There is a uniform electric field $E$ in the region between two parallel plates. A charged particle with a specific charges enters the region with a speed $u$ , in a direction parallel to the plates. The length of each plate is $L$ . The time taken to come out of the plates is $t$ and it suffers a vertical displacement $Δ$ . Also as the particle crosses the plates, its direction of motion changes by an angle $θ$ , then

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$θ = \tan^{- 1} (\frac{E s t}{u})$

$θ = \tan^{- 1} (\frac{E s L}{u^{2}})$

$Δ = \frac{E s L^{2}}{2 u^{2}}$

$t = \frac{L}{u}$

answer is A, B, C, D.

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Detailed Solution

$a = \frac{q E}{m} = s E$

Since, L=u t

$\Rightarrow t = \frac{L}{u}$

Also, $Δ = \frac{1}{2} a t^{2}$

$\Rightarrow Δ = \frac{1}{2} (s E) \frac{L^{2}}{u^{2}} \Rightarrow Δ = \frac{E s L^{2}}{2 u^{2}} \tan θ = \frac{v_{y}}{v_{x}} = \frac{a t}{u} \Rightarrow \tan θ = \frac{E s t}{u} = \frac{E s L}{u^{2}}$

Hence, (1), (2), (3) and (4) are correct.

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