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PHYS219 Fall Semester 2014
Lecture 4:
Gauss Law & Potential Energy
Dimitrios Giannios
Purdue University
Help Center Info (see syllabus link)
Remember: Homework 1 is due this Friday,
10:30am before the recitation!
Electric Flux and Gauss’ Law
• Gauss’ Law can be used to find the electric field of a
complex charge distribution
• Easier than treating it as a collection of point charge
and using superposition
• To use Gauss’ Law, a quantity called electric flux is
needed
• The electric flux is equal the product of the electric
field that passes through a particular surface and
the area of the surface
• The electric flux is denoted by E
The idea of an electric flux
How do you account
(in a very general mathematical way) for the E fields
that fill space?
How many E-field
lines penetrate a small
surface with area dA?
Definition of the electric flux,
E||=|E|sin
E =|E|cos
E
Must be able to visualize
the components of E
n̂
DEFINE
E =dA x E
= dA E cos θ
[units]: Nm2/C
Area dA
Special cases:
Gauss’ Law
Think about closed surfaces: two possibilities
E ,total≈
q
dA
dA
What can be shown:
For ANY closed surface
ΦE,total=closed surfacedA E cos θ = 4 π k qenclosed
dA
when no net charge
lies inside the surface
define k
1
4 π ε0
ε0 =8.85 x 10-12 C2 / m2 N
qenclosed
Gauss ' s Law:
E ,total
ε0
Applications of Gauss Law
Calculate the electric field for charge configurations
with some symmetry:
Gauss’ Law: Point Charge
• Choose a Gaussian
surface
• Want a surface that will
make the calculation as
easy as possible
• Choose a surface that
matches the symmetry
of the problem
• For a point charge, the
field lines have a
spherical symmetry
Gauss’ Law: Point Charge (more)
• Since the field is perpendicular to the area, the flux is
•
•
•
•
the product of the field and the area: E = E Asphere
Asphere is the area of the Gaussian sphere
With a radius of r, Asphere = 4  r2
Therefore, E = 4  r2 E
From Gauss’ Law,
FE = 4p r E =
2
q
eo
and
E=
q
4pe o r 2
• This agrees with the result from Coulomb’s Law
Applications of Gauss law will follow in next lectures
Part II: Electric Potential Energy
• Electric forces can do work on a charged object
• Work is related to changes in electric potential energy
• Analogous in many ways to gravitational potential energy
• Conservation of energy will be revisited
• The ideas of forces, work, and energy will be extended
to electric forces and systems
Electric Potential Energy
• A point charge in an electric field experiences a
force:
• Assume the charge moves a distance Δx
• The work done by the electric force on the charge is
W = FE Δx
• The electric force is conservative, so the work done
is independent of the path
• If the electric force does an amount of work W on a
charged particle, there is an accompanying change
in electric potential energy
Electric Potential Energy, cont.
• The change in electric
potential energy is
Δ(PE)E = - W = - FE Δx =
-q E Δ x
• The change in potential
energy depends on the
endpoints of the motion,
but not on the path
taken
Δ(PEE) + W=0
Notation: PE=Potential Energy (J)
KE=Kinetic Energy (J)
W=work (J)
Gravitational Potential
Energy (PEG)
Electrostatic Potential
Energy (PEE)
end
end
Δr
Δh
begin
+
m
q1 +
Conventions:
near earth’s surface, PEG=0
for orbiting masses, PEG=0 at infinity
earth
+ q2
begin
“fixed”
PEE=0 (by definition) when
separation is at infinity
“fixed”
Electrostatic potential energy is the energy stored in a
system of charges. It depends on the relative positions
(not the motion) of the charges in the system.
Gravity and Electrostatics
What is the same, what’s different between
What’s the Same: PE can be converted to KE and vice
versa. PE can be converted to W and vice versa.
What’s Different:
• Gravity is attractive: Gravitational Potential Energy
is negative (when set to 0 at infinite distance
• Electrostatic Potential Energy both positive and
negative (depending on the sign of charges)
Bottom Line: The situation for
electrostatics is similar to gravity, except
you can’t see the hills and valleys.
Need an imagination!
Review: Gravitational Analogy
(Potential Energy is Stored Energy)
PEG increases
Startin
g
position
m
Reference
Level
Hill
Δ(PEG)+W=0
Work done
on the ball
It’s basically
Conservation of
Energy
Hole
PEG decreases
Δ(PEG)+W=0
Work done
by the ball
CASE I: Change in Electrostatic Potential
Energy (case of like charges)
end
PEE
Must push “+” charge
with external force
Notes:
•The change in
PEE is a positive Q
number
•Charges are
stationary at
beginning and end
(KE=0)
•Same diagram
if both charges
are negative
begin
+ +
PEE increases, W is negative
r
+ q
An amount of
work W is
required
Δ(PEE) + W=0
The work done by the electric
force is negative
Electrostatic Potential Energy
(case of unlike charges)
CASE II: Change in PEE
Note that the
change in PEE is a
positive number
Q
Must push “+” charge
with external force
- +
PE increases
Charges are
stationary at
beginning and end
(KE=0)
begin
PEE increases, W is negative
r
end
+
q
An amount of
work W is
required
Δ(PEE) + W=0
13e electric
The work is done by th
force is negative
Electrostatic Potential Energy (more)
Now return the charge to original
position. What is the change in
Electrostatic Potential Energy?
PEE
Charges are
stationary at
beginning and end
(KE=0)
Q
Must hold back “+”
charge with external
force
r
+
- +
PE
decreases
begin
+
q
Work will be
recovered
Δ(PEE) + W=0
end
PEE decreases, W is positive
We say work is done by the
electric fields of the charges
Making the Discussion Quantitative
(point charges ONLY)
ri
rf
Move charge q2 from ri to rf
What is the PEE stored in the system?
ri infinity
PEE,i = 0
Convention: PEE is zero
at infinite distance
0
PEE
D(PEE ) = PEE,f - PEE,i
separation
kq1q2 kq1q2
=
rf
ri
PEE,f
kq1q2
=
rf
0
NOTE
It is important to use the signs
associated with each charge!!!
When the signs of the two charges
are opposite, the potential energy
of the pair is negative.
When the signs of the two charges
are the same, the potential energy
of the pair is positive.
kq1q2
PEE =
r
An example
-2 C
kq1q2
PEE =
r
3 C
-
+
PEE (r
Electrostatic
Potential Energy
PEe
2.5 m
q1
-
2.5m) = -2.16 x1010 J
What does the minus sign mean??
q2
+
r
Extracting PE from a system wrt infinite
separation indicates a preference toward
confinement, like digging a hole on the beach
in gravitational situation.
PEelec and Superposition
• The results for two point charges can be extended
by using the superposition principle
• If there is a collection of point charges, the total
potential energy is the sum of the potential energies
kq1q2 kq2q3 kq1q3
of each pair of charges
PEE =
+
+
r12
r23
r13
question: how many
potential terms do
you have for 4
charges?
1+2+3 = 6 terms
(n-1) x n/2 for n charges