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UNIT 3:-MAGNETOSTATIC
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Summary: Lecture 5 The Electric Field is related to
Coulombs Force by Thus knowing the field we can
calculate the force on a charge The Electric Field is a
vector
field Using superposition we thus
find Field lines illustrate the strength
& direction of the Electric field
!
ii
i
iQr
rE
||4
12
0TI
0Q
FE !
EF Q!
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Today Electric Flux
Gausss Law
Examples of using Gausss Law Properties of Conductors
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Electric Flux
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Electric Flux:
Fora constant field perpendicularto a surface A
A
Electric Flux is
defined as
A||E!*
E
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Electric Flux:
Electric Flux is
defined as
Ucos|| AE!*A
E
U
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Electric Flux:
A||E!*
||EwV
AA ||EwV
Numberof flux lines *wN
A
E
Field line
density
Field line density
Area
FLUX
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Quiz What is the electric flux through a cylindrical
surface? The electric field, E, is uniform and
perpendicular to the surface. The cylinder has radius
r and length L A) E 4/3 T r3 L
B) E r L
C) E T r2 L
D) E 2 T r L
E) 0
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Gausss Law
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Flux through a sphere from a point
charge
2
10 ||4
1||
rE
Q
TI
!
2
12
10
||4||4
1r
rT
TI
v!*Q
0I
Q!*
r1
The electric field
around a point charge
Thusthe
flux on a
sphere is E
Area
AreaE
Cancelling
we get
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Now we change the
radius of sphere
2
20 ||4
1
|| rE
Q
TI!
2
22
20
2 ||4||4
1r
rT
TI
v!*Q
0
2I
Q!*
r2
0
12I
Q!*!*
Flux through a sphere from a
point charge
2
10 ||4
1||
rE
Q
TI
!
212
10
||4||4
1 rr
T
TI
v!* Q
0I
Q!*
r1
The electric field
around a point charge
Thusthe
flux on asphere is E
Area
AreaE
Cancelling
we get
The flux is
the same
as before
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Flux lines & Flux*wN Nw*
and numberof lines passing
through each sphere isthe same
In factthe numberof flux
lines passingthrough any
surface surroundingthis
charge isthe same
1*
2*
s*
even when a line
passes in and out
ofthe surface it
crossesoutonce
more than in
outin
out
012 I
Q
S
!*!*!*
Just what we would expect because the
numberof field lines passingthrough each
sphere isthe same
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Principle of superposition:What is the flux from two
charges?
s*
Q1
Q2
0
2
0
1
II
QQS !*
!* 0Ii
S
Q
Ingeneral
GausssLaw
Forany
surface
Since the flux isrelated tothe
numberof field lines passing
through a surface the total flux is
the total from each charge
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Quiz
-Q/I0 0 +Q/I0 +2Q/I0
*1
*2
*3
2*Q1
1*
3*
What flux is passingthrough each of
these surfaces?
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What is Gausss Law?
GausssLaw issometimes easierto use than
CoulombsLaw, especially ifthere is lotsofsymmetry inthe
problem
GausssLaw doesnottell us anythingnew, it
is NOT a new law of physics, but another
wayof expressing CoulombsLaw
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Examples of using Gausss Law
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Example of using Gausss Law 2
Whatsthe field around a charged spherical
shell?
Q
0
outI
Q!*
Again considersphericalsurface centred on
charged shell
2
04
1||
r
QE
TI
!
0!E
Outside
out*in
* So as e.g. 1
Inside
0in !*
charge withinsurface = 0
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Properties of Conductors
Using GausssLaw
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Properties of Conductors
1. E is zero within the conductor2. Any net charge, Q, is
distributed on surface
(surface charge density W=Q/A)
3. E immediately outside isB
to surface4. W is greatest where the radius of curvature
is smaller
Fora conductorin electrostatic equilibrium
1W2W 211 WW ""
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1. E is zero within conductorIf there is a field in the
conductor, then the freeelectrons would feel a force and be
accelerated. They would then move and since
there are charges moving the conductor
would not be in electrostatic equilibrium
Thus E=0
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2. Any net charge, Q, is distributed
on surface
qi
Assurface can be drawn
arbitrarily close tosurface of
conductor, all net charge must
be distributed onsurface
Considersurface S below surface of conductor
Since we are in a conductorin
equilibrium, rule 1says E=0, thus*=0
!!* 0/IqEAGausssLaw
! 0/ 0IiqthusSo, net charge within
the surface is zero
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3. E immediately outside is B to
surfaceConsidera small cylindrical surface atthe surface
ofthe conductor
I/qEA !!*GausssLaw
cylinderissmall enough that E is constant
EB
E||If E||>0 it would cause surface charge q to move thusit
would not be in electrostatic equilibrium, thus E||=0
thus IAqE /!
IW /!B
E
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Summary
Electric Flux
Gausss Law
Examples of using
Gausss Law
Isolated charge Charged shell
Line of charge
Uniform sphere
Properties of
Conductors E is zero within the conductor
Any net charge, Q, isdistributed on surface
(surface charge density
W=Q/A)
E immediately outside is B to
surface
W is greatest where the radius
of curvature is smaller
Ucos|| AE!*
!*0I
iS
Q
