Question

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)
• B. $\tan ^{-1}$ (2)
• C. $\tan ^{-1}\left(\frac{1}{3}\right)$
• D. $\tan ^{-1}$ (3)

Answer 1

Answer: (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 $
$\Rightarrow \theta=\tan ^{-1}(2)$