MCQ
A uniform electric field $E =(8\,m / e ) V / m$ 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 $2\,m / s$. The angle of the deviation $(\theta)$ of the path of the electron as it comes out of the field will be........
  • A
    $\tan ^{-1} (4)$
  • $\tan ^{-1}(2)$
  • C
    $\tan ^{-1}\left(\frac{1}{3}\right)$
  • D
    $\tan ^{-1} (3)$

Answer

Correct option: B.
$\tan ^{-1}(2)$
b
$a _{ y }=\frac{ F _{ y }}{ m }=\frac{ e ( E )}{ m }=\frac{ e \left(\frac{8\,m }{ e }\right)}{ m }=8\,m / s ^{2}$

$s _{ x }= u _{ x } t$

$1=2 \times t$

$t =\frac{1}{2} sec$

$v _{ y }= u _{ y }+ a _{ y } t$

$v _{ y }=0+8 \times \frac{1}{2}$

$v _{ y }=4\,m / s$

$\tan \theta=\frac{ v _{ y }}{ v _{ x }}=\frac{4}{2}=2 \Rightarrow \theta=\tan ^{-1}(2)$

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