Version One – Homework 1 – Juyang Huang – 24018 – Jan 16 ...

Version One – Homework 1 – Juyang Huang – 24018 – Jan 16, 2008 4
The electric field strength E ∝ 1 r 2 , so
tan θ = F 1,3
F 1,2
( )
θ = tan −1 F1,3
F 1,2
( 1.34813 × 10
= tan −1 −5 )
N
2.99585 × 10 −6 N
= 77.4712 ◦
E A
E B
=
keywords:
1
r 2 A
1
r 2 B
= r2 B
r 2 A
=
(2 r)2
r 2 = 4 .
below the negative x-axis. From the positive
x-axis, the angle is
−180 ◦ + 77.4712 ◦ = −102.529 ◦ .
Two Charge Field
23:13, trigonometry, multiple choice, > 1 min,
wording-variable.
008 (part 1 of 3) 10 points
Two point-charges at fixed locations produce
an electric field as shown below.
keywords:
AP B 1993 MC 68
23:07, trigonometry, multiple choice, < 1 min,
fixed.
007 (part 1 of 1) 10 points
The diagram shows an isolated, positive
charge Q, where point B is twice as far away
from Q as point A.
A
X
B
+Q A B
Y
0 10 cm 20 cm
The ratio of the electric field strength at
point A to the electric field strength at point
B is
1. E A
E B
= 8 1 .
2. E A
E B
3. E A
E B
= 2 1 .
4. E A
E B
= 1 1 .
5. E A
E B
= 1 2 .
Explanation:
= 4 1 . correct
Let : r B
= 2 r A
.
A negative charge placed at point X would
move
1. toward charge B. correct
2. toward charge A.
3. along an equipotential plane.
Explanation:
The electric field runs from a positive potential
to a negative potential, so it points
from a positive charge to a negative charge.
Therefore the charge B is positive. A negative
charge will move toward a positive potential,
which creates lower potential energy and a
higher kinetic energy.
009 (part 2 of 3) 10 points
The electric field at point X is