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QJEEM21Ph35f3: JEE Main 2021 Physics - Class 12 JEEM21Ph35f3 Solution: A particle of mass 1 mg and charge q is lying at the mid-point…

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.

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JEE Main 2021 (Online) 25th July Morning Shift Physics - Electrostatics Question 123 English Explanation

Net force on free charged particle

F=kq2(d+x)2−kq2(d−x)2F = {{k{q^2}} \over {{{(d + x)}^2}}} - {{k{q^2}} \over {{{(d - x)}^2}}}

F=−kq2[4dx(d2−x2)2]F = - k{q^2}\left[ {{{4dx} \over {{{({d^2} - {x^2})}^2}}}} \right]

a=−4kq2dm(xd4)a = - {{4k{q^2}d} \over m}\left( {{x \over {{d^4}}}} \right)

a=−(4kq2md3)xa = - \left( {{{4k{q^2}} \over {m{d^3}}}} \right)x

So, angular frequency

ω=4kq2md3\omega = \sqrt {{{4k{q^2}} \over {m{d^3}}}}

ω=4×9×109×101×10−6×13\omega = \sqrt {{{4 \times 9 \times {{10}^9} \times 10} \over {1 \times {{10}^{ - 6}} \times {1^3}}}}

ω=6×108\omega = 6 \times {10^8} rad/sec

ω=6000×105\omega = 6000 \times {10^5} rad/sec

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