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JEE PHYSICS MCQ: ELECTRIC FIELDS AND CHARGES

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A particle of mass

\text{m}

and charge

\text{q}

is thrown perpendicular to an electric field of intensity

\text{E}

with an initial velocity

\text{v}.

The particle moves a distance

x

perpendicular to the field and a distance

\text{y}

along the direction of the field. If

\text{y}=\alpha \text{x}^2

then the

\alpha

is given by:

View answer and explanation

Correct option: D

\frac{\mathrm{qE}}{2 \mathrm{mv}^2}

Option D (

\frac{\mathrm{qE}}{2 \mathrm{mv}^2}

) is correct because it is derived from the equations of motion under constant acceleration due to the electric field.

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