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Physics 2112
Unit 4: Gauss’ Law
Today’s Concepts:
A) Conductors
B) Using Gauss’ Law
Unit 4, Slide 1
Use #1: Determining E without calculus
E from infinite line of charge.
We did this before with calculus.
Remember?

E x ( P )  k
f
h * sec  * sin 
2
 d  ( h * tan  )
0
2
h
2

Let’s do it again using Guass Law
Unit 4, Slide 2
Use #2: Determine Charge on Surfaces
E  0 inside any conductor at equilibrium
(If E ≠0, then charge feels force and moves!)
Excess charge on conductor only on surface
at equilibrium
E0
Why?
 Apply Gauss’ Law
 Take Gaussian surface to be just inside conductor surface



Q enc
E  dA  0 
surface
o
Q enc  0
Unit 4, Slide 3
Gauss’ Law + Conductors + Induced Charges
How Does This Work?
Charges in conductor move to surfaces to make
Qenclosed  0.
We say charge is induced on the surfaces of
conductors

 Q

E  d A  en c
su rfa ce
o
ALWAYS TRUE!
Unit 4, Slide 4
Example 4.1
An infinite line of charge with linear
density λ1 = -5.1μC/m is positioned
along the axis of a neutral conducting
shell of inner radius a = 2.7 cm and
outer radius b = 5.0 cm and infinite
length.
A) What is λb, the linear charge density on the outer
surface of the conducting shell ?
B) What is Ex(R), the electric field at point R, located a
distance dR = 0.9 cm from the origin and making an angle
of 30o with respect to the y-axis as shown?
Unit 4, Slide 5
Example 4.2
xR
An infinite line of charge with linear
density λ1 = -5.1μC/m is positioned
along the axis of a thick conducting
shell of inner radius a = 2.7 cm and
outer radius b = 5.0 cm and infinite
length. The conducting shell is
uniformly charged with a linear charge
density λ 2 = +2.0 μC/m.
C) What is Ex(R), the electric field at point R, located a
distance dR = 7.0 cm from the origin and making an angle
of 30o with respect to the y-axis as shown?
Unit 4, Slide 6
Gauss’ Law
  Q
enc
E

d
A


ALWAYS TRUE!
0
In cases with symmetry can pull E outside and get
E 
Q enc
A 0
In General, integral to calculate flux is difficult…. and not useful!
To use Gauss’ Law to calculate E, need to choose surface carefully!
1) Want E to be constant and equal to value at location of interest
OR
2) Want E dot A  0 so doesn’t add to integral
Unit 4, Slide 7
CheckPoint: Charged Conducting Sphere & Shell 1
A positively charged solid conducting sphere is contained
within a negatively charged conducting spherical shell as
shown. The magnitude of the total charge on each sphere is
the same. Which of the following statements best describes
the electric field in the region between the spheres?
A. The field points radially outward
B. The field points radially inward
C. The field is zero
Unit 4, Slide 8
Gauss’ Law Symmetries

  Q
E  d A  enc
ALWAYS TRUE!
0
In cases with symmetry can pull E outside and get E 
Spherical
A  4 r
E 
2
Q enc
4 r  0
2
Cylindrical
Planar
A  2  rL
A  2 r
E 

2 r  0
E 
Q enc
A 0
2

2 0
Unit 4, Slide 9
CheckPoint: Gaussian Surface Choice
You are told to use Gauss' Law to calculate the electric field at a distance R away from
a charged cube of dimension a. Which of the following Gaussian surfaces is best suited
for this purpose?
A.
B.
C.
D.
E.
a sphere of radius R+1/2a
a cube of dimension R+1/2a
a cylinder with cross sectional radius of R+1/2a and arbitrary length
This field cannot be calculated using Gauss' law
None of the above
Electricity & Magnetism Lecture 4, Slide 10
CheckPoint: Charged Sphericlal Shell
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?
A.
B.
C.
D.
ρ/εo
zero
ρ(b3-a3)/(3εor2)
none of the above
Electricity & Magnetism Lecture 4, Slide 11
Example 4.3
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 a< r < b?
Electricity & Magnetism Lecture 4, Slide 12
Quick Review: Infinite Sheet of Charge
Unit 4, Slide 13
CheckPoint: Infinite Sheets of Charge
In both cases shown below, the colored lines represent
positive (blue) and negative (red) charged planes. The
magnitudes of the charge per unit area on each plane is the
same. In which case is the magnitude of the electric field at
point P bigger?
A. Case A
B. Case B
C. They are the same
Unit 4, Slide 14
Example 4.4
y
r2
Point charge 3Q at center of
neutral conducting shell of inner
radius r1 and outer radius r2.
neutral
conductor
3Q
r1
x
What is E for the three regions of:

r < r1

r1< r < r2

r2 < r
?
Electricity & Magnetism Lecture 4, Slide 15
Calculation
y
r2
Point charge 3Q at center of neutral
conducting shell of inner radius r1 and outer
radius r2.
neutral
conductor
3Q
r1
  Q
enc
E

d
A


0
x
What is E everywhere?
r < r1
r > r2
r1 < r < r2
E 
1
3Q
4  0 r
2
E 0
Electricity & Magnetism Lecture 4, Slide 16

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