04. Electric flux.pdf
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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.
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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
-
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.
-
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 ?