Version 027/AABCD – midterm 01a – turner – (56705)
1
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001
10.0 points
P
.
x
y
++++
−−−−
L
.
x
y
++++++
−−−−−−
M
.
x
y
+++++
+++++
+
+
+
+
+
−
−
−
−
−
G
.
x
y
−−−−
++++
S
.
x
y
++++
++++
±or which confguration(s) does the total
electric feld vector at the origin have non
zero components in the
x
direction as well as
the
y
direction (
i.e.
,
both
x
and
y
components
are nonzero)?
1.
Confgurations
P
,
G
and
L
only
2.
Confgurations
P
and
S
only
3.
Confgurations
S
and
L
only
4.
Confguration
P
only
5.
Confgurations
S
,
M
and
L
only
6.
Confguration
S
only
7.
Confguration
M
only
8.
Confguration
G
only
correct
9.
Confgurations
P
,
S
and
L
only
10.
Confgurations
P
and
L
only
Explanation:
Basic Concepts:
Δ
E
=
k
Δ
q
r
2
ˆ
r
and
E
=
s
Δ
E
.
Symmetry oF the confguration will
cause some component oF the electric feld to
be zero.
Solution:
Confguration
P
: It is anti
symmetric about the
y
axis (opposite sign
oF charges), so the electric feld has no
y

component.
x
y
P
Confguration
S
: It is symmetric by a ro
tation oF 180
◦
, so the electric felds generated
by these two pieces have opposite directions;
thereFore the total feld is zero.
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2
x
y
++++
++++
S
Confguration
G
: It is antisymmetric by
rotation oF a 180
◦
, so the total feld has non
zero components in both
x
and
y
directions,
just like the feld generated by just one piece
oF charge.
x
y
−−−−
++++
G
Confguration
M
: It is symmetric about
the
x
axis, so the
y
component oF the total
feld must vanish.
x
y
+++++
+++++
+
+
+
+
+
−
−
−
−
−
M
Confguration
L
: It is symmetric about the
y
axis, so the
x
component oF the total feld
must vanish.
x
y
++++++
−−−−−−
L
002
10.0 points
Three identical point charges, each oF mass
150 g and charge +
q
, hang From three strings,
as in the fgure.
9
.
8 m
/
s
2
10
.
7 cm
150 g
+
q
150 g
+
q
38
◦
150 g
+
q
IF the lengths oF the leFt and right strings
are each 10
.
7 cm, and each Forms an an
gle oF 38
◦
with the vertical, determine the
value oF
q
.
The acceleration oF gravity
is 9
.
8 m
/
s
2
, and the Coulomb constant is
8
.
98755
×
10
9
N
·
m
2
/
C
2
.
1. 0.415744
2. 0.66606
3. 0.517167
4. 1.1998
5. 1.07128
6. 0.750419
7. 1.14671
8. 0.825952
9. 1.04714
10. 0.479555
Correct answer: 0
.
66606
μ
C.
Explanation:
Let :
θ
= 38
◦
,
m
= 150 g = 0
.
15 kg
,
L
= 10
.
7 cm = 0
.
107 m
,
g
= 9
.
8 m
/
s
2
,
and
k
e
= 8
.
98755
×
10
9
N
·
m
2
/
C
2
.
Consider the Forces acting on the charge on
the right. There must be an electrostatic Force
F
acting on this charge, keeping it balanced
against the Force oF gravity
mg
. The electro
static Force is due to the other two charges
and is thereFore horizontal.
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 Spring '10
 Lam
 Electric charge, midterm 01a, Version 027/AABCD

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