Version One – Homework 1 – Juyang Huang – 24018 – Jan 16 ...

Version One – Homework 1 – Juyang Huang – 24018 – Jan 16, 2008 14
y
a
E
× (0.133632 N m 2 /C)
= 1.1832 × 10 −12 C .
z
c
What is the magnitude of the net charge
enclosed by the surface
Correct answer: 1.1832 × 10 −12 C.
Explanation:
Let : a = b = 0.4 m ,
c = 0.36 m ,
α = 3 N/C , and
β = 2 N/(C m 2 ) .
The electric field throughout the region is
directed along the x-axis and the direction of
d A ⃗ is perpendicular to its surface. Therefore,
⃗E is parallel to d A ⃗ over the four faces of
the surface which are perpendicular to the
yz plane, and E ⃗ is perpendicular to d A ⃗ over
the two faces which are parallel to the yz
plane. That is, only the left and right sides
of the right rectangular parallel piped which
encloses the charge will contribute to the flux.
The net electric flux through the cube is
∫
∫
∆Φ = E x d A ⊥ − E x d A ⊥
right side
b
a
left side
= a b [ α + β(a + c) 2 − α − β a 2]
= a b β (2 a c + c 2 )
= a b c β (2 a + c)
= (0.4 m) (0.4 m) (0.36 m)
× [2 N/(C m 2 )] [2 (0.4 m) + 0.36 m]
= 0.133632 N m 2 /C ,
so the enclosed charge is
q = ɛ 0 ∆Φ
= [8.85419 × 10 −12 C 2 /(N m 2 )]
x
030 (part 2 of 2) 10 points
What is the sign of the charge enclosed in the
surface
1. positive correct
2. negative
3. Cannot be determined
Explanation:
Since there is more flux coming out of the
surface than going into the surface, the sign
of the enclosed charge must be positive.
Flux Through a Loop 01
24:01, calculus, numeric, > 1 min, normal.
031 (part 1 of 1) 10 points
A 40 cm diameter loop is rotated in a uniform
electric field until the position of maximum
electric flux is found. The flux in this position
is measured to be 520000 N · m 2 /C.
What is the electric field strength
Correct answer: 4.13803 × 10 6 N/C.
Explanation:
Let : r = 20 cm = 0.2 m and
By Gauss’ law,
∮
Φ =
Φ = 520000 N · m 2 /C .
⃗E · d ⃗ A
The position of maximum electric flux will be
that position in which the plane of the loop is
perpendicular to the electric field; i.e., when
⃗E · d ⃗ A = E dA. Since the field is constant,
E =
Φ = E A = Eπ r 2
Φ
π r 2
= 520000 N · m2 /C
π (0.2 m) 2
= 4.13803 × 10 6 N/C .