A uniform electric field E = 8 m e V m - 1 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 enters the fiel

Question

A uniform electric field E = 8 m e V m - 1 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 enters the field symmetrically between the plates with a speed of 2 m s - 1 . The angle of the deviation θ of the path of the electron as it comes out of the field will be _____ .

TestHub · Accepted Answer

Time taken by the electron to cross will be, $t = \frac{1}{v_{x}} = \frac{1}{2} s$ .
Acceleration of the electron along $y$ axis
$a_{g} = \frac{F}{m} = \frac{E e}{m} = \frac{8 m}{e} \times \frac{e}{m} = 8 m s^{- 2}$
$v_{y}$ of the electron when it has crossed the capacitor will be,
$v_{y} = a_{y} t = 8 \times \frac{1}{2} = 4 m s^{- 1}$
Therefore,
$\tan θ = \frac{v_{y}}{v_{x}} = \frac{4}{2} = 2 \Rightarrow θ = \tan^{- 1} (2)$

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