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Physics 21
Fall, 2009
Solution to HW3
Coulomb’s Law Tutorial
This problem asks you to fnd
the Couloumb Forces exerted by particles 1, 2, and 3 on par
ticle 0, as shown in the diagram. (a) Consider the two posi
tively charged particles, one oF charge
q
0
(particle 0) fxed at
the origen, and another oF charge
q
1
(particle 1) fxed on the
y
axis at (0
,d
1
,
0). What is the net Force
F
1on0
on
particle
0
due to
particle 1? (b) Now consider the third, negatively
charged, particle, whose charge is
−
q
2
(particle 2). Particle
2i
sfxedonthe
y
axis at position (0
2
,
0). What is the
new net Force
on
particle 0,
from
particle 1 and particle 2?
(c) Particle 0 experiences a repulsion
from
particle 1 and an
attraction
toward
particle 2. ±or certain values oF
d
1
and
d
2
, the repulsion and attraction should balance each other,
resulting in no net Force. ±or what ratio
d
1
/d
2
is there no
net Force on particle 0? (d) Now consider the Fourth charged
particle, particle 3, with positive charge
q
3
,fxedinthe
yz

plane at (0
2
2
). What is the net Force
F
3on0
on particle
0 due
solely
to this charge?
(a) ±rom the expression For
F
2on1
on the Equation Sheet,
we can write
F
=
1
4
π±
0
q
0
q
1
(
r
0
−
r
1
)

r
0
−
r
1

3
.
r
0
−
r
1
points From the charge point (particle 1 at
r
1
=
d
1
ˆ
j
)
to the feld point (particle 0 at
r
0
= 0, the origin). Then
r
0
−
r
1
=
−
d
1
ˆ
j
and

r
0
−
r
1

=
d
1
,
and so the fnal answer is
F
=
−
1
4
0
q
0
q
1
d
2
1
ˆ
j
.
(b) The third charged particle on the
y
axis (particle 2)
introduces an additional Force on particle 0 that adds to the
Force From particle 1 via superposition. Since the charge is
−
q
2
, the additional Force is
F
2on0
=
1
4
0
q
0
(
−
q
2
)(
r
0
−
r
2
)

r
0
−
r
2

3
=
1
4
0
q
0
(
−
q
2
)
³
−
d
2
ˆ
j
´
d
3
2
=
1
4
0
q
0
q
2
d
2
2
ˆ
j
.
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 Spring '07
 Hickman
 Force

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