Download Mid-Term Exam Solutions, Electromagnetic Theory I

Mid-Term Exam Solutions, Electromagnetic Theory I
Dr. Christopher S. Baird, Fall 2014
University of Massachusetts Lowell
PART I: Multiple Choice (30 points).
Circle the one best answer to each question.
1. A non-uniform line charge λ(z) extends infinitely in the z direction. The line charge is located in
cylindrical coordinates at ρ=a , ϕ=0 . What is the expression for the volume charge density ρch (x) of
this distribution in terms of cylindrical coordinates?
(a) ρch (x)=δ( z− z 0)δ(ρ−a) δ(ϕ−ϕ0 )
(b) ρch (x)=λ( z )δ(ρ−a )δ (ϕ)
(c) ρch (x)=λ(z ) δ(ρ−a )
(d) ρch (x)=q δ(ρ−a )
δ( ϕ)
ρ
δ( ϕ)
ρ sin ϕ
2. A large chunk of conducting material has a right-angle wedge cut into it
as shown in the diagram and carries a net positive charge. How does the
surface charge density at the two points compare?
(a) σ1 > σ2
(b) σ1 < σ2
(c) σ1 = σ2
(d) none of the above
σ1
σ2
3. A hollow circular cylinder of height h is aligned with the z axis. Its bottom and top faces are held at
zero potential, and its curved side is held at the potential V(z). The solution for the potential inside the
cylinder will contain what types of functions?
(a) ln
(b) Nν where ν ≠ 0
(c) Jν where ν ≠ 0
(d) J0
4. How do we find the Green's function for a particular problem (Dirichlet boundary conditions)?
(a) Place a Gaussian surface around half the charges and solve for the potential on the surface.
(b) Solve for the potential of the simpler problem of a unit point charge near a grounded conducting
surface.
(c) Place image charges that mirror the real charges of the original problem and then maximize the
energy density everywhere. The final energy density becomes the Green's function.
(d) Place image charges at points of symmetry and then maximize the energy density everywhere. The
final energy density becomes the Green's function.
5. A hollow cube has one corner at the origin and the far corner at (a, a, a). All sides of the cube are
held at zero potential except the z = 0 face which is held at +V and the z = a face which is held
at -V. What is the potential inside the cube?
∞
(a) Φ=−
∑
n , m=odd
(b) Φ=−
∞
∑
n , m=odd
(c) Φ=
∞
∑
n ,m=odd
(d) Φ=−
)]
(
( ) ( )
[
]
[
)]
(
2z
sinh π √ n2 +m2
−1
2
a
16 V
nπ x
mπ y
cos
cos
a
a
n m π2
sinh π √ n 2+m2
2
( ) ( )
[
[
]
)]
(
2z
cosh π √ n2 +m 2
−1
2
a
16 V
cos ( n x ) cos ( mπ )
n m π2
cosh π √ n 2+m2
2
∞
∑
[
2z
sinh π √ n 2+m 2
−1
2
a
16 V
nπ x
mπ y
sin
sin
a
a
n m π2
sinh π √ n 2+m 2
2
n , m=odd
[
]
[
(
)]
2z
J 0 π √ n2+m2
−1
2
a
16 V
nπ x
mπ y
sin
sin
a
a
n m π2
J 0 π √ n2+m2
2
( ) ( )
[
]
6. What best describes the distribution of the surface charge density on the surface of a perfect
conductor?
(a) it collects in holes and away from sharp points.
(b) it is always uniform and constant.
(c) it has the same overall distribution pattern as the electric field strength on the surface.
(d) it has an arbitrary distribution independent of the electric field.
PART II: Diagram Problem (20 points)
As shown in the diagram, a grounded conducting sphere has two points charges placed near it (ignore
the red dot until part 5).
1. Draw and label the appropriate image charges.
2. Draw the actual electric field lines everywhere they exist.
3. What is the total charge on the sphere?: 0
4. Consider that the origin is placed at the sphere's center and the z axis is lined up with the charges. If
we were to solve for the potential outside the sphere using a series solution in spherical coordinates,
what types of functions would the solution contain (circle one)?
(a) spherical harmonics
(b) Legendre polynomials
(c) Bessel functions
(d) real-valued exponentials
5. A small negatively-charged test charge is placed at rest at the location of the red dot and then let go.
Draw a thick arrow on the red dot to show the direction in which it begins moving.
+q
-q'
+q'
-q
PART III: Diagram Problem (20 points)
A hollow box is uniform and is so tall in the dimension going into the page that this dimension can be
ignored. Two opposing sides of the box are held at an electric potential of +V and the other two sides
are held at zero potential, as shown in the diagram.
1. Draw the electric field lines in the box using solid arrows.
2. Draw the equipotential lines that pass through the exact center of the box (if any) using dashed lines.
3. Mark with a large dot on the diagram the location inside the box where the field energy density is
zero.
4. Could we use the Relaxation Method to solve this problem numerically? YES
Φ = +V
Φ=0
Φ=0
Φ = +V
PART IV: Work Problem (30 Points)
A hollow sphere of radius a is centered at the origin and is held at the potential V sin(2 θ) sin(ϕ) . Find
the potential everywhere inside the sphere.
8π
(Y +Y 2,−1) .
Note that sin(2 θ)sin (ϕ)=i
15 2,1
√
SOLUTION:
∞
Φ=∑
l
∑ ( Alm r l+ Blm r−l −1 )Y lm (θ , ϕ)
l=0 m=−l
We need a finite potential at the origin, so Blm = 0, leaving:
∞
Φ=∑
l
∑
l=0 m=−l
Alm r l Y lm( θ , ϕ)
Apply the last boundary condition:
Vi
√
l
∞
8π
(Y 2,1+Y 2,−1)=∑ ∑ Alm al Y lm(θ , ϕ)
15
l =0 m=−l
Because the spherical harmonics form an orthogonal set of functions and each term on the left is one of
these functions, all Alm are zero except A2,1 and A2,-1.
Vi
√
8π
(Y 2,1+Y 2,−1)= A2,1 a 2 Y 2,1 (θ , ϕ)+A2,−1 a 2 Y 2,−1 (θ , ϕ)
15
A2,1 =A2,−1=a−2 V i
√
8π
15
The final solution is:
2
2
Φ=A2,1 r Y 2,1 (θ , ϕ)+A2,−1 r Y 2,−1 (θ , ϕ)
2
√ ( )(
8π r
Φ=V i
15 a
2
()
Y 2,1 (θ , ϕ)+Y 2,−1 (θ , ϕ) )
r
Φ=V
sin( 2θ)sin( ϕ)
a