Post on 24-Jun-2018
Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 1
Summary: Applications of Gauss’Law
1. Field outside of a uniformly charged
sphere of radius a:
2. An infinite, uniformly charged plane
with charge density σ:
3. Field inside a univormly charged sphere
of charge density %:
4. A long coaxial wire:
Suggested Reading:
Griffiths: Chapter 2, Sections 2.1.1 - 2.1.4, pages
58-64; Sections 2.2.1-2.2.3, pages 65-757.
Weber and Arfken, Chapter 1, Section 1.13,
pages 82-85.
Wangsness: Chapters 3 and 4, pages 40-67.
Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 2
Applications of Gauss’ Law: A uni-formly charged sphere
To find the electric field outside of some uniformly
charged sphere, volume charge density %, radius r0,
centred at the origin, simply construct a spherical
“Gaussian surface” concentric with the centre of
the sphere, with a radius R > r0. Then, by Gauss’
Law: ∫∫S
~E · d ~A =Q
ε0
where ‘Q’ is the total charge enclosed, i.e.,
Q =
∫∫∫V
%dV
Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 3
By symmetry the electric field at the surface of
this sphere must have the same magnitude, ER
and must point normal to the surface (any con-
tributions to components tangential to the surface
will have equal and opposite compensating contri-
butions from elsewhere in the sphere). Thus∫∫S
~E·d ~A = |ER|∫ 2π
0
dφ
∫ π
0
(r·r)R2 sin θdθ = 4πERR2 =Q
ε0
So that the electric field everywhere on this spher-
ical surface of radius R is just
~E(R) =Q
4πε0
r
R2
which is precisely the electric field that would arise
from a point charge ‘Q’ located at the origin.
Infinite uniformly charged plane:
A surface “S” extends to infinity in both the x and
y directions at z = 0. It is covered with a uniform
Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 4
charge density σ. Use Gauss’ Law to determine
the electric field on either side of the plane.
We construct a small “pill-box”shaped Gaussian
surface S’ which penetrates the surface S (as shown
in the figure above). By symmetry we know that
the electric field, whatever its magnitude, must
point perpendicular to the surface at any point
(think about it!). Thus, in evaluating the total
flux through the Gaussian surface S’ we need only
worry about the top and bottom surfaces which
are parallel to S. Since the charge density on the
surface is uniform, the total charge enclosed by the
Gaussian pill-box is just
σ × cross-sectional area of the pill-box
(Note: if we like we can shrink the sides of the
pill-box which are perpendicular to S down to an
infinitesimal height). It is also obvious from sym-
metry that the value of ~E(x, y, z) cannot depend
on either x or y. Finally, as we increase the height
of the box above the plane, the flux through the
top (or bottom) surface remains the same. Thus,~E(x, y, z) is independent of z as well (except that
below the plane it points in the −z direction while
above the plane it points in the +z direction. Gauss’
Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 5
Law states that
Φ =
∫∫S′
~E · d ~A =Qencl
ε0
Since ~E = E0 z (by symmetry) we can take it
outside the integral and
Φ =
∫∫S′
~E · d ~A = E0 × 2A =σ A
ε0
where “A” is the area of the top surface of the pill-
box. The electric field above the plane then is just
~E = E0 z =σ
2ε0z
while below the plane it has the same magnitude
but points in the opposite direction.
We can now make the following observations:
• For a point charge at the origin, | ~E| ∝ 1/r2
• For an infinite line of charge, | ~E| ∝ 1/r
• For an infinite charged plane,| ~E| ∝ constant
Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 6
Problem 2.12 Use Gauss’ law tofind ~E inside a uniformly chargedsphere having charge density %
Φ =
∫∫S′
~E · d ~A =Qencl
ε0
By symmetry no one point on the surface “S ′”
can be any different than any other point. Thus,~E(r < a) = Er r, where “a” is the radius of the
charged sphere.
Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 7
The charge enclosed by S ′ is simply
Qencl =
∫∫∫S′
%dV = %
∫ 2π
0
∫ π
0
∫ r
0
r2 sin θdrdθdφ
=4πr3
3%
Thus,
~E(r < a) =%
ε0
4πr3/3
4πr2=
%
ε0
r
3Note: The charge outside of the Gaussian surface
S ′ does not contribute to the electric field on the
surface S ′.
What if the Gaussian surface S ′ was not centred
at the origin?
Gauss’ Law still gives the net flux through the
Gaussian surface as just Qencl/ε0. However, we
cannot easily evaluate the surface integral because
Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 8
the electric field on this surface depends on posi-
tion (i.e., we can’t use symmetry to argue that it
is constant).
Problem 2.16 A long co-axial ca-ble consists of an inner cylindri-cal rod (radius a), uniformly posi-tively charged with volume chargedensity %, and a concentric outercylindrical tube, of radius b > a,with a negative surface charge den-sity σ which ensures that charge isbalanced (i.e., total charge is pre-cisely 0). Use Gauss’ law to find~E everywhere
To find the electric field outside of the coaxial ca-
ble, construct a cylindrical Gaussian surface of length
L concentric to the axis of the cable but with a ra-
dius r > b
Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 9
By symmetry, ~E must point radially outward from
the axis of the coaxial cable and can only be a
function of r.
For r ≥ b:
Qencl = 0 and by Gauss’ Law∫∫S
~E · d ~A = |E|∫∫
S′dA =
Qencl
ε0= 0
Here, S ′ is the outer curved surface of the Gaussian
cylinder but does not include the ends, for which
there would be no contribution. Therefore, since∫∫S′ dA = 2πr · L 6= 0 we conclude that outside
Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 10
the coaxial cable
|E| = 0, for r ≥ b
For a ≤ r ≤ b:
Construct a cylindrical Gaussian surface with ra-
dius r which now lies in between the inner conduc-
tor and the outer conductor
Now,
Qencl
ε0=
∫∫S
~E · d ~A = |E|∫∫
S′dA = |E| · 2πr · L
=1
ε0
∫∫∫V
%dV =%
ε0· πa2L
so that
|E| =%a2
2ε0r, radially outward if % > 0
For r ≤ a:
Construct a cylindrical Gaussian surface inside the
inner electrode (recall that it is a solid conductor
having volume charge density %).
Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 11
Using the same arguments, we find that
Qencl
ε0=
∫∫S
~E · d ~A = |E|∫∫
S′dA = |E| · 2πr · L
=1
ε0
∫∫∫V
%dV =%
ε0· πr2L
Giving
|E| =%r
2ε0radially outward if r ≤ a
Summarizing, the electric field for all values of r
is: