Question

There is a uniform electric field E in the region between two parallel plates. A charged particle with a specific charge s enters the region with a speed u, in a direction parallel to the plates. The length of each plate is L. The time taken to come out of the plates is t and it suffers a vertical displacement. Also as y the particle crosses the plates, its direction of motion changes by an angle $\theta$, then

• A. $y=\frac{Es{L}^2}{4u^2}$
• B. $t=2\frac{L}{u}$
• C. $\theta =ta{n}^{-1}\left(\frac{EsL}{2u^2}\right)$
• D. $\theta =co{t}^{-1}\left(\frac{EsL}{u^2}\right)$

Answer 1

The correct option is C $\theta =ta{n}^{-1}\left(\frac{EsL}{2u^2}\right)$

$a=\frac{qE}{m}=sE$
Since $L=ut\Rightarrow t=\frac{L}{u}$
Also, $y=\frac{1}{2}a{t}^2$
$\Rightarrow y=\frac{1}{2}\left(sE\right)\frac{L^2}{u^2}\Rightarrow y=\frac{Es{L}^2}{2u^2}$