Question

A stream of a positively charged particles having $\frac{q}{ m }=2 \times 10^{11} \frac{ C }{ kg }$ and velocity $\vec{v}_0=3 \times 10^7 \hat{i} m / s$ is deflected by an electric field $1.8 kV / m$. The electric field exists in a region of $10 cm$ along $x$ direction. Due to the electric field, the deflection of the charge particles in the $y$ direction is___ $mm$.

Answer 1

$ a =\frac{ F }{ m }=\frac{ qE }{ m }=\left(2 \times 10^{11}\right)\left(1.8 \times 10^3\right) $
$ =3.6 \times 10^{14} m / s ^2 $
$ \text { Time to cross plates }=\frac{ d }{ v } $
$ t =\frac{0.10}{3 \times 10^7} $
$ y =\frac{1}{2} at ^2=\frac{1}{2}\left(3.6 \times 10^{14}\right)\left(\frac{0.01}{9 \times 10^{14}}\right) $
$=0.2 \times 0.01$
$ =0.002 \,m $
$=2 \,mm $