chapter 22 Lecture 2 - Santa Monica Collegehomepage.smc.edu/kocharian_armen/Physics22/chapter 22...
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Chapter 22 Gauss’s Law Lecture 4 Dr. Armen Kocharian
Transcript of chapter 22 Lecture 2 - Santa Monica Collegehomepage.smc.edu/kocharian_armen/Physics22/chapter 22...
Chapter 22
Gauss’s LawLecture 4
Dr. Armen Kocharian
Field Due to a Plane of Charge
E must be perpendicular to the plane and must have the same magnitude at all points equidistant from the planeChoose a small cylinder whose axis is perpendicular to the plane for the gaussian surface
Φ = = =
= ⇒ =
in2
22
Eo o
o o
q σAEAε ε
σA σEA Eε ε
Field Due to a Plane of Charge, cont
E is parallel to the curved surface and there is no contribution to the surface area from this curved part of the cylinderThe flux through each end of the cylinder is EA and so the total flux is 2EA
Field Due to a Plane of Charge, final
The total charge in the surface is σAApplying Gauss’s law
Note, this does not depend on rTherefore, the field is uniform everywhere
22E
o o
σA σEA and Eε ε
Φ = = =
Electrostatic Equilibrium
When there is no net motion of charge within a conductor, the conductor is said to be in electrostatic equilibrium
Properties of a Conductor in Electrostatic Equilibrium
The electric field is zero everywhere inside the conductorIf an isolated conductor carries a charge, the charge resides on its surfaceThe electric field just outside a charged conductor is perpendicular to the surface and has a magnitude of σ/εoOn an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature is the smallest
Property 1: Einside = 0Consider a conducting slab in an external field EIf the field inside the conductor were not zero, free electrons in the conductor would experience an electrical forceThese electrons would accelerateThese electrons would not be in equilibriumTherefore, there cannot be a field inside the conductor
Property 1: Einside = 0, cont.Before the external field is applied, free electrons are distributed throughout the conductorWhen the external field is applied, the electrons redistribute until the magnitude of the internal field equals the magnitude of the external fieldThere is a net field of zero inside the conductorThis redistribution takes about 10-15s and can be considered instantaneous
Property 2: Charge Resides on the SurfaceChoose a gaussian surface inside but close to the actual surfaceThe electric field inside is zero (prop. 1)There is no net flux through the gaussian surfaceBecause the gaussian surface can be as close to the actual surface as desired, there can be no charge inside the surface
Property 2: Charge Resides on the Surface, cont
Since no net charge can be inside the surface, any net charge must reside onthe surfaceGauss’s law does not indicate the distribution of these charges, only that it must be on the surface of the conductor
Property 3: Field’s Magnitude and DirectionChoose a cylinder as the gaussian surfaceThe field must be perpendicular to the surface
If there were a parallel component to E, charges would experience a force and accelerate along the surface and it would not be in equilibrium
Property 3: Field’s Magnitude and Direction, cont.
The net flux through the gaussian surface is through only the flat face outside the conductor
The field here is perpendicular to the surface
Applying Gauss’s law
Eo o
σA σEA and Eε ε
Φ = = =
Conductors in Equilibrium, example
The field lines are perpendicular to both conductorsThere are no field lines inside the cylinder
Example
( )
( )
21 1
0
1
4 ,
00
in
in
qE A E r
qE r a
πε
= =
=
= ≤
E EAΦ =
( )
( )
22 2
0
2 2 20
4
222
4
in
in
e
qE A E r
q Qk QQE a r b
r r
πε
πε
= =
=
= = ≤ ≤
Find the field in regions 1, 2, 3 and 4
Find in region 1 Field in region 2
Example
( )( )
( )
24 4
0
4 2 20
4 ,
2
4
in
in
e
qE A E r
q Q Q Qk QQE r c
r r
πε
πε
= =
= + − =
= = ≥
E EAΦ =
Find the field in regions 1, 2, 3 and 4
( )
( )
23 3
0
3
4 ,
00
in
in
qE A E r
qE b r c
πε
= =
=
= ≤ ≤
Field in region 3 Field in region 4
Example
( )1 0E r a= ≤
( )2 22 ek Q
E a r br
= ≤ ≤
Electrical field is zero inside conductors
Field exist outside of conductors
( )3 0E b r c= ≤ ≤
( )4 2 204
ek QQE r cr rπε
= = ≥
Derivation of Gauss’s LawWe will use a solid angle, ΩA spherical surface of radius r contains an area element ΔAThe solid angle subtended at the center of the sphere is defined to be 2
ArΔ
Ω =
Some Notes About Solid Angles
A= 4πr2 and r2 have the same units, so Ω is a dimensionless ratioWe give the name steradian to this dimensionless ratioThe total solid angle subtended by a sphere is 4π steradians
Derivation of Gauss’s Law, cont.
Consider a point charge, q, surrounded by a closed surface of arbitrary shapeThe total flux through this surface can be found by evaluating E.ΔA for each small area element and summing over all the elements
Derivation of Gauss’s Law, final
The flux through each element is
Relating to the solid angle
where this is the solid angle subtended by ΔA
The total flux is
( ) 2coscosE e
A θE θ A k qr
ΔΦ = ⋅ Δ = Δ =E A
2cosA θr
ΔΔΩ =
2cos
E e eo
dA θ qk q k q dr ε
Φ = = Ω =∫ ∫
Derivation of Gauss’s Law, cont.
(a) in 3 2q Q Q Q= + − = + in 0q >
.
The charge distribution is spherically symmetric and directed radially outward.
in2 2
2e ek q k QE
r r= = r c≥
Insulating sphere 3Q within conducting shell -Q. Prob. 55
(d) in 0q =
.
Since all points within this region are located inside conducting material, E=0.
b r c< <
in 00 0E Ed qΦ = ⋅ = ⇒ =∈ Φ =∫E A
.
(f) in 3q Q= +
in2 2
3e ek q k QE
r r= = a r b≤ <
Directed radially outward.
Insulating sphere 3Q within conducting shell –Q, cont.
.
Directed radially outward.
33
in 3 343
3 43
3Q r
q V r Qa a
⎛ ⎞+ ⎛ ⎞= = = +⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ρ π
π
(h)
3in
2 2 3 33 3e ee
k q k r rE Q k Q
r r a a
⎛ ⎞= = + =⎜ ⎟⎝ ⎠
Insulating sphere 3Q within conducting shell –Q, cont.
Problem
Inner surface density
(a) The inner charge is q+
q−on the inner surface of the conductor, where its surface density is:
24a
q
a
−=σ
π.
(b) The outer surface carries charge
Q q+
24b
Q q
b
+=σ
π
.
Outer surface density
Derivation of Gauss’s Law –Notes
This result matches the earlier result for the equation for Gauss’s lawThis derivation was independent of the shape of the closed surfaceThis derivation was independent of the position of the charge within the surface
Problem
E EAΦ =
ds Rd= θ
2 2 sinr R=π π θ
( ) ( ) ( )2 2 20
0 0
2 2 sin 2 sin 2 cos 2 1 cosA rds R Rd R d R Rθ θ θ
π π θ θ π θ θ π θ π θ= = = = − = −∫ ∫ ∫
( ) ( )22
0 0
12 1 cos 1 cos
4 2EQ Q
RR
Φ = ⋅ − = −∈ ∈
π θ θπ
Find the flux through the circular cap
Electric FluxField lines penetrating an area A perpendicular to the fieldThe product of EA is the flux, ΦIn general:ΦE = E A sin θ
Chapter 24
Assessment Test
A uniformly charged rod has a finitelength L. The rod is symmetric under rotations about the axis and under reflection in any plane containing the axis. It is not symmetric under translations or under reflections in a plane perpendicular to the axis other than the plane that bisects the rod. Which field shape or shapes match the symmetry of the rod?
A. a and dB. c and eC. b onlyD. e onlyE. none of the above
A uniformly charged rod has a finitelength L. The rod is symmetric under rotations about the axis and under reflection in any plane containing the axis. It is not symmetric under translations or under reflections in a plane perpendicular to the axis other than the plane that bisects the rod. Which field shape or shapes match the symmetry of the rod?
A. a and dB. c and eC. b onlyD. e onlyE. none of the above
This box contains
A. a net positive charge.B. no net charge. C. a net negative charge.D. a positive charge. E. a negative charge.
This box contains
A. a net positive charge.B. no net charge. C. a net negative charge.D. a positive charge. E. a negative charge.
The total electric flux through this box is
A. 0 Nm2/C.B. 1 Nm2/C.C. 2 Nm2/C.D. 4 Nm2/C.E. 6 Nm2/C.
The total electric flux through this box is
A. 0 Nm2/C.B. 1 Nm2/C.C. 2 Nm2/C.D. 4 Nm2/C.E. 6 Nm2/C.
These are two-dimensional cross sections through three-dimensional closed spheres and a cube. Rank order, from largest to smallest, the electric fluxes a to e through surfaces a to e.
A. Φa > Φc > Φb > Φd > ΦeB. Φb = Φe > Φa = Φc = ΦdC. Φe > Φd > Φb > Φc > ΦaD. Φb > Φa > Φc > Φe > ΦdE. Φd = Φe > Φc > Φa = Φb
These are two-dimensional cross sections through three-dimensional closed spheres and a cube. Rank order, from largest to smallest, the electric fluxes a to e through surfaces a to e.
Which Gaussian surface would allow you to use Gauss’s law to determine the electric field outside a uniformly charged cube?
A. A sphere whose center coincides with the center of the charged cube.
B. A cube whose center coincides with the center of the charged cube and which has parallel faces.
C. Either A or B.D. Neither A nor B.
Which Gaussian surface would allow you to use Gauss’s law to determine the electric field outside a uniformly charged cube?
A. A sphere whose center coincides with the center of the charged cube.
B. A cube whose center coincides with the center of the charged cube and which has parallel faces.
C. Either A or B.D. Neither A nor B.
Chapter 24Reading Quiz
The amount of electric field passing through a surface is called
A. Electric flux.B. Gauss’s Law.C. Electricity.D. Charge surface density.E. None of the above.
The amount of electric field passing through a surface is called
A. Electric flux.B. Gauss’s Law.C. Electricity.D. Charge surface density.E. None of the above.
Gauss’s law is useful for calculating electric fields that are
A. due to point charges.B. uniform.C. symmetric.D. due to continuous charges.
Gauss’s law is useful for calculating electric fields that are
A. due to point charges.B. uniform.C. symmetric.D. due to continuous charges.
Gauss’s law applies to
A. lines.B. flat surfaces.C. spheres only.D. closed surfaces.
Gauss’s law applies to
A. lines.B. flat surfaces.C. spheres only.D. closed surfaces.
The electric field inside a conductor in electrostatic equilibrium is
A. uniform.B. zero.C. radial.D. symmetric.
The electric field inside a conductor in electrostatic equilibrium is
A. uniform.B. zero.C. radial.D. symmetric.
