Question

A small ball of mass $m$ and charge $+q$ tied with a string of length $l$, rotating in a vertical circle under gravity and a uniform horizontal electric field $E$ as shown. The tension in the string will be minimum for

• A. $\theta ={\tan }^{-1}\left(\frac{qE}{mg}\right)$
• B. $\theta =\pi$
• C. $\theta =0^{\circ }$
• D. $\theta =\pi +{\tan }^{-1}\left(\frac{qE}{mg}\right)$

Answer 1

The correct option is D $\theta =\pi +{\tan }^{-1}\left(\frac{qE}{mg}\right)$


Tension in the string is minimum at the point diagonally opposite to equilibrium position the particle in vertical circle.
Let's start first to calculate equilibrium position from the FBD.

At position $\text{A}$, from equilibrium of forces,

$T\sin \alpha =qE.......\left(1\right)$

and, $T\cos \alpha =mg........\left(2\right)$

from Eqs.(1) and (2)

$\Rightarrow \tan \alpha =\frac{qE}{mg}$

$\Rightarrow \alpha ={\tan }^{-1}\left(\frac{qE}{mg}\right)$

Minimum tension will occur at diametrically opposite point $B$. So,

$\theta =\pi +\alpha =\pi +{\tan }^{-1}\left(\frac{qE}{mg}\right)$.

Hence, option $\left(d\right)$ is the correct answer.