Question 89 of 311

A uniform electric field $E=\left(\dfrac{8m}{e}\right)V{m}^{-1}$ 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 $2m{s}^{-1}$. The angle of the deviation $\left(\theta \right)$ of the path of the electron as it comes out of the field will be _____ .

A${\tan }^{-1}\left(4\right)$B${\tan }^{-1}\left(2\right)$C${\tan }^{-1}\left(\dfrac{1}{3}\right)$D${\tan }^{-1}\left(3\right)$

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Answer: (B) ${\tan }^{-1}\left(2\right)$

Time taken by the electron to cross will be, $t=\dfrac{1}{{v}_{x}}=\dfrac{1}{2}s$.
Acceleration of the electron along $y$ axis
${a}_{g}=\dfrac{F}{m}=\dfrac{Ee}{m}=\dfrac{8m}{e}\times \dfrac{e}{m}=8m{s}^{-2}$
${v}_{y}$ of the electron when it has crossed the capacitor will be,
${v}_{y}={a}_{y}t=8\times \dfrac{1}{2}=4m{s}^{-1}$
Therefore,
$\tan \theta =\dfrac{{v}_{y}}{{v}_{x}}=\dfrac{4}{2}=2 \Rightarrow \theta ={\tan }^{-1}\left(2\right)$