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Q.

A uniform electric field E = (8m/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

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

d

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

answer is B.

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Detailed Solution

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$a_{y} = \frac{F_{y}}{m} = \frac{e (\frac{8 m}{e})}{m} = 8 m / s^{2}$

$s_{x} = u_{x} t \Rightarrow 1 = 2 \times t \Rightarrow t = \frac{1}{2} \sec$

$v_{y} = u_{y} + a_{y} t \Rightarrow v_{y} = 0 + 8 \times \frac{1}{2} \Rightarrow v_{y} = 4 m / s$

$\Rightarrow \tan (θ) = \frac{v_{y}}{v_{x}} = \frac{4}{2} = 2 \Rightarrow θ = \tan^{- 1} (2)$

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A uniform electric field E = (8m/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 .........