ECE106 Lecture Notes - Lecture 5: Net Force
6 Sep 2018
School
Department
Course
Professor
19
∵
the electron does not move
∴
Σ
F
˙
=
0
(2.11)
Note
that
direction
of
F
˙
e
and
F
˙
g
are
anti-par
allel.
W
e
ha
v
e
F
e
=
mg
(2.12)
kq2
d2 =meg
d
2
=
=
kq
2
m
e
g
8.99 × 109 × (1.6 × 10−19)2
9.11 × 10−31 × 9.8
d
=5
.
077
=
5
.
08 (m)
Example 3 Find the charge
q
on each sphere (point mass) so the system is in equilibrium.
q
=?
Figure 2.6: Example 3
Because the system is in equilibrium, the net force applied on the sphere should equals zero. If
we consider the forces on the left sphere,
F
˙
x
=
(
T
sin(
θ
/
2)
−
F
e
)
x
ˆ
=
0
F
˙
y
=
(
T
cos(
θ
/
2)
−
mg
)
y
ˆ
=
0
(2.13)
kq
2
F
e
=
mg
×
tan(
θ/
2) (2.14)
(2
L
sin(
θ/
2))
2
=
mg
tan(
θ/
2)
.
4
mgL
2
sin
3
(
θ/
2)
Example 4 Two charges
Q
1 and
Q
2 are fixed in space as shown, where would a third have to be
placed so that the net force on it is zero?
4 m
Q
1
=
−
7 nC
Q
2
=
12 nC
Figure 2.7: Example 4
1.
it should be apparent that the third charge will have to be placed along the line connecting
Q
1
and
Q
2
. (We are not saying between
Q
1
and
Q
2
, but along the line).
k cos(θ/2)
Fe
L
T
q
=?
θ
mg
.
Σ
q
=
Document Summary
Note that direction of f e and f g are anti-parallel. 9. 11 10 31 9. 8 d =5. 077 = 5. 08 (m) Example 3 find the charge q on each sphere (point mass) so the system is in equilibrium. Because the system is in equilibrium, the net force applied on the sphere should equals zero. If we consider the forces on the left sphere, F x = (t sin( /2) fe)x = 0. F y = (t cos( /2) mg)y = 0. Figure 2. 7: example 4: it should be apparent that the third charge will have to be placed along the line connecting. Q1 and q2. (we are not saying between q1 and q2, but along the line).
