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Physics 202, Lecture 4

Today’s Topics

More on Gauss’s Law

Electric Potential (Ch. 23) Electric Potential Energy and Electric Potential Electric Potential and Electric Field

Gauss’s Law: Review

!E =!Eid!A"" =

qencl

#0

Use it to obtain E field for highlysymmetric charge distributions.

Method: evaluate flux over carefully chosen “Gaussian surface”

spherical cylindrical planar

(point charge, uniform sphere, spherical shell,…)

(infinite uniform line of charge or cylinder…)

(infinite uniform sheetof charge,…)

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Spherical Symmetry: Examples

Spherical symmetry: choose a spherical Gaussian surface, radius r

Then:

Example 1 (last lecture): uniformly charged sphere (radius a, charge Q)

E =qencl

4!"0r2

!Eid!A"! = E4"r2 =

qencl

#0

!Er>a

=Q

4!"0r2r

!Er<a

=Qr

4!"0a3r

Example 2: Thin Spherical ShellFind E field inside/outside a uniformly charged thin

spherical shell, surface charge density σ, total charge q

Gaussian Surfacefor E(outside)

Gaussian Surfacefor E(inside)

Er<a

= 0Er>a

=q

4!"0r2

qencl = 0qencl = q

Trickier examples: set up(holes, conductors,…)22.29-31, 22.54 (board)

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Cylindrical symmetry: ExamplesCylindrical symmetry.

Example: infinite uniform line of charge

!

r

!

E indep of ,

z

z !

Gaussian surface: cylinder of length L

, in radial direction

!Eid!A = E(r)2!rL =

qencl

"0

"#

!E(r) =

!

2"#0rr

Similar examples: infinite uniform cylinder, cylindrical shell

Text examples (board): 22.34-35

E =qencl

2!rL"0

Then:

Planar Symmetry: Examples

Planar symmetry. Example: infinite uniform sheet of charge.

x

y

!

E indep of x,y, in z direction

z

Gaussian surface: pillbox, area of faces=A

!Eid!A = 2EA =

qencl

!0

"" =#A

!0

!E =

!

2"0

z

!

Examples: multiple charged sheets, infinite slab (hmwk)

Text example (board): 22.24 (parallel plate)

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Electric Potential Energy and Electric Potential

Kinetic Energy (K). Potential Energy U: for conservative forces(can be defined since work done by F is path-independent)

If only conservative forces present in system,conservation of mechanical energy: constant

Conservative forces: Springs: elastic potential energy Gravity: gravitational potential energy Electrostatic: electric potential energy (analogy with gravity)

Nonconservative forces Friction, viscous damping (terminal velocity)

Review: Conservation of Energy (particle)

K =1

2mv

2

U(x, y, z)

K +U =

U = kspringx22

Electric Potential EnergyGiven two positive charges q and q0: initially very far apart,

choose

To push particles together requires work (they want to repel).Final potential energy will increase!

If q fixed, what is work needed to move q0 a distance r from q?

Note: if q negative, final potential energy negativeParticles will move to minimize their final potential energy!

q q0U

i= 0

!U =U f "Ui = !W

q q0

!W =!Fus id!l

"

r

# = $!Fe id!l

"

r

# = $kqq

0

%r 2d %r =

"

r

#kqq

0

r

q q0r

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Electric Potential EnergyElectric potential energy between two point charges:

U is a scalar quantity, can be + or - convenient choice: U=0 at r= ∞ SI unit: Joule (J)

Electric potential energy for system of multiple charges:sum over pairs:

r

q0

qU(r) =kq

0q

r

U(r) =kqiqj

rijj

!i< j

!

Text example: 23.54This is the work required to assemble charges.

Electric PotentialCharge q0 is subject to Coulomb force in electric field E.

Work done by electric force:

W =

!Fid!l = q

0i

f

!!Eid!l =

i

f

! " #U

independent of q0

!V "!U

q0

= #!Eid!l

A

B

$ = VB #VA

Electric Potential Difference:

Often called potential V, but meaningful only as potential difference

Customary to choose reference point V=0 at r = ∞ (OK for localized charge distribution)

Units: Volts (1 V = 1 J/C)

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Electric Potential and Point Charges

For point charge q shown below, what is VB-VA ?

A

B

rA

rB

q

Equipotentials:lines of constant potential

VB !VA = ! E(r)dr = !A

B

" kqdr

r2

A

B

" = kq1

rB!1

rA

#

$%&

'(

Many point charges: superposition

V (r) =kq

r

V (r) = kqi

rii

!

Potential of point charge:

independent of path b/w A and B!

Electric Potential: Continuous Distributions

Two methods for calculating V:

1. Brute force integration (next lecture)

2. Obtain from Gauss’s law and definition of V:

V (r) = !!Eid!l

ref

r

"

Examples (today or next lecture) : 23.19, parallel plate

dV = kdq

r,V = dV!

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Obtaining the Electric Field Fromthe Electric Potential

Three ways to calculate the electric field: Coulomb’s Law Gauss’s Law Derive from electric potential

Formalism

dV = !

!Eid!l = !Exdx ! Eydy ! Ezdz

Ex= !

"V

"x , E

y= !

"V

"y ,E

z= !

"V

"z or

!E = !#V

!V = "

!Eid!l

A

B

#