Question

Two small particles $A$ and $B$ having masses $m =0.5 kg$ each and charge $q_{1}=\left(-\frac{155}{18} \mu C \right)$ and $q_{2}=(+100 \mu C )$ respectively, are connected at the ends of a non-conducting, flexible and inextensible string of length $r=0.5 m$. Particle $A$ is fixed and $B$ is whirled along a vertical circle with centre at $A$. If a vertically upward electric field of strength $E=1.1 \times 10^{5} NC ^{-1}$ exists in the space, the minimum velocity of the particle $B$ at the highest position is (in $m / s )$ so that it may just complete the circle. $\left(g=10 ms ^{-2}\right.$ ) (Approximate the answer to the nearest integer.)

Answer 1

$F_{A B}=\frac{K q_{1} q_{2}}{r^{2}}=31N=F_{1} \\ F_{E O}=F_{2}=q_{2}E=11N \\ W=5$
Tension can be zero at lowest point
$F_{1}+F_{2}-W=\frac{1}{r}mV_{m i n}^{2} \\ V_{m i n}=\sqrt{37}$
$\frac{1}{2}mV_{B m i n .}^{2}=\frac{1}{2}mv_{m i n}^{2}+F_{2}\left(\right.2r\left.\right)-W\left(\right.2r\left.\right)$
$V_{B m i n .}=\sqrt{61}m/s$