Lecture 6
Gauss’s Law and
Applications
Ch (27) in the book of
Randall D. Knight, Physics for Scientists and Engineers, 3rd edition.
Flux and Charge
Consider the flux through a spherical surface of
radius R that surrounds a point charge q located at
its center.
This result is independent of how big we choose
to make the spherical surface. A very large or a
very small surface will have the same flux, which
depends only on the size of the charge q at the
center of the sphere.
Essentially, q/e0 is the number of field lines
attached to charge q, and e0 is the charge
producing one unit of flux.
2
Flux is Independent of Surface Shape and Radius
What about the flux when the surface has an arbitrary non-spherical shape?
We can approximate such a surface by a set of spherical pieces. Summing
over these gives Fe=q/e0.
We conclude that the flux through an arbitrary surface
always depends only on the charge q enclosed by the surface.
3
Charge Outside the Surface
What about the flux through a surface from any
charges not enclosed by the surface?
There is no net flux because entering and exiting
fluxes cancel (see Figure below).
4
Multiple Charges
So far we have considered only
the flux from a single point charge.
What about the case of multiple
charges?
F e   E  dA
  E1  dA   E2  dA   E3  dA 
 F1  F 2  F 3 

q1
e0

q2
e0

q3
e0

Qin  q1  q2  q3 
Qin
Therefore, Fe   E  dA 
e0
5
Gauss’s Law
The previous arguments lead to the conclusion that
the correspondence between the charge Qin enclosed
by a surface and the net flux Fe through that surface
is a general result. It is called Gauss’s Law, and is
usually written as:
F e   E  dA 
Qin
e0
Johann Carl Friedrich Gauss
(1777 – 1855)
Gauss’s Law is the first of four master equations, collectively called
Maxwell’s Equations, that together constitute a “unified field theory” of
electromagnetism.
In essence, Gauss’s Law says that diverging field lines from a point indicate
the presence of an electric charge at that point, and that this charge can be
“detected” by surrounding the point with a surface and observing the flux
through the surface.
6
Gaussian Spherical Geometry
Spherical distributions of charge
have spherical symmetry, so the
electric field must also have
spherical symmetry.
This means that the electric
field can have only radial
components and can depend only
on radius. If a concentric
spherical Gaussian surface is
constructed around the charge
distribution, the E-field and dA
surface vectors will be parallel,
and the magnitude of the E-field
will be constant over the surface.
Therefore, Fe = E·dA = EA,
greatly simplifying the flux
calculation.
7
Example: E Inside a hollow spherical shell of Charge
What is the electric field Ei at some
point x inside a hollow spherical shell of
radius R with total charge Q distributed
uniformly over the shell?
xr
R
Q
Solution
Put a concentric spherical Gaussian surface through
x, so that it has radius r<R. From the spherical
symmetry of the charge distribution, the electric field
Ei on this surface must be strictly radial (in spherical
coordinates).
Then Ei and dA are parallel everywhere on the
surface, so Fe=Ei(4pr2)=Qin/e0. But Qin=0 because all
the charge lies outside the Gaussian sphere.
Therefore, Ei=0.
In other words, the electric field anywhere inside
a uniform spherical shell of charge is zero.
8
Example: The E-Field Outside a Charged Spherical Shell
What is the electric field Eout at some
point x outside a hollow spherical shell of
radius R with total charge Q distributed
uniformly over the shell?
Solution
x
r
R
Q
Put a concentric spherical Gaussian surface
through x, so that it has radius r>R. From the
spherical symmetry of the charge distribution,
the electric field Eout on this surface must be
radially outward (in spherical coordinates).
Then Eout and dA are parallel everywhere on the surface, so Fe = EoutA =
Eout(4pr2)=Qin/e0, with Qin=Q because all of the charge lies inside the Same result we
Gaussian sphere. Therefore, Eout = (Qin/e0)/(4pr2) = (1/4pe0)Q/r2.
got in Lect. 4,
page 11
In other words, the electric field anywhere outside a uniform
spherical shell of charge Q is the same as the electric field for a point
charge Q at the center of the sphere.
9
Summary: The E-Field of
a Spherical Shell of Charge
xr
out
x
rin
R
Q
Eoutside  (1/ 4pe 0 )Q / r 2
Einside  0
Esurface  (1/ 4pe 0 )Q / R 2
  / e0
Same result we got
in Lect. 4, page 11
10
Stop To Think 27.4
These are two-dimensional cross sections through three-dimensional closed
spheres and a cube. Rank in order, from largest to smallest, the electric fluxes Fu to Fy
through surfaces u to y.
(u)
(v)
(w)
What is the ranking of the electric
fluxes for these systems?
(a) Fx = Fy > Fu = Fv = Fw ;
(b) Fx = Fy = Fu > Fv = Fw ;
(c) Fx = Fy > Fu = Fv > Fw ;
(d) Fv = Fy > Fu = Fw = Fx ;
(x)
(y)
Solution
(d) Fv = Fy > Fu = Fw = Fx ;
The flux through a closed surface depends only
on the amount of enclosed charge, not the size
or shape of the surface.
(e) Fv = Fy > Fu > Fw > Fx .
11
Example: The E-Field Outside a Solid Sphere of Charge
What is the electric field Eout at some
point x outside a solid sphere of radius R
with total charge Q distributed uniformly in
the volume?
Solution
Put a concentric spherical Gaussian surface
through x, so that it has radius r>R. From the
spherical symmetry of the charge distribution,
the electric field Eout on this surface must be
radially outward (in spherical coordinates).
x
+ +R
++ +
+ +
r
Q
Then Eout and dA are parallel everywhere on the surface, so Fe = EoutA =
Eout(4pr2)=Qin/e0, with Qin=Q because all of the charge lies inside the
Gaussian sphere. Therefore, Eout = (Qin/e0)/(4pr2) = (1/4pe0)Q/r2.
In other words, the electric field anywhere outside a solid uniform
sphere of charge Q is the same as the electric field for a point charge
Q at the center of the sphere and for a spherical shell of charge Q.
12
Example: The E-Field Inside a Solid Sphere of Charge
What is the electric field Ein at some
point x inside a solid sphere of radius R with
total charge Q distributed uniformly in the
volume?
Solution
More details is in page 796
of the book
+ +R
xr
+ +
++
Put a concentric spherical Gaussian surface
through x, so that it has radius r<R. From the
spherical symmetry of the charge distribution,
the electric field Ein on this surface must be
radially outward (in spherical coordinates).
Then Ein and dA are parallel everywhere on the surface, so Fe = Ein A =
Ein(4pr2)=Qin/e0, with Qin=QVG / V0 = Q(r/R)3 because only part of the charge
lies inside the Gaussian sphere. (VG is the volume of the spherical Gaussian
surface VG = 4/3 pr3 and V0 = 4/3 pR3 )
Therefore, Eout = (Qin/e0)/(4pr2) = (1/4pe0)Q(r/R)3/r2 = (1/4pe0)Qr/R3.
In other words, the electric field anywhere inside a solid uniform
sphere of charge Q is proportional to the distance r from the center
of the sphere and is equal to zero at the center.
Q
13
Summary: The E-Field of
a Solid Sphere of Charge
xr
out
x
rin
R
Q
Eoutside  (1/ 4pe 0 )Q / r 2
Einside  (1/ 4pe 0 )Qr / R3
Esurface  (1/ 4pe 0 )Q / R2
Compare the above results with page 10
14
Example: The E-Field Outside a Long Charged Cylinder
What is the electric field Ecyl at some
point x outside a long charged cylinder of
radius R and charge per unit length l
distributed uniformly through the cylinder?
Solution
x
Put a coaxial cylindrical Gaussian surface
of length L through x, so that it has radius
r>R. From the cylindrical symmetry of the
charge distribution, the electric field Ecyl
anywhere on the Gaussian cylinder must be radially outward (in cylindrical
coordinates).
Therefore, Ecyl||dA on the curved wall of the Gaussian cylinder and
Ecyl^dA on the circular end caps.
The enclosed charge is Qin=lL, so Fe = EwallAwall + EtopAtop+ EbottomAbottom =
EcylAwall = Ecyl(2prL)=Qin/e0 Therefore,
Ecyl = (Qin/e0)/(2prL) = (1/4pe0)2l/r
15
Example: The E-Field Inside a Long Charged Cylinder
What is the electric field Ecyl at some
point x inside a long charged cylinder of
radius R and charge per unit length l
distributed uniformly through the
cylinder?
Solution
R
x
r
+ + + + + +
+ + + + + +
+
+
+
+
+
+
Section of long cylinder
Put a coaxial cylindrical Gaussian surface
of length L through x, so that it has radius
r<R. From the cylindrical symmetry of the
charge distribution, the electric field Ein
anywhere on the Gaussian cylinder must be
radially outward (in cylindrical coordinates).
Therefore, Ein||dA on the curved wall of the Gaussian cylinder and Ein^dA on
the circular end caps. The Gaussian cylinder encloses only part of the charge, so
Qin=lL VG / V0 = lL(r/R)2, (VG is the volume of the cylindrical Gaussian surface
VG = L pr2 and V0 = L pR2 )
so Fe = EwallAwall+ EtopAtop+ EbottomAbottom = EinAwall = Ein(2prL)=Qin/e0.
Therefore, Ein = (Qin/e0)/(2prL) = (1/4pe0)2l(r/R)2/r = (1/4pe0)2lr/R2.
16
Summary: The E-Field of a
Long Charged Cylinder
outside
Ecylinder

in solid
cylinder
E
1 2l
4pe 0 r
~
1
r
1 2l r

~r
2
4pe 0 R
in hollow
Ecylinder
0
on surface
cylinder
E
1 2l

4pe 0 R
17
Example: The E-Field Outside a Plane of Charge
What is the electric field Eplane at
some point x that is a distance L from
an infinite plane with charge per unit
area  distributed uniformly through
the plane?
L
x
Solution
Put a cylindrical Gaussian surface of length
2L and cap area A through x, so that it extends
symmetrically through the plane. From the
rotational and translational symmetry of the charge distribution, the
electric field Eplane anywhere on the Gaussian cylinder must be outward
perpendicular to the plane and must not depend on where the axis of the
cylinder passes through the plane. Therefore, Eplane ^dA on the curved wall
of the Gaussian cylinder and Eplane || dA is constant on the circular end caps.
The enclosed charge is Qin=A, so Fe = EtopA + EbottomA = A(Etop+ Ebottom) =
2EplaneA=Qin/e0= A/e0. Therefore, Eplane = /2e0.
18
Conductors in Electrostatic Equilibrium
The electric field must be zero
inside any conductor in electrostatic
equilibrium (charges are all
stationary) If this weren’t true,
the electric field would cause the
charge carriers to move and thus
violate the assumption that all the
charges are at rest
Now suppose that we place a
Gaussian surface just inside the
“skin” of a conducting object. The
flux through this Gaussian surface is
Fe= 0 because E= 0. But Fe= Qin/e0, so
Qin= 0 also.
Therefore, all of the charge present
in a charged conductor resides on the
outside surface of the conductor.
19
The Surface Field of a Conductor
Suppose that the electric field at the surface
of a conductor had a component tangent to the
surface. Then mobile electric charges at the
surface would feel a force and would move in
response to the field, until equilibrium was
established and the tangential field vanished.
Therefore, the field at the surface of a
conductor must be perpendicular to the surface,
with no tangential components.
Suppose that a charged conductor has local
surface charge density . Put a perpendicular
cylinder of end-cap area A through this region of
the surface. The field on the inner end cap is
zero, and the field is perpendicular to the area
vectors on the curved wall of the cylinder, so Fe =
Esurf A Qin/e0A/e0. Therefore, Esurf = / e0.
(Note that this matches the surface field of a
sphere.)
20
Charges and Fields Inside a Conductor
Suppose there is a cavity in the interior of a
conductor.
Can any of the net charge of the conductor reside
on the surface of the cavity?
Place a Gaussian surface in the conductor just
outside the cavity surface. There E = 0 so Fe= 0, so
Qin = 0. Therefore, there is no charge or field
within or at the surface of the cavity.
Now suppose we force the issue by placing a
point charge q in the interior cavity of a neutral
conductor. Using the same Gaussian surface, Qin= 0
as before. This means that a compensating charge
-q must be present on the interior cavity surface,
to cancel the field of the interior point charge, and
to compensate for that, a charge q must be present
on the outer surface of the conductor.
21
Screening E-Fields
Suppose that we wish to create a region
where the electric field is zero within a region
containing a strong electric field.
All we have to do is surround the region
with a neutral conductor, and the conducting
properties will ”screen” the region, insuring an
electric field of zero. The charges on the
outside of the inserted conductor will
distribute themselves so as to “stop” the
incoming field lines on one side of the
conductor and “restart” them on the other
side.
The “price” paid for such screening is that
the exterior electric field may be very
distorted by the presence of the inserted
conductor.
22
Summaries
Knight,
Chapters 25, 26 and 27
Summary: Chapter 25
24
Charges
Fields
Not really true…
semi-conductors
Not always true…
Only if object
is polarizable
25
Chapter 26 Summary
The E field exists at each point in space
Non-zero values of the E-field are induced by
electric charges
26
27
Chapter 27 Summary
28
29