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AP Physics C
Mrs. Coyle
Part I
 Electric Flux
 Gauss’s Law
 Gauss’s Law allows us to calculate the electric field in
an alternative way to what we did so far.
 It makes the calculation easier!
 It utilizes the electric field lines.
 It utilizes electric flux.
 It utilizes symmetry.
 Electric field
lines due
individual point
charges
Is there symmetry?
Electric Field Lines of an Infinitely Long Cylinder
Is there symmetry?
Infinitely Long Sheet
Is there a charge enclosed in this box?
Is there a charge enclosed in this box?
Is there a charge enclosed in this box?
Gaussian Surface
A Gaussian surface is an
imaginary closed surface
around a charge.
Electric Flux, F:
The number of electric (flux) field lines which pass
through a given cross-sectional area A.
This is a side view of a ring:
Which has the largest electric flux?
Does flux depend on direction?
Answer:1
Area vector A is perpendicular to
the surface A and has a magnitude
equal to the area A.
Electric Flux through a Gaussian Surface
 Flux is a dot product and scalar
 Units:
[F ]=
N 2
m
C
[A]= area m2
 q: angle formed between E and the normal to the surface.
 The circle on the integral means “surface integral” over the
entire closed surface.
When E is perpendicular to the loop:
F= EA
Why?
Prob # 1: Finding net flux
An electric field with a magnitude of 4 N/C is applied along
the x axis.
Calculate the electric flux through a 0.50m cube that lies on
the origin and on xyz axis.
Strategy:
-Add the EA’s for each of the six sides.
-If flux lines are exiting a surface
the flux is (+).
Ans: 0
Gauss’s Law: the total electric flux
through a closed surface is proportional
to the enclosed charge.
-This constant is used with Gauss’s law for vacuum (free space).
Note: the dot product
Note:
 The “Gaussian” surface can have any shape. You choose
the surface.
 The flux is positive for exiting field lines, and negative
for entering.
 The flux due to a charge qenclosed is the same regardless
of the shape or size of the surface.
 If the surface contains no charge, the flux through it is
zero. That means that every field line that enters the
surface will also exit the surface.
More Notes on Gauss’s Law
 The normal area vector is taken to point from the
inside of the closed surface out.
 Gauss’s Law is also known as Maxwell’s first equation.
Prob. 2: The Electric Field due to a point charge. Prove
E=kq/r2 at a point at a distance r from the point charge.
Hint: The surface area of a sphere is 4pr2
What do you think would be your answer for
E, if you chose a different Gaussian surface?
 You can pick any surface, but the math is easier if you
pick a surface with the same symmetry as the charge.
Prob #13

Part II- Applications of Gauss’s Law for
Continuous Charge Distributions
Prob. 3: Show E= q /(2 Aε0) for a
rectangular metal slab positively charged
with a uniform surface charge density.
Problem 4: Field Due to a Spherically Symmetric
Charge Distribution of an Insulator
Outside the sphere( r>a)
Outside the sphere
Ans:
E  ke
Q
r2
Problem 4: Field Due to a Spherically Symmetric Charge
Distribution of an Insulator Inside the sphere( r<a)
 qin < Q and since the
volume charge density is
uniform:
qin
Q
r3

 qin  Q 3
3
3
4__p a
4__p r
a
3
3
Problem 4: Field Due to a Spherically
Symmetric Charge Distribution (Insulator)
Problem 5: Field Due to a Thin Spherical Shell
 For r > a, the enclosed charge is Q and E = keQ / r2
 For r < a, the charge inside the surface is 0 and E = 0
Problem 5: Field due to a Line of Charge
Problem 5: Field Due to a Line of Charge
End view
 The flux through the
ends of the gaussian
cylinder is 0. Why?
Problem 5: Field Due to a Line of Charge
qin
F E   E  dA   EdA 
εo
λ
E  2πr  
εo
λ
λ
E
 2ke
2πεo r
r