Transcript Applications of Gauss` Law to Charged Insulators

Applications of Gauss’ Law
to Charged Insulators
AP Physics C
Montwood High School
R. Casao
• Gauss’ law is useful when there is a high
degree of symmetry in the charge
distribution.
• The surface should always be chosen so
that it has the same symmetry as that of
the charge distribution.
• Electric Field due to
a Point Charge:
For a point charge, choose
a spherical gaussian
surface of radius R that is
centered on the point
charge.
Electric Field Due to a Point Charge
• The electric field of a positive point charge is
directed radially outward by symmetry and is
normal to the surface at every point.
– E is parallel to the area vector at each point. The
angle between the area vector and the electric field
vector is zero.
– E is constant everywhere on the surface of the
gaussian sphere and can be pulled out in front of the
integral sign.
– Applying Gauss’ Law:


Φ  E  dA  E  dA
– Integrating over the surface area of the sphere gives
us the area, 4·p·r2.
Electric Field Due to a Point Charge
• The equation for the electric field of a point
charge is:
k Q
E 2
r
• The flux equation becomes:
Φ
k Q
r
2
4πr  4πk Q
2
• Remember that Q refers to the entire
charge located inside the sphere (Qin or
qin).
Spherically Symmetric Charge Distribution
• Consider an insulating sphere of radius R
having a uniform charge density r and a
total positive charge Q.
• Charge will remain in its position when on
an insulating sphere, therefore, the charge
on the insulating sphere will be uniformly
distributed throughout the volume of the
sphere (spherically symmetric).
• For determining the electric field within the
insulating sphere of radius R, (for situations
in which r < R):
Spherically Symmetric
Charge Distribution
• Apply a spherical gaussian
surface of radius r, centered on
the center of the insulating
sphere.
• The amount of charge enclosed
within the gaussian sphere will be
less than the total charge Q.
• The symmetric charge distribution
allows for a proportional
relationship to be set up between
the total charge Q, total volume V,
enclosed charge qin, and the
volume of the gaussian sphere.
Spherically Symmetric Charge Distribution
• Volume of a sphere: V  4  π  R 3
3
• Relationship:
q in
q in
Q
Q


3
4 πR
4  π  r3
V Vgauss
3
3
• Solve for qin:
q in
3
4
3
Q πr
Qr
3


4  π  R3
R3
3
• The electric field is constant in magnitude
everywhere on the spherical Gaussian surface and
is parallel to the area vector at each point. The
angle between the area vector and the electric field
vector is zero.
Spherically Symmetric Charge Distribution
• The area vector is parallel to the electric field
vector at the surface of the spherical Gaussian
surface.
• Applying Gauss’ law:


Φ  E  dA  E  dA  E  A  cos θ
cos θ  1
sphere A  4  π  r 2
4  π  k  q in  E  4  π  r
E
4  π  k  q in
4 π  r2

2
k  q in
r2
Φ  E  A  E  4 π  r2
k Q  r3 k  Q  r
 2 3 
r
R
R3
Spherically Symmetric Charge Distribution
• As r increases to R, the magnitude of the
electric field will increase as more charge is
included within the volume of the spherical
Gaussian surface.
• The maximum electric field
will be obtained when r = R.
• For a spherical Gaussian
surface of radius r  R, we
can consider the charged
insulating sphere as a point
charge.
Spherically Symmetric Charge Distribution


Φ  E  dA  E  dA  E  A  cos θ
cos θ  1 sphere A  4  π  r 2
Φ  4  π  k  q in
E
4  π  k  q in
Φ  E  4 π  r2
4  π  k  q in  E  4  π  r 2

k  q in

k Q
4 π  r2
r2
r2
• Conclusion: the electric field E inside a uniformly
charged solid insulating sphere varies linearly with
the radius until reaching the surface. The electric
field E outside the sphere varies inversely with 1/r2.
The diagram on the next slide illustrates this
relationship.
Spherically Symmetric Charge Distribution
Thin Spherical Shell
• Consider a thin
spherical shell of
radius R with a total
charge Q distributed
uniformly over its
surface.
• For points outside the
shell, r  R: the
spherical Gaussian
surface will enclose
the thin spherical
shell, so qin = Q.
Thin Spherical Shell
• The area vector is parallel to the electric field
vector at the surface of the spherical Gaussian
surface. The angle between the area vector and
the electric field vector is zero.


Φ  E  dA  E  dA  E  A  cos θ
cos θ  1 sphere A  4  π  r 2
Φ  4  π  k  q in
E
4  π  k  q in
4πr
2
Φ  E  4 π  r2
4  π  k  q in  E  4  π  r

k  q in
r
2

k Q
r
2
2
Thin Spherical Shell
• For points inside the thin spherical shell,
r < R:
– The charge lies on the surface of the thin
spherical shell.
– No charge is found within the shell, so qin = 0
and the electric field and flux within the thin
spherical shell is also 0.
Cylindrically Symmetric Charge Distribution
• Consider a uniformly charged cylindrical rod
of length L.
Cylindrically Symmetric Charge Distribution
Cylindrically Symmetric Charge Distribution
• Instead of the shortened Gaussian cylinder
indicated in the diagrams, I use a Gaussian
cylinder that is the same length as the charged
cylindrical rod.
• The electric field lines are perpendicular to the
charged cylindrical rod and are directed outward in
all directions.
• The electric field is perpendicular to the Gaussian
cylinder and is parallel to the area vector at those
points where it passes through the Gaussian
cylinder. The angle between the area vector and
the electric field vector is zero.
Cylindrically Symmetric Charge Distribution
• The equation for the area of a cylinder is:
A = 2·p·r·L
• Applying Gauss’ law:


Φ  E  dA  E  dA  E  A  cos θ
cos θ  1 cylinder A  2  π  r  L Φ  E  2  π  r  L
Φ  4  π  k  q in
4  π  k  q in  E  2  π  r  L
4  π  k  q in 2  k  q in 2  k  Q 2  k  λ
E



2πr L
rL
rL
r
Nonconducting Plane Sheet of Charge
• Consider a plane sheet of charge with a uniform
charge density s.
Nonconducting Plane Sheet of Charge
• The diagrams show a small cylinder through the
plane sheet of charge; I use a rectangle that
encloses the entire sheet of charge. The faces of
the rectangle will have the same area as the plane
sheet of charge.
• The electric field E is perpendicular to the plane
sheet of charge and the area vector for the
rectangular face on either side of the plane sheet
of charge is in the same direction as the electric
field vector.
– The angle between the area vector and the
electric field vector is 0.
Nonconducting Plane Sheet of Charge
• The electric field lines near the top of the
plane of charge will run parallel to the sides
of the rectangle and will not result in any flux
through the rectangular surface.
• There are two surfaces of the rectangle
through which the electric field lines pass.
• Applying Gauss’ law:


cos θ  1
  2EA
Φ  2  E  dA  2  E  dA  2  E  A  cos θ
Nonconducting Plane Sheet of Charge
4  π  k  q in  2  E  A
4  π  k  q in 4  π  k  q in 2  π  k  q in
E


2A
2A
A
Q
σ
q in  Q
E  2πk σ
A
• Textbooks will have:
σ
E
2  εo
Nonconducting Plane Sheet of Charge
• Here is the conversion:
1
k
4  π  εo
1
so
 4 πk
εo
σ
4 πk σ
E

 2 πk σ
2  εo
2
Two Parallel Nonconducting Sheets
•
•
•
•
The situation is different if you bring two nonconducting sheets of charge
close to each other.
In this case, the charges cannot move, so there is no shielding, but now we
can use the principle of superposition.
The electric field on the left due to the positively charged sheet is canceled
by the electric field on the left of the negatively charged sheet, so the field
there is zero.
The electric field on the right due to the negatively charged sheet is
canceled by the electric field on the right of the positively charged sheet.

The electric field in between
the two plates is equal to
2·E.