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2. Electric Potential and Capacitance
hard
A particle of mass $m$ and charge $q$ is kept at the top of a fixed frictionless sphere. $A$ uniform horizontal electric field $E$ is switched on. The particle looses contact with the sphere, when the line joining the center of the sphere and the particle makes an angle $45^o$ with the vertical. The ratio $\frac{qE}{mg}$ is :-
A
$\frac{3}{{3 + 2\sqrt 2 }}$
B
$\frac{{3 + 2\sqrt 2 }}{3}$
C
$\frac{3}{{3 - 2\sqrt 2 }}$
D
$\frac{{3 - 2\sqrt 2 }}{3}$
Solution
$m g \cos \theta-q E \sin \theta=\frac{m v^{2}}{R}$ ………$(1)$
Applying work energy theorem
$\frac{1}{2} m v^{2}=m g R(1-\cos \theta)+q E R \sin \theta$ ……..$(2)$
Solving $(1)$ and $(2)$
$\frac{q E}{m g}=\frac{3-2 \sqrt{2}}{3}$
Standard 12
Physics
