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JEE Main 2013 (9 Apr, Shift 1) · Motion of Charges in an Electric Field

Motion of Charges in an Electric Field question from JEE Main 2013 (9 Apr, Shift 1)

Q39JEE Main 2013 · 9 APR · SHIFT 1Easy

Motion of Charges in an Electric Field

A uniform electric field $\vec{E}$ exists between the plates of a charged condenser. A charged particle enters the space between the plates and perpendicular to $\vec{E}$ . The path of the particle between the plates is a:

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Answer: (c)

Along the plates (perpendicular to $\vec{E}$ ): no force, so $x = ut$ .
Along $\vec{E}$ : constant acceleration, so $y = \frac{1}{2}\frac{qE}{m}t^2$ .
Eliminating $t$ : $y = \frac{qE}{2mu^2}x^2$ , a parabola.
So the answer is (c) parabola.

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