HW 2 solution - song(shs546 Homework 2 weathers(22202 This...

song (shs546) – Homework 2 – weathers – (22202) 1
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before answering.
001 10.0points
Four point charges are placed at the four cor-
ners of a square, where each side has a length
a . The upper two charges have identical pos-
itive charge, and the two lower charges have
charges of the same magnitude as the first two
but opposite sign. That is, q
1
= q
2
= q and
q
3
= q
4
= − q where q > 0.
q
4
q
2
q
3
q
1
P
j
i
Determine the direction of the electric field
at the center, point P.
1.i
2. − i
3.j
4. 0
5. −
1
√
2
( i − j )
6.
1
√
2
( i − j )
7.
1
√
2
( i + j )
8. − jcorrect
9. −
1
√
2
( i + j )
Explanation:
The direction is already clear: all the x -
components cancel, and the lower charges at-
tract and the top ones repel, so the answer is
− j .
002(part1of2)10.0points
Three positive charges are arranged as shown
at the corners of a rectangle.
+
+
+
0 . 321 m
0 . 963 m
4 . 4 nC
2 . 2 nC
2 . 35 nC
Find the magnitude of the electric field
at the fourth corner of the rectangle. The
Coulomb constant is 8 . 98755 × 10
9
N m
2
/ C
2
.
Correct answer: 207 . 862 N / C.
Explanation:
r
1
r
2
θ
ϕ
Q 2
Q 1 Q 3
E
1
E
2
E
3
Let : q
1
= 2 . 2 nC = 2 . 2 × 10
− 9
C ,
q
2
= 4 . 4 nC = 4 . 4 × 10
− 9
C ,
q
3
= 2 . 35 nC = 2 . 35 × 10
− 9
C ,
r
2
= 0 . 963 m ,
r
1
= 0 . 321 m , and
k
e
= 8 . 98755 × 10
9
N m
2
/ C
2
.
The length of the diagonal is
r
3
=
radicalBig
r
2
1
+ r
2
2
=
radicalBig
(0 . 963 m) 2 + (0 . 321 m) 2
= 1 . 01509 m .
The electric field produced by the charge q
1
is along the positive y -axis, so
E
1 ,x
= 0
E
1 ,y
= E
1
= k
c
q
1
r
2
1
.
The electric field produced by the charge q
2
is along the negative x -axis, so
E
2 ,x
= − E
2
= − k
c
q
2
r
2
2
E
2 ,y
= 0 .

song (shs546) – Homework 2 – weathers – (22202) 2
The electric field produced by the charge q
3
is upward and to the left along the diagonal of
the rectangle, and is directed away from the
charge since q
3
is positive, so
E
3
= k
e
q
3
r
2
3
.
Consider the direction of E
3
cos ϕ =
r
2
r
3
and sin ϕ =
r
1
r
3
Thus
E
3 ,x
= − k
e
q
3
r
2
3
sin ϕ = − k
e
q
3
r
2
r
3
3
E
3 ,y
= k
C
q
3
r
2
3
cos ϕ = k
e
q
3
r
1
r
3
3
, so that
E
net,x
= E
2 ,x
+ E
3 ,x
= − k
e
q
2
r
2
2
− k
e
q
3
r
2
r
3
3
= −
(
8 . 98755 × 10
9
N m
2
/ C
2
)
×
4 . 4 × 10
− 9
C
(0 . 963 m)
2
−
(
8 . 98755 × 10
9
N m
2
/ C
2
)
×
(2 . 35 × 10
− 9
C) (0 . 963 m)
(1 . 01509 m)
3
= − 62 . 0879 N / C , and
E
net,y
= E
1 ,y
+ E
3 ,y
= k
e
q
1
r
2
1
+ k
e
q
3
r
1
r
3
3
=
(
8 . 98755 × 10
9
N m
2
/ C
2
)
×
2 . 2 × 10
− 9
C
(0 . 321 m) 2
+
(
8 . 98755 × 10
9
N m
2
/ C
2
)
×
(2 . 35 × 10
− 9
C) (0 . 321 m)
(1 . 01509 m)
3
= 198 . 373 N / C , so that
E
net
=
radicalBig
E
net,x
2
+ E
net, 4
2
E
net
=
bracketleftbig
( − 62 . 0879 N / C)
2
+(198 . 373 N / C)
2
bracketrightbig
1 / 2
= 207 . 862 N / C .
003(part2of2)10.0points
What is the direction of this electric field (as
an angle measured from the positive x -axis,
with counterclockwise positive)?
Correct answer: 107 . 379
◦
.
Explanation:
tan θ =
E
net,y
E
net,x
θ = tan
− 1
parenleftbigg
E
net,y
E
net,x
parenrightbigg
= tan
− 1
parenleftbigg
198 . 373 N / C
− 62 . 0879 N / C
parenrightbigg
= 107 . 379
◦
.