ID: B
1
UCSD Physics 2B
Unit Exam 1B
Charges, Force & Fields
Answer Section
MULTIPLE CHOICE
1. ANS: B
τ
=
p
×
E
=
pE
sin
θ
=
35
×
10
−
9
C
⋅
m
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
2.2
×
−
3
N
C
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
sin165
°=+
2.0
×
−
11
N
⋅
m
TOP:
ELECTRIC DIPOLE TORQUE
2. ANS: A
To apply the line/cylinder formula, we need the linear charge density
λ
. Pick some length
L
for a sample
section of cylinder. Then the lateral surface area of that section will be
A
=
2
π
RL
and the total charge is the area times the surface charge density
Q
=
2
RL
σ
.
Now
=
Q
L
=
2
RL
L
=
2
R
so that
E
=
2
k
r
=
2
k
2
R
r
=
4
k
R
r
=
49
.0
0
×
9
Nm
2
/
C
2
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
3.14
()
19.00
×
−
2
m
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
27.0
×
−
9
C
/
m
2
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
31.0
×
−
2
m
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
=
1.87
×
3
N
=
kN
C
The
much
easier way is to apply Gauss's Law from scratch:
Charged Cylinder
Q
=
2
RL
Gaussian Cylinder Area
A
=
2
rL
E
=
Q
ε
0
A
=
2
RL
2
rL
0
=
R
r
0
=
19
31
27.0
×
−
9
8.85
×
−
12
Ê
Ë
Á
Á
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
˜
˜
Note the extra advantage of not having to convert centimeters to meters since the units cancel
TOP:
ELECTRIC FIELD CYLINDER
3. ANS: A
This is one of the basic facts about charges  from grade school, I think.
TOP: CHARGE CONCEPT
4. ANS: A
The electric field inside any ideal conductor, even if it’s hollow & empty inside, is always zero.
TOP:
CONDUCTOR FIELD CONCEPT
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentID: B
2
5. ANS: D
The electric field from a point charge is
E
=
kQ
r
2
. Outside a charged spherical shell, the field looks the
same as if the whole charge were located in a point at the center. Halfway between the shells, the
distance from their center is
3
2
R
. Hence the field is
E
=
r
2
=
3
2
R
Ê
Ë
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
This is the end of the preview.
Sign up
to
access the rest of the document.
 Summer '09
 Lever
 Charge, Force, Electric charge

Click to edit the document details