JEE PYQ: Electrostatics Question 5
Question 5 - 2021 (24 Feb Shift 2)
Two electrons each are fixed at a distance $2d$. A third charge proton placed at the midpoint is displaced slightly by a distance $x$ ($x \ll d$) perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency: ($m$ = mass of charged particle)
(1) $\left(\frac{q^2}{2\pi\varepsilon_0 md^3}\right)^{1/2}$
(2) $\left(\frac{2q^2}{\pi\varepsilon_0 md^3}\right)^{1/2}$
(3) $\left(\frac{2\pi\varepsilon_0 md^3}{q^2}\right)^{1/2}$
(4) $\left(\frac{q^2}{4\pi\varepsilon_0 md^3}\right)^{1/2}$
Show Answer
Answer: (1)
Solution
Restoring force on proton: $F_r = \frac{2Ke^2 x}{(d^2+x^2)^{3/2}}$. For $x \ll d$: $F_r \approx \frac{2Ke^2}{d^3}x = kx$. Angular frequency $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{2Ke^2}{md^3}} = \left(\frac{q^2}{2\pi\varepsilon_0 md^3}\right)^{1/2}$.
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