Two forces each of magnitude q are fixed
Answers
Answer:
Explanation:
For a small displacement x perpendicular to the line joining the charges
the potential energy , U ( r ) =
4πϵ
0
r
Q
, of the displaced charge is now :
U(x)=2×−
4πϵ
0
a
2
+x
2
Q
The force vector acting on the particle is in the negative derivative w.r.t x of this conservative potential function.
Form symmetry we know that it will act "perpendicular to the line joining the charges".
F
= -
dx
du
=
2πϵ
0
Q
(a
2
+x
2
)
2
1
x
By taylor expansion, for small x we have:
(a
2
+x
2
)
2
3
x
=
a
3
x
(1+
a
2
x
2
)
−
2
3
In other words,
F
linearises as:
F
= −
2a
3
πϵ
0
Q
x,
and it is this linearise restorative force that characterises the harmonic oscillator.
i.e. x+
2πma
3
ϵ
o
Q
x=0 can be written in the standrad SHM wave form x+ω
2
x=0
Therefore;
T=
w
2π
= 2π
Q
2ma
3
πϵ
0
Explanation:
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