homework 06-solutions - marout(ssm979 homework 06...

marout (ssm979) – homework 06 – Turner – (56725) 1
This print-out should have 9 questions.
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001 10.0 points
A cubic box of side a , oriented as shown, con-
tains an unknown charge. The vertically di-
rected electric field has a uniform magnitude
E at the top surface and 2 E at the bottom
surface.
a
E
2 E
How much charge Q is inside the box?
1. Q
encl
= 0
2. Q
encl
= 2 ǫ
0
E a
2
3. Q
encl
= 3 ǫ
0
E a
2
4. Q
encl
= ǫ
0
E a
2
correct
5. Q
encl
=
1
2
ǫ
0
E a
2
6. insufficient information
Explanation:
Electric flux through a surface S is, by con-
vention, positive for electric field lines going
out of the surface S and negative for lines
going in.
Here the surface is a cube and no flux goes
through the vertical sides.
The top receives
Φ
top
= - E a
2
(inward is negative) and the bottom
Φ
bottom
= 2 E a
2
.
The total electric flux is
Φ
E
= - E a
2
+ 2 E a
2
= E a
2
.
Using Gauss’s Law, the charge inside the
box is
Q
encl
= ǫ
0
Φ
E
= ǫ
0
E a
2
.
002 10.0 points
A closed surface with dimensions a = b =
0 . 385 m and c = 0 . 693 m is located as in
the figure. The electric field throughout the
region is nonuniform and given by
vector
E = ( α +
β x
2
)ˆ ı where x is in meters, α = 3 N / C, and
β = 2 N / (C m
2
).
E
y
x
z
a
c
b
a
What is the magnitude of the net charge
enclosed by the surface?
Correct answer: 2 . 6612 × 10
- 12
C.
Explanation:
Let : a = b = 0 . 385 m ,
c = 0 . 693 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
vector
A is perpendicular to its surface. Therefore,
vector
E