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

Version One – Homework 1 – Juyang Huang – 24018 – Jan 16, 2008 2
The force exerted on the particle is
F = k e
|q 1 | |q 2 |
r 2
‖⃗a‖ = k e
‖⃗q‖ ‖ ⃗ Q‖
m d 2
keywords:
= m a
= k e
∣ ∣ 5 × 10 −5 C ∣ ∣ ∣ ∣−2 × 10 −5 C ∣ ∣
(0.05 kg) (0.5 m 2 )
= 719 m/s 2 .
Hanging Charges
23:05, trigonometry, numeric, > 1 min, normal.
004 (part 1 of 1) 10 points
Two identical small charged spheres hang in
equilibrium with equal masses as shown in
the figure. The length of the strings are equal
and the angle (shown in the figure) with the
vertical is identical.
The acceleration of gravity is 9.8 m/s 2
and the value of Coulomb’s constant is
8.98755 × 10 9 N m 2 /C 2 .
q
m
θ
a
L
q
m
From the right triangle in the figure above,
we see that
sin θ = a L .
Therefore
a = L sin θ
= (0.15 m) sin(5 ◦ )
= 0.0130734 m .
The separation of the spheres is r = 2 a =
0.0261467 m . The forces acting on one of the
spheres are shown in the figure below.
T cos θ
F
e
θ
mg
T
θ
T sin θ
Because the sphere is in equilibrium, the
resultant of the forces in the horizontal and
vertical directions must separately add up to
zero:
∑
Fx = T sin θ − F e = 0
0.15 m
∑
Fy = T cos θ − m g = 0 .
5 ◦
0.03 kg 0.03 kg
Find the magnitude of the charge on each
sphere.
Correct answer: 4.4233 × 10 −8 C.
Explanation:
Let : L = 0.15 m ,
m = 0.03 kg ,
θ = 5 ◦ .
and
From the second equation in the system
above, we see that T =
m g , so T can be
cos θ
eliminated from the first equation if we make
this substitution. This gives a value
F e = m g tan θ
= (0.03 kg) ( 9.8 m/s 2) tan(5 ◦ )
= 0.0257217 N ,
for the electric force.
From Coulomb’s law, the electric force between
the charges has magnitude
|F e | = k e
|q| 2
r 2 ,