Verified Solution • 2025–26 Session moderate

QJEEM20Phd447: JEE Main 2020 Physics - Class 12 JEEM20Phd447 Solution: A particle of mass m and charge q is released from rest in a…

A particle of mass m and charge q is released from rest in a uniform electric field. If there is no other force on the particle, the dependence of its speed v on the distance x travelled by it is correctly given by (graphs are schematic and not drawn to scale)

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/8e2b32d741adfe8f1f9609150421f0125fa74699cb4473f49cfbcafeeeafb9a4.webp
  • B https://ik.imagekit.io/novatra/photos/caf9fa562638352d9bd3850260f7478e91f4623ea28b32e71c99a742047a297f.webp
  • C https://ik.imagekit.io/novatra/photos/9f096f1b84ba27ebbe3c8b673ba2655bcb1518bb4d0ee91f28587f26e639f2df.webp
  • D https://ik.imagekit.io/novatra/photos/6d4abf0e1f4c2ced3366eccf73596e0759d5c5edae1a24aba4c05992d647826d.webp

Step-by-Step Verified Solution

As F = qE

⇒\Rightarrow ma = qE

⇒\Rightarrow a = qEm{{qE} \over m}

Also we know, v2 = u2 + 2as

⇒\Rightarrow v2 = 0 + 2 (qEm)\left( {{{qE} \over m}} \right) s

⇒\Rightarrow v = 2qEmx{\sqrt {{{2qE} \over m}} \sqrt x }

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