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 $1.8kV/m$. The electric field exists in a region of $10cm$ along $x$ direction. Due to the electric field, the deflection of the charge particles in the $y$ direction is___ $mm$.

Answer 1

Answer: 2

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