Physics 122B
Electricity and Magnetism
Lecture 7
Gauss’s Law and Applications
April 09, 2007
Martin Savage
Lecture 7 Announcements
 Lecture Homework #2 has been posted on the Tycho system.
It is due on Wednesday, April 11 , at 10 PM.
 On Friday, April 13 we will have Exam 1, covering Knight, Chapters
25-27. You should bring with you to the Exam a good calculator, a
Scantron sheet, and one sheet of 8.5”x11” paper on which you have
written anything you wish on both sides.
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Hints on Preparing for Exam 1
 Review Chapters 25, 26 and 27, the lectures on the
Web, and the important concepts featured in the
Chapter Summaries.
 Fill in identification on Scantron sheet before the
Exam.
 Make your 2-sided page of notes and focus on
understanding the formulas you write there.
 Make sure you know how to do all of the Tycho
homework problems.
 Practice for the exam by working problems at the
ends of chapters 25-27.
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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.
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Flux is Independent of
Surface Shape and Radius
What about
the flux when the
surface has an
arbitrary nonspherical 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.
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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.
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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 
Therefore, F e   E  dA 
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Qin
e0
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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.
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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.
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Example:
E Inside a Sphere 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
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.
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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?
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
Gaussian sphere. Therefore, Eout = (Qin/e0)/(4pr2) = (1/4pe0)Q/r2.
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.
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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
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Question 1
(u)
(v)
(w)
What is the ranking of the electric
fluxes for these systems?
(x)
(y)
(a) Fx = Fy > Fu = Fv = Fw ;
(b) Fx = Fy = Fu > Fv = Fw ;
(c) Fx = Fy > Fu = Fv > Fw ;
(d) Fv = Fy > Fu = Fw = Fx ;
(e) Fv = Fy > Fu > Fw > Fx .
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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?
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
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.
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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?
R
Q
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 = EinA =
Ein(4pr2)=Qin/e0, with Qin=QVG/V0=Q(r/R)3 because only part of the charge lies
inside the Gaussian sphere. 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.
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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 / R 3
Esurface  (1/ 4pe 0 )Q / R 2
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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?
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
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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?
R
x
r
+
+
+
+
+
+
+ + + + + +
Put a coaxial cylindrical Gaussian surface
of length L through x, so that it has radius
Section of long cylinder
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=lLVG/V0=lL(r/R)2, 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.
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Summary: The E-Field of a
Long Charged Cylinder
outside
Ecylinder

in solid
cylinder
E
2l
4pe 0 r
1
~
1
r
1 2l r

~r
2
4pe 0 R
in hollow
Ecylinder
0
on surface
cylinder
E
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2l

4pe 0 R
1
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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
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.
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Conductors and Gauss’s Law
The electric field must be zero inside
any conductor in electrostatic equilibrium
because charges are free to move in a
conductor, and any internal field would
cause the mobile charges to rearrange
themselves until the field disappeared.
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.
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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 =
EsurfA Qin/e0A/e0. Therefore, Esurf = /e0. (Note
that this matches the surface field of a sphere.)
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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.
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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.
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End of Lecture 7
 Lecture Homework #2 has been posted on the Tycho system.
It is due on Wednesday, April 11 , at 10 PM.
 On Friday, April 13 we will have Exam 1, covering Knight, Chapters
25-27. You should bring with you to the Exam a good calculator, a
Scantron sheet, and one sheet of 8.5”x11” paper on which you have
written anything you wish on both sides.
7/19/2016
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