Transcript Conductors in Electrostatic Equilibrium

Conductors in
Electrostatic
Equilibrium
AP Physics C
Montwood High School
R. Casao
•
•
•
•
A good electrical conductor contains electrons
that are not bound to any atom and are free to
move within the conductor.
If the electric field E is not zero in some area, the
electrons there feel a force F = E·q and start to
move.
The electrons adjust their positions until the
force on every electron is zero.
When there is no net motion of charges within a
conductor, the conductor is in electrostatic
equilibrium.
•
Properties of a conductor in electrostatic
equilibrium:
– The electric field is zero everywhere inside
the material.
– Any excess charge on an isolated conductor
must be entirely on its surface.
– The electric field just outside a charged
conductor is perpendicular to the conductor’s
surface and has magnitude 2··k· (or /eo).
– On an irregularly shaped conductor, charge
tends to accumulate at locations where the
radius of curvature is the smallest (at sharp
points).
Gaussian
surface
is
labeled
A
Under electrostatic conditions, any excess
charge resides entirely on the surface of a
solid conductor.
• Generally draw a Gaussian surface to
satisfy two conditions:
–Condition 1: electric field constant
over the surface by symmetry.
–Condition 2: the dot product of
EdA = E·A·cos q because E and A
are parallel (q = 0°).
• Proof that electric field just outside a
charged conductor is 2··k· (or
/eo).
• Consider a section of the surface small
enough to neglect any curvature and take
the section to be flat.
• Imagine a tiny cylindrical Gaussian surface
to be embedded in the section. One end is
fully inside the conductor, the other
end is fully outside the
conductor and the
cylinder is
perpendicular to the
conductor’s surface.
• E at and just outside the conductor’s surface must
be perpendicular to the surface. If not, a
component of E would exist in the direction of the
conductor’s surface that would exert forces on the
surface charges and cause them to move.
• There is no flux thru
the internal end of the
cylinder because no
electric field lines pass
through the inner surface
of the Gaussian cylinder
because the electric field
E inside the conductor is
zero.
• There is no flux thru the curved surface of the
cylinder because E = 0 N/C within the
conductor and externally E is parallel to the
curved portion of the Gaussian cylinder.
• The only flux thru the Gaussian cylinder occurs thru the
external end of the
cylinder where E is
perpendicular to A.
The charge enclosed
(qin) by the Gaussian
surface lies on the
conductor’s surface
in an area A;  = Q/A
so qin = ·A.
• The magnitude of E just outside a
conductor is proportional to the surface
charge density at that location on the
conductor.
• If the charge on the conductor is
positive, E is directed away from the
conductor.
• If the charge on the conductor is
negative, E is directed toward the
conductor.
Spherical Gaussian surfaces around (a)
positive and (b) negative point charge.
Q
σ
A
q in
Q
proportional relationship : 
A 2 πr L

Φ  E  dA Φ  4  π  k  q in  E  A  cos θ
2πr LQ
cos θ  1 q in 
 2πr Lσ
A
4  π  k  q in 4  π  k  2  π  r  L  σ
E

 4πk σ
A
2πr L
σ
E  4πk σ 
εo
Off-Centered Charge Distribution
• Consider a –5 mC point charge located
closer to one side of an electrically neutral
shell.
• A charge of +5 mC will lie on the inner wall of the shell; the
E inside the shell must be 0 N/C; qin = -5 mC + 5 mC = 0 C.
• If the point charge were centered inside the shell, the
positive charge would be uniformly distributed along the
inner wall. Since the point charge is off-center, the
distribution of positive charge tends to collect on the section
of the wall nearest the -5 mC.
• Because the shell is electrically neutral, its inner wall can
only have a charge of +5 mC if electrons with a total charge
of -5 mC leave the inner wall and move to the outer wall.
• There they spread out uniformly because the shell is
spherical and because the distribution of positive charge on
the inner wall cannot produce an electric field in the shell to
affect the distribution of charge on the outer wall.
Example 24.7 A
Sphere Inside a
Spherical Shell
Example 24.7 A Sphere Inside a Spherical Shell
• Consider a solid conducting sphere of
radius a surrounded by a conducting
spherical shell of inner radius b and outer
radius c.
• The net charge on the solid conducting
sphere is +2·Q. The electric field E from
the solid conducting sphere radiates
outward from the sphere of radius a and
would be uniform.
• The net charge on the conducting spherical
shell is –Q. The electric field E from the
conducting spherical shell radiates inward
and would be uniform.
• The entire system is in electrostatic
equilibrium.
• The spherical nature of the objects
satisfies condition 1 (electric field E is
constant over the surface by
symmetry).
• Region 1: the charge 2·Q will lie on
the outer surface of the solid
conducting sphere of radius a.
• Draw a spherical Gaussian surface of
radius r < a.
• No electric field E can exist within the
conductor.
– The negative charges from the outer
conducting spherical shell attracts the
positive charge on the solid conducting
sphere to the surface.
– The repulsion of the positive charges on
the solid conducting sphere also
contributes to the positioning of the
positive charge on the surface of the solid
conducting sphere.
– No charge lies within the solid sphere,
therefore, qin = 0 C.


Φ  E  dA  E  dA  E  A  cos θ
Φ  4  π  k  q in
cos θ  1
4  π  k  q in  E  A  cos θ
A  4 π  r2
4  π  k  q in 4  π  k  q in k  q in
E


A
4 π  r2
r2
q in  0
E  0 N/C
• Region 2: the charge 2·Q will lie on the
outer surface of the solid conducting
sphere of radius a.
• Draw a spherical Gaussian surface of
radius r  a.
• Charge located inside the spherical
Gaussian surface of radius r is 2·Q;
therefore, qin = 2·Q
• Spherical Gaussian surface satisfies
condition 1: electric field constant over
the surface by symmetry.
• Spherical Gaussian surface satisfies
condition 2: the dot product of EdA =
E·A·cos q because E and A are
parallel (q = 0°).


Φ  E  dA  E  dA  E  A  cos θ
Φ  4  π  k  q in
cos θ  1
A  4πr
2
4  π  k  q in  E  A  cos θ
4  π  k  q in 4  π  k  q in k  q in
E


2
2
A  cos θ
4πr
r
k 2Q
2k Q
q in  2  Q
E
E

2
2
r
r
• Region 3: inside the conducting spherical
shell which is in electrostatic equilibrium:
– The negative charges on this conductor are
located on the inner surface of the conducting
spherical shell because they are attracted to the
positive charge located on the solid conducting
sphere.
– The charge on the inner surface of the spherical
shell must be -2·Q to balance the
+2·Q
charge located on the solid sphere. Therefore,
a charge of +Q must reside on the outer surface
of the conducting shell, bringing the net charge
to –Q; (-2·Q +Q = -Q).
– Qnet = Qinner surface + Qouter surface
– The distribution of negative charge on the inner
surface and the positive charge on the outer
surface of the shell means that there is no
charge located within the spherical shell itself
because all the charge is located on the
surface.
– Draw a spherical Gaussian surface of radius
b  r < c.
– The spherical Gaussian surface satisfies
condition 1 and condition 2; the electric field E
can be considered to be constant over the
surface by symmetry and the dot product of
EdA = E·A·cos q because E and A are parallel
(q = 0°).


Φ  E  dA  E  dA  E  A  cos θ
Φ  4  π  k  q in
cos θ  1
4  π  k  q in  E  A  cos θ
A  4πr
2
4  π  k  q in 4  π  k  q in k  q in
E


2
2
A
4 πr
r
q in  0
E  0 N/C
• Region 4: lies outside the conducting
spherical shell.
– A spherical Gaussian surface can be drawn to
enclose the conducting spherical shell and the
solid conducting sphere.
– The radius of the spherical Gaussian surface is r
 c.
– The spherical Gaussian surface satisfies
condition 1 and condition 2; the electric field E
can be considered to be constant over the
surface by symmetry and the dot product of
EdA = E·A·cos q because E and A are parallel
(q = 0°).


Φ  E  dA  E  dA  E  A  cos θ
Φ  4  π  k  q in
4  π  k  q in  E  A  cos θ
cos θ  1
A  4πr
4  π  k  q in 4  π  k  q in k  q in
E


2
2
A
4 πr
r
q in  2  Q  2  Q  Q  Q
2
E
k Q
r
2
as if it were a point charge