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Electrostatics question

2020 · 2 Sep · Shift 1 · Q54
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Electrostatics question

2020 · 2 Sep · Shift 1 · Q54

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A charged particle (mass m and charge q) moves along X-axis with velocity V0. When it passes through the origin it enters a region having uniform electric field E→=−Ej^\overrightarrow E = - E\widehat jE=−Ej​ which extends upto x = d. Equation of path of electron in the region x > d is JEE Main 2020 (Online) 2nd September Morning Slot Physics - Electrostatics Question 173 English
  1. A
    y = qEdmV02(x−d){{qEd} \over {mV_0^2}}\left( {x - d} \right)mV02​qEd​(x−d)
  2. B
    y = qEdmV02(d2−x){{qEd} \over {mV_0^2}}\left( {{d \over 2} - x} \right)mV02​qEd​(2d​−x)
  3. C
    y = qEdmV02x{{qEd} \over {mV_0^2}}xmV02​qEd​x
  4. D
    y = qEd2mV02x{{qE{d^2}} \over {mV_0^2}}xmV02​qEd2​x
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