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Gradient – Curl – Faraday’s Law
Problem 1:
Use Stokes’ Theorem to evaluate
S
dS
V n
i)
2 2 2
yz xz xy
V x y z x e x y e y z e z , and S is the hemisphere
2 2 2
x y z 4, z 0,
oriented upward.
ii)
2
V x y z , , xyz ˆ x xy y ˆ x yz z ˆ
, and S consists of the top and the four sides (but not
the bottom) of the cube with vertices
, oriented outward.
Solution:
i)
S
dS
V n
C
d
V r
=
2 2
2 2 2 2
0 0
x y 2 d 2 4cos sin 4sin cos d
x y
2
3 3
0
8 cos sin 0
ii)
S
dS
V n
C
d
V r
1 1 1 1
1 1
1 1
1 1 1 1 1 1
y y
z z
x x
xyz dx xy dy xyz dx xy dy
Gradient – Curl – Faraday’s Law
Problem 2:
Use Stokes’ Theorem to evaluate
C
d
V r
. In each case, C is oriented counterclockwise as
viewed from above.
i)
2 2 2
V x y z , , x y x ˆ y z y ˆ z x z ˆ , and C is the triangle with vertices
1, 0, 0 , 0,1, 0 , and 0, 0, 0 .
ii)
xy
V x y z yz x xz y e z , and C is the circle
2 2
x y 16, z 5.
Solution:
i)
C
d
V r
S
dS
V n
1 1
0 0
x
z
S S
dS dS ydydx
V z V
ii)
C
d
V r
S
dS
V n
2 4
5
0 0
z z
S S
dS dS z d d
V z V
Gradient – Curl – Faraday’s Law
Problem 4:
A possible electric field is given as
E 2 y x 2 x sin y y .
a) Determine
E
and use it to decide whether integrals of the form
C
d
E
are
independent of path.
b) If
C
d
E
is path independent, find a potential such that E .
c) Evaluate the line integral
C
d
E
where C is the path from (0, 0, 0) to (2, / 2,0) both in
cylindrical and rectangular coordinate.
Solution:
2 2 sin 0
x y z
y x y
x y z
E z ,
C
d
E
is path independent
y x z
x y z
E x y z
x
x
y y x C
x
2 sin cos 2
y
y
x y y y x C
y
z
z
C
z
cos y 2 x y C
If points are interpreted in cylindrical coordinates:
2 2
0 0 0
2 2 sin sin 1 cos 2 1.
x
y x y dy ydy
x y y
If points are interpreted in rectangular coordinates:
2 / 2 / 2
0 2
0 0 0
2 2 sin 0 4 sin 2 1 7.
y x
y dx x y dy y dy
Gradient – Curl – Faraday’s Law
Problem 5:
Two students propose the following two different flux densities as solutions to an electrostatics
problem:
2
1 0
D ρ ,
2 2 2
2 0 0
D sin ρ 3 z cos z ˆ .
a) Show that the volume charge densities that correspond to these two fields are identical.
b) Choose one of the following equivalent statements and, showing all work, use it to
determine which of the flux densities in part a is a valid electric field:
i)
C
d
E
ii)
E 0
iii)
E
c) Assume that the potential associated with the charge distribution in part a depends on
alone, that is, we may write the potential as
. Show that setting E results
in a simple differential equation for the potential. Solve this equation for the unknown
potential by integrating the equation, using the valid field from parts a and b. You may
assume any convenient reference point.
v
D
,
2 3
1 0 0
3 C/m
D
2 2 2 2 2 3
2 0 0 0 0 0
sin 3 z cos 3 sin 3 cos 3 C/m
z
D
1
2
z
ρ φz z
E
2 2
2
2 2 2
/ / / 6 cos sin 3 cos 2 sin cos ˆ 0
sin 0 3 cos
z z z
z
ρ φz z
E ρ φz z
2
1 0
D ρ is the valid field.
3
2
C
z
E ρ φz z
Gradient – Curl – Faraday’s Law
Problem 7:
A slab of charge has a uniform charge density 0
v
At
1 1 1
, and.
x
x x E E There is no
variation of E in the
y and z directions. Find the potential
x
within the slab.
x
y
z
1
x
2
x
v 0
1 1
, E
Solution:
Since
1 1
0 0
0 0
or
x
E x
x v v
v x
E x
dE
= dE dx
dx
D
0
1 1 1 2
0
for.
v
x
E x x E x x x
Now, E
1 1
0 0
1 1 1 1
0 0
or
x
v v
x
x
d
E x x E dV x x E dx
dx
Hence,
2
0
1 1 1 1 1 2
0
for.
v
x x x E x x x x x
