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A stream of a positively charged particles having $\frac{ q }{ m }=2 \times 10^{11} \frac{ C }{ kg }$ and velocity $\overrightarrow{ v }_0=3 \times 10^7 \hat{ i ~ m} / s$ is deflected by an electric field $1.8 \hat{ j } kV / m$. 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 $...........mm$
$2$
$4$
$0.5$
$9$
Solution
$a =\frac{ F }{ m }=\frac{ qE }{ m }=\left(2 \times 10^{11}\right)\left(1.8 \times 10^3\right)$
$=3.6 \times 10^{14} m / s ^2$
$\text { Time to cross plates }=\frac{ d }{ v }$
$t =\frac{0.10}{3 \times 10^7}$
$y =\frac{1}{2} at ^2=\frac{1}{2}\left(3.6 \times 10^{14}\right)\left(\frac{0.01}{9 \times 10^{14}}\right)$
$=0.2 \times 0.01$
$=0.002\,m$
$=2\,mm$
