Question

A point object having charge $q$ and mass $m$ kept on the top of a frictionless and insulated hemisphere. When a horizontal electric field is applied, the object starts sliding down and loses contact from the surface at an angle $\theta =si{n}^{-1}\frac{3}{5}$ from the vertical. The ratio of force of gravity and the force due to interaction with the field acting on this object is

• A. $\frac{9}{2}$
• B. $\frac{7}{2}$
• C. $\frac{9}{7}$
• D. $\frac{7}{5}$

Answer 1

The correct option is A $\frac{9}{2}$

At an angle $\alpha$ from the vertical,


FBD of the object is


Writing force balance equation perpendicular to the surface,

$mgcos\alpha -N-qEsin\alpha =\frac{m{v}^2}{R}$

At the point where it loses contact with the surface,

$N=0$

Given at this point, $\alpha =\theta =si{n}^{-1}\left(\frac{3}{5}\right)$
Then $sin\theta =\frac{3}{5}cos\theta =\frac{4}{5}$
$\Rightarrow mgcos\theta -qEsin\theta =\frac{m{v}^2}{R}$

$\Rightarrow mg\left(\frac{4}{5}\right)-qE\left(\frac{3}{5}\right)=\frac{m{v}^2}{R}$ ---- (1)

Using Work-Energy theorem,

Work done by electric field + Work done by gravity = Change in kinetic energy


$qE\left(Rsin\theta \right)+mg(R-Rcos\theta )=\frac{1}{2}m{v}^2$

$qE\left(R\frac{3}{5}\right)+mg(R-R\frac{4}{5})=\frac{1}{2}m{v}^2$ - (2)

Dividing (1) by (2)

$\frac{\frac{4}{5}Rmg-\frac{3}{5}RqE}{\frac{3}{5}qER+mg(R-R(\frac{4}{5})}=2$

$\Rightarrow \frac{4}{5}mg-\frac{3}{5}qE=\frac{6}{5}qE+\frac{2}{5}mg$

$\Rightarrow \frac{2}{5}mg=\frac{9}{5}qE$
$\Rightarrow \frac{mg}{qE}=\frac{9}{2}$