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STUDENT NO. ___________________
PHY481 Exam 1
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Notations and Formulas
Coordinates:
Unit vectors:
Cartesian
î , ĵ, k̂
Cylindrical
r̂, k̂
Spherical
r̂, θ̂, φ̂
Position vector x =
xî + yĵ + zk̂
rr̂ + zk̂
rr̂
Einstein notation for Cartesian coordinates (repeated indices are summed over)
F ⋅ G = FiGi ; F × G = ε ijk F j Gk ; ε ijk ε klm = δ ilδ jm − δ imδ jl (corrected)
∇⋅F =
∂Fi
∂xi
; ( ∇ × F )i = ε ijk
∂Fk
∂x j
Divergence and curl in curvilinear coordinates
1 ⎛ ∂ ( h2 h3 F1 ) ∂ ( h1h3 F2 ) ∂ ( h1h2 F3 ) ⎞
∇⋅F =
+
+
h1h2 h3 ⎜⎝ ∂u1
∂u2
∂u3 ⎟⎠
∂ hF ⎞
⎛∂ h F
( ∇ × F )i = 1 ⎜ ( k k ) − ( j j ) ⎟
hh
∂u
∂u
j k
⎝
j
k
⎠
Integrals
1
NAME _____________________
STUDENT NO. ___________________
1) [15 pts] Consider a vector function F(x) = î × x , and
let S be the unit square in the y-z plane, as shown in the
figure
a) [5 pts] Calculate the line integrals
∫ F ⋅ d for each
of the four segments of the unit square.
b) [5 pts] Calculate the loop integral 
∫ F ⋅ d
z
F3 = yk̂ − ĵ
d  3 = ĵ dy
1
F2 = k̂ − zĵ
F4 = 0k̂ − zĵ
d 2 = k̂dz
d 4 = k̂ dz
C
c) [5 pts] Verify that this loop integral agrees with
Stokes’s theorem: 
∫ (∇ × F ) ⋅ n̂ dA = ∫ F ⋅ d .
S
C
F1 = yk̂ − 0 ĵ
x
d 1 = ĵdy
ANS.
F(x) = î × (xî + yĵ + zk̂)
= yk̂ − zĵ
1) d = ĵdy, z = 0, 2) d = k̂dz, y = 1
3) d = ĵdy, z = 1,
a)
4) d = k̂dz, y = 0
1
1
0
0
0
0
1
1
∫1F ⋅ d = −0 ∫ dy = 0 ; ∫2 F ⋅ d = 1∫ dz =1;
∫3 F ⋅ d = −1∫ dy = 1; ∫4 F ⋅ d = 0 ∫ dz = 0
b) ∫ F ⋅ d = 2
C
⎛ ∂F ∂Fy ⎞ ⎛ ∂Fx ∂Fz ⎞
⎛ ∂Fy ∂Fx ⎞
+ ĵ ⎜⎝
−
−
c) ∇ × F = î ⎜ z −
⎟⎠ + k̂ ⎜
⎟
⎟ = î(2) + ĵ( 0 ) + k̂(0)
⎝ ∂y
⎝ ∂x
∂z ⎠
∂z
∂x
∂y ⎠
n̂ = î; dA = dxdy ;
∫ (∇ × F ) ⋅ n̂ dA = 2
S
∴
∫ (∇ × F ) ⋅ n̂ dA = ∫ F ⋅ d = 2
S
C
2
1
y
NAME _____________________
STUDENT NO. ___________________
2) [12 pts] Consider electric field and potential determined by a charge density ρ(x) .
Answer the following questions clearly identifying any vector functions or variables.
a) [3 pts] The scalar potential V (x) is determined by what integral over the source
position x ′ .
1
V (x) =
4πε 0
3
ρ( x ′ )d x ′
∫ x − x′
b) [3 pts] A charge density ρ(x) and the resulting electric field E(x) are related by
what partial differential equation?
∇ ⋅ E(x) =
ρ(x)
ε0
c) [3 pts] An electric field E(x) and the scalar potential V (x) are related by what
differential equation?
E(x) = −∇V (x)
d) [3 pts] The electric field E(x) is determined by what integral over the source
position x ′
E(x) =
1
4πε 0
x − x′
∫ x − x′ 3 ρ(x′)d
3
3
x′
NAME _____________________
STUDENT NO. ___________________
3) [13 pts] The integral form of Gauss’s law relates the electric flux through a closed
surface S to the charge enclosed by that surface.
a) [2 pts] What is the integral form of Gauss’s law?
b) [3 pts] Use Gauss’s law and symmetry to determine the electric field E(x) above
and below an infinite sheet of charge with a constant charge density σ .
c) [8 pts] Using a spherical Gaussian surface, again determine the electric field E(x) of
an infinite sheet of charge with constant charge density σ .
a)
∫ E ⋅ dA =
qencl
ε0
S
θ
R
E = E0k̂
n̂ = r̂
n̂ = k̂
σ
σ
n̂ = −k̂
E = −E0k̂
b)
ET (x) = E0k̂; E B (x) = − E0k̂ and n̂T = k̂; n̂ B = −k̂
∫ E ⋅ dA = ∫ E0k̂ ⋅k̂ dA + ∫ ( − E0k̂ ) ⋅( −k̂ dA) = qencl
S
T
ε0 = σ A ε0
B
2E0 A = σ A ε 0
E0 = σ 2ε 0
c)
2
n̂ = r̂; dA = R sin θ dθ dφ ; qencl = σπ R
2
∫ E ⋅ dA = ∫ E0k̂ ⋅r̂ dA + ∫ ( − E0k̂ ) ⋅(k̂ ⋅r̂ dA)
S
T
B
⎡
⎤
π
2
= 2π E0 R ⎢ ∫ dθ cosθ sin θ − ∫ dθ cosθ sin θ ⎥
⎢0
⎥
π
⎣
⎦
2
1
0
⎤
2⎡
= 2π E0 R ⎢ ∫ xdx − ∫ xdx ⎥
⎣0
⎦
1
1
⎡ 2 ⎤
x ⎥
2
2
= 2π E0 R ⎢ 2
= 2π E0 R
⎢⎣ 2 0 ⎥⎦
σ
2
2
2π E0 R = qencl ε 0 = σπ R / ε 0 ∴ E0 =
2ε 0
π
2
4
NAME _____________________
STUDENT NO. ___________________
4) [20 pts] Consider a semicircle of charge with a linear charge density λ .
a) [4 pts] Determine the total charge on the semicircle.
b) [6 pts] Display the integral that will determine all components of the electric field
E(x) on the z axis, AND draw on the figure the corresponding vector quantities.
c) [6 pts] Determine the E(x) (vector) on the z axis.
d) [4 pts] For z >> R , show that the z-component of field approaches the field of a
point charge with a value equal to the total charge on the semicircle.
a)
3
q = ∫ ρ( x ′ )d x ′ = ∫ λ d =
1
4πε 0
x − x′
π 2
∫
Rdφ = π Rλ
−π 2
∫ x − x′ 3 ρ(x′)d
3
b)
E(x) =
c)
x − x ′ = zk̂ − R ( î cos φ + ĵsin φ ) = ( z + R
3
3
2
)
2 32
zk̂ − R ( î cos φ + ĵsin φ )
λ Rdφ
∫
( z 2 + R2 )3 2
2
π 2
⎞
1 ⎛
λR
λR
(
)
=
π
z
k̂
+
î
cos
φ
+
ĵsin
φ
d
φ
⎜
⎟
∫
2 32
4πε 0 ⎝ ( 2
⎠
( z 2 + R2 )3 2 −π 2
z +R )
1
λR
(π zk̂ − 2Rî )
=
2 32
4πε 0 ( 2
z +R )
1
πλ Rz
q
z
Ez (x) =
=
3
2
2
2 32
4πε 0 ( 2
4πε 0 3
z +R )
z 1+ ( R z)
q
Ez (x) ≈
R z → 0 (i.e., z >> R)
2
4πε 0 z
This is field is the same as that of the point charge q .
E(x) =
d)
x′
1
4πε 0
(
5
)
NAME _____________________
STUDENT NO. ___________________
5) [10 pts] Use the Levi-Chivita tensor to determine an identity for ∇ ⋅ ( A × B ) involving
∇ × A and ∇ × B .
Let C = A × B where Ci = ε ijk Aj Bk
∇ ⋅ (C) =
∂Ci
∂xi
=
∂Aj
∂B
∂
ε ijk Aj Bk = Bk ε ijk
+ Aj ε ijk k
∂xi
∂xi
∂xi
(
)
Make the replacements ε kij = ε ijk
∂Aj
( ) ∂x
∇ ⋅ ( C ) = Bk ε kij
i
(
+ Aj −ε jik
and − ε jik = ε ijk
∂B
) ∂xk
i
= Bk ( ∇ × A )k − Aj ( ∇ × B ) j = B ⋅ ( ∇ × A ) − A ⋅ ( ∇ × B )
6
NAME _____________________
STUDENT NO. ___________________
6) [15 pts] Consider a spherical shell with a
surface charge density σ (x) = σ 0 cosθ , where θ
is the standard polar angle.
a) [8pts] Determine the dipole moment p of this
charge distribution
b) [2pts] State why the answer to a) makes sense.
Hint: the dipole moment is a vector.
c) [5 pts] Display an expression that represents
the dominate term in the potential V (r,θ ) of
this charge distribution in terms of the dipole
moment p at a large distance from the shell.
a)
r̂ = k̂ cosθ + sin θ ( î cos φ + jsin φ )
3
p = ∫ rρ(x)d x ;
p= R
3
∫ r̂ (σ 0 cosθ ) sinθ dθdφ
π
2π
⎡π
⎤
2
3
2
= 2π R σ 0 ⎢ ∫ k̂ cos θ sin θ dθ ⎥ + R σ 0 ∫ cosθ sin θ dθ ∫ dφ ( î cos φ + jsin φ )
⎣0
⎦
0
0
3
−1
3
⎡ x3 ⎤
4π R σ 0
= 2π R σ 0k̂ ∫ −x dx = 2π R σ 0k̂ ⎢ − ⎥ =
k̂
⎣ 3 ⎦1
3
1
3
b)
−1
2
3
Positive charge is displaced in the +z direction and negative charge in the –z
direction, yielding a dipole moment in the +z direction.
2
Extra: The total charge in the upper half sphere, qT = π R σ 0 . Writing p = qT ”d”,
the effective charge spacing ”d” = 4R 3 .
Instead, with a constant charge density σ 0 (upper) and −σ 0 (lower) the total
2
charge in the upper half sphere is qT = 2π R σ 0 and the dipole moment is
3
p = π R σ 0 . The effective charge spacing ”d” = R / 2 , less than half the previous
case.
c)
From far away this will look like a dipole along the positive z axis. Therefore, the
dominate term will be the dipole term:
p ⋅ r̂
V (r,θ ) =
2
4πε 0 r
7
NAME _____________________
STUDENT NO. ___________________
7. [15 pts]
∞
∫ δ (ax + b) f (x)dx =
a) [10 pts] Prove
−∞
f (− b a)
3
(d x should be dx, answer OK)
a
nd
x ′ = ax + b; dx = dx ′ a ; 2 integral is f at x ′ = 0
∞
∫ δ (ax + b) f (x)dx =
−∞
∞
b) [5 pts] Prove
∫δ
3
∞
1
f (− b a)
δ ( x ′ ) f ( x ′ a − b a)dx ′ =
∫
a −∞
a
−1
3
(Ax + b) f (x)d x = f (−A b) / det A ,
−∞
where det A is the Jacobian of the transformation x ′ = Ax + b .
3
3
d x ′ = Det A d x
∞
3
∫ δ (Ax + b) f (x)d x =
−∞
∞
∫ δ (x′) f (A
−1
−1
3
x ′ − A b) d x ′ Det A
−∞
−1
= f (−A b) Det A
8