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JEE Main 2020 (8 Jan, Shift 2) · Motion of Charges in an Electric Field

Motion of Charges in an Electric Field question from JEE Main 2020 (8 Jan, Shift 2)

Q32JEE Main 2020 · 8 JAN · SHIFT 2Easy

Motion of Charges in an Electric Field

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)

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Answer: (b)

Constant force $qE$ gives constant acceleration $a = \frac{qE}{m}$ .
From rest: $v^2 = 2ax = \frac{2qE}{m}x$ , so $v \propto \sqrt{x}$ .
The $v$ – $x$ graph is a parabola opening along $x$ : steep at the origin, then flattening.
So the answer is (b).

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