Transcript Electric field of a ball of charge Q

Hw: All Chapter 5 problems and exercises
Average 75
Median 78
521 – 79
522 – 76
523 – 73.3
524 – 76.5
525 – 76
526 – 69.7
Test 1 results
40
35
30
25
20
Series1
15
10
5
0
1
>90
2
3
>80
>70
4
>60
5
6
>50
<50
Outline
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Applications of Gauss’s Law
The single Fixed Charge
Field of a sphere of charge
Field of a spherical shell
A Line of Charge
Conductors and Insulators
The electric field of a conductor
The field in the cavity of a conductor;
Faraday’s Cage
Solid conducting sphere with charge Q
A
E
r A E0
rA E
V
A
r
1
Q
40 r 2
1 Q
rA V 
40 A
1
Q
40 A
r
rA V 
1 Q
40 r
Electric field of a ball of charge
Q
1
Q
rR E
40 r 2
1
rQ
rR E
3
40 R
Electric field outside of a charged sphere is exactly the
same as the electric field produced by a point charge,
located at the center of the sphere, with charge equal to the
total charge on the sphere.
Electric field of a spherical shell
Q
The field outside the shell is like that of a point charge,
while the field everywhere inside the shell is zero.
Electric field of a line of charge


E
20 r
A Charged, Thin Sheet of Insulating Material
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
E
2 0
Conductors and insulators
Charges reside at the surface of the conductor
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Conductor
E=0
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What have we learned about
conductors?
• There is no electric field inside a conductor
• Net charge can only reside on the surface
of a conductor
• Any external electric field lines are
perpendicular to the surface (there is no
component of electric field that is tangent
to the surface).
• The electric potential within a conductor is
constant
Electric field near a surface of a conductor
l
a
 
 E  dS 
 EdS  Ea
cap
a
Ea 
0

E
0
Two parallel conducting plates


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l
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a
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+
d
 
 E  dS 
 EdS  Ea
cap
a
Ea 
0

E
0
(the total field at any point
between the plates)
An Apparent Contradiction
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
E
0
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-
An Apparent Contradiction


+
2
+

E
2 0

E
0
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+
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2
E

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
?
2 0
Near the surface of any conductor in electrostatics

E
0
1) There is a conducting spherical shell, inner radius A
and outer radius B. If you put a charge Q on it, find the
charge density everywhere.
2) There is a conducting spherical shell, inner radius A and
outer radius B. A charge Q is put at the center. If you put a
charge Q2 on the shell, find the charge density
everywhere.
A sphere of radius A has a charge Q uniformly spread
throughout its volume. Find the difference in the electric
potential, in other words, the voltage difference, between
the center and a point 2A from the center.
There is a conducting spherical shell, inner radius A and
outer radius B. A charge Q1 is put at the center. If you now
put charge -2Q1 on the shell, find the charge density at r=A
and r=B.

r2
 
Vr2  Vr1    E  dr  0

r1
since

E  0 inside the conductor.

For any two points r1
and

r2
inside the conductor
Vr1  Vr2
The conductor’s surface is an equipotential.
Equipotential Surfaces
An equipotential surface is a surface on which
the electric potential V is the same at every
point.
Because potential energy does not change as a test charge moves over
an equipotential surface, the electric field can do no work on such a

charge. So, electric field must be perpendicular
to the surface at every
point so that the electric force qE
is always perpendicular to the
displacement of a charge moving on the surface.
Field lines and equipotential surfaces are always
mutually perpendicular.
Method of images: What is a force on the point charge near a conducting plate?
Equipotential surface
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--
The force acting on the positive charge is exactly the same as it would
be with the negative image charge instead of the plate.
a
The point charge feels a force towards the plate with a magnitude:
1
2
q
F
40 (2a) 2
Method of images: A point charge near a conducting plane.

E ?
Equipotential surface
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--
P
r
a
E  
1
aq
40 (a 2  r 2 )3 2
1
aq
E  
40 (a 2  r 2 )3 2
2
aq
E
40 (a 2  r 2 ) 3 2
Equilibrium in electrostatic field: Earnshaw’s theorem
There are NO points of stable equilibrium in any electrostatic field!
How to prove it? Gauss’s Law will help!
Imaginary surface
surrounding P
P
If the equilibrium is to be a stable one, we require that if we move the
charge away from P in any direction, there should be a restoring force
directed opposite to the displacement. The electric field at all nearby points
must be pointing inward – toward the point P. But that is in violation of
Gauss’ law if there is no charge at P.
Thomson’s atom
1899
If charges cannot be held stably, there cannot be matter made up of
static point charges (electrons and protons) governed only by the laws
of electrostatics. Such a static configuration would collapse!
Capacitors
Consider two large metal plates which are parallel to each other
and separated by a distance small compared with their width.
y
Area A
L








The field between plates is






 
 
V

E
0

 [V (top)  V (bottom)]   E y dy 
dy   L
0
0
0
0
L
L


 A
QL
 [V (top)  V (bottom)]   L  
L
0
0 A
0 A
QL
V 
A 0
The capacitance is:
A 0
Q
Q
C


QL
V
L
A 0
Have a great day!