Physics 202, Lecture 3
Conductors And Electrostatic Equilibrium
Today’s Topics
Conductors: charges (electrons) able to move freely !
Charges redistribute when subject to E field.
Charge redistribution " electrostatic equilibrium.
!! Conductors in Electrostatic Equilibrium
!! Gauss’s Law
E
E
- - - - - - -
- - - - -
E=0
+ + + + + + +
E
Initial
"
Conductors And Electrostatic Equilibrium
1. E field is always zero inside the conductor.
2. All charges reside on the surface of conductor.
3. E field outside conductor is perpendicular to the
surface, has magnitude
(show later using Gauss’s Law) !
E field is also zero inside any cavity within the conductor.
transient, <10-16s
" equilibrium
(right after E applied)
Electric Flux
Electric flux !E through a surface:
(component of E-field " surface) X (surface area)
Flux proportional to
# E- field lines
penetrating surface
For Flat Surface
! !
!E = E • A
For Arbitrary Surface
The above properties are valid regardless of the
shape of and the total charge on the conductor!
!E =
"
surface
! !
E • dA
Karl Friedrich Gauss
Why Perpendicular Component?
Normal vector to surface makes angle # with respect to field:
!
E = E || sˆ + E " nˆ
Gauss’s Law
Net electric flux through any closed surface (“Gaussian
surface”) equals the total charge enclosed inside the
closed surface divided by the permittivity of free space.
Component || surface
Component " surface
!
A = Anˆ
! Only " component
‘goes through’ surface
!!
$E = EA cos #
! !
!E = E • A
$E =0 if E parallel A
!! $E = EA (max) if E " A
!!
!!
!
Flux SI units are N!m2/C
electric flux
qencl : all charges enclosed
regardless of positions
! ! q
! E = " E idA = encl
#0
%0: permittivity constant k =
1
4 "#0
!
Using Gauss’ Law
Choose a closed (Gaussian) surface such that the
surface integral is trivial.
Use symmetry arguments:
1. Direction. Choose a Gaussian surface such that E is clearly
either parallel or perpendicular to each piece of surface
2. Magnitude. Choose a surface such that E is known to
have the same value at all points on the surface
Then:
! !
Eid
"! A =
The Gauss and Coulomb Laws
"!Suppose we have a point-like
charge Q. Let’s compute the
electric field at a distance, r,
from the charge:
1.! Define sphere of radius r
around point-like charge Q
2.! Apply Gauss’s Law
! !
# EdA =
! !
qencl
"! EdA = E "! dA = EA = " 0
qencl Q
=
"0
"0
# EdA = E # dA = E 4$r
E 4 $r 2 =
Given qencl, can solve for E (at surface), and vice versa
E=
!
Q
"0
Q
Q
=k 2
4 $"0 r 2
r
2
Gaussian
Surface
Conductors and Gauss’s Law
Use Gauss’s Law to understand properties
of conductors in electrostatic equilibrium:
More examples: three general cases
Method: evaluate flux over carefully chosen “Gaussian surface”:
Works for highly symmetric charge distributions:
1. Choose a Gaussian surface inside
(as close to the surface as desired)
spherical
No net flux through the Gaussian surface
(Ein=0). Any net charge must reside on the
surface (cannot be inside!)
cylindrical
planar
E=0
2. E field always perp to surface:
choose Gaussian surface (“pillbox”)
! E = EA =
qencl # A
=
"0
"0
E=
!
"0
(point charge, uniform
sphere, spherical shell,…)
Spherical Symmetry: Examples
(infinite uniform line of
charge or cylinder…)
(infinite uniform sheet
of charge or slab,…)
Example: Uniform Charge Sphere. Solution:
(see board)
Spherical symmetry: choose a spherical Gaussian surface, radius r
E.g. uniformly charged sphere radius a and total charge Q,
Note: same form as
a point charge
Then:
!
!
"! EidA = E4" r
2
qencl
E=
4!" 0 r 2
=
qencl
#0
where, qencl =
!
!
r3
Q if r < a
a3
Q if r > a
inside
outside
How not to apply Gauss’s Law
Two charges +2Q and –Q are placed at locations
shown. Find the electric field at point P.
1. Draw a Gaussian surface passing through P
2. Apply Gauss’s law:
!
!
"! EidA =
P
qencl
"0
EP=?
r
qencl = 2Q + (!Q) = Q
3. Surface integral:
! !
2
Eid
"! A = 4" r E
!!
E=
+2Q
-Q
Q
4!" 0 r 2
Is this correct? No! Which step is wrong? Last step