Question

Two small particles $A$ and $B$ having masses $m=0.5kg$ 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 $r=0.5m$. Particle $A$ is fixed and $B$ is whirled along a vertical circle with centre at $A$. If a vertically upward electric field of strength $E=1.1\times 10^{5}N{C}^{-1}$ 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. $(g=10m{s}^{-2}$ ) (Approximate the answer to the nearest integer.)

Answer 1

Answer: 8

$$\begin{gathered} F_{AB}=\frac{K{q}_1{q}_2}{r^2}=31N=F_1 \\ {F}_{EO}=F_2=q_{2}E=11N \\ W=5 \end{gathered}$$
Tension can be zero at lowest point
$$\begin{gathered} F_1+F_2-W=\frac{1}{r}m{V}_{min}^2 \\ {V}_{min}=\sqrt{37} \end{gathered}$$
$\frac{1}{2}m{V}_{Bmin.}^2=\frac{1}{2}m{v}_{min}^2+F_2(2r)-W(2r)$
$V_{Bmin.}=\sqrt{61}m/s$