JEE MAIN - Physics (2021 - 25th July Morning Shift - No. 23)
A particle of mass 1 mg and charge q is lying at the mid-point of two stationary particles kept at a distance '2 m' when each is carrying same charge 'q'. If the free charged particle is displaced from its equilibrium position through distance 'x' (x < < 1 m). The particle executes SHM. Its angular frequency of oscillation will be ____________ $$\times$$ 105 rad/s if q2 = 10 C2.
Answer
6000
Explanation
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Net force on free charged particle
$$F = {{k{q^2}} \over {{{(d + x)}^2}}} - {{k{q^2}} \over {{{(d - x)}^2}}}$$
$$F = - k{q^2}\left[ {{{4dx} \over {{{({d^2} - {x^2})}^2}}}} \right]$$
$$a = - {{4k{q^2}d} \over m}\left( {{x \over {{d^4}}}} \right)$$
$$a = - \left( {{{4k{q^2}} \over {m{d^3}}}} \right)x$$
So, angular frequency
$$\omega = \sqrt {{{4k{q^2}} \over {m{d^3}}}} $$
$$\omega = \sqrt {{{4 \times 9 \times {{10}^9} \times 10} \over {1 \times {{10}^{ - 6}} \times {1^3}}}} $$
$$\omega = 6 \times {10^8}$$ rad/sec
$$\omega = 6000 \times {10^5}$$ rad/sec
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