Ash (bka258) – HW2 – Tsoi – (60250)
1
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001
(part 1 of 2) 10.0 points
Consider a square with side
a
. Four charges
+
q
, +
q
,
−
q
, and +
q
are placed at the corners
A
,
B
,
C
, and
D
, respectively.
+
+
−
+
D
C
A
B
a
The magnitude of the electric field at
D
due
to the charges at
A
,
B
, and
C
is given by
1.
bardbl
vector
E
bardbl
= 3
k q
a
2
2.
bardbl
vector
E
bardbl
=
√
2
k q
a
2
3.
bardbl
vector
E
bardbl
=
5
4
k q
a
2
4.
bardbl
vector
E
bardbl
=
9
4
k q
a
2
5.
bardbl
vector
E
bardbl
=
5
2
k q
a
2
6.
bardbl
vector
E
bardbl
=
3
4
k q
a
2
7.
bardbl
vector
E
bardbl
= 2
k q
a
2
8.
bardbl
vector
E
bardbl
=
7
2
k q
a
2
9.
bardbl
vector
E
bardbl
=
k q
a
2
10.
bardbl
vector
E
bardbl
=
3
2
k q
a
2
correct
Explanation:
The magnitudes of the electric fields at
D
due to
A
and
C
are
E
A
=
E
C
=
k q
a
2
since they are at a distance
a
from
d
, whereas
E
B
=
k q
(
a
√
2)
2
=
k q
2
a
2
since
B
is at a distance
√
2
a
from
d
.
292
.
53
◦
E
C
E
A
E
B
E
A
+
E
C
E
A
+
E
B
+
E
C
+
D
As for directions,
vector
E
A
(downwards since C
is positive) and
vector
E
C
(right since
A
is negative)
are at right angles and can be added with
Pythagoras theorem:
bardbl
vector
E
A
+
vector
E
C
bardbl
=
k q
a
2
√
1 + 1 =
√
2
k q
a
2
This resultant vector points 45
◦
down to the
right. The vector
vector
E
B
points at 135
◦
down to
the left.
The angle between these is 90
◦
, so
we apply Pythagoras’ theorem again, now to
all three:
bardbl
(
vector
E
A
+
vector
E
C
) +
vector
E
B
bardbl
=
radicalBigg
parenleftBig
√
2
parenrightBig
2
+
parenleftbigg
1
2
parenrightbigg
2
k q
a
2
,
which means
bardbl
vector
E
bardbl
=
radicalbigg
9
4
k q
a
2
=
3
2
k q
a
2
.
002
(part 2 of 2) 10.0 points
The polar angle of the corresponding electric
field vector at
D
is within the range
1.
315
◦
≤
θ <
360
◦
2.
180
◦
≤
θ <
225
◦
3.
0
◦
≤
θ <
45
◦
4.
90
◦
≤
θ <
135
◦
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Ash (bka258) – HW2 – Tsoi – (60250)
2
5.
45
◦
≤
θ <
90
◦
6.
270
◦
≤
θ <
315
◦
correct
7.
225
◦
≤
θ <
270
◦
8.
135
◦
≤
θ <
180
◦
Explanation:
The
x
components of the vectors are
E
Cx
= 0
E
Bx
=
−
1
√
2
k q
2
a
2
E
Ax
=
E
A
=
k q
a
2
,
and the
y
components are
E
Cy
=
−
E
C
=
−
k q
a
2
E
By
=
−
1
√
2
k q
2
a
2
E
Ay
= 0
.
Thus the components of the total vector are
E
x
=
−
1
√
2
k q
2
a
2
+
k q
a
2
= +
parenleftBigg
1
−
√
2
4
parenrightBigg
k q
a
2
E
y
=
−
k q
a
2
−
1
√
2
k q
2
a
2
=
−
parenleftBigg
1 +
√
2
4
parenrightBigg
k q
a
2
Which means the tangent of the angle
α
be
tween the resultant vector and the horizontal
is
tan
α
=
−
parenleftBigg
1 +
√
2
4
parenrightBigg
k
q
a
2
+
parenleftBigg
1
−
√
2
4
parenrightBigg
k
q
a
2
=
−
√
2 +
1
2
√
2
−
1
2
.
Therefore the angle is
α
=
−
64
.
4712
◦
+
n
(180
◦
)
We see from the figure that the resultant must
point down, so we must have
n
= 2 to obtain
the right range
α
=
295
.
529
◦
.
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 Spring '08
 ERSKINE/TSOI
 Charge, Electrostatics, Work, Electric charge

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