Electrostatics Question 2
Question 2 - 24 January - Shift 1
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 $1.8 \hat{j k V} / m$. 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 $mm$.
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Answer: (2)
Solution:
Formula: Electric force
$ \begin{aligned} & \xrightarrow[V_0=3 \times 10^{7} m / s]{ }, E=1.8 \times 10^{3} N / m. \\ & \longmapsto \iota=10 cm \square \\ & a=\frac{F}{m}=\frac{qE}{m}=(2 \times 10^{11})(1.8 \times 10^{3}) \\ & =3.6 \times 10^{14} m / s^{2} \end{aligned} $
Time to cross plates $=\frac{d}{V}$
$t=\frac{0.10}{3 \times 10^{7}}$
$y=\frac{1}{2} at^{2}=\frac{1}{2}(3.6 \times 10^{14})(\frac{0.01}{9 \times 10^{14}})$
$=0.2 \times 0.01$
$=0.002 m$
$=2 mm$
