Gauss’ Law
c
a
b
Phys 122 Lecture 5
Overview of Lecture
•
•
•
Gauss' Law: Motivation & Definition
Coulomb's Law as a consequence of Gauss' Law
Charges on Conductors:
• Where are they?
•
Applications of Gauss' Law
•
•
•
•
•
Infinite Line of Charge
Infinite Sheet of Charge
Two infinite sheets of charge
Shortcuts
Uniform Charged Sphere
• The rather challenging Checkpoint Questions
PreLecture Thoughts
Overwhelming feeling:
• Found this one (Gauss’ Law
Prelecture) to extremely confusing.
Gauss' Law
The net electric flux through any closed surface is
proportional to the charge enclosed by that surface.
– TRUE always, but it not always easy to use
– Use when physical situation exhibits SYMMETRY
– Choose a closed surface to make integral trivial
– Direction: surface such that E is either parallel or perpendicular each
piece of surface
– Magnitude: chose surface such that E has same value at all points on
the surface when E is perpendicular to the surface.
CheckPoint
D) The field cannot be calculated using Gauss’ Law
E) None of the above
THE CUBE HAS NO GLOBAL SYMMETRY!
THE FIELD AT THE FACE OF THE CUBE IS NOT PERPENDICULAR OR PARALLEL
3D
2D
1D
POINT
LINE PLANE SPHERICAL
CYLINDRICAL
PLANAR
Example: Rediscovering Coulomb’s Law
What is E field around a point charge?
Symmetry  E field of point charge is radial
and spherically symmetric
Draw a sphere of radius R centered on the charge.
E is normal to every point on that surface

E has same value at every point on surface
 can take E outside of the integral!
Gauss' Law 
We are free to choose the surface in such problems…we
call this a “Gaussian” surface
E
R
+Q
CheckPoint
A charged spherical insulating
shell has inner radius a and outer
radius b. The charge density on the
shell is ρ.
What is the magnitude of the E-field at a distance r away from the center of the
shell where r < a?
CheckPoint
What is direction of field OUTSIDE the red sphere?
Gauss’ Law: Field depends on enclosed charge
Let’s try another: Uniform charged sphere
What is the magnitude of the electric
field due to a solid sphere of radius a
with uniform charge density  (C/m3)?
r
a

• Outside sphere: (r > a)
•
Spherical symmetry centered on the center of the sphere of charge
•
Choose Gaussian surface to be hollow sphere of radius r

Gauss’
Law
Uniform charged sphere
• Outside sphere: (r > a)
a
r

• Inside sphere: (r < a)
•
Spherical symmetry, centered on the center of the sphere of
charge.
•
Choose Gaussian surface to be sphere of radius r.
Gauss’
Law
But,
E
Thus:
a
r
How is charge carried on macroscopic objects?
Assume (for now) that only two kinds of objects in the world:
Insulators.. Once charged, the charges CANNOT MOVE.
Plastics, glass, and other “bad conductors of electricity” are good
examples of insulators.
Conductors.. Here, the charges ARE FREE TO MOVE.
Metals are good examples of conductors.
How do the charges move in a
conductor?
DEMO: Faraday Pail
Charge the inside, all of this charge
moves to the outside.
Recall Conductors Charges Free to Move
Claim: E  0 inside any conductor at equilibrium Charges in conductor move to make E field zero inside. (Induced charge distribution). If E ≠0, then charge would experience a force and move!
Claim: Excess charge on conductor is only on surface
To demonstrate this, we apply Gauss’ Law
-- Take Gaussian surface to be just inside conductor surface
-- E  0 everywhere inside conductor

 
E  dA  0
surface
-- Gauss’ Law:
  Qenc
 E  dA 
surface
o
Qenc  0
E0
Charges on a Conductor
We have just seen that the charges always move to the surface of a
conductor.
– Gauss’ Law tells us that …
– E = 0 inside a conductor when in equilibrium
– If E  0, then the charges would have forces on them and
they would move !
Therefore the charge on a conductor must only reside on the
surface(s)
++++++++++++
++++++++++++
Infinite conducting plane
+
+
+
+
+
+
+
+
Conducting sphere
Clicker
A Q = -3 C charge is surrounded by
an uncharged conducting spherical
shell (in yellow)
2
X
Compare the electric field at point X to the
one you would find if the conducting shell
was removed.
(a) Eshell < ENoShell
(b) Eshell = ENoShell
1
Q
(c) Eshell > ENoShell
Select a sphere passing through the point X as the Gaussian surface.
How much charge does it enclose?
Answer: Q, whether or not the uncharged shell is present.
(The field at point X is determined only by the objects with NET CHARGE.)
Clicker
A Q = -3 C charge is surrounded by
an uncharged conducting spherical
shell (in yellow)
2
r1
r2
Q
What is the value of the surface charge density
1 on the inner surface of the conducting shell?
(a) 1 = -Q (b) 1 = +Q
1
ொ
(c) 1 = 0 (d) 1 = ସగ௥ మ
భ
(e) 1 =
ିொ
ସగ௥భమ
Inside the conductor, we know the field: E = 0
Select a Gaussian surface inside the conductor
Since E = 0 on this Gaussian surface, the total enclosed charge must be 0
Therefore 1 must be positive, to cancel the charge Q=-3 C
Infinite Line of Charge
Symmetry  E field must be  to
line and can only depend on
distance from line
y
Er
Er
CHOOSE Gaussian surface to be a
cylinder of radius r and length h
+ + +++++++ + +++++++++++++ + + + + + +
aligned with the x-axis.
Apply Gauss' Law:
h
On the ends,
On the barrel,
and

x
Clicker
A line charge  C/m is placed along the axis of
an uncharged conducting cylinder of inner
radius ri = a, and outer radius ro = b as shown.
What is the value of the charge density
o (C/m2) on the outer surface of the
cylinder?
a

(c)
(b)
(a)
b
View end on:
Draw Gaussian tube which surrounds only the outer edge
o
b
0
Infinite sheet of charge

Symmetry:
direction of E = x-axis
CHOOSE Gaussian surface to be a
cylinder whose axis is aligned with
the x-axis.
Apply Gauss' Law:
A
x
E
E
On the barrel,
On the ends,
The charge enclosed = A
Therefore, Gauss’ Law

Conclusion: An infinite plane sheet of charge creates a CONSTANT
electric field .
Two Infinite Sheets
(into screen)
Field outside the sheets must be
zero. Two ways to see:
Superposition
Gaussian surface
encloses zero charge
Field inside sheets is NOT zero:
Superposition
Gaussian surface
encloses non-zero chg
0
+
E=0 
+
+
+
+
A
+
+
+
+
A+
+
+
 E=0
E
CheckPoint
In both cases shown below, the colored lines represent
positive (blue) and negative (red) charged planes.
In which case is E at point P the biggest?
A) A B) B C) the same
Superposition:
P
P
NET

Case A



Case B
Remember: infinite planes produce uniform fields that do not vary with
distance. The two additional plates have cancelling fields at point P.
A long thin wire has a uniform positive charge density of 2.5 Cm. Concentric with the wire is a long thick conducting cylinder, with inner radius 3 cm, and outer radius 5 cm. The conducting cylinder has a net linear charge density of 4 Cm.
What is the linear charge density of the induced charge on the inner surface of the conducting cylinder iand on the outer surface o?
i :
2.5 Cm
 o:
6.5 Cm
4 Cm
0
2.5 Cm
2.5 Cm
2.5 Cm
1.5 Cm
0
4 Cm
A B C D E
o
i
A long thin wire has a uniform positive charge density of 2.5 Cm. Concentric with the wire is a long thick conducting cylinder, with inner radius 3 cm, and outer radius 5 cm. The conducting cylinder has a net linear charge density of 4 Cm.
What is the linear charge density of the induced charge on the inner surface of the conducting cylinder iand on the outer surface o?
i :
2.5 Cm
 o:
6.5 Cm
4 Cm
0
2.5 Cm
2.5 Cm
2.5 Cm
1.5 Cm
0
4 Cm
A B C D E
o
i
Gauss’ Law tips (look at offline)
• Gauss’ Law is ALWAYS VALID!!
If you have (a) spherical, (b) cylindrical, or (c) planar symmetry AND:
If you know the charge (RHS), you can calculate the electric
field (LHS)
If you know the field (LHS, usually because E=0 inside
conductor), you can calculate the charge (RHS).
• Spherical Symmetry: Gaussian surface = Sphere of radius r
LHS:
RHS: q = ALL charge inside radius r
• Cylindrical Symmetry: Gaussian surface = Cylinder of radius r
LHS:
RHS: q = ALL charge inside radius r, length L
• Planar Symmetry: Gaussian surface = Cylinder of area A
LHS:
RHS: q = ALL charge inside cylinder=A
Smart Clicker: Charge in Cavity of Conductor
A particle with charge Q is placed in the center of an uncharged conducting hollow sphere. How much charge will be induced on the inner and outer surfaces of the sphere?
Qouter
A) inner = Q, outer = Q
B) inner = Q/2 , outer = Q/2
Qinner
C) inner = 0, outer = 0
Q
D) inner = Q/2, outer = Q/2
E) inner = Q, outer = Q
Since E = 0 in conductor

  Q
E  dA  enc
 Gauss’ Law:surface
o
Qenc  0
Smart Clicker: Infinite Cylinders
A long thin wire has a uniform positive charge density of 2.5 Cm. Concentric with the wire is a long thick conducting cylinder, with inner radius 3 cm, and outer radius 5 cm. The conducting cylinder has a net linear charge density of 4 Cm.
What is the linear charge density of the induced charge on the inner surface of the conducting cylinder iand on the outer surface o?
i :
2.5 Cm
o:
6.5 Cm
4 Cm
0
2.5 Cm
2.5 Cm
2.5 Cm
1.5 Cm
0
4 Cm
o
A B C D E
• E = 0 in conducting cylinder means net charge
enclosed = 0
• Thus,  = -2.5 C/m to cancel wire
• If NET is -4 C/m and inside is -2.5 C/m, then outside
has to be -1.5 C/m to add up
• Follow up for you: What is field outside based on?
i