

Eykz

jy

ix

VE

ˆ)ˆ0ˆˆ0(

)ˆˆˆ(

−=++−=

∂

∂+

∂

∂+

∂

∂−=∇−=

rr

Given E or V in some region of space, can find the other.

Cylindrical and spherical symmetry cases:For E radial case and r is distancefrom point (spherical) or axis (cylindridal):

r

VEr

∂

∂−=

Example: E of point charge:

2

0

2

0

0

4)

1)(

4(

)4

(

r

q

r

q

r

q

rr

VEr

πεπε

πε

=−

−=

∂

∂−=

∂

∂−=

)ˆˆ3ˆ(2

)ˆ2ˆ6ˆ2(

)}3({)ˆˆˆ( )222

kzjyixA

kAzjAyiAx

zyxAkz

jy

ix

VE

+−−=

+−−=

+−∂

∂+

∂

∂+

∂

∂−=∇−=

rr

This graph shows the electric

potential at various points

along the x-axis.

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

A B C D

UI7PF7: Clicker Problem

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

( ) 323 xxxV −=

1

UI7ACT1: Clicker Problem

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

( ) 323 xxxV −=

VE ∇−=rr

We know V(x) “everywhere”

To obtain Ex “everywhere”, use

x

VEx

∂

∂−=

236 xxEx +−=

( ) 012122 =+−=xE

1

UI7ACT1: Clicker Problem

allows us to calculate the potential function V everywhere (keep

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

0q

WVV AB

AB ≡− ⇒⇒⇒⇒ ∫ •−=−B

A

AB ldEVVrr

If we know the electric field E everywhere,

VE ∇−=rr

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/Take Home Message

(but remember we measure potential differences with a voltmeter)