Physics - Electrostatics Question with Solution | TestHub

Angular velocity of B wrt A at the instant when three particles become collinear will be

$$\omega =\frac{{\mathrm{v}}_{\mathrm{B}}-{\mathrm{v}}_{\mathrm{A}}}{\ell }=\frac{3\mathrm{v}}{\ell }=\mathrm{q}\sqrt{\frac{3\mathrm{k}\ell }{2\mathrm{m}}}$$

If T be the tension in string AB , then

$$\begin{array}{rlr}& \mathrm{T}-\frac{{\mathrm{kq}}^2}{{\ell }^2}-\frac{{\mathrm{kq}}^2}{(2\ell {)}^2}=\mathrm{m}{\omega }^2\ell & \Rightarrow \mathrm{T}=\frac{11{\mathrm{kq}}^2}{4{\ell }^2}\stackrel{\mathrm{T}}{B}\cdot \mathrm{A}\cdot \mathrm{C}\end{array}$$

Let ${\mathrm{x}}_0=$ Maximum displacement of B . Then maximum displacement of $\mathrm{A}=2{\mathrm{x}}_0($ as ${\mathrm{v}}_{\mathrm{A}}=2\mathrm{v})$ At the time of maximum displacement, particles again form an equilateral triangle. The positions of B and C wrt A are as shown in figure.

$$\begin{array}{rlr}& \text{ Also, }\mathrm{AD}=\frac{\sqrt{3}}{2}\ell & \therefore {\mathrm{x}}_0+2{\mathrm{x}}_0=2\times \mathrm{AD}\Rightarrow {\mathrm{x}}_0=\ell /\sqrt{3}\end{array}$$