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A uniform electric field $E = (12 m / e)$ V/m is created between two parallel plates of length 1 m as shown in figure. (where m=mass of electron and e= charge of electron). An electron centers the field symmetrically between the plates with a speed of 2 m/s. The angle of the deviation $(θ)$ of the path of the electron as it comes out of the field will be_____

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$\tan^{- 1} (4)$

$\tan^{- 1} (2)$

$\tan^{- 1} (\frac{1}{3})$

$\tan^{- 1} (3)$

answer is D.

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Detailed Solution

We have $a_{y} = \frac{F_{y}}{m} = \frac{c (E)}{m} = \frac{e (\frac{12 m}{e})}{m} = 12 m / s^{2}$
And $s_{x} = u_{x} t$
$\Rightarrow 1 - 2 \times t \Rightarrow t = \frac{1}{2} \sec$
And $v_{y} = u_{y} + a_{y} t$

$\Rightarrow v_{y} = 0 + 12 \times \frac{1}{2} \Rightarrow v_{y} = 6 m / s$

$\tan θ = \frac{v_{y}}{v_{x}} = \frac{6}{2} = 3 \Rightarrow θ = \tan^{- 1} (3)$

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