- Let Q = charge on A & B Separated by distance d
q = charge on c displaced $\bot$ to -AB
So, force on $0=\overline{\text{F}}_\text{AB}+\overline{\text{F}}_\text{BO}$
But $\text{F}_\text{AO}\cos\theta=\text{F}_\text{BO}\cos\theta$
So, force on ‘0’ in due to vertical component.
$\overline{\text{F}}=\text{F}_\text{AO}\sin\theta+\text{F}_\text{BO}\sin\theta$ $|\text{F}_\text{AO}|=|\text{F}_\text{BO}|$
$=2\frac{\text{KQq}}{\Big(\frac{\text{d}}{2^2+\text{x}^2}\big)}\sin\theta$
$\text{F}=\frac{2\text{KQq}}{\Big(\frac{\text{d}}{2}\Big)^2+\text{x}^2}\sin\theta$
$=\frac{4\times2\times2\text{KQq}}{(\text{d}^2+4\text{x}^2)}\times\frac{\text{x}}{\Big[\big(\frac{\text{d}}{2}\big)^2+\text{x}^2\Big]^{\frac{1}{2}}}$
$=\frac{2\text{kQq}}{\Big[\big(\frac{\text{d}}{2}\big)^2+\text{x}^2\Big]^{\frac{3}{2}}}\text{x}$ = Electric force $\Rightarrow\text{F}\propto\text{x}$
- When x << d
$\text{F}=\frac{2\text{kQq}}{\Big[\big(\frac{\text{d}}{2}\big)^2+\text{x}^2\Big]^{\frac{3}{2}}}\text{x}$ x << d
$\Rightarrow\text{F}=\frac{2\text{kQq}}{\Big(\frac{\text{d}^2}{4}\Big)^{\frac{3}{2}}}\text{x}\Rightarrow\text{F}\propto\text{x}$
$\text{a}=\frac{\text{F}}{\text{m}}=\frac{1}{\text{m}}\Bigg[\frac{2\text{kQqx}}{\Big[\big(\frac{\text{d}^2}{4}\big)+\ell^2}\Bigg]$
So time period $\text{T}=2\pi\sqrt{\frac{\ell}{\text{g}}}$
$=2\pi\sqrt{\frac{\ell}{\text{a}}}$
