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Chapter 4 Gauss’s Law
4.1 Electric Flux............................................................................................................. 1 4.2 Gauss’s Law............................................................................................................. 2 Example 4.1: Example 4.2: Example 4.3: Example 4.4: Infinitely Long Rod of Uniform Charge Density ................................ 7 Infinite Plane of Charge.................................................................

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Chapter 4Gauss’s Law
4.1 Electric Flux.............................................................................................................14.2 Gauss’s Law.............................................................................................................2Example 4.1: Infinitely Long Rod of Uniform Charge Density................................7Example 4.2: Infinite Plane of Charge.......................................................................9Example 4.3: Spherical Shell...................................................................................11Example 4.4: Non-Conducting Solid Sphere...........................................................134.3 Conductors.............................................................................................................14Example 4.5: Conductor with Charge Inside a Cavity............................................17Example 4.6: Electric Potential Due to a Spherical Shell........................................184.4 Force on a Conductor.............................................................................................214.5 Summary................................................................................................................234.6 Appendix: Tensions and Pressures........................................................................24
Animation 4.1:
Charged Particle Moving in a Constant Electric Field..................25
Animation 4.2
: Charged Particle at Rest in a Time-Varying Field........................26
Animation 4.3
: Like and Unlike Charges Hanging from Pendulums.....................284.7 Problem-Solving Strategies...................................................................................294.8 Solved Problems....................................................................................................314.8.1 Two Parallel Infinite Non-Conducting Planes................................................314.8.2 Electric Flux Through a Square Surface.........................................................324.8.3 Gauss’s Law for Gravity.................................................................................344.8.4 Electric Potential of a Uniformly Charged Sphere.........................................344.9 Conceptual Questions............................................................................................364.10 Additional Problems............................................................................................364.10.1 Non-Conducting Solid Sphere with a Cavity................................................364.10.2 P-N Junction..................................................................................................364.10.3 Sphere with Non-Uniform Charge Distribution...........................................374.10.4 Thin Slab.......................................................................................................374.10.5 Electric Potential Energy of a Solid Sphere..................................................384.10.6 Calculating Electric Field from Electrical Potential.....................................380
Gauss’s Law
4.1
Electric Flux
In Chapter 2 we showed that the strength of an electric field is proportional to the numberof field lines per area. The number of electric field lines that penetrates a given surface iscalled an “electric flux,” which we denote as
E
Φ
. The electric field can therefore bethought of as the number of lines per unit area.
Figure 4.1.1
Electric field lines passing through a surface of area
A
.Consider the surface shown in Figure 4.1.1. Let
ˆ
A
=
An
be defined as the
area vector
having a magnitude of the area of the surface,
A
, and pointing in the normal direction,. If the surface is placed in a uniform electric field
E
ˆ
n
that points in the same directionas , i.e., perpendicular to the surface
A
, the flux through the surface is
ˆ
n
ˆ
E
A EA
Φ = ⋅ = ⋅ =
EAEn
(4.1.1)On the other hand, if the electric field
E
makes an angle
θ
with (Figure 4.1.2), theelectric flux becomes
ˆ
n
n
cos
E
EA E A
θ
Φ = ⋅ = =
EA
(4.1.2)where is the component of
E
n
ˆ
E
= ⋅
En
perpendicular to the surface.
Figure 4.1.2
Electric field lines passing through a surface of area
A
whose normal makesan angle
θ
with the field.1
Note that with the definition for the normal vector , the electric flux is positive if the electric field lines are leaving the surface, and negative if entering the surface.
ˆ
n
E
Φ
In general, a surface
S
can be curved and the electric field
E
may vary over the surface.We shall be interested in the case where the surface is
closed
. A closed surface is asurface which completely encloses a volume. In order to compute the electric flux, wedivide the surface into a large number of infinitesimal area elements
ˆ
i i i
A
∆ = ∆
An
, asshown in Figure 4.1.3. Note that for a closed surface the unit vector is chosen to pointin the
outward
normal direction.ˆ
i
n
Figure 4.1.3
Electric field passing through an area element
i
∆
A
, making an angle
θ
withthe normal of the surface.The electric flux through is
i
∆
A
cos
E i i i i
E A
θ
∆Φ = ⋅∆ = ∆
EA
(4.1.3)The total flux through the entire surface can be obtained by summing over all the areaelements. Taking the limit and the number of elements to infinity, we have
0
i
∆ →
A
0
lim
E iiiS
A
d
∆ →
Φ = ⋅ = ⋅
d
∑∫∫
EAEA
(4.1.4)where the symbol
S
∫∫
denotes a double integral over a
closed
surface
S
. In order toevaluate the above integral, we must first specify the surface and then sum over the dotproduct .
d
⋅
EA
4.2
Gauss’s Law
Consider a positive point charge
Q
located at the center o a sphere of radius
r
, as shownin Figure 4.2.1. The electric field due to the charge
Q
isf
r
20
ˆ(/4)
Q r
πε
=
E
, which points2
in the radial direction. We enclose the charge by an imaginary sphere of radius
r
calledthe “Gaussian surface.”
Figure 4.2.1
A spherical Gaussian surface enclosing a charge
Q
.ates, a small surface area element on the sphere is given by (Figure.2.2)In spherical coordin4
2
ˆsin
d r d d
θ θ φ
=
Ar
(4.2.1)
Figure 4.2.2
A small area element on the surface of a sphere of radius
r
.Thus, the net electric flux through the area element is
( )
2
20
4
E
r
0
1sin =sin4
Q Qd d EdA r d d d d
θ θ φ θ θ φ πε
⎛ ⎞Φ = ⋅ = =⎜ ⎟
EA
(4.2.2)he total flux through the entire surface is
πε
⎝ ⎠
T
20000
4
E S
sin =
Q Qd d d
π π
θ θ φ πε ε
Φ = ⋅ =
∫∫ ∫ ∫
EA
(4.2.3)us
r
has a surface area
The same result can also be obtained by noting that a sphere of radi
2
4
A r
π
=
, and since the magnitude of the electric field at any point on the sphericalsurface is
20
/4
E Q r
πε
=
, the electric flux through the surface is3

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