A uniform electric field $E=\frac{8m}{e} Vm^{-1}$ 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 $2ms^{-1}$. The angle of the deviation $\theta$ of the path of the electron as it comes out of the field will be _____ .

Question

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

A uniform electric field $E=\frac{8m}{e} Vm^{-1}$ is created between two parallel plates of length $1m$ as…

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