Physics - Electrostatics Question with Solution | TestHub

By symmetry and momentum conservation, with $\theta$ the half-angle at M: ${\dot{y}}_M=-\frac{\ell \sin \theta \dot{\theta }}{2}$

M moves by $\frac{\ell }{2}(\cos 30^{\circ }-\cos 90^{\circ })=\frac{\sqrt{3}\ell }{4}$ to the mean (collinear) position, and equally beyond: amplitude $=\frac{\sqrt{3}\ell }{4}$ (A correct).

Energy: $m{\ell }^2{\dot{\theta }}^2\left(1-\frac{{\sin }^2\theta }{2}\right)+\frac{k{q}^2}{2\ell \sin \theta }=\frac{k{q}^2}{\ell }$. At $\theta =90^{\circ }$: $v_M=\frac{\ell \dot{\theta }}{2}=\frac{q}{4}\sqrt{\frac{1}{\pi {\epsilon }_{0}m\ell }}$ (B correct).

At the straight line each m moves with speed $\ell \dot{\theta }/2$, equal to M's speed (C correct).

Newton's law for either m at the straight line gives $T=\frac{q}{16\pi {\epsilon }_0{\ell }^2}(4Q+5q)$ (D correct).