Physics 202, Lecture 4
Today’s Topics

Review: Gauss’s Law

Electric Potential (Ch. 25-Part I)
 Electric Potential Energy and Electric Potential
 Electric Potential and Electric Field

Next Tuesday: Electric Potential (Ch. 25-Part II)

Homework #1 due tomorrow (9/14) at 10 PM
Homework #2 (now on WebAssign) due 9/24 at 10 PM
Gauss’s Law: Review
! !
! E = " EidA =
#q
in
$0
Fundamental equation of electrostatics
(equivalent to Coulomb’s Law)
Can use it to obtain E for highly symmetric charge distributions.
Method: evaluate flux over carefully chosen “Gaussian surface”:
spherical
cylindrical
(point chg, uniform
(infinite uniform line of
sphere, spherical shell,…) charge or cylinder…)
planar
(infinite uniform sheet
of charge,…)
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Gauss’s Law: Examples
1. Spherical symmetry (last lecture).
2. Cylindrical symmetry. Example: infinite uniform line of charge.
z
Symmetry: E indep of
!
!
3. Planar symmetry.
Example: infinite uniform sheet of charge.
!
!
, in radial direction
Gaussian surface: cylinder of length L
! !
q
$L
"# EidA = E(r)2! rL = "in0 = " 0
!
!
E(r) ==
r̂
2"# 0 r
r
y
z, !
z
x
Symmetry: E indep of x,y, in z direction
Gaussian surface: pillbox, area of faces=A
! !
!
q
#A
!
E=
ẑ
" EidA = 2EA = !in0 = ! 0
2" 0
Exercise: try for E field just outside of a conductor
Electric Potential Energy and Electric Potential
Review: Conservation of Energy (particle)
 Kinetic Energy (K)
K=
1 2
mv
2
 Potential Energy U: conservative forces
(work independent of path)
U(x, y, z)
 If only conservative forces present in system,
conservation of mechanical energy: K + U = constant
 Examples of conservative forces:
 Springs: elastic potential energy
U = kspring x 2 2
 Gravity: gravitational potential energy
 Electrostatic: electric potential energy (today)
 Examples of nonconservative forces
 Friction, viscous damping (terminal velocity)
2
Electric Potential Energy (I)
Compare with gravitational force (Ch. 13):
!
mm
F12 = !G 1 2 2 r̂12
r
W=
! ! Gm1m2 Gm1m2
Fid
! path s = rf " ri
Gravitational Potential energy:
U =!
path independent!
Gm1m2
r
 Electric Force:
!
qq
F12 = ke 1 2 2 r̂12
r
Electric
U=
Potential Energy
ke q1q2
r
Electric Potential Energy (II)
Given two positive charges q and q0:
q0
q
Initially charges very far apart: U i = 0
(we are free to define the potential energy zero somewhere)
To push particles together requires work (they want to repel).
q
q0
Final potential energy will increase! !U = U f " U i = !W
Now, suppose q is fixed at the origin. What is work required to
move q0 from infinity to a distance r away from q?
q
q0
r
! !
! !
k qq
k qq
!W = # Fus ids = $ # Fe ids = $ # e 2 0 dr % = e 0
r%
r
"
"
"
r
r
r
Note: if q negative, final potential energy negative
Particles will move to minimize their final potential energy!
3
Electric Potential Energy: Summary
 Electric potential energy between two point charges:
U(r) =
ke q0 q
r
q
r
 U is a scalar quantity, can be + or  convenient choice: U=0 at r= ∞
 SI unit: Joule (J)
q0
 Electric potential energy for system of multiple charges:
sum over pairs:
ke qi q j
U(r) = ! !
rij
i< j j
Integral if continuous distribution
Example: Three Charge system
 What is work required to assemble the three charge
system as shown? (q1=q2 =q3=Q)
q3
Answer: ke 3Q2/a (see board)
a
q1
a
a
q2
 What if q1=q2=Q but q3=-Q ?
Answer: -keQ2/a
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Electric Potential Energy:
Charge In An Electric Field
 Charge q0 is subject to Coulomb force in electric field E:
!
!
F = q0 E
 Work done by electric force:
W=
!
i
f
! !
f !
!
Fids = q0 ! Eids = " #U
i
!U = U f " U i = "q0 #
i
f
! !
Eids
independent of q0
Electric Potential Difference
 Electric Potential Energy: q0 In a Generic E. Field
!U = U B " U A = "q0 #
system potential energy
B
A
! !
Eids = q0 !V
test
source
 Electric Potential Difference
B !
!U
!
!V "
= # $ Eids = VB # VA
A
q0
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Properties of the Electric Potential
 Results from conservative nature of the electric force
 associated with source field only (indep. of test charge)
 units: J/C ≡ Volt (V)
 often called potential, but meaningful only as
potential difference VB-VA.
A convenient point (! , earth..) typically chosen as “ground” 
ΔV=V-(VA=0) =V
 scalar quantity (no vector operations necessary!)
 related to electric potential energy by ΔU=q0ΔV
Exercise 2: E. Potential and Point Charges
In the configuration shown, find VB-VA
B
rB
Answer:
VB-VA = ke(q/rB-q/rA)
A
rA
q
(See board)
6
Exercise 1: Uniform E. Field
In the uniform electric field shown:
1. Find potential at B,C,D,G
2. If a charge +q is placed at B,
what is the potential energy UB?
3.If now a –q is at B, what is UB ?
4. If a -q is initially at rest at G,
will it move to A or B?
C
D
B
A
VA=0
G
d
5. What is the kinetic energy when it reaches A?
A Picture to Remember
+q
-q
higher V
E
lower V
 Field lines always point towards lower electric potential
 In an electric field:
 positive charges are always subject to a force in the
direction of field lines, towards lower V
 negative charge is always subject a force in the
opposite direction of field lines, towards higher V
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Obtaining the Electric Field From
the Electric Potential
 Three ways to calculate the electric field
 Coulomb’s Law E=ΣEi
 Gauss’s Law
 Derive from electric potential 
 Formalism
B !
!
!V = " # Eids
A
! !
dV = ! Eid s = !Ex dx ! Ey dy ! Ez dz
"V
Ex = !
"x
"V
, Ey = !
"y
"V
,Ez = !
"z
or
!
E = !#V
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