Question

Two positive point charges each of magnitude $10C$ are fixed at positions $A$ and $B$ at a separation $2d=6m.A$ negatively charged particle of mass $m=90g$ and charge of magnitude $10\times 10^{-6}C$ is revolving in a circular path of radius $4m$ in the plane perpendicular to the line $AB$ and bisecting the line $AB$. Neglect the effect of gravity. Find the angular velocity of the particle.

• A. 200 rad/s
• B. 400 rad/s
• C. 250 rad/s
• D. 100 rad/s

Answer 1

Answer: (B)

Net force on $-q$ towards the centre,

$F=\left(2F_1\cdot \sin \theta \right)$
$=2\cdot \frac{kQ\cdot q}{d^2+R^2}\times \frac{R}{\sqrt{d^2+R^2}}$
For particle to move in circle, $F=m{\omega }^{2}R$
$\Rightarrow \frac{2kQqR}{{\left(d^2+R^2\right)}^{3/2}}=m{\omega }^{2}R$
$\Rightarrow \omega =\sqrt{\frac{2kQq}{m{\left(d^2+R^2\right)}^{3/2}}}$
$=\sqrt{\frac{2\times 9\times 10^9\times 10\times 10\times 10^{-6}}{90\times 10^{-3}\times {\left(3^2+4^2\right)}^{3/2}}}=400rad/s$