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 $18 kV / m$ 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

Correct Answer: 2

The correct answer is 2.

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