JEE MAIN - Physics (2019 - 8th April Evening Slot - No. 2)
An electric dipole is formed by two equal and
opposite charges q with separation d. The
charges have same mass m. It is kept in a
uniform electric field E. If it is slightly rotated
from its equilibrium orientation, then its angular
frequency $$\omega$$ is :-
$$\sqrt {{{qE} \over {md}}} $$
$$\sqrt {{{qE} \over {2md}}} $$
$$\sqrt {{{qE} \over {-2md}}} $$
$$\sqrt {{{2qE} \over {md}}} $$
Explanation
_8th_April_Evening_Slot_en_2_1.png)
Moment of inertia$$(I) = m{\left( {{d \over 2}} \right)^2} \times 2 = {{m{d^2}} \over 2}$$
Now by $$\tau = l\alpha $$
$$(qE)(d\,\sin \theta ) = {{m{d^2}} \over 2}.\alpha $$
$$\alpha = \left( {{{2qE} \over {md}}} \right)\sin \theta $$ for small $$\theta $$
$$ \Rightarrow \alpha = \left( {{{2qE} \over {md}}} \right)\theta $$
$$ \Rightarrow $$ Angular frequency $$\omega = \sqrt {{{2qE} \over {md}}} $$
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