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

The correct answer is 2.

a=mF​=mqE​=(2×1011)(1.8×103)
=3.6×1014m/s2
Time to cross plates =vd​
t=3×1070.10​
y=21​at2=21​(3.6×1014)(9×10140.01​)
=0.2×0.01
=0.002m
=2mm