A uniform electric field E = (8 m/e) V/m is created between two… | JEE Main 2022 (28 Jul, Shift 2)

Q22JEE Main 2022 · 28 JUL · SHIFT 2Medium

Motion of Charges in an Electric Field

A uniform electric field $E = (8\,m/e)\text{ V/m}$ is created between two parallel plates of length $1\text{ m}$ as shown in figure, (where $m =$ mass of electron and e = charge of electron). An electron enters the field symmetrically between the plates with a speed of $2\text{ m/s}$ . The angle of the deviation $(\theta)$ of the path of the electron as it comes out of the field will be _________.

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Answer: (b)

Acceleration across the plates: $a = \frac{eE}{m} = 8\text{ m/s}^2$ .
Time inside: $t = \frac{L}{u} = \frac{1}{2} = 0.5\text{ s}$ .
$v_y = at = 4\text{ m/s}$ , $v_x = 2\text{ m/s}$ , so $\tan\theta = \frac{4}{2} = 2$ .
So the answer is (b) $\tan^{-1}(2)$ .

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Motion of Charges in an Electric Field question from JEE Main 2022 (28 Jul, Shift 2)

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