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Q.

A uniform electric field $E = (8 m / e) v / m$ is created between two parallel plates of length 1m 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 2m/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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a

$\tan^{- 1} 4$

b

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

c

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

d

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

answer is B.

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Detailed Solution

$a_{y} = \frac{\sqrt{y}}{m} = \frac{e (E)}{m} = \frac{e (\frac{8 m}{e})}{m} = 8 m / s^{2} S_{x} = u_{x} t \Rightarrow 1 = 2 \times t \Rightarrow t = \frac{1}{2} \sec U \sin g v_{y} = u_{y} + a_{y} t v_{y} = 0 + 8 \times \frac{1}{2} = 4 m / s \tan θ = \frac{v_{y}}{v_{x}} = \frac{4}{2} = 2 \Rightarrow θ = \tan^{- 1} (2)$

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