Electric Flux and Gauss

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Electric Flux and Gausss Law
Gausss law states that the electric flux through
a closed surface is proportional to the charge
enclosed by the surface
Note A charge outside the surface will
contribute nothing to the total flux. This is the
essence of Gausss Law.
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Gausss Law in differential form
Gausss law in its integral form
Using divergence theorem
Qenc in terms of charge density ?
So Gausss law becomes
Since this holds for any volume, the integrands
must be equal
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Gauss Law and Coulomb law
The angle ? between E and dA is zero at any
point on the surface, we can re-write Gauss Law
as
A spherical Gaussian surface centered on a point
charge q
E has the same value at all points on the surface
E can be moved out,
Integral is the sum of surface area
Coulombs Law
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Applications of Gausss Law
Gausss law can be used to find the electric
field in systems with simple configurations.
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Applications of Gauss Law
Gausss law is always true, but it is not always
useful. In what conditions Gausss Law can not
be used ? If charge density is not uniform, or
our chosen Gaussian surface (closed) is such that
E is not in the same direction as da at every
point through it would still have been true that
the flux of E is (1/?0)q Why We could not pull
?E? out of the integral. When we can use Guass
Law? Symmetry is crucial
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• Criteria for calculation of E using Guasss Law
• Construct a closed surface called Gaussian
  surface near the charge distribution, over which
• magnitude of E have constant value for uniform
  charge distributions.
• angle (?) between E and normal to this surface
  have constant value for symmetrical charge
  distributions and symmetrical surfaces so that
  ECos? can be taken outside the integral sign.

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• There are only three kinds of symmetry which we
  can use for calculation of E
• Spherical symmetry Make your Gaussian surface a
  concentric sphere.
• Cylindrical symmetry Make your Gaussian surface
  a coaxial cylinder.
• Plane symmetry Use a Gaussian pillbox which
  straddles the surface.

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Using Cylindrical Symmetry
Question Evaluate the electric field that arises
from an infinite line of charge.
Pick a point P for evaluation of E.
Understand the symmetry
The only variable on which E may depend is R, the
distance from the line of charge. There is no
angular dependence because the line is
cylindrically symmetric, i.e. it does not matter
at which angle about the line you view it.
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There is no axial dependence because the line of
charge is infinitely long, i.e. there is no
preferred position along the line. The
direction of E is directly away from the line,
perpendicular to it at all locations. There is
no azimuthally or axial components because the
line is infinite and has no extent in the
transverse directions.
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Construct the Gaussian surface
A cylindrical tube having a radius equal to the
distance between the point of evaluation and the
line of charge. This surface can be divided in 3
parts.
Surface 2 The electric field anywhere on this
surface has the same direction as the
infinitesimal area vector and has a constant
value everywhere on the surface. Surface 1 and
3 Since a Gaussian surface must be closed, the
tube is then capped with flat endcaps, 1 and 3.
The field vector is perpendicular to the area
vector so there is no flux through the endcaps.
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Examine the Gaussian surface
The flux through the endcap 1 is zero since A1 is
perpendicular to E everywhere on this surface.
Similarly, the flux through surface 3 is zero.
Since the electric field is everywhere constant
on surface 2 and points in the same direction as
the surface vector at any point, the flux through
surface 2 will be the field magnitude E times the
area of the tube wall.
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Evaluate the electric flux through the Gaussian
surface
Evaluate the charge enclosed by the Gaussian
surface
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Apply Gauss's law for the result
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What is the electric field from an infinitely
long wire with linear charge density of 100
nC/m at a point 10cm away from it ?

.
R 10 cm
Ey
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Line of Charge by Direct Integration 1
Problem Evaluate the electric field that arises
from an infinite line of charge.
Evaluate the contribution from the infinitesimal
charge element
Considering a positive test charge at point P,
the direction of the electric field is shown. The
strength of the electric field contribution from
the infinitesimal charge shown is proportional to
dq and inversely proportional to the square of
the distance separating point P and the charge
element.
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Exploit symmetry as appropriate
The horizontal components of the two
contributions will balance. We need only
consider the vertical contribution to the
electric field by each infinitesimal charge.
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Substituting these terms, dEy becomes
Setting up the integrating over the charge
distributing
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Expressing the integrand in terms of theta and
switching the limits of integration
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Problem 2 (a)Electric field inside a conductor.
Ans E0 inside the a conductor. Why
If we put a conductor in an external field
(E0). Due to this electric field electrons will
start to move in the opposite direction of the
electric field. After some time they will reach
to the end of the material, the charges pile up
plus on the right hand side and minus on the left
hand side. These induced charges produce a
field of its own in the opposite direction to
E0. The field of the induced charges tends to
cancel off the original field.
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(b) Charge inside a conductor? Gausss Law If
E0 so ? is also zero. (c) Any net charge
resides on the surface. If we place a Gaussian
surface just inside the surface of the charged
conductor, E 0 at all points on the Gaussian
surface. According to Gauss Law the net charge
enclosed the Gaussian surface must also be zero.
Therefore the excess charged must be outside
the Gaussian surface and must lie on the actual
surface of the conductor.
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(d) A conductor is an equipotential Let a and b
are two points inside or on the surface of a
given conductor hence
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(e) E is perpendicular to the surface, just
outside a conductor

• Think an electric field at some arbitrary angle
  at the surface of a conductor.
• There is a component perpendicular to the
  surface, so charges will move in this direction
  until they reach the surface, and then, since
  they cannot leave the surface, they stop.
• There is also a component parallel to the
  surface, so there will be forces on charges in
  this direction.

Since they are free to move, they will move to
nullify any parallel component of E. In a very
short time, only the perpendicular component is
left.
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(f) E inside the cavity of a conductor( if cavity
has no charge)
Zero If E0 within an empty cavity, there is no
charge on the surface of the cavity. Using this
concept you can make a metal car during the
thunderstorm If lightning strikes, you will not
be electrocuted.
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(f) E just outside the conductor
Consider a section of the surface that is small
enough to neglect any curvature and assume the
section is flat.
Embed a tiny cylindrical Gaussian surface with
one end cap full inside the conductor and the
other fully outside and cylinder is perpendicular
to the surface.
The electric field just outside the surface must
also be perpendicular to that surface otherwise
surface charges would be subject to motion.
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We now sum the flux through the Gaussian
surface. There is no flux through the internal
endcap. Why ?? There is no flux through the
curved surface. Why ?? The only flux through the
Gaussian surface is that through the external
endcap where E is perpendicular. Assuming a cap
area of A, the flux through the cap is EA. The
charge qenc enclosed by the Gaussian surface lies
on the conductors surface area A. If ? is the
charge per unit area , the qenc is equal to ? A.
Therefore Gauss Law becomes
?oEA ? A
E ? / ?o (conducting
surface)
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Using Planar Symmetry
Find the electric field of an infinite plane
carries a uniform surface charge ?.
Draw a Gaussian pillbox extending equal distances
above and below the plane. From symmetry , E
must be perpendicular to the surface and endcaps.
The surface charge is positive so E emanates
from the surface. Since the field lines do not
piece the curved surface there is no flux.
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Using Gausss law
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Using Planar Symmetry
A cross-section of a thin, infinite conducting
plate having (a) positive charge (b) negative
charge uniform charge densities (?1) on both the
surfaces. Find the field away from the sheets.
(a)
away from the plate.
For figure (b) plate is a identical plate with
negative charge. In this case E points inward.
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• Problem Two infinite parallel planes carry equal
  but opposite uniform charge densities ?2?. Find
  the field in each of the three regions
• To the left of the both
• Between them
• To the right of the both

Left plate produces a field (1/?0)?, points away
from it, the left in region (i), and to the right
in regions (ii) and (iii) Right plate produces a
field (1/?0)?, points toward it, the right in
region (i) and (ii) to the left in region (iii)
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In (i) and (iii) regions field due to both
planes will cancel each other. So E0 In (ii)
region points toward right and amount of
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Using Spherical Symmetry
Q. Find the field (a) inside and (b) outside of a
uniformly charged spherical shell of radius R.
Ans.(a)Since fields lines can not cross, the
spherical symmetry requires E field to be in
the radially direction and uniform on the
Gaussian surface, therefore Gauss Law
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Ans(b) Outside the hollow sphere
We define a Gaussian surface around the sphere
for which rgtR.
Again since fields lines can not cross, the
spherical symmetry requires E field to be in
the radially direction and uniform on the
Gaussian surface, therefore Gauss Law becomes
Same as point charge
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Summarized result Using Gauss Law, we are able
to prove two theorems for a spherical shell. A
shell of uniform charge attracts and repels a
charged particle that is outside the shell as if
all shells charge were concentrated at the
center. If a charged particle is located inside a
shell of uniform charge, there is no net electric
force on the particle from the shell.
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Q. Find the field outside of a uniformly charged
solid sphere of radius R and total charge q.
Ans Draw a spherical (Gaussian) surface at
radius rgtR. Gausss law says for this surface
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Symmetry allows us to extract E from under the
integral sign because E points radially outward,
as does da so drop the dot product.
Also the magnitude of E is constant over the
Gaussian surface, so it comes outside the
integral
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Q. Find the field (a) outside and (b) inside of a
uniformly charged solid sphere of radius R with
uniform charge density ?(C/m3). Ans (a) Outside
(agtR)
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Ans (b) inside (Rgta) Still we have spherical
symmetry centered on the center of the
charge. Choose a Gaussian surface sphere of
radius a
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Inside the sphere
Outside the sphere