Download Gauss`s Law 3.1 Quiz

Physics 102
Conference 3
Gauss’s Law
Conference 3
Physics 102
General Physics II
Monday, February 10th, 2014
3.1
Quiz
Problem 3.1
A spherical shell of radius R has charge Q spread uniformly over its surface.
Find the electric field inside and outside the shell.
We’ll use Gauss’s law – for r < R, the charge enclosed by a (fictitious) Gaussian
sphere is zero, so that E(r) 4 π r2 = 0 and E = 0. For r > R, the charge
enclosed by a Gaussian sphere is Q, and E(r) 4 π r2 = Q/0 so that E(r) =
Q
.
4 π 0 r 2
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3.1. QUIZ
Conference 3
Problem 3.2
An infinite line of charge has constant λ (charge-per-unit-length). Find the
electric field a distance s from the line of charge.
Once again, Gauss’s law – for a Gaussian cylinder that encloses the line of charge,
and has radius s, length `, the charge enclosed is: λ `, and then E(s) 2 π s ` =
λ `/0 gives E = 2 πλ0 s .
Problem 3.3
An infinite sheet of charge has constant charge-per-unit-area σ, find the electric
field a height z above the sheet.
For a Gaussian “pillbox” with top surface area A and height z above the plane
(and extending to −z below), the charge enclose is σ A, and Gauss’s law read:
E(z) A 2 = σA/0 – then E(z) = 2σ0 .
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3.2. FLUX
3.2
Conference 3
Flux
Flux is defined in terms of a surface and an electric field. Given a surface with
vector area element da (normal to the surface) and an electric field E, the flux
through the surface is defined to be:
Z
Φ = E · da.
(3.1)
Problem 3.4
A point charge sits at the back corner of a cube. Find the electric flux through
the face of the cube shown below.
Find flux of electric field through this face
Charge q sits at the back left corner
Think of the larger cube that encloses the point charge, shown below:
The flux through this large cube, which encloses the point charge, is q0 by
Gauss’s law. The shaded face is 1/24 of the total flux, so the flux through that
face is just Φ = 24q0 .
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3.2. FLUX
Conference 3
Problem 3.5
A sphere of radius R has charge-per-unit-volume ρ = α r for constant α. What
are the units of α? Find the electric field inside and outside the sphere.
In both cases, the electric field depends only on the distance to the origin, and
points away from the center of the sphere. What changes in the application
of Gauss’s law is the amount of charge enclosed by the Gaussian sphere: For a
Gaussian sphere of radius r < R, we have
Z r
r4
4 π r̄2 (α r̄) dr̄ = 4 π
Qenc =
α.
(3.2)
4
0
H
H
The integral of E over the surface is: E(r) · da = E(r) da = E(r) 4 π r2 .
Putting the two together in Gauss’s law gives:
I
Qenc
E · da =
0
(3.3)
α
π r4
2
E(r) 4 π r =
0
2
so that we end up with E(r) = α4 r0 . α must be a charge/ `4 so that α r is a
charge-per-unit-volume. Outside the sphere, our Gaussian surface encloses all
of the charge, so we can take the integral in (3.2) out to R, giving:
Qenc = 4 π
R4
α,
4
and then E(r) 4 π r2 = Qenc /0 gives E(r) =
expected.
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α R4
.
4 0 r 2
(3.4)
These match at r = R as
3.2. FLUX
Conference 3
Problem 3.6
There is an infinite slab with charge-per-unit-volume ρ0 extending from z = −d
to z = d. At the “center” of the slab, we carve out a hollow sphere of radius
d – find the electric field a height h above the center of the slab, directly over
the hollow sphere.
ẑ
E =?
h
d
ŷ
x̂
z=0
empty
⇢0
d
The electric field above the infinite slab, with no hole cut out, is, from Gauss’s
Law (applied to a cylinder of height 2 z centered at the x − y plane, with top
area A):
I
Qenc
E · da =
0
ρ0 2 d A
(3.5)
E(z) 2 A =
0
ρ0 d
E(z) =
0
The electric field above a uniformly charged sphere with constant −ρ0 is Es (h) =
−ρ0 34 π d3
.
4 π 0 h2
The superposition of these two configurations gives the target configuration, so the superposition of these two fields is the electric field:
E=
ρ0 d
ρ0 d3
ẑ −
ẑ.
0
3 0 h2
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(3.6)
3.2. FLUX
Conference 3
Problem 3.7
An infinite line charge has constant charge-per-unit-length λ. Surrounding the
line charge is a cylindrical shell of radius R, and carrying a constant chargeper-unit-area σ. Given λ, what must σ be in order to get zero electric field for
all points outside the cylindrical shell? For that σ, what is the electric field in
between the line of charge and the shell?
=?
R
The electric field outside of a wire with constant λ is E = 2 πλ0 s – this is also
the electric field in between the line charge and the cylindrical shell (it does not
depend on σ at all). The electric field outside a cylinder with constant σ is, via
Gauss’s law: E(s) 2 π s ` = σ 2 π R `/0 giving E(s) = σ0Rs . The total electric
field outside the configuration is, then,
1
λ
E(s) =
σR+
.
(3.7)
0 s
2π
Then σ = − 2 πλ R is the charge density on the cylindrical surface that will ensure
that the electric field outside the entire configuration is zero. Note that it has
the correct dimensions to be a surface charge.
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