A small ball of mass $m=0.6kg$ carrying a charge $q=80\mu C$ is connected to a light thread of length $l=30cm.$ The other end of the thread is fixed. A horizontal electric field $E={10}^{5}N.C^{-1}$  exists in the space. The ball is given a speed and it moves in a vertical circle. Find the speed (in m/s) of the ball at the top point of the circle if it is just able to complete the circle without the string getting slack at any point $\left[g=10m{s}^{-2}\right].$

(i) Apart from the string tension, the ball experiences $T_A=0$two forces $mg$ and $qE.$ Resultant of these two forces is $F=\sqrt{{\left(mg\right)}^2+{\left(qE\right)}^2}$

The particle is moving in an environmental where we can regard the effective free fall acceleration to be

$g_{eff}=\frac{F}{m}=\sqrt{g^2+{\left(\frac{qE}{m}\right)}^2}$

Direction of $g_{eff}$ makes an angle $\theta$ with vertical where $\tan \theta =\frac{qF}{mg}$

(ii) Tension will be maximum and minimum when the particle (on its circular path) lies on the diameter along the direction of $g_{eff}.$

$g_{eff}=\sqrt{g^2+{\left(\frac{qE}{m}\right)}^2}$ $=\sqrt{{10}^2+{\left(\frac{80\times {10}^{-6}\times {10}^5}{0.6}\right)}^2}$ $=\frac{50}{3}m{s}^{-2}$

Direction of $g_{eff}$  makes an angle $\theta$  with the vertical where $\tan \theta =\frac{qE}{mg}=\frac{4}{3}\Rightarrow \theta =53°$

Tension will be maximum when the ball is at B and it will be minimum when the ball is at A For completing the circle, tension at A should be     $T_A\ge 0$

In circle case,  $\Rightarrow \frac{m{v}^2}{l}=m{g}_{eff}\Rightarrow v=\sqrt{l{g}_{eff}}$   $\Rightarrow v=\sqrt{0.3\times \frac{50}{3}}=\sqrt{5}m{s}^{-1}$

To get the speed at C we will use conservation of energy between points A and C

$K_A+U_{gA}+U_{eA}=K_B+U_{gC}+U_{eC}$    $\Rightarrow \frac{1}{2}m{v}^2+\left(U_{eA}-U_{eC}\right)=\frac{1}{2}m{u}^2+\left(U_{gC}-U_{gA}\right)$

$\Rightarrow \frac{1}{2}m{v}^2+\left(V_A-V_C\right)q=\frac{1}{2}m{u}^2+mgl\left(1-\cos \theta \right)$

$\Rightarrow \frac{1}{2}m{v}^2+\left(E.l.\sin \theta \right)q=\frac{1}{2}m{u}^2+mgl\left(1-\cos \theta \right)$

$\Rightarrow \frac{1}{2}\times 0.6\times 5+{10}^5\times 0.3\times \frac{4}{5}\times 80\times {10}^{-6}=\frac{1}{2}\times 0.6\times u^2+0.6\times 10\times 0.3\times \left(1-\frac{3}{5}\right)$

$\Rightarrow u=3m{s}^{-1}$