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Electricity & Magnetism
Lecture 6: Gauss’s Law
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Summary: Lecture 5• The Electric Field is related to
Coulomb’s 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
i
ii
iQr
rE ˆ
||4
12
0
0Q
FE
EF Q
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Today
• Electric Flux
• Gauss’s Law
• Examples of using Gauss’s Law
• Properties of Conductors
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Electric Flux
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Electric Flux: Field Perpendicular
For a constant field perpendicular to a surface A
A
Electric Flux is defined as
A|| E
E
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Electric Flux:Non perpendicular
For a constant field NOT perpendicular to a surface A
Electric Flux is defined as
cos|| AEA
E
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Electric Flux:Relation to field lines
A|| E
|| E
AA || E
Number of flux lines N
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 r3 L
– B) E r L
– C) E r2 L
– D) E 2 r L
– E) 0
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Gauss’s Law
Relates flux through a closed surface to
charge within that surface
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Flux through a sphere from a point charge
210 ||4
1||
rE
Q
212
10
||4||4
1r
r
Q
0Q
r1
The electric field around a point charge
Thus the flux on a sphere is E × Area
AreaE
Cancelling we get
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Now we change the radius of sphere
220 ||4
1||
rE
Q
222
202 ||4
||4
1r
r
Q
02
Q
r2
012
Q
Flux through a sphere from a point charge
210 ||4
1||
rE
Q
212
10
||4||4
1r
r
Q
0Q
r1
The electric field around a point charge
Thus the flux on a sphere is E ×Area
AreaE
Cancelling we get
The flux is the same as before
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Flux lines & Flux
N N
and number of lines passing through each sphere is the same
In fact the number of flux lines passing through any surface surrounding this charge is the same
1
2
s
even when a line passes in and out of the surface it crosses out once more than in
outin
out
012
QS
Just what we would expect because the number of field lines passing through each sphere is the same
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Principle of superposition:What is the flux from two charges?
s
Q1
Q2
0
2
0
1
S
0i
S
Q
In general
Gauss’s Law
For any
surface
Since the flux is related to the number of field lines passing through a surface the total flux is the total from each charge
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Quiz
-Q/0 0 +Q/0 +2Q/0
1
2
3
2Q1
1
3
What flux is passing through each of these surfaces?
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What is Gauss’s Law?
Gauss’s Law is sometimes easier to use than Coulomb’s Law, especially if there is lots of symmetry in the problem
Gauss’s Law does not tell us anything new, it is NOT a new law of physics, but another way of expressing Coulomb’s Law
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Examples of using Gauss’s Law
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Example of using Gauss’s Law 1oh no! I’ve just forgotten Coulomb’s Law!
r2
Q
0Q
Not to worry I remember Gauss’s Law
By symmetry E is to surface
consider spherical surface centred on charge
0
||Q
AE 0
24||
QrE
200
2 4
1
4
1||
r
rE
0
24
1
rF F=qE
q
Phew!
Using the Symmetry
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Example of using Gauss’s Law 2What’s the field around a charged spherical
shell?
Q
0out
Q
Again consider spherical surface centred on charged shell
204
1||
r
QE
0E
Outside
outinSo as e.g. 1
Inside
0in
charge within surface = 0
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Examples
Gauss’s Law and a line of
charge
Gauss’s Law and a uniform
sphere
Gauss’s Law around a point
charge
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Quiz
• In a model of the atom the nucleus is a uniform ball of +ve charge of radius R. At what distance is the E field strongest?– A) r = 0– B) r = R/2– C) r = R– D) r = 2 R– E) r = 1.5 R
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Properties of Conductors
Using Gauss’s Law
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Properties of Conductors
1. E is zero within the conductor
2. Any net charge, Q, is distributed on surface (surface charge density =Q/A)
3. E immediately outside is to surface
4. is greatest where the radius of curvature is smaller
For a conductor in electrostatic equilibrium
12
211
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1. E is zero within conductor
If there is a field in the conductor, then the free electrons 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
As surface can be drawn arbitrarily close to surface of conductor, all net charge must be distributed on surface
Consider surface S below surface of conductor
Since we are in a conductor in equilibrium, rule 1 says E=0, thus =0
0/qEAGauss’s Law
0/ 0iqthusSo, net charge within the surface is zero
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3. E immediately outside is to surface
Consider a small cylindrical surface at the surface of the conductor
/qEA Gauss’s Law
q
cylinder is small enough that E is constant
E
E||
If E|| >0 it would cause surface charge q to move thus it would not be in electrostatic equilibrium, thus E|| =0
thus AqE /
/E
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Summary: Lecture 6
• Electric Flux
• Gauss’s Law
• Examples of using Gauss’s Law– Isolated charge
– Charged shell
– Line of charge
– Uniform sphere
• Properties of Conductors– E is zero within the
conductor
– Any net charge, Q, is distributed on surface (surface charge density =Q/A)
– E immediately outside is to surface
is greatest where the radius of curvature is smaller
cos|| AE
0i
S
Q