Electrostatics Part 1

The arrangement of charges is shown belowAs we know that,
Coulomb's force between two charges. i.e. q1 and q2,
F=‌

1

4πε0

‌

q1q2

r2

=‌

1

4πε0

‌

q1q2

(d2+x2)

. . . (i)
Here, q1=q2=q
Force in SHM,F=mω2x ...(ii)
Since, in order to have SHM+q should move downwards and force responsible for this will be only
F′=F‌sin‌θ+F‌sin‌θ=2F‌sin‌θ. . . (iii)
Using Eqs. (ii) and (iii), we get
2F‌sin‌θ=mω2x
⇒‌‌‌

2

4πε0

‌

q2

(d2+x2)

‌sin‌θ=mω2x
⇒‌‌‌

2

4πε0

‌

q2

(d2+x2)

⋅‌

x

(d2+x2)1∕2

=mω2x
⇒‌‌ω=(‌

1

2πε0

‌

q2

(d2+x2)3∕2m

)1∕2
As, x<<d
∴‌‌ω=(‌

1

2πε0

‌

q2

md3

)1∕2