villa (pv5936) – HW02 – keto – (55825)
This print-out should have 9 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 (part 1 of 4) 10.0 points
Consider four charges of equal magnitude q
on the corners of a square with sides of length
a.
A
B
−
+
y
O
a
x
−
+
D
C
What is the magnitude of the electric field
at the center O?
ke q
1. EO = 2
a
2. EO = 0
√ ke q
2 2
a
√ ke q
=4 2 2
a
1 ke q
= √
2 2 a2
1 ke q
=√
2 a2
ke q
=3 2
a
√ ke q
= 2 2
a
1 ke q
= √
4 2 a2
√ ke q
=2 2 2
a
3. EO = 3
4. EO
5. EO
6. EO
7. EO
8. EO
9. EO
10. EO
002 (part 2 of 4) 10.0 points
Find the direction of the electric field.
~ =0
1. Undetermined since E
1
2. √ (~i − ~j)
2
1
3. ~j
4. −~i
1
5. √ (−~i + ~j)
2
6. ~i
1
7. − √ (~i + ~j)
2
8. −~j
1
9. √ (~i + ~j)
2
003 (part 3 of 4) 10.0 points
What is the magnitude of the electric field at
C due to the charges at A, B, and D?
1 ke q
√
3 2 a2
√ ke q
2. EC = 2 2 2
a
ke q
3. EC = 3 2
a
√ ke q
4. EC = 2 2
a
ke q
5. EC = 2
a
9 ke q
6. EC =
4 a2
3 ke q
7. EC =
2 a2
√ ke q
8. EC = 3 2 2
a
√ ke q
9. EC = 4 2 2
a
7 ke q
10. EC = √
4 2 a2
1. EC =
004 (part 4 of 4) 10.0 points
villa (pv5936) – HW02 – keto – (55825)
A
+
−
1. Aligned with the negative x-axis
B
y
D
α
2. Along the 225◦ direction in quadrant I
x
O
−
+
2
3. Along the 45◦ direction in quadrant I
4. Along the 60◦ direction in quadrant I
C
5. Along the 135◦ direction in quadrant I
Find tan α, where α as the angle between
the horizontal and the electric field at C due
to the three charges at A, B, and D.
6. Along the 30◦ direction in quadrant I
7. Aligned with the positive y-axis
1
1. tan α = √
2 2+1
√
2. tan α = 3
√
3. tan α = 2
√
4. tan α = 2 2 + 1
√
2 2−1
5. tan α = √
2 2+1
√
6. tan α = 2 2 − 1
8. Aligned with the positive x-axis
9. Zero with undefined direction
10. Aligned with the negative y-axis
006 (part 1 of 3) 10.0 points
Two point charges at fixed locations produce an electric field as shown.
7. tan α = 1
1
8. tan α = √
2
√
2 2+1
9. tan α = √
2 2−1
1
10. tan α = √
2 2−1
A
X
005 10.0 points
Consider symmetrically placed rectangular
insulators with uniformly charged distributions of equal magnitude as shown.
y
++
++
++
++
++
++
B
Y
How would a negative charge placed at
point X move?
1. Toward charge B
x
−−−−−−
What is the direction of the net field at the
origin?
2. Along an equipotential plane
3. Toward charge A
007 (part 2 of 3) 10.0 points
The electric field at point X is
villa (pv5936) – HW02 – keto – (55825)
1. the same as that the field at point Y .
2. stronger than the field at point Y .
3. weaker than the field at point Y .
008 (part 3 of 3) 10.0 points
Estimate the ratio of the magnitude of
charge A to the magnitude of charge B.
Your answer must be within ± 5.0%
009 10.0 points
Three identical charges (q = +8.7 µC) are
along a circle with a radius of 7.7 m at angles
of 51.0◦ , 171.0◦ , and 291.0◦ , as shown.
q
+
171.0◦
q +
51.0◦
b
291.0◦
+
q
What is the resultant electric field at the
center? The value of the Coulomb constant is
8.99 × 109 N · m2 /C2 .
1. 0.0883705 N/C at 111◦
2. 10.1575 N/C at 291◦
3. 0.0114767 N/C at 111◦
4. None of these
5. 0.0114767 N/C at 291◦
6. 0.0883705 N/C at 291◦
7. 10.1575 N/C at 111◦
8. 0 N/C
3