Download Surface charge on a conductor

Lecture-7
Conductors
Field At the Surface of a Conductor
•

E

E||

E
•
•
•
•
September 26, 2007
Imagine an electric field at some
arbitrary angle at the surface of a
conductor.
There is a component perpendicular to
the surface, so charges will move in this
direction until they reach the surface,
and then, since they cannot leave the
surface, they stop.
There is also a component parallel to
the surface, so there will be forces on
charges in this direction.
Since they are free to move, they will
move to nullify any parallel component
of E.
In a very short time, only the
perpendicular component is left.
Field Inside a Conductor
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•
•
•
•
September 26, 2007
We can use Gauss’ Law to show that the inside of a
conductor must have no net charge.
Take an arbitrarily shaped conductor, and draw a gaussian
surface just inside.
Physically, we expect that there is no electric field inside,
since otherwise the charges would move to nullify it.
Since E = 0 everywhere inside, E must be zero also on the
gaussian surface, hence there can be no net charge
inside.
Hence, all of the charge must be on the surface (as
discussed in the previous slide).
Field Inside a Conductor
•
•
•
•
•
•
We can use Gauss’ Law to show that the inside of a
conductor must have no net charge.
Take an arbitrarily shaped conductor, and draw a gaussian
surface just inside.
Physically, we expect that there is no electric field inside,
since otherwise the charges would move to nullify it.
Since E = 0 everywhere inside, E must be zero also on the
gaussian surface, hence there can be no net charge
inside.
Hence, all of the charge must be on the surface (as
discussed in the previous slide).
If we make a hole in the conductor, and surround the hole
with a gaussian surface, by the same argument there is
no E field through this new surface, hence there is no net
charge in the hole.
Field Inside a Conductor
•
•
•
if you try to deposit charge on the
inside of the conductor...
The charges all move to the outside
and distribute themselves so that
the electric field is everywhere
normal to the surface.
This is NOT obvious, but Gauss’ Law
allows us to show this!
There are two ideas here
• Electric field is zero inside conductors
• Because that is true, from Gauss’
Law, cavities in conductors have E = 0
September 26, 2007
A Charge Inside a Conductor
Q.
Spherical
cavity
What will happen when we add a
charge inside a conductor?
A.
B.
C.
D.
Conducting
sphere
September 26, 2007
Positive point
charge
E.
E field is still zero in the cavity.
E field is not zero in the cavity, but it is
zero in the conductor.
E field is zero outside the conducting
sphere.
E field is the same as if the conductor
were not there (i.e. radial outward
everywhere).
E field is zero in the conductor, and
negative (radially inward) outside the
conducting sphere.
Use Gauss’ Law to Find Out
 
 0  E  dA  qenc
Is E = 0 in the cavity?
No, because there is charge enclosed (Gauss’ Law).
Gaussian Surface
Is E = 0 in the conductor?
Yes, because as before, if there were an electric
field in the conductor, the charges would move in
response (NOT Gauss’ Law).
If we enlarge the gaussian surface so that it is
inside the conductor, is there any net charge
enclosed?
It looks like there is, but there cannot be, because
Gauss’ Law says E = 0 implies qenc = 0!
How do we explain this?
There must be an equal and opposite charge
induced on the inner surface.
September 26, 2007
E Field of Charge In Conductor
This negative charge acts with the inner charge to
make the field radial inside the cavity.
This negative charge cannot appear out of nowhere.
Where does it come from?
It comes from the outer surface (electrons drawn
inward, attracted to the positive charge in the
center). Therefore, it leaves positive charge
behind.
The net positive charge that appears conductor is
exactly the same as the original charge in the
center, so what do the field lines look like?
By spherical symmetry, the positive shell of charge
acts like a point charge in the center, so field is the
same as the field of the original point charge.
September 26, 2007
Back to the Previous Question
Q.
Spherical
cavity
What will happen when we add a
charge inside a conductor?
A.
B.
C.
D.
Conducting
sphere
September 26, 2007
Positive point
charge
E.
E field is still zero in the cavity.
E field is not zero in the cavity, but it is
zero in the conductor.
E field is zero outside the conducting
sphere.
E field is the same as if the conductor
were not there (i.e. radial outward
everywhere).
E field is zero in the conductor, and
negative (radially inward) outside the
conducting sphere.
E Field of Charge In Conductor
What happens when we move the inner charge
off-center?
It induces an off-center charge distribution on the
Inner wall.
Note that the field lines distorted, so they remain
perpendicular to the inner wall. What happens to
the outer positive charge distribution?
Draw a gaussian surface inside the conductor to
find out.
The net charge enclosed is zero, so E = 0, which
we already knew because it is inside the
conductor. The inner charge is shielded by the
induced charge distribution, so the outer charges
will be evenly distributed.
September 26, 2007
Electrostatic Boundary Conditions
E
σ 2ε0
Discontinuity
For an infinite plane carrying uniform
surface charge 
E
y
Discontinuity
 σ 2ε0
R
For a spherical shell of radius R,
carrying uniform surface charge 
r
Electrostatic Boundary Conditions
From Gauss’ law

S


E above
 E below

1
σ
ε0
1
1
E  ds  Qenc  A
0
0
E

above
E

above
AE
E

below

below
1
A  A
0
1
 
0
Conclusion: The normal component of E is discontinuous by an
amount σ/ε0 at any boundary.
Electrostatic Boundary Conditions
The line integral of the static electric field E around a
closed path is zero

E  dl  0
s
E
||
above
lE
E
||
above
E
||
below
l0
||
below
The tangential component of E by contrast is always
continuous
Electrostatic Boundary Conditions
The potential is continuous
across any boundary.
b
Vabove  Vbelow    E  dl
a
As the path length shrinks to zero i.e
Vabove  Vbelow
Basic properties
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E=0 inside a conductor.
=0 inside a conductor.
Any net charge resides on the surface.
A conductor is an equi-potential.
E is perpendicular to the surface, just outside
a conductor.
• Note, none of this is true for insulators.
Summary
• Electric flux is the amount of electric field passing through a
closed surface.
• Flux is positive when electric field is outward, and negative
when electric field is inward through the closed surface.
• Gauss’ Law states that the electric flux is proportional to the
net charge enclosed by the surface, and the constant of
proportionality is 0. In symbols, it is    q
0
enc
 
 0  E  dA  qenc
• There are three geometries we typically deal with:
Geometry
Charge Density
Gaussian surface
Linear
l = q/L
Cylindrical, with axis
along line of charge
E
l
20 r
Sheet or
Plane
 = q/A
Cylindrical, with axis
along E.
E

0
Spherical
 = q/V
Spherical, with center
on center of sphere
E
September 26, 2007
Electric field
Line of Charge
Conducting
q
40 r
2
rR
E

2 0
Nonconducting
 q 
r
E  
3 
4

R
0


r<R
Q.
A metal sphere of radius R, carrying charge q is surrounded by a thick concentric
metal shell. The shell carries no net charge.
(a) Find the surface charge density at R, a and b
(b) Find the potential at the centre, using infinity as the reference point
(c) If outer surface is grounded, then find out the answer of (a) and (b)
Answer:
a
q
b
(a) R 
R
q
q
q
,



,


a
b
4R 2
4a 2
4b2
(b) V 0 
(c)
1 q q q
   
4πε0  b R a 
1 q q q
=0


V
0

   
b
4πε0  b
R
a
Surface charge on a conductor


σ
E above  E below  nˆ
ε0
Electrostatic boundary condition:
=> Field outside a conductor
 σ
E  nˆ
ε0
Force on a charge element dq placed in an external field E(e) :
F  dq Ee  dq E


On a volume charge distribution : F   ρ E dτ
 τ 
On a surface charge distribution : F   σ E da
s
“ BUT E is discontinuous across a surface charge distribution”
On the surface charge the force per unit area :

 1 

f  σ E above  E below
2

Force on a conductor
Force (per unit area) on the conductor surface:

1 2
f 
σ nˆ
2 ε0
Outward electrostatic Pressure on the conductor surface :
1
P  ε0 E 2
2
Two large metal plates (each of area A) are held a distance d apart.
Suppose we put a charge Q on each plate, what is the electrostatic
pressure on the plates ?
2
Answer:
P
Q
2 ε 0 A2
Q.5 consider two potential fields V1 = y and V2 =y + ex sin y.
(a) Is 2V1 = 0? (b) Is 2 V2 = 0? (c) Is V1 = 0 at y = 0?
(d) Is V2 = 0 at y = 0 ? (e) Is V1 = π at y = π ? (f) Is V2 = π
at y = π ? (g) Are V1 and V2 identical ?
Ans: (a) yes (b) yes (c) yes (d) yes (e) yes (f) yes (g) No