Gauss

1
Chapter 24

• Gausss Law

2
Electric Flux

• Electric flux is the product of the magnitude of
  the electric field and the surface area, A,
  perpendicular to the field
• FE EA

3
Electric Flux

• The electric flux is proportional to the number
  of electric field lines penetrating some surface
• The field lines may make some angle ? with the
  perpendicular to the surface
• Then
• FE EA cos ?
• The flux is a maximum (zero) when the surface is
  perpendicular (parallel) to the field

4
Electric Flux

• If the field varies over the surface, F EA cos
  ? is valid for only a small element of the area
• In the more general case, look at a small area
  element
• In general, this becomes

5
Electric Flux

• The surface integral means the integral must be
  evaluated over the surface in question
• The value of the flux depends both on the field
  pattern and on the surface
• SI units N.m2/C

6
Electric Flux, Closed Surface

• For a closed surface, by convention, the A
  vectors are perpendicular to the surface at each
  point and point outward
• (1) ? lt 90o, F gt 0
• (2) ? 90o, F 0
• (3) 180o gt ? gt 90o, F lt 0

7
Electric Flux, Closed Surface

• The net flux through the surface is proportional
  to the number of lines leaving the surface minus
  the number entering the surface

8
Electric Flux, Closed Surface

• Example flux through a cube
• The field lines pass perpendicularly through two
  surfaces and are parallel to the other four
  surfaces
• Side 1 F E l2
• Side 2 F E l2
• For the other sides, F 0
• Therefore, Ftotal 0

9
Chapter 24Problem 5

• A pyramid with horizontal square base, 6.00 m on
  each side, and a height of 4.00 m is placed in a
  vertical electric field of 52.0 N/C. Calculate
  the total electric flux through the pyramids
  four slanted surfaces.

10
Gauss Law

• Gauss Law electric flux through any closed
  surface is proportional to the net charge Q
  inside the surface
• eo 8.85 x 10-12 C2/Nm2 permittivity of free
  space
• The area in F is an imaginary Gaussian surface
  (does not have to coincide with the surface of a
  physical object)

11
Gauss Law

• A positive point charge q is located at the
  center of a sphere of radius r
• The magnitude of the electric field everywhere on
  the surface of the sphere is E keq / r2
• Asphere 4pr2

12
Gauss Law

• Gaussian surfaces of various shapes can surround
  the charge (only S1 is spherical)
• The electric flux is proportional to the number
  of electric field lines penetrating these
  surfaces, and this number is the same
• Thus the net flux through any closed surface
  surrounding a point charge q is given by q/eo and
  is independent of the shape of the surface

13
Gauss Law

• If the charge is outside the closed surface of an
  arbitrary shape, then any field line entering the
  surface leaves at another point
• Thus the electric flux through a closed surface
  that surrounds no charge is zero

14
Gauss Law

• Since the electric field due to many charges is
  the vector sum of the electric fields produced by
  the individual charges, the flux through any
  closed surface can be expressed as
• Although Gausss law can, in theory, be solved to
  find for any charge configuration, in
  practice it is limited to symmetric situations
• One should choose a Gaussian surface over which
  the surface integral can be simplified and the
  electric field determined

15
Field Due to a Spherically Symmetric Charge
Distribution

• For r gt a
• For r lt a

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Field Due to a Spherically Symmetric Charge
Distribution

• Inside the sphere, E varies linearly with r (E ?
  0 as r ? 0)
• The field outside the sphere is equivalent to
  that of a point charge located at the center of
  the sphere

17
Chapter 24Problem 18

• A solid sphere of radius 40.0 cm has a total
  positive charge of 26.0 1µC uniformly distributed
  throughout its volume. Calculate the magnitude of
  the electric field (a) 0 cm, (b) 10.0 cm, (c)
  40.0 cm, and (d) 60.0 cm from the center of the
  sphere.

18
Electric Field of a Charged Thin Spherical Shell

• The calculation of the field outside the shell is
  identical to that of a point charge
• The electric field inside the shell is zero

19
Field Due to a Plane of Charge

• The uniform field must be perpendicular to the
  sheet and directed either toward or away from the
  sheet
• Use a cylindrical Gaussian surface
• The flux through the ends is EA and there is no
  field through the curved part of the surface
• Surface charge density s Q / A

20
Conductors in Electrostatic Equilibrium

• When no net motion of charge occurs within a
  conductor, the conductor is said to be in
  electrostatic equilibrium
• An isolated conductor has the following
  properties
• Property 1 The electric field is zero everywhere
  inside the conducting material
• If this were not true there were an electric
  field inside the conductor, the free charge there
  would move and there would be a flow of charge
  the conductor would not be in equilibrium

21
Conductors in Electrostatic Equilibrium

• When no net motion of charge occurs within a
  conductor, the conductor is said to be in
  electrostatic equilibrium
• An isolated conductor has the following
  properties
• Property 2 Any excess charge on an isolated
  conductor resides entirely on its surface
• The electric field (and thus the flux) inside is
  zero whereas the Gaussian surface can be as close
  to the actual surface as desired, thus there can
  be no charge inside the surface and any net
  charge must reside on the surface

22
Conductors in Electrostatic Equilibrium

• When no net motion of charge occurs within a
  conductor, the conductor is said to be in
  electrostatic equilibrium
• An isolated conductor has the following
  properties
• Property 3 The electric field just outside a
  charged conductor is perpendicular to the surface
  and has a magnitude of s/eo
• If this was not true, the component along the
  surface would cause the charge to move no
  equilibrium

23
Conductors in Electrostatic Equilibrium

• When no net motion of charge occurs within a
  conductor, the conductor is said to be in
  electrostatic equilibrium
• An isolated conductor has the following
  properties
• Property 4 On an irregularly shaped conductor,
  the charge accumulates at locations where the
  radius of curvature of the surface is smallest
• Proof see Chapter 25

24
Answers to Even Numbered Problems Chapter 24
Problem 64 For r lt a ?/2??0r radially
outward. For a lt r lt b ? ?? (r2?a2)/2??0r
radially outward. For r gt b ? ??
(b2?a2)/2??0r radially outward.