Question 87 of 311

A stream of a positively charged particles having $\dfrac{q}{m}=2\times {10}^{11}C{kg}^{-1}$ and velocity ${\vec{v}}_{0}=3\times {10}^{7}\hat{i}m{s}^{-1}$ is deflected by an electric field $1.8\hat{j}kV{m}^{-1}$. 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 _____ $mm$.

Your answerCheck answer

Show answer and solution▾

Answer: 2

Acceleration of the particles due to electric field, $a=\dfrac{F}{m}=\dfrac{qE}{m}=\left(2\times {10}^{11}\right)\left(1.8\times {10}^{3}\right)$
$=3.6\times {10}^{14}m{s}^{-2}$
Time to cross plates $=\dfrac{d}{v}$
$\Rightarrow t=\dfrac{0.10}{3\times {10}^{7}}$
Now deflection,
$y=\dfrac{1}{2}{at}^{2} =\dfrac{1}{2}\left(3.6\times {10}^{14}\right){\left(\dfrac{0.10}{3\times {10}^{7}}\right)}^{2}$
$=0.002m$
$=2mm$