As we know that, Coulomb's force between two charges. i.e. q1 and q2, F=4πε01r2q1q2=4πε01(d2+x2)q1q2. . . (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⇒4πε02(d2+x2)q2sinθ=mω2x⇒4πε02(d2+x2)q2⋅(d2+x2)1/2x=mω2x⇒ω=(2πε01(d2+x2)3/2mq2)1/2 As, x≪d∴ω=(2πε01md3q2)1/2