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Electric flux Think of air blowing in through a window. How
much air comes through the window depends upon the speed of the
air, the direction of the air, and the area of the window. We might
call this air that comes through the window the "air flux".
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Electric flux
The field is uniform
The plane is perpendicular to the field
E E A
We will define the electric flux for an electric field that is
perpendicular to an area as
= E A
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Flux for a uniform electric field passing
through an arbitrary plane
The field is uniform
The plane is not perpendicular to the field
E E A E Acos
n
A AnE E A
-
n
A
E
E
General flux definition
E E n A E Acos
The field is not uniform
The surface is not perpendicular to the field
A If the surface is made up of a mosaic of N little surfaces
N
E i i i
i 1
E n A
E E ndA E dA
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Electric Flux, General
In the more general
case, look at a small
area element
In general, this
becomes
cosE i i i i iE A E A
0surface
limi
E i iA
E A d
E AIf we let the area of each element approach zero, then the
number of elements approaches infinity and the sum is replaced by
an integral.
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Electric Flux Calculations
The surface integral means the integral
must be evaluated over the surface in
question
In general, the value of the flux will depend
both on the field pattern and on the
surface
The units of electric flux will be N.m2/C2
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Electric Flux, Closed Surface
The vectors Ai
point in different
directions
At each point, they
are perpendicular to
the surface
By convention, they
point outward
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Closed surfaces
+ Q 3Q+ +Q
EE ndA E dA
n or A point in the direction outward from the closed
surface
Negative Flux(total charge negative) Zero Flux
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Flux from a point charge through a closed sphere
EE ndA E dA
E n E n E
2
kqE r E constant
r
E sphereE ndA E dA E A
2
E 2
o
kq q4 r 4 kq
r
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Gausss Law
Gauss asserts that the proceeding calculation for the flux from
a point charge is true for any charge distribution!!!
enclosedE enclosed
o
qE dA 4 kq
This is true so long as Q is the charge enclosed by the surface
of integration.
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Example
3 2cos 3.50 10 0.350 0.700 cos0 858 N m CE EA
An electric field with a magnitude of 3.50 kN/C is applied along
the x axis. Calculate the electric flux through a rectangular plane
0.350 m wide and 0.700 m long assuming that
(a) the plane is parallel to the yz plane; (b) the plane is
parallel to the xy plane; (c) the plane contains the y axis, and
its normal
makes an angle of 40.0 with the x axis.
90.0 0E
3 23.50 10 0.350 0.700 cos40.0 657 N m CE
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Example
The following charges are located inside a submarine: 5.00 C,
9.00 C, 27.0 C, and 84.0 C.
(a) Calculate the net electric flux through the hull of
the submarine. (b) Is the number of electric field lines leaving
the
submarine greater than, equal to, or less than the number
entering it?
6 2 2in12 2 2
0
5.00 C 9.00 C 27.0 C 84.0 C6.89 10 N m C
8.85 10 C N mE
q
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Example A point charge Q = 5.00 C is located at the center of a
cube of edge L = 0.100 m. In addition, six other identical point
charges having q = 1.00 C are positioned symmetrically around Q as
shown in Figure below. Determine the electric flux through one face
of the cube.
-
0
6 22
12 2one face0
6
6 5.00 6.00 10 C N m18.8 kN m C
6 6 8.85 10 C
E total
E
Q q
Q q
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Problem
An electric field is given by , where sign(x) equals 1 if x <
0, 0 if x = 0, and +1 if x > 0. A cylinder of length 20 cm and
radius 4.0 cm has its center at the origin and its axis along the x
axis such that one end is at x = +10 cm and the other is at x = 10
cm. (a) What is the electric flux through each end? (b) What is the
electric flux through the curved surface of the cylinder? (c) What
is the electric flux through the entire closed surface? (d) What is
the net charge inside the cylinder?
( )(300 / )E sign x N C i
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The field at both circular faces of the cylinder is parallel to
the outward vector normal to the surface, so the flux is just EA.
There is no flux through the curved surface because the normal to
that surface is perpendicular to Er. The net flux through the
closed surface is related to the net charge inside by Gausss
law.
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APPLICATION OF GAUSSS LAW TO CHARGED INSULATORS
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Ways of choosing the gaussian surface
to determine the electric field
Ways of choosing the gaussian surface over which the surface
integral given by Equation can be simplified and the electric field
determined
EE ndA E dA
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In choosing the surface, we should always take advantage of the
symmetry of the charge distribution so that we can remove E from
the integral and solve for it. The goal in this type of calculation
is to determine a surface that satisfies one or more of the
following conditions:
1. The value of the electric field can be argued by symmetry to
be constant over the surface. 2. The dot product in Equation can be
expressed as a simple algebraic product E dA because E and dA are
parallel. 3. The dot product in Equation is zero because E and dA
are perpendicular.
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Select a surface Try to imagine a surface where the electric
field is constant everywhere. This is accomplished if the surface
is equidistant from the charge. Try to find a surface such that the
electric field and the normal to the surface are either
perpendicular or parallel.
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L
r
Example a line of charge
total charge
total length
E
1. Find the correct closed surface 2. Find the charge inside
that closed surface
enclosedE
o
qE dA
-
0
0
(2 )
2
insideqEA
LE rL
kE
r
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Example
A uniformly charged, straight filament 7.00 m in length has a
total positive charge of 2.00 C. An uncharged cardboard cylinder
2.00 cm in length and 10.0 cm in radius surrounds the filament at
its center, with the filament as the axis of the cylinder. Using
reasonable approximations, find (a) the electric field at the
surface of the cylinder and (b) the total electric flux through the
cylinder.
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(2 )
charge over length of cylinder=7
(2 )7
27
4
enclosedE
o
E cylinder
enclosedr
o
enclosed
enclosedr
o o
r
o
qE dA
E ndA E dA E A
qE rL
Qq xL
q Q LE rL
Q
kE
r r
9 2 2 62 8.99 10 N m C 2.00 10 C 7.00 m20.100 m
ekE
r
51.4 kN C , radially outw ardE
cos 2 cos0E EA E r 4 25.14 10 N C 2 0.100 m 0.0200 m 1.00 646 N
m CE
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Example a charged plane
+Q
total charge
total area
1. Find the correct closed surface 2. Find the charge inside
that closed surface
enclosedE
o
QE dA
E
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Example a charged plane
+Q
total charge
total area
1. Find the correct closed surface(cylinder) 2. Find the charge
inside that closed surface
enclosedE
o
QE dA
E0
0
0
2
2
2
insideqEA
AEA
E
-
r
Example a solid sphere of charge
+Q uniformly distributed total charge
total volume
a
Inside the charged sphere:
r a
1. Find the correct closed surface 2. Find the charge inside
that closed surface
enclosedE
o
qE dA
E
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Example a solid sphere of charge
+Q uniformly distributed total charge
total volume
a
EOutside the charged sphere:
r a
1. Find the correct closed surface 2. Find the charge inside
that closed surface
enclosedE
o
qE dA
r
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A non-conducting sphere of radius 6.00 cm has a uniform volume
charge density of 450 nC/m3. (a) What is the total charge on the
sphere? Find the electric field at the following distances from the
spheres center: (b) 2.00 cm, (c) 5.90 cm, (d) 6.10 cm, and (e) 10.0
cm. Ans: (a)0.407 nc (b)339 N/C1KN/C(d)983N/C(e)366 N/C (e)
PROBLEM
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A sphere of radius R has volume charge density = B/r for r <
R , where B is a constant and = 0 for r > R. (a) Find the total
charge on the sphere. (b) Find the expressions for the electric
field inside and
outside the charge distribution (c) Sketch the magnitude of the
electric field as a function
of the distance r from the spheres center.
PROBLEM
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CONDUCTORS IN ELECTROSTATIC EQUILIBRIUM
Insulators, like the previous charged sphere, trap excess charge
so it cannot move.
Conductors have free electrons not bound to any atom. The
electrons are free to move about within the material. If excess
charge is placed on a conductor, the charge winds up on the surface
of the conductor. Why?
The electric field inside a conductor is always zero.
The electric field just outside a conductor is perpendicular to
the conductors surface and has a magnitude, /o
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Einside = 0, cont.
Before the external field is applied, free electrons are
distributed throughout the conductor
When the external field is applied, the electrons redistribute
until the magnitude of the internal field equals the magnitude of
the external field
There is a net field of zero inside the conductor
This redistribution takes about 10-15s and can be considered
instantaneous
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Charge Resides on the Surface
Choose a gaussian surface
inside but close to the actual
surface
The electric field inside is
zero (prop. 1)
There is no net flux through
the gaussian surface
Because the gaussian
surface can be as close to
the actual surface as
desired, there can be no
charge inside the surface
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Charge Resides on the Surface,
cont
Since no net charge can be inside the
surface, any net charge must reside on
the surface
Gausss law does not indicate the
distribution of these charges, only that it
must be on the surface of the conductor
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Fields Magnitude and Direction
Choose a cylinder as the gaussian surface
The field must be perpendicular to the surface
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Fields Magnitude and Direction,
cont.
The net flux through the gaussian surface
is through only the flat face outside the
conductor
The field here is perpendicular to the surface
Applying Gausss law
E
o o
A EA and E
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Conductors in Equilibrium,
example
The field lines are
perpendicular to
both conductors
There are no field
lines inside the
cylinder
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r
Example a solid conducting sphere of
charge surrounded by a conducting shell
+2Q on inner sphere -Q on outer shell
a
Find the Electric Field:
r a
1. Find the correct closed surface 2. Find the charge inside
that closed surface
enclosedE
o
qE dA
E
b
c
r c
a r b
b r c
At inner surface of shell, gaussian surface Since E=0,
Qenc=0=charge on inner sphere+charge on inner surface of shell
Therefore charge on inner shell=-2Q
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Example
Consider a thin spherical shell of radius 14.0 cm with a total
charge of 32.0 C distributed uniformly on its surface. Find the
electric field
(a) 10.0 cm and (b) 20.0 cm from the center of the charge
distribution.
9 6
2 2
8.99 10 32.0 107.19 M N C
0.200
ekQE
r
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Example P24.43
A square plate of copper with 50.0-cm sides has no net charge
and is placed in a region of uniform electric field of 80.0 kN/C
directed perpendicularly to the plate. Find
(a) the charge density of each face of the plate and (b) the
total charge on each face.
4 12 7 28.00 10 8.85 10 7.08 10 C m
277.08 10 0.500 CQ A 71.77 10 C 177 nCQ
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Example
A long, straight wire is surrounded by a hollow metal cylinder
whose axis coincides with that of the wire. The wire has a charge
per unit length of , and the cylinder has a net charge per unit
length of 2. From this information, use Gausss law to find (a) the
charge per unit length on the inner and outer surfaces of the
cylinder and (b) the electric field outside the cylinder, a
distance r from the axis.
0
2 3 6 3 radially outw ard
2
e ek k
Er r r
in0 q inq 3
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Recall
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Del
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Grad
Vector operator acts on a scalar field to generate a vector
field
Example
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Div
Vector operator acts on a vector field to generate a scalar
field
Example
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Curl
Vector operator acts on a vector field to generate a vector
field
Example
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Gausss Theorem (differential form)
Applying the Divergence Theorem
The integral and differential form of Gausss Theorem
Qin
Gauss' divergence theorem relates triple integrals and surface
integrals.
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del operator
x y zx y z
: sin cosd d
x xdx dx
Examples of derivative operators:
scalar
vector
This is a vector operator.
: sin cos
sin sin cos
sin sin cos
d dx x x x x
dx dx
d dx x x x x x x
dx dx
d dx y x x y x z x
dx dx
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Example
, , sin 3E x y z x x y y z xy
, , sin 3E x y z x y z x x y y z xyx y z
, , sin 3E x y z x y xyx y z
, , cos 3 0 3 cosE x y z x x
, ,E x y zFind
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Example: Problem If the electric field in some region is given
(in spherical coordinates) by the expression
where A and B are constants, what is the charge density ?