Question

The time when will the velocity of the bead vanish for the first time is

• A. $\sqrt{\frac{{\pi }^{2}m{R}^3}{k2Qq}}$
• B. $\sqrt{\frac{{\pi }^{2}m{R}^3}{kQq}}$
• C. $\sqrt{\frac{{\pi }^{2}m{R}^3}{k4Qq}}$
• D. $\sqrt{\frac{{2\pi }^{2}m{R}^3}{kQq}}$

Answer 1

The correct option is A $\sqrt{\frac{{\pi }^{2}m{R}^3}{k2Qq}}$

Let's first calculate the force on the -q to determine whether it is in SHM or not.

From the FBD shown, it can be seen that vertical components of force will be cancelled out and horizontal components will be added. So, total force in horizontal direction can be written as:

$F_{total}=2\frac{KQ(-q)}{(\sqrt{R^2+x^2}{)}^2}cos\theta$

$F_{total}=2\frac{KQ(-q)}{(\sqrt{R^2+x^2}{)}^2}\frac{x}{\left(\sqrt{R^2+x^2}\right)}$

$F_{total}=2\frac{KQ(-q)x}{(\sqrt{R^2+x^2}{)}^{\frac{3}{2}}}$

It is given that x is very small as compared to R, so $R^2+x^2\approx R^2$. Therefore:

$F_{total}=-\frac{2KQqx}{R^3}x$

This resembles the equation of SHM, also note that SHM starts from one of the extreme position, hence it's velocity will become zero again at other extreme, which will take time half of the time period:

$t=\frac{T}{2}=\frac{2\pi }{2}\sqrt{\frac{m}{k}}=\sqrt{\frac{{\pi }^{2}m{R}^3}{2KQq}}$