Question

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

$F =\left(2 F_{1} \cdot \sin \theta\right)$
$=2 \cdot \frac{k Q \cdot q}{d^{2}+R^{2}} \times \frac{R}{\sqrt{d^{2}+R^{2}}} $
For particle to move in circle, $F=m \omega^{2} R$
$\Rightarrow \frac{2 k Q q R}{\left(d^{2}+R^{2}\right)^{3 / 2}}=m \omega^{2} R$
$\Rightarrow \omega=\sqrt{\frac{2 k Q q}{m\left(d^{2}+R^{2}\right)^{3 / 2}}}$
$=\sqrt{\frac{2 \times 9 \times 10^{9} \times 10 \times 10 \times 10^{-6}}{90 \times 10^{-3} \times\left(3^{2}+4^{2}\right)^{3 / 2}}}=400\,rad / s$