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JEE Main 2018 · Physics · Electrostatics · medium

A uniformly charged thin ring of radius 0.09 m carries a charge of +1 muC. A particle of mass 1 g and charge -1 muC is released from rest on the axis…

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A uniformly charged thin ring of radius 0.09 m0.09\ \mathrm{m} carries a charge of +1 μC+1\ \mu\mathrm{C}. A particle of mass 1 g1\ \mathrm{g} and charge −1 μC-1\ \mu\mathrm{C} is released from rest on the axis of the ring at a distance 3\sqrt3 times the ring radius from its centre. Taking 14πε0=9×109 N m2C−2\dfrac{1}{4\pi\varepsilon_0}=9\times10^9\ \mathrm{N\,m^2C^{-2}} and ignoring gravity, the speed of the particle as it passes through the centre of the ring is
  1. A. 10 m s−110\ \mathrm{m\,s^{-1}}
  2. B. 14.1 m s−114.1\ \mathrm{m\,s^{-1}}
  3. C. 5 m s−15\ \mathrm{m\,s^{-1}}
  4. D. 7.07 m s−17.07\ \mathrm{m\,s^{-1}}
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Answer: (A) 10 m s−110\ \mathrm{m\,s^{-1}}

The potential on the axis at distance xx is V=kQR2+x2V=\dfrac{kQ}{\sqrt{R^2+x^2}}. At the start, x=3Rx=\sqrt3R, so Vi=kQ2RV_i=\dfrac{kQ}{2R}; at the centre, Vf=kQRV_f=\dfrac{kQ}{R}. The negative charge is attracted inward and gains kinetic energy 12mv2=∣q∣(Vf−Vi)=k∣q∣Q2R\tfrac12mv^2=|q|(V_f-V_i)=\dfrac{k|q|Q}{2R}. Thus v2=kQ∣q∣mR=9×109×10−1210−3×0.09=100v^2=\dfrac{kQ|q|}{mR}=\dfrac{9\times10^9\times10^{-12}}{10^{-3}\times0.09}=100, so v=10 m s−1v=10\ \mathrm{m\,s^{-1}}. Taking the start at infinity would give 14.114.1.
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