Question

There exists a region of electric field with potential difference of 25 V as shown in figure. An electron enters this region at speed of v = 3 $\times {10}^6$ m/s at angle of $\alpha$ and emerges out of region at angle $\beta$. The $\frac{sin\alpha }{sin\beta }$ is [Electric field exist in given region only]
$e=1.6\times {10}^{-}19,m=9.1\times {10}^{-}31$ (2 decimal places)

Answer 1

$usin\alpha =v\Rightarrow \frac{sin\alpha }{sin\beta }=\frac{v}{u}$
$\frac{1}{2}m{u}^2+qV=\frac{1}{2}m{v}^2\Rightarrow \frac{v}{u}=\sqrt{1+\frac{2qV}{m{u}^2}}=1.406$